3.4.59 \(\int \frac {x^{3/2} (A+B x^2)}{(a+b x^2)^2} \, dx\)

Optimal. Leaf size=284 \[ -\frac {(A b-5 a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{3/4} b^{9/4}}+\frac {(A b-5 a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{3/4} b^{9/4}}-\frac {(A b-5 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{3/4} b^{9/4}}+\frac {(A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{3/4} b^{9/4}}-\frac {\sqrt {x} (A b-5 a B)}{2 a b^2}+\frac {x^{5/2} (A b-a B)}{2 a b \left (a+b x^2\right )} \]

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Rubi [A]  time = 0.21, antiderivative size = 284, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {457, 321, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {(A b-5 a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{3/4} b^{9/4}}+\frac {(A b-5 a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{8 \sqrt {2} a^{3/4} b^{9/4}}-\frac {(A b-5 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{3/4} b^{9/4}}+\frac {(A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{4 \sqrt {2} a^{3/4} b^{9/4}}-\frac {\sqrt {x} (A b-5 a B)}{2 a b^2}+\frac {x^{5/2} (A b-a B)}{2 a b \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^(3/2)*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

-((A*b - 5*a*B)*Sqrt[x])/(2*a*b^2) + ((A*b - a*B)*x^(5/2))/(2*a*b*(a + b*x^2)) - ((A*b - 5*a*B)*ArcTan[1 - (Sq
rt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(4*Sqrt[2]*a^(3/4)*b^(9/4)) + ((A*b - 5*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt
[x])/a^(1/4)])/(4*Sqrt[2]*a^(3/4)*b^(9/4)) - ((A*b - 5*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sq
rt[b]*x])/(8*Sqrt[2]*a^(3/4)*b^(9/4)) + ((A*b - 5*a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]
*x])/(8*Sqrt[2]*a^(3/4)*b^(9/4))

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 457

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> -Simp[((b*c - a*d
)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*b*e*n*(p + 1)), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b
*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] &&
LeQ[-1, m, -(n*(p + 1))]))

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rubi steps

\begin {align*} \int \frac {x^{3/2} \left (A+B x^2\right )}{\left (a+b x^2\right )^2} \, dx &=\frac {(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}+\frac {\left (-\frac {A b}{2}+\frac {5 a B}{2}\right ) \int \frac {x^{3/2}}{a+b x^2} \, dx}{2 a b}\\ &=-\frac {(A b-5 a B) \sqrt {x}}{2 a b^2}+\frac {(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}+\frac {(A b-5 a B) \int \frac {1}{\sqrt {x} \left (a+b x^2\right )} \, dx}{4 b^2}\\ &=-\frac {(A b-5 a B) \sqrt {x}}{2 a b^2}+\frac {(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}+\frac {(A b-5 a B) \operatorname {Subst}\left (\int \frac {1}{a+b x^4} \, dx,x,\sqrt {x}\right )}{2 b^2}\\ &=-\frac {(A b-5 a B) \sqrt {x}}{2 a b^2}+\frac {(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}+\frac {(A b-5 a B) \operatorname {Subst}\left (\int \frac {\sqrt {a}-\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 \sqrt {a} b^2}+\frac {(A b-5 a B) \operatorname {Subst}\left (\int \frac {\sqrt {a}+\sqrt {b} x^2}{a+b x^4} \, dx,x,\sqrt {x}\right )}{4 \sqrt {a} b^2}\\ &=-\frac {(A b-5 a B) \sqrt {x}}{2 a b^2}+\frac {(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}+\frac {(A b-5 a B) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {a} b^{5/2}}+\frac {(A b-5 a B) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}+x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {a} b^{5/2}}-\frac {(A b-5 a B) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}+2 x}{-\frac {\sqrt {a}}{\sqrt {b}}-\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} a^{3/4} b^{9/4}}-\frac {(A b-5 a B) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{a}}{\sqrt [4]{b}}-2 x}{-\frac {\sqrt {a}}{\sqrt {b}}+\frac {\sqrt {2} \sqrt [4]{a} x}{\sqrt [4]{b}}-x^2} \, dx,x,\sqrt {x}\right )}{8 \sqrt {2} a^{3/4} b^{9/4}}\\ &=-\frac {(A b-5 a B) \sqrt {x}}{2 a b^2}+\frac {(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}-\frac {(A b-5 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{3/4} b^{9/4}}+\frac {(A b-5 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{3/4} b^{9/4}}+\frac {(A b-5 a B) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{3/4} b^{9/4}}-\frac {(A b-5 a B) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{3/4} b^{9/4}}\\ &=-\frac {(A b-5 a B) \sqrt {x}}{2 a b^2}+\frac {(A b-a B) x^{5/2}}{2 a b \left (a+b x^2\right )}-\frac {(A b-5 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{3/4} b^{9/4}}+\frac {(A b-5 a B) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{3/4} b^{9/4}}-\frac {(A b-5 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{3/4} b^{9/4}}+\frac {(A b-5 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{8 \sqrt {2} a^{3/4} b^{9/4}}\\ \end {align*}

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Mathematica [A]  time = 0.37, size = 353, normalized size = 1.24 \begin {gather*} \frac {\frac {2 \sqrt {2} (5 a B-A b) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{a^{3/4}}+\frac {2 \sqrt {2} (A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{a^{3/4}}-\frac {\sqrt {2} A b \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{a^{3/4}}+\frac {\sqrt {2} A b \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{a^{3/4}}-\frac {8 A b^{5/4} \sqrt {x}}{a+b x^2}+\frac {8 a \sqrt [4]{b} B \sqrt {x}}{a+b x^2}+5 \sqrt {2} \sqrt [4]{a} B \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )-5 \sqrt {2} \sqrt [4]{a} B \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )+32 \sqrt [4]{b} B \sqrt {x}}{16 b^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^(3/2)*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

(32*b^(1/4)*B*Sqrt[x] - (8*A*b^(5/4)*Sqrt[x])/(a + b*x^2) + (8*a*b^(1/4)*B*Sqrt[x])/(a + b*x^2) + (2*Sqrt[2]*(
-(A*b) + 5*a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/a^(3/4) + (2*Sqrt[2]*(A*b - 5*a*B)*ArcTan[1 + (
Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/a^(3/4) - (Sqrt[2]*A*b*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt
[b]*x])/a^(3/4) + 5*Sqrt[2]*a^(1/4)*B*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x] + (Sqrt[2]*A*
b*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/a^(3/4) - 5*Sqrt[2]*a^(1/4)*B*Log[Sqrt[a] + Sqrt
[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(16*b^(9/4))

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IntegrateAlgebraic [A]  time = 0.65, size = 167, normalized size = 0.59 \begin {gather*} \frac {(5 a B-A b) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{4 \sqrt {2} a^{3/4} b^{9/4}}-\frac {(5 a B-A b) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{4 \sqrt {2} a^{3/4} b^{9/4}}+\frac {\sqrt {x} \left (5 a B-A b+4 b B x^2\right )}{2 b^2 \left (a+b x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(x^(3/2)*(A + B*x^2))/(a + b*x^2)^2,x]

[Out]

(Sqrt[x]*(-(A*b) + 5*a*B + 4*b*B*x^2))/(2*b^2*(a + b*x^2)) + ((-(A*b) + 5*a*B)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(S
qrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(4*Sqrt[2]*a^(3/4)*b^(9/4)) - ((-(A*b) + 5*a*B)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(
1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(4*Sqrt[2]*a^(3/4)*b^(9/4))

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fricas [B]  time = 1.01, size = 725, normalized size = 2.55 \begin {gather*} \frac {4 \, {\left (b^{3} x^{2} + a b^{2}\right )} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {a^{2} b^{4} \sqrt {-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}} + {\left (25 \, B^{2} a^{2} - 10 \, A B a b + A^{2} b^{2}\right )} x} a^{2} b^{7} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {3}{4}} + {\left (5 \, B a^{3} b^{7} - A a^{2} b^{8}\right )} \sqrt {x} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {3}{4}}}{625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}\right ) + {\left (b^{3} x^{2} + a b^{2}\right )} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {1}{4}} \log \left (a b^{2} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {1}{4}} - {\left (5 \, B a - A b\right )} \sqrt {x}\right ) - {\left (b^{3} x^{2} + a b^{2}\right )} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {1}{4}} \log \left (-a b^{2} \left (-\frac {625 \, B^{4} a^{4} - 500 \, A B^{3} a^{3} b + 150 \, A^{2} B^{2} a^{2} b^{2} - 20 \, A^{3} B a b^{3} + A^{4} b^{4}}{a^{3} b^{9}}\right )^{\frac {1}{4}} - {\left (5 \, B a - A b\right )} \sqrt {x}\right ) + 4 \, {\left (4 \, B b x^{2} + 5 \, B a - A b\right )} \sqrt {x}}{8 \, {\left (b^{3} x^{2} + a b^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="fricas")

[Out]

1/8*(4*(b^3*x^2 + a*b^2)*(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a
^3*b^9))^(1/4)*arctan((sqrt(a^2*b^4*sqrt(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^
3 + A^4*b^4)/(a^3*b^9)) + (25*B^2*a^2 - 10*A*B*a*b + A^2*b^2)*x)*a^2*b^7*(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 15
0*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(3/4) + (5*B*a^3*b^7 - A*a^2*b^8)*sqrt(x)*(-(625*B^4*
a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(3/4))/(625*B^4*a^4 - 500*A
*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)) + (b^3*x^2 + a*b^2)*(-(625*B^4*a^4 - 500*A*B^3*a
^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(1/4)*log(a*b^2*(-(625*B^4*a^4 - 500*A*B^3*a
^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(1/4) - (5*B*a - A*b)*sqrt(x)) - (b^3*x^2 +
a*b^2)*(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(1/4)*log
(-a*b^2*(-(625*B^4*a^4 - 500*A*B^3*a^3*b + 150*A^2*B^2*a^2*b^2 - 20*A^3*B*a*b^3 + A^4*b^4)/(a^3*b^9))^(1/4) -
(5*B*a - A*b)*sqrt(x)) + 4*(4*B*b*x^2 + 5*B*a - A*b)*sqrt(x))/(b^3*x^2 + a*b^2)

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giac [A]  time = 0.43, size = 283, normalized size = 1.00 \begin {gather*} \frac {2 \, B \sqrt {x}}{b^{2}} - \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a b^{3}} - \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{8 \, a b^{3}} - \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a b^{3}} + \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} B a - \left (a b^{3}\right )^{\frac {1}{4}} A b\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{16 \, a b^{3}} + \frac {B a \sqrt {x} - A b \sqrt {x}}{2 \, {\left (b x^{2} + a\right )} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="giac")

[Out]

2*B*sqrt(x)/b^2 - 1/8*sqrt(2)*(5*(a*b^3)^(1/4)*B*a - (a*b^3)^(1/4)*A*b)*arctan(1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4
) + 2*sqrt(x))/(a/b)^(1/4))/(a*b^3) - 1/8*sqrt(2)*(5*(a*b^3)^(1/4)*B*a - (a*b^3)^(1/4)*A*b)*arctan(-1/2*sqrt(2
)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x))/(a/b)^(1/4))/(a*b^3) - 1/16*sqrt(2)*(5*(a*b^3)^(1/4)*B*a - (a*b^3)^(1/4)*A
*b)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a*b^3) + 1/16*sqrt(2)*(5*(a*b^3)^(1/4)*B*a - (a*b^3)^(1/
4)*A*b)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a*b^3) + 1/2*(B*a*sqrt(x) - A*b*sqrt(x))/((b*x^2 +
a)*b^2)

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maple [A]  time = 0.02, size = 323, normalized size = 1.14 \begin {gather*} -\frac {A \sqrt {x}}{2 \left (b \,x^{2}+a \right ) b}+\frac {B a \sqrt {x}}{2 \left (b \,x^{2}+a \right ) b^{2}}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 a b}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, A \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 a b}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, A \ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{16 a b}-\frac {5 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{8 b^{2}}-\frac {5 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, B \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{8 b^{2}}-\frac {5 \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, B \ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {a}{b}}}\right )}{16 b^{2}}+\frac {2 B \sqrt {x}}{b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(B*x^2+A)/(b*x^2+a)^2,x)

[Out]

2*B/b^2*x^(1/2)-1/2/b*x^(1/2)/(b*x^2+a)*A+1/2/b^2*x^(1/2)/(b*x^2+a)*B*a+1/8/b*(a/b)^(1/4)/a*2^(1/2)*A*arctan(2
^(1/2)/(a/b)^(1/4)*x^(1/2)+1)+1/8/b*(a/b)^(1/4)/a*2^(1/2)*A*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)+1/16/b*(a/b)
^(1/4)/a*2^(1/2)*A*ln((x+(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2))/(x-(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2)))
-5/8/b^2*(a/b)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)-5/8/b^2*(a/b)^(1/4)*2^(1/2)*B*arctan(2^(1
/2)/(a/b)^(1/4)*x^(1/2)-1)-5/16/b^2*(a/b)^(1/4)*2^(1/2)*B*ln((x+(a/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2))/(x-(a
/b)^(1/4)*2^(1/2)*x^(1/2)+(a/b)^(1/2)))

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maxima [A]  time = 2.49, size = 250, normalized size = 0.88 \begin {gather*} \frac {{\left (B a - A b\right )} \sqrt {x}}{2 \, {\left (b^{3} x^{2} + a b^{2}\right )}} + \frac {2 \, B \sqrt {x}}{b^{2}} - \frac {\frac {2 \, \sqrt {2} {\left (5 \, B a - A b\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} {\left (5 \, B a - A b\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} {\left (5 \, B a - A b\right )} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (5 \, B a - A b\right )} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}}{16 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(B*x^2+A)/(b*x^2+a)^2,x, algorithm="maxima")

[Out]

1/2*(B*a - A*b)*sqrt(x)/(b^3*x^2 + a*b^2) + 2*B*sqrt(x)/b^2 - 1/16*(2*sqrt(2)*(5*B*a - A*b)*arctan(1/2*sqrt(2)
*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + 2*sqrt
(2)*(5*B*a - A*b)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sq
rt(a)*sqrt(sqrt(a)*sqrt(b))) + sqrt(2)*(5*B*a - A*b)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a)
)/(a^(3/4)*b^(1/4)) - sqrt(2)*(5*B*a - A*b)*log(-sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(3/
4)*b^(1/4)))/b^2

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mupad [B]  time = 0.37, size = 744, normalized size = 2.62 \begin {gather*} \frac {2\,B\,\sqrt {x}}{b^2}-\frac {\sqrt {x}\,\left (\frac {A\,b}{2}-\frac {B\,a}{2}\right )}{b^3\,x^2+a\,b^2}+\frac {\mathrm {atan}\left (\frac {\frac {\left (A\,b-5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,b^2-10\,A\,B\,a\,b+25\,B^2\,a^2\right )}{b}-\frac {\left (A\,b-5\,B\,a\right )\,\left (8\,A\,a\,b^2-40\,B\,a^2\,b\right )}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}+\frac {\left (A\,b-5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,b^2-10\,A\,B\,a\,b+25\,B^2\,a^2\right )}{b}+\frac {\left (A\,b-5\,B\,a\right )\,\left (8\,A\,a\,b^2-40\,B\,a^2\,b\right )}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}}{\frac {\left (A\,b-5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,b^2-10\,A\,B\,a\,b+25\,B^2\,a^2\right )}{b}-\frac {\left (A\,b-5\,B\,a\right )\,\left (8\,A\,a\,b^2-40\,B\,a^2\,b\right )}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}\right )}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}-\frac {\left (A\,b-5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,b^2-10\,A\,B\,a\,b+25\,B^2\,a^2\right )}{b}+\frac {\left (A\,b-5\,B\,a\right )\,\left (8\,A\,a\,b^2-40\,B\,a^2\,b\right )}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}\right )}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}}\right )\,\left (A\,b-5\,B\,a\right )\,1{}\mathrm {i}}{4\,{\left (-a\right )}^{3/4}\,b^{9/4}}+\frac {\mathrm {atan}\left (\frac {\frac {\left (A\,b-5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,b^2-10\,A\,B\,a\,b+25\,B^2\,a^2\right )}{b}-\frac {\left (A\,b-5\,B\,a\right )\,\left (8\,A\,a\,b^2-40\,B\,a^2\,b\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}\right )}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}+\frac {\left (A\,b-5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,b^2-10\,A\,B\,a\,b+25\,B^2\,a^2\right )}{b}+\frac {\left (A\,b-5\,B\,a\right )\,\left (8\,A\,a\,b^2-40\,B\,a^2\,b\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}\right )}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}}{\frac {\left (A\,b-5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,b^2-10\,A\,B\,a\,b+25\,B^2\,a^2\right )}{b}-\frac {\left (A\,b-5\,B\,a\right )\,\left (8\,A\,a\,b^2-40\,B\,a^2\,b\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}-\frac {\left (A\,b-5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (A^2\,b^2-10\,A\,B\,a\,b+25\,B^2\,a^2\right )}{b}+\frac {\left (A\,b-5\,B\,a\right )\,\left (8\,A\,a\,b^2-40\,B\,a^2\,b\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}\right )\,1{}\mathrm {i}}{8\,{\left (-a\right )}^{3/4}\,b^{9/4}}}\right )\,\left (A\,b-5\,B\,a\right )}{4\,{\left (-a\right )}^{3/4}\,b^{9/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^(3/2)*(A + B*x^2))/(a + b*x^2)^2,x)

[Out]

(2*B*x^(1/2))/b^2 - (x^(1/2)*((A*b)/2 - (B*a)/2))/(a*b^2 + b^3*x^2) + (atan((((A*b - 5*B*a)*((x^(1/2)*(A^2*b^2
+ 25*B^2*a^2 - 10*A*B*a*b))/b - ((A*b - 5*B*a)*(8*A*a*b^2 - 40*B*a^2*b))/(8*(-a)^(3/4)*b^(9/4)))*1i)/(8*(-a)^
(3/4)*b^(9/4)) + ((A*b - 5*B*a)*((x^(1/2)*(A^2*b^2 + 25*B^2*a^2 - 10*A*B*a*b))/b + ((A*b - 5*B*a)*(8*A*a*b^2 -
40*B*a^2*b))/(8*(-a)^(3/4)*b^(9/4)))*1i)/(8*(-a)^(3/4)*b^(9/4)))/(((A*b - 5*B*a)*((x^(1/2)*(A^2*b^2 + 25*B^2*
a^2 - 10*A*B*a*b))/b - ((A*b - 5*B*a)*(8*A*a*b^2 - 40*B*a^2*b))/(8*(-a)^(3/4)*b^(9/4))))/(8*(-a)^(3/4)*b^(9/4)
) - ((A*b - 5*B*a)*((x^(1/2)*(A^2*b^2 + 25*B^2*a^2 - 10*A*B*a*b))/b + ((A*b - 5*B*a)*(8*A*a*b^2 - 40*B*a^2*b))
/(8*(-a)^(3/4)*b^(9/4))))/(8*(-a)^(3/4)*b^(9/4))))*(A*b - 5*B*a)*1i)/(4*(-a)^(3/4)*b^(9/4)) + (atan((((A*b - 5
*B*a)*((x^(1/2)*(A^2*b^2 + 25*B^2*a^2 - 10*A*B*a*b))/b - ((A*b - 5*B*a)*(8*A*a*b^2 - 40*B*a^2*b)*1i)/(8*(-a)^(
3/4)*b^(9/4))))/(8*(-a)^(3/4)*b^(9/4)) + ((A*b - 5*B*a)*((x^(1/2)*(A^2*b^2 + 25*B^2*a^2 - 10*A*B*a*b))/b + ((A
*b - 5*B*a)*(8*A*a*b^2 - 40*B*a^2*b)*1i)/(8*(-a)^(3/4)*b^(9/4))))/(8*(-a)^(3/4)*b^(9/4)))/(((A*b - 5*B*a)*((x^
(1/2)*(A^2*b^2 + 25*B^2*a^2 - 10*A*B*a*b))/b - ((A*b - 5*B*a)*(8*A*a*b^2 - 40*B*a^2*b)*1i)/(8*(-a)^(3/4)*b^(9/
4)))*1i)/(8*(-a)^(3/4)*b^(9/4)) - ((A*b - 5*B*a)*((x^(1/2)*(A^2*b^2 + 25*B^2*a^2 - 10*A*B*a*b))/b + ((A*b - 5*
B*a)*(8*A*a*b^2 - 40*B*a^2*b)*1i)/(8*(-a)^(3/4)*b^(9/4)))*1i)/(8*(-a)^(3/4)*b^(9/4))))*(A*b - 5*B*a))/(4*(-a)^
(3/4)*b^(9/4))

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sympy [A]  time = 84.94, size = 984, normalized size = 3.46 \begin {gather*} \begin {cases} \tilde {\infty } \left (- \frac {2 A}{3 x^{\frac {3}{2}}} + 2 B \sqrt {x}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {- \frac {2 A}{3 x^{\frac {3}{2}}} + 2 B \sqrt {x}}{b^{2}} & \text {for}\: a = 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {9}{2}}}{9}}{a^{2}} & \text {for}\: b = 0 \\- \frac {\sqrt [4]{-1} A a^{\frac {5}{4}} b \sqrt [4]{\frac {1}{b}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} + \frac {\sqrt [4]{-1} A a^{\frac {5}{4}} b \sqrt [4]{\frac {1}{b}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} - \frac {2 \sqrt [4]{-1} A a^{\frac {5}{4}} b \sqrt [4]{\frac {1}{b}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{b}}} \right )}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} - \frac {\sqrt [4]{-1} A \sqrt [4]{a} b^{2} x^{2} \sqrt [4]{\frac {1}{b}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} + \frac {\sqrt [4]{-1} A \sqrt [4]{a} b^{2} x^{2} \sqrt [4]{\frac {1}{b}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} - \frac {2 \sqrt [4]{-1} A \sqrt [4]{a} b^{2} x^{2} \sqrt [4]{\frac {1}{b}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{b}}} \right )}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} - \frac {4 A a b \sqrt {x}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} + \frac {5 \sqrt [4]{-1} B a^{\frac {9}{4}} \sqrt [4]{\frac {1}{b}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} - \frac {5 \sqrt [4]{-1} B a^{\frac {9}{4}} \sqrt [4]{\frac {1}{b}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} + \frac {10 \sqrt [4]{-1} B a^{\frac {9}{4}} \sqrt [4]{\frac {1}{b}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{b}}} \right )}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} + \frac {5 \sqrt [4]{-1} B a^{\frac {5}{4}} b x^{2} \sqrt [4]{\frac {1}{b}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} - \frac {5 \sqrt [4]{-1} B a^{\frac {5}{4}} b x^{2} \sqrt [4]{\frac {1}{b}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} + \frac {10 \sqrt [4]{-1} B a^{\frac {5}{4}} b x^{2} \sqrt [4]{\frac {1}{b}} \operatorname {atan}{\left (\frac {\left (-1\right )^{\frac {3}{4}} \sqrt {x}}{\sqrt [4]{a} \sqrt [4]{\frac {1}{b}}} \right )}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} + \frac {20 B a^{2} \sqrt {x}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} + \frac {16 B a b x^{\frac {5}{2}}}{8 a^{2} b^{2} + 8 a b^{3} x^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(B*x**2+A)/(b*x**2+a)**2,x)

[Out]

Piecewise((zoo*(-2*A/(3*x**(3/2)) + 2*B*sqrt(x)), Eq(a, 0) & Eq(b, 0)), ((-2*A/(3*x**(3/2)) + 2*B*sqrt(x))/b**
2, Eq(a, 0)), ((2*A*x**(5/2)/5 + 2*B*x**(9/2)/9)/a**2, Eq(b, 0)), (-(-1)**(1/4)*A*a**(5/4)*b*(1/b)**(1/4)*log(
-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(8*a**2*b**2 + 8*a*b**3*x**2) + (-1)**(1/4)*A*a**(5/4)*b*(1/b)**
(1/4)*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(8*a**2*b**2 + 8*a*b**3*x**2) - 2*(-1)**(1/4)*A*a**(5/4
)*b*(1/b)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/b)**(1/4)))/(8*a**2*b**2 + 8*a*b**3*x**2) - (-1)**(1/4)
*A*a**(1/4)*b**2*x**2*(1/b)**(1/4)*log(-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(8*a**2*b**2 + 8*a*b**3*x
**2) + (-1)**(1/4)*A*a**(1/4)*b**2*x**2*(1/b)**(1/4)*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(8*a**2*
b**2 + 8*a*b**3*x**2) - 2*(-1)**(1/4)*A*a**(1/4)*b**2*x**2*(1/b)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/
b)**(1/4)))/(8*a**2*b**2 + 8*a*b**3*x**2) - 4*A*a*b*sqrt(x)/(8*a**2*b**2 + 8*a*b**3*x**2) + 5*(-1)**(1/4)*B*a*
*(9/4)*(1/b)**(1/4)*log(-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(8*a**2*b**2 + 8*a*b**3*x**2) - 5*(-1)**
(1/4)*B*a**(9/4)*(1/b)**(1/4)*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(8*a**2*b**2 + 8*a*b**3*x**2) +
10*(-1)**(1/4)*B*a**(9/4)*(1/b)**(1/4)*atan((-1)**(3/4)*sqrt(x)/(a**(1/4)*(1/b)**(1/4)))/(8*a**2*b**2 + 8*a*b
**3*x**2) + 5*(-1)**(1/4)*B*a**(5/4)*b*x**2*(1/b)**(1/4)*log(-(-1)**(1/4)*a**(1/4)*(1/b)**(1/4) + sqrt(x))/(8*
a**2*b**2 + 8*a*b**3*x**2) - 5*(-1)**(1/4)*B*a**(5/4)*b*x**2*(1/b)**(1/4)*log((-1)**(1/4)*a**(1/4)*(1/b)**(1/4
) + sqrt(x))/(8*a**2*b**2 + 8*a*b**3*x**2) + 10*(-1)**(1/4)*B*a**(5/4)*b*x**2*(1/b)**(1/4)*atan((-1)**(3/4)*sq
rt(x)/(a**(1/4)*(1/b)**(1/4)))/(8*a**2*b**2 + 8*a*b**3*x**2) + 20*B*a**2*sqrt(x)/(8*a**2*b**2 + 8*a*b**3*x**2)
+ 16*B*a*b*x**(5/2)/(8*a**2*b**2 + 8*a*b**3*x**2), True))

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