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

Optimal. Leaf size=298 \[ -\frac {(5 a B+3 A b) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(5 a B+3 A b) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{7/4} b^{9/4}}-\frac {(5 a B+3 A b) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(5 a B+3 A b) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{7/4} b^{9/4}}-\frac {\sqrt {x} (5 a B+3 A b)}{16 a b^2 \left (a+b x^2\right )}+\frac {x^{5/2} (A b-a B)}{4 a b \left (a+b x^2\right )^2} \]

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Rubi [A]  time = 0.21, antiderivative size = 298, 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, 288, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {(5 a B+3 A b) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(5 a B+3 A b) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{64 \sqrt {2} a^{7/4} b^{9/4}}-\frac {(5 a B+3 A b) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(5 a B+3 A b) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{32 \sqrt {2} a^{7/4} b^{9/4}}-\frac {\sqrt {x} (5 a B+3 A b)}{16 a b^2 \left (a+b x^2\right )}+\frac {x^{5/2} (A b-a B)}{4 a b \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*x^(5/2))/(4*a*b*(a + b*x^2)^2) - ((3*A*b + 5*a*B)*Sqrt[x])/(16*a*b^2*(a + b*x^2)) - ((3*A*b + 5*a
*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(7/4)*b^(9/4)) + ((3*A*b + 5*a*B)*ArcTan[1 +
(Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(32*Sqrt[2]*a^(7/4)*b^(9/4)) - ((3*A*b + 5*a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1
/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(7/4)*b^(9/4)) + ((3*A*b + 5*a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4
)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(64*Sqrt[2]*a^(7/4)*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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 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 )^3} \, dx &=\frac {(A b-a B) x^{5/2}}{4 a b \left (a+b x^2\right )^2}+\frac {\left (\frac {3 A b}{2}+\frac {5 a B}{2}\right ) \int \frac {x^{3/2}}{\left (a+b x^2\right )^2} \, dx}{4 a b}\\ &=\frac {(A b-a B) x^{5/2}}{4 a b \left (a+b x^2\right )^2}-\frac {(3 A b+5 a B) \sqrt {x}}{16 a b^2 \left (a+b x^2\right )}+\frac {(3 A b+5 a B) \int \frac {1}{\sqrt {x} \left (a+b x^2\right )} \, dx}{32 a b^2}\\ &=\frac {(A b-a B) x^{5/2}}{4 a b \left (a+b x^2\right )^2}-\frac {(3 A b+5 a B) \sqrt {x}}{16 a b^2 \left (a+b x^2\right )}+\frac {(3 A b+5 a B) \operatorname {Subst}\left (\int \frac {1}{a+b x^4} \, dx,x,\sqrt {x}\right )}{16 a b^2}\\ &=\frac {(A b-a B) x^{5/2}}{4 a b \left (a+b x^2\right )^2}-\frac {(3 A b+5 a B) \sqrt {x}}{16 a b^2 \left (a+b x^2\right )}+\frac {(3 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 )}{32 a^{3/2} b^2}+\frac {(3 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 )}{32 a^{3/2} b^2}\\ &=\frac {(A b-a B) x^{5/2}}{4 a b \left (a+b x^2\right )^2}-\frac {(3 A b+5 a B) \sqrt {x}}{16 a b^2 \left (a+b x^2\right )}+\frac {(3 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 )}{64 a^{3/2} b^{5/2}}+\frac {(3 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 )}{64 a^{3/2} b^{5/2}}-\frac {(3 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 )}{64 \sqrt {2} a^{7/4} b^{9/4}}-\frac {(3 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 )}{64 \sqrt {2} a^{7/4} b^{9/4}}\\ &=\frac {(A b-a B) x^{5/2}}{4 a b \left (a+b x^2\right )^2}-\frac {(3 A b+5 a B) \sqrt {x}}{16 a b^2 \left (a+b x^2\right )}-\frac {(3 A b+5 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(3 A b+5 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(3 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 )}{32 \sqrt {2} a^{7/4} b^{9/4}}-\frac {(3 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 )}{32 \sqrt {2} a^{7/4} b^{9/4}}\\ &=\frac {(A b-a B) x^{5/2}}{4 a b \left (a+b x^2\right )^2}-\frac {(3 A b+5 a B) \sqrt {x}}{16 a b^2 \left (a+b x^2\right )}-\frac {(3 A b+5 a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(3 A b+5 a B) \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{32 \sqrt {2} a^{7/4} b^{9/4}}-\frac {(3 A b+5 a B) \log \left (\sqrt {a}-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(3 A b+5 a B) \log \left (\sqrt {a}+\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {b} x\right )}{64 \sqrt {2} a^{7/4} b^{9/4}}\\ \end {align*}

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Mathematica [A]  time = 0.43, size = 389, normalized size = 1.31 \begin {gather*} \frac {-\frac {2 \sqrt {2} (5 a B+3 A b) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{a^{7/4}}+\frac {2 \sqrt {2} (5 a B+3 A b) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{a^{7/4}}-\frac {3 \sqrt {2} A b \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{a^{7/4}}+\frac {3 \sqrt {2} A b \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{a^{7/4}}-\frac {5 \sqrt {2} B \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{a^{3/4}}+\frac {5 \sqrt {2} 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^2+a b x^2}-\frac {32 A b^{5/4} \sqrt {x}}{\left (a+b x^2\right )^2}-\frac {72 \sqrt [4]{b} B \sqrt {x}}{a+b x^2}+\frac {32 a \sqrt [4]{b} B \sqrt {x}}{\left (a+b x^2\right )^2}}{128 b^{9/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-32*A*b^(5/4)*Sqrt[x])/(a + b*x^2)^2 + (32*a*b^(1/4)*B*Sqrt[x])/(a + b*x^2)^2 - (72*b^(1/4)*B*Sqrt[x])/(a +
b*x^2) + (8*A*b^(5/4)*Sqrt[x])/(a^2 + a*b*x^2) - (2*Sqrt[2]*(3*A*b + 5*a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x
])/a^(1/4)])/a^(7/4) + (2*Sqrt[2]*(3*A*b + 5*a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/a^(7/4) - (3*
Sqrt[2]*A*b*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/a^(7/4) - (5*Sqrt[2]*B*Log[Sqrt[a] - S
qrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/a^(3/4) + (3*Sqrt[2]*A*b*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sq
rt[x] + Sqrt[b]*x])/a^(7/4) + (5*Sqrt[2]*B*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/a^(3/4)
)/(128*b^(9/4))

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IntegrateAlgebraic [A]  time = 0.74, size = 191, normalized size = 0.64 \begin {gather*} -\frac {(5 a B+3 A b) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{32 \sqrt {2} a^{7/4} b^{9/4}}+\frac {(5 a B+3 A b) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{32 \sqrt {2} a^{7/4} b^{9/4}}+\frac {-5 a^2 B \sqrt {x}-3 a A b \sqrt {x}-9 a b B x^{5/2}+A b^2 x^{5/2}}{16 a b^2 \left (a+b x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*A*b*Sqrt[x] - 5*a^2*B*Sqrt[x] + A*b^2*x^(5/2) - 9*a*b*B*x^(5/2))/(16*a*b^2*(a + b*x^2)^2) - ((3*A*b + 5*
a*B)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])])/(32*Sqrt[2]*a^(7/4)*b^(9/4)) + ((3*A*b +
5*a*B)*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/(32*Sqrt[2]*a^(7/4)*b^(9/4))

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fricas [B]  time = 1.20, size = 806, normalized size = 2.70 \begin {gather*} \frac {4 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (-\frac {625 \, B^{4} a^{4} + 1500 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 540 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{9}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {a^{4} b^{4} \sqrt {-\frac {625 \, B^{4} a^{4} + 1500 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 540 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{9}}} + {\left (25 \, B^{2} a^{2} + 30 \, A B a b + 9 \, A^{2} b^{2}\right )} x} a^{5} b^{7} \left (-\frac {625 \, B^{4} a^{4} + 1500 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 540 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{9}}\right )^{\frac {3}{4}} - {\left (5 \, B a^{6} b^{7} + 3 \, A a^{5} b^{8}\right )} \sqrt {x} \left (-\frac {625 \, B^{4} a^{4} + 1500 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 540 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{9}}\right )^{\frac {3}{4}}}{625 \, B^{4} a^{4} + 1500 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 540 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}\right ) + {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (-\frac {625 \, B^{4} a^{4} + 1500 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 540 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{9}}\right )^{\frac {1}{4}} \log \left (a^{2} b^{2} \left (-\frac {625 \, B^{4} a^{4} + 1500 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 540 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{9}}\right )^{\frac {1}{4}} + {\left (5 \, B a + 3 \, A b\right )} \sqrt {x}\right ) - {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )} \left (-\frac {625 \, B^{4} a^{4} + 1500 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 540 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{9}}\right )^{\frac {1}{4}} \log \left (-a^{2} b^{2} \left (-\frac {625 \, B^{4} a^{4} + 1500 \, A B^{3} a^{3} b + 1350 \, A^{2} B^{2} a^{2} b^{2} + 540 \, A^{3} B a b^{3} + 81 \, A^{4} b^{4}}{a^{7} b^{9}}\right )^{\frac {1}{4}} + {\left (5 \, B a + 3 \, A b\right )} \sqrt {x}\right ) - 4 \, {\left (5 \, B a^{2} + 3 \, A a b + {\left (9 \, B a b - A b^{2}\right )} x^{2}\right )} \sqrt {x}}{64 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} 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)^3,x, algorithm="fricas")

[Out]

1/64*(4*(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2)*(-(625*B^4*a^4 + 1500*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 540*A
^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^9))^(1/4)*arctan((sqrt(a^4*b^4*sqrt(-(625*B^4*a^4 + 1500*A*B^3*a^3*b + 1350*A^
2*B^2*a^2*b^2 + 540*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^9)) + (25*B^2*a^2 + 30*A*B*a*b + 9*A^2*b^2)*x)*a^5*b^7*(-
(625*B^4*a^4 + 1500*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 540*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^9))^(3/4) - (5*B
*a^6*b^7 + 3*A*a^5*b^8)*sqrt(x)*(-(625*B^4*a^4 + 1500*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 540*A^3*B*a*b^3 + 8
1*A^4*b^4)/(a^7*b^9))^(3/4))/(625*B^4*a^4 + 1500*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 540*A^3*B*a*b^3 + 81*A^4
*b^4)) + (a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2)*(-(625*B^4*a^4 + 1500*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 540*
A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^9))^(1/4)*log(a^2*b^2*(-(625*B^4*a^4 + 1500*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^
2 + 540*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^9))^(1/4) + (5*B*a + 3*A*b)*sqrt(x)) - (a*b^4*x^4 + 2*a^2*b^3*x^2 + a
^3*b^2)*(-(625*B^4*a^4 + 1500*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 540*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7*b^9))^(1
/4)*log(-a^2*b^2*(-(625*B^4*a^4 + 1500*A*B^3*a^3*b + 1350*A^2*B^2*a^2*b^2 + 540*A^3*B*a*b^3 + 81*A^4*b^4)/(a^7
*b^9))^(1/4) + (5*B*a + 3*A*b)*sqrt(x)) - 4*(5*B*a^2 + 3*A*a*b + (9*B*a*b - A*b^2)*x^2)*sqrt(x))/(a*b^4*x^4 +
2*a^2*b^3*x^2 + a^3*b^2)

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giac [A]  time = 0.52, size = 298, normalized size = 1.00 \begin {gather*} \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} B a + 3 \, \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 )}{64 \, a^{2} b^{3}} + \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} B a + 3 \, \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 )}{64 \, a^{2} b^{3}} + \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} B a + 3 \, \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 )}{128 \, a^{2} b^{3}} - \frac {\sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} B a + 3 \, \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 )}{128 \, a^{2} b^{3}} - \frac {9 \, B a b x^{\frac {5}{2}} - A b^{2} x^{\frac {5}{2}} + 5 \, B a^{2} \sqrt {x} + 3 \, A a b \sqrt {x}}{16 \, {\left (b x^{2} + a\right )}^{2} a 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)^3,x, algorithm="giac")

[Out]

1/64*sqrt(2)*(5*(a*b^3)^(1/4)*B*a + 3*(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^2*b^3) + 1/64*sqrt(2)*(5*(a*b^3)^(1/4)*B*a + 3*(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^2*b^3) + 1/128*sqrt(2)*(5*(a*b^3)^(1/4)*B*a + 3*(a*b^3)^(1/4)*A*b)*l
og(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^2*b^3) - 1/128*sqrt(2)*(5*(a*b^3)^(1/4)*B*a + 3*(a*b^3)^(1/
4)*A*b)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^2*b^3) - 1/16*(9*B*a*b*x^(5/2) - A*b^2*x^(5/2) +
5*B*a^2*sqrt(x) + 3*A*a*b*sqrt(x))/((b*x^2 + a)^2*a*b^2)

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maple [A]  time = 0.02, size = 334, normalized size = 1.12 \begin {gather*} \frac {3 \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 )}{64 a^{2} b}+\frac {3 \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 )}{64 a^{2} b}+\frac {3 \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 )}{128 a^{2} 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 )}{64 a \,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 )}{64 a \,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 )}{128 a \,b^{2}}+\frac {\frac {\left (A b -9 B a \right ) x^{\frac {5}{2}}}{16 a b}-\frac {\left (3 A b +5 B a \right ) \sqrt {x}}{16 b^{2}}}{\left (b \,x^{2}+a \right )^{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)^3,x)

[Out]

2*(1/32*(A*b-9*B*a)/a/b*x^(5/2)-1/32*(3*A*b+5*B*a)/b^2*x^(1/2))/(b*x^2+a)^2+3/64/b/a^2*(a/b)^(1/4)*2^(1/2)*A*a
rctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)+3/128/b/a^2*(a/b)^(1/4)*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)))+3/64/b/a^2*(a/b)^(1/4)*2^(1/2)*A*arctan(2^(1/2)/(a/b)^(1/
4)*x^(1/2)+1)+5/64/b^2/a*(a/b)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1)+5/128/b^2/a*(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)))+5/64/b^2/
a*(a/b)^(1/4)*2^(1/2)*B*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)+1)

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maxima [A]  time = 2.38, size = 280, normalized size = 0.94 \begin {gather*} -\frac {{\left (9 \, B a b - A b^{2}\right )} x^{\frac {5}{2}} + {\left (5 \, B a^{2} + 3 \, A a b\right )} \sqrt {x}}{16 \, {\left (a b^{4} x^{4} + 2 \, a^{2} b^{3} x^{2} + a^{3} b^{2}\right )}} + \frac {\frac {2 \, \sqrt {2} {\left (5 \, B a + 3 \, 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 + 3 \, 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 + 3 \, 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 + 3 \, 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}}}}{128 \, a 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)^3,x, algorithm="maxima")

[Out]

-1/16*((9*B*a*b - A*b^2)*x^(5/2) + (5*B*a^2 + 3*A*a*b)*sqrt(x))/(a*b^4*x^4 + 2*a^2*b^3*x^2 + a^3*b^2) + 1/128*
(2*sqrt(2)*(5*B*a + 3*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 + 3*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))) + sqrt(2)*(5*B*a + 3*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 + 3*A*b)*log(-sqrt(2)*a^
(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(3/4)*b^(1/4)))/(a*b^2)

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mupad [B]  time = 0.47, size = 799, normalized size = 2.68 \begin {gather*} -\frac {\frac {\sqrt {x}\,\left (3\,A\,b+5\,B\,a\right )}{16\,b^2}-\frac {x^{5/2}\,\left (A\,b-9\,B\,a\right )}{16\,a\,b}}{a^2+2\,a\,b\,x^2+b^2\,x^4}+\frac {\mathrm {atan}\left (\frac {\frac {\left (3\,A\,b+5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,b^2+30\,A\,B\,a\,b+25\,B^2\,a^2\right )}{64\,a^2\,b}-\frac {\left (3\,A\,b^2+5\,B\,a\,b\right )\,\left (3\,A\,b+5\,B\,a\right )}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}\right )\,1{}\mathrm {i}}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}+\frac {\left (3\,A\,b+5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,b^2+30\,A\,B\,a\,b+25\,B^2\,a^2\right )}{64\,a^2\,b}+\frac {\left (3\,A\,b^2+5\,B\,a\,b\right )\,\left (3\,A\,b+5\,B\,a\right )}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}\right )\,1{}\mathrm {i}}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}}{\frac {\left (3\,A\,b+5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,b^2+30\,A\,B\,a\,b+25\,B^2\,a^2\right )}{64\,a^2\,b}-\frac {\left (3\,A\,b^2+5\,B\,a\,b\right )\,\left (3\,A\,b+5\,B\,a\right )}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}\right )}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}-\frac {\left (3\,A\,b+5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,b^2+30\,A\,B\,a\,b+25\,B^2\,a^2\right )}{64\,a^2\,b}+\frac {\left (3\,A\,b^2+5\,B\,a\,b\right )\,\left (3\,A\,b+5\,B\,a\right )}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}\right )}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}}\right )\,\left (3\,A\,b+5\,B\,a\right )\,1{}\mathrm {i}}{32\,{\left (-a\right )}^{7/4}\,b^{9/4}}+\frac {\mathrm {atan}\left (\frac {\frac {\left (3\,A\,b+5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,b^2+30\,A\,B\,a\,b+25\,B^2\,a^2\right )}{64\,a^2\,b}-\frac {\left (3\,A\,b^2+5\,B\,a\,b\right )\,\left (3\,A\,b+5\,B\,a\right )\,1{}\mathrm {i}}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}\right )}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}+\frac {\left (3\,A\,b+5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,b^2+30\,A\,B\,a\,b+25\,B^2\,a^2\right )}{64\,a^2\,b}+\frac {\left (3\,A\,b^2+5\,B\,a\,b\right )\,\left (3\,A\,b+5\,B\,a\right )\,1{}\mathrm {i}}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}\right )}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}}{\frac {\left (3\,A\,b+5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,b^2+30\,A\,B\,a\,b+25\,B^2\,a^2\right )}{64\,a^2\,b}-\frac {\left (3\,A\,b^2+5\,B\,a\,b\right )\,\left (3\,A\,b+5\,B\,a\right )\,1{}\mathrm {i}}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}\right )\,1{}\mathrm {i}}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}-\frac {\left (3\,A\,b+5\,B\,a\right )\,\left (\frac {\sqrt {x}\,\left (9\,A^2\,b^2+30\,A\,B\,a\,b+25\,B^2\,a^2\right )}{64\,a^2\,b}+\frac {\left (3\,A\,b^2+5\,B\,a\,b\right )\,\left (3\,A\,b+5\,B\,a\right )\,1{}\mathrm {i}}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}\right )\,1{}\mathrm {i}}{64\,{\left (-a\right )}^{7/4}\,b^{9/4}}}\right )\,\left (3\,A\,b+5\,B\,a\right )}{32\,{\left (-a\right )}^{7/4}\,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)^3,x)

[Out]

(atan((((3*A*b + 5*B*a)*((x^(1/2)*(9*A^2*b^2 + 25*B^2*a^2 + 30*A*B*a*b))/(64*a^2*b) - ((3*A*b^2 + 5*B*a*b)*(3*
A*b + 5*B*a))/(64*(-a)^(7/4)*b^(9/4)))*1i)/(64*(-a)^(7/4)*b^(9/4)) + ((3*A*b + 5*B*a)*((x^(1/2)*(9*A^2*b^2 + 2
5*B^2*a^2 + 30*A*B*a*b))/(64*a^2*b) + ((3*A*b^2 + 5*B*a*b)*(3*A*b + 5*B*a))/(64*(-a)^(7/4)*b^(9/4)))*1i)/(64*(
-a)^(7/4)*b^(9/4)))/(((3*A*b + 5*B*a)*((x^(1/2)*(9*A^2*b^2 + 25*B^2*a^2 + 30*A*B*a*b))/(64*a^2*b) - ((3*A*b^2
+ 5*B*a*b)*(3*A*b + 5*B*a))/(64*(-a)^(7/4)*b^(9/4))))/(64*(-a)^(7/4)*b^(9/4)) - ((3*A*b + 5*B*a)*((x^(1/2)*(9*
A^2*b^2 + 25*B^2*a^2 + 30*A*B*a*b))/(64*a^2*b) + ((3*A*b^2 + 5*B*a*b)*(3*A*b + 5*B*a))/(64*(-a)^(7/4)*b^(9/4))
))/(64*(-a)^(7/4)*b^(9/4))))*(3*A*b + 5*B*a)*1i)/(32*(-a)^(7/4)*b^(9/4)) - ((x^(1/2)*(3*A*b + 5*B*a))/(16*b^2)
- (x^(5/2)*(A*b - 9*B*a))/(16*a*b))/(a^2 + b^2*x^4 + 2*a*b*x^2) + (atan((((3*A*b + 5*B*a)*((x^(1/2)*(9*A^2*b^
2 + 25*B^2*a^2 + 30*A*B*a*b))/(64*a^2*b) - ((3*A*b^2 + 5*B*a*b)*(3*A*b + 5*B*a)*1i)/(64*(-a)^(7/4)*b^(9/4))))/
(64*(-a)^(7/4)*b^(9/4)) + ((3*A*b + 5*B*a)*((x^(1/2)*(9*A^2*b^2 + 25*B^2*a^2 + 30*A*B*a*b))/(64*a^2*b) + ((3*A
*b^2 + 5*B*a*b)*(3*A*b + 5*B*a)*1i)/(64*(-a)^(7/4)*b^(9/4))))/(64*(-a)^(7/4)*b^(9/4)))/(((3*A*b + 5*B*a)*((x^(
1/2)*(9*A^2*b^2 + 25*B^2*a^2 + 30*A*B*a*b))/(64*a^2*b) - ((3*A*b^2 + 5*B*a*b)*(3*A*b + 5*B*a)*1i)/(64*(-a)^(7/
4)*b^(9/4)))*1i)/(64*(-a)^(7/4)*b^(9/4)) - ((3*A*b + 5*B*a)*((x^(1/2)*(9*A^2*b^2 + 25*B^2*a^2 + 30*A*B*a*b))/(
64*a^2*b) + ((3*A*b^2 + 5*B*a*b)*(3*A*b + 5*B*a)*1i)/(64*(-a)^(7/4)*b^(9/4)))*1i)/(64*(-a)^(7/4)*b^(9/4))))*(3
*A*b + 5*B*a))/(32*(-a)^(7/4)*b^(9/4))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \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)**3,x)

[Out]

Timed out

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