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

Optimal. Leaf size=255 \[ \frac {\sqrt [4]{a} (A b-a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} b^{9/4}}-\frac {\sqrt [4]{a} (A b-a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} b^{9/4}}+\frac {\sqrt [4]{a} (A b-a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} b^{9/4}}-\frac {\sqrt [4]{a} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} b^{9/4}}+\frac {2 \sqrt {x} (A b-a B)}{b^2}+\frac {2 B x^{5/2}}{5 b} \]

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Rubi [A]  time = 0.20, antiderivative size = 255, 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 = {459, 321, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} \frac {2 \sqrt {x} (A b-a B)}{b^2}+\frac {\sqrt [4]{a} (A b-a B) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} b^{9/4}}-\frac {\sqrt [4]{a} (A b-a B) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{2 \sqrt {2} b^{9/4}}+\frac {\sqrt [4]{a} (A b-a B) \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} b^{9/4}}-\frac {\sqrt [4]{a} (A b-a B) \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )}{\sqrt {2} b^{9/4}}+\frac {2 B x^{5/2}}{5 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(A*b - a*B)*Sqrt[x])/b^2 + (2*B*x^(5/2))/(5*b) + (a^(1/4)*(A*b - a*B)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/
a^(1/4)])/(Sqrt[2]*b^(9/4)) - (a^(1/4)*(A*b - a*B)*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(Sqrt[2]*b^(
9/4)) + (a^(1/4)*(A*b - a*B)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*b^(9/4)) -
(a^(1/4)*(A*b - a*B)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(2*Sqrt[2]*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 459

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

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

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Mathematica [A]  time = 0.20, size = 208, normalized size = 0.82 \begin {gather*} \frac {40 \sqrt {x} (A b-a B)+\frac {5 \sqrt {2} \sqrt [4]{a} (A b-a B) \left (\log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )-\log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )\right )}{\sqrt [4]{b}}+\frac {10 \sqrt {2} \sqrt [4]{a} (A b-a B) \left (\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )-\tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right )\right )}{\sqrt [4]{b}}+8 b B x^{5/2}}{20 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(40*(A*b - a*B)*Sqrt[x] + 8*b*B*x^(5/2) + (10*Sqrt[2]*a^(1/4)*(A*b - a*B)*(ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x]
)/a^(1/4)] - ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)]))/b^(1/4) + (5*Sqrt[2]*a^(1/4)*(A*b - a*B)*(Log[Sqr
t[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x] - Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*
x]))/b^(1/4))/(20*b^2)

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IntegrateAlgebraic [A]  time = 0.19, size = 158, normalized size = 0.62 \begin {gather*} -\frac {\left (a^{5/4} B-\sqrt [4]{a} A b\right ) \tan ^{-1}\left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{\sqrt {2} b^{9/4}}+\frac {\left (a^{5/4} B-\sqrt [4]{a} A b\right ) \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt {2} b^{9/4}}+\frac {2 \sqrt {x} \left (-5 a B+5 A b+b B x^2\right )}{5 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.29, size = 660, normalized size = 2.59 \begin {gather*} -\frac {20 \, b^{2} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {1}{4}} \arctan \left (\frac {\sqrt {b^{4} \sqrt {-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}} + {\left (B^{2} a^{2} - 2 \, A B a b + A^{2} b^{2}\right )} x} b^{7} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {3}{4}} + {\left (B a b^{7} - A b^{8}\right )} \sqrt {x} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {3}{4}}}{B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}\right ) + 5 \, b^{2} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (b^{2} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {1}{4}} - {\left (B a - A b\right )} \sqrt {x}\right ) - 5 \, b^{2} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {1}{4}} \log \left (-b^{2} \left (-\frac {B^{4} a^{5} - 4 \, A B^{3} a^{4} b + 6 \, A^{2} B^{2} a^{3} b^{2} - 4 \, A^{3} B a^{2} b^{3} + A^{4} a b^{4}}{b^{9}}\right )^{\frac {1}{4}} - {\left (B a - A b\right )} \sqrt {x}\right ) - 4 \, {\left (B b x^{2} - 5 \, B a + 5 \, A b\right )} \sqrt {x}}{10 \, 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),x, algorithm="fricas")

[Out]

-1/10*(20*b^2*(-(B^4*a^5 - 4*A*B^3*a^4*b + 6*A^2*B^2*a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9)^(1/4)*arctan(
(sqrt(b^4*sqrt(-(B^4*a^5 - 4*A*B^3*a^4*b + 6*A^2*B^2*a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9) + (B^2*a^2 -
2*A*B*a*b + A^2*b^2)*x)*b^7*(-(B^4*a^5 - 4*A*B^3*a^4*b + 6*A^2*B^2*a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9)
^(3/4) + (B*a*b^7 - A*b^8)*sqrt(x)*(-(B^4*a^5 - 4*A*B^3*a^4*b + 6*A^2*B^2*a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^
4)/b^9)^(3/4))/(B^4*a^5 - 4*A*B^3*a^4*b + 6*A^2*B^2*a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)) + 5*b^2*(-(B^4*a^5
- 4*A*B^3*a^4*b + 6*A^2*B^2*a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9)^(1/4)*log(b^2*(-(B^4*a^5 - 4*A*B^3*a^
4*b + 6*A^2*B^2*a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9)^(1/4) - (B*a - A*b)*sqrt(x)) - 5*b^2*(-(B^4*a^5 -
4*A*B^3*a^4*b + 6*A^2*B^2*a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9)^(1/4)*log(-b^2*(-(B^4*a^5 - 4*A*B^3*a^4*
b + 6*A^2*B^2*a^3*b^2 - 4*A^3*B*a^2*b^3 + A^4*a*b^4)/b^9)^(1/4) - (B*a - A*b)*sqrt(x)) - 4*(B*b*x^2 - 5*B*a +
5*A*b)*sqrt(x))/b^2

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giac [A]  time = 0.41, size = 263, normalized size = 1.03 \begin {gather*} \frac {\sqrt {2} {\left (\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 )}{2 \, b^{3}} + \frac {\sqrt {2} {\left (\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 )}{2 \, b^{3}} + \frac {\sqrt {2} {\left (\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 )}{4 \, b^{3}} - \frac {\sqrt {2} {\left (\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 )}{4 \, b^{3}} + \frac {2 \, {\left (B b^{4} x^{\frac {5}{2}} - 5 \, B a b^{3} \sqrt {x} + 5 \, A b^{4} \sqrt {x}\right )}}{5 \, b^{5}} \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),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 299, normalized size = 1.17 \begin {gather*} \frac {2 B \,x^{\frac {5}{2}}}{5 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 )}{2 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 )}{2 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 )}{4 b}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, B a \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )}{2 b^{2}}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, B a \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )}{2 b^{2}}+\frac {\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, B 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 )}{4 b^{2}}+\frac {2 A \sqrt {x}}{b}-\frac {2 B a \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),x)

[Out]

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

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maxima [A]  time = 2.24, size = 235, normalized size = 0.92 \begin {gather*} \frac {{\left (\frac {2 \, \sqrt {2} {\left (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 (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 (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 (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}}}\right )} a}{4 \, b^{2}} + \frac {2 \, {\left (B b x^{\frac {5}{2}} - 5 \, {\left (B a - A b\right )} \sqrt {x}\right )}}{5 \, 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),x, algorithm="maxima")

[Out]

1/4*(2*sqrt(2)*(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)*(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))) + sqrt(2)*(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)*(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)))*a/b^2 + 2/5*(B*b*x^(5/2) - 5*(B*a - A*b)*sqrt(x))/b^2

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mupad [B]  time = 0.38, size = 789, normalized size = 3.09 \begin {gather*} \sqrt {x}\,\left (\frac {2\,A}{b}-\frac {2\,B\,a}{b^2}\right )+\frac {2\,B\,x^{5/2}}{5\,b}-\frac {{\left (-a\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}-\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )}{2\,b^{9/4}}\right )\,1{}\mathrm {i}}{2\,b^{9/4}}+\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}+\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )}{2\,b^{9/4}}\right )\,1{}\mathrm {i}}{2\,b^{9/4}}}{\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}-\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )}{2\,b^{9/4}}\right )}{2\,b^{9/4}}-\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}+\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )}{2\,b^{9/4}}\right )}{2\,b^{9/4}}}\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{b^{9/4}}-\frac {{\left (-a\right )}^{1/4}\,\mathrm {atan}\left (\frac {\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}-\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{2\,b^{9/4}}\right )}{2\,b^{9/4}}+\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}+\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{2\,b^{9/4}}\right )}{2\,b^{9/4}}}{\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}-\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{2\,b^{9/4}}\right )\,1{}\mathrm {i}}{2\,b^{9/4}}-\frac {{\left (-a\right )}^{1/4}\,\left (A\,b-B\,a\right )\,\left (\frac {16\,\sqrt {x}\,\left (A^2\,a^2\,b^2-2\,A\,B\,a^3\,b+B^2\,a^4\right )}{b}+\frac {{\left (-a\right )}^{1/4}\,\left (32\,A\,a^2\,b^2-32\,B\,a^3\,b\right )\,\left (A\,b-B\,a\right )\,1{}\mathrm {i}}{2\,b^{9/4}}\right )\,1{}\mathrm {i}}{2\,b^{9/4}}}\right )\,\left (A\,b-B\,a\right )}{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),x)

[Out]

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

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sympy [A]  time = 16.60, size = 393, normalized size = 1.54 \begin {gather*} \begin {cases} \tilde {\infty } \left (2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}\right ) & \text {for}\: a = 0 \wedge b = 0 \\\frac {\frac {2 A x^{\frac {5}{2}}}{5} + \frac {2 B x^{\frac {9}{2}}}{9}}{a} & \text {for}\: b = 0 \\\frac {2 A \sqrt {x} + \frac {2 B x^{\frac {5}{2}}}{5}}{b} & \text {for}\: a = 0 \\\frac {\sqrt [4]{-1} A \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 b} - \frac {\sqrt [4]{-1} A \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 b} + \frac {\sqrt [4]{-1} A \sqrt [4]{a} \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 )}}{b} + \frac {2 A \sqrt {x}}{b} - \frac {\sqrt [4]{-1} B a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}} \log {\left (- \sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 b^{2}} + \frac {\sqrt [4]{-1} B a^{\frac {5}{4}} \sqrt [4]{\frac {1}{b}} \log {\left (\sqrt [4]{-1} \sqrt [4]{a} \sqrt [4]{\frac {1}{b}} + \sqrt {x} \right )}}{2 b^{2}} - \frac {\sqrt [4]{-1} B a^{\frac {5}{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 )}}{b^{2}} - \frac {2 B a \sqrt {x}}{b^{2}} + \frac {2 B x^{\frac {5}{2}}}{5 b} & \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),x)

[Out]

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

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