3.3.31 \(\int \frac {x^{3/2}}{(b x^2+c x^4)^3} \, dx\)

Optimal. Leaf size=264 \[ -\frac {165 c^{7/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}+\frac {165 c^{7/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}-\frac {165 c^{7/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{19/4}}+\frac {165 c^{7/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{19/4}}+\frac {55 c}{16 b^4 x^{3/2}}-\frac {165}{112 b^3 x^{7/2}}+\frac {15}{16 b^2 x^{7/2} \left (b+c x^2\right )}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2} \]

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Rubi [A]  time = 0.23, antiderivative size = 264, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 10, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {1584, 290, 325, 329, 211, 1165, 628, 1162, 617, 204} \begin {gather*} -\frac {165 c^{7/4} \log \left (-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}+\frac {165 c^{7/4} \log \left (\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {b}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}-\frac {165 c^{7/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{19/4}}+\frac {165 c^{7/4} \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}+1\right )}{32 \sqrt {2} b^{19/4}}+\frac {55 c}{16 b^4 x^{3/2}}+\frac {15}{16 b^2 x^{7/2} \left (b+c x^2\right )}-\frac {165}{112 b^3 x^{7/2}}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(b*x^2 + c*x^4)^3,x]

[Out]

-165/(112*b^3*x^(7/2)) + (55*c)/(16*b^4*x^(3/2)) + 1/(4*b*x^(7/2)*(b + c*x^2)^2) + 15/(16*b^2*x^(7/2)*(b + c*x
^2)) - (165*c^(7/4)*ArcTan[1 - (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(19/4)) + (165*c^(7/4)*ArcTan
[1 + (Sqrt[2]*c^(1/4)*Sqrt[x])/b^(1/4)])/(32*Sqrt[2]*b^(19/4)) - (165*c^(7/4)*Log[Sqrt[b] - Sqrt[2]*b^(1/4)*c^
(1/4)*Sqrt[x] + Sqrt[c]*x])/(64*Sqrt[2]*b^(19/4)) + (165*c^(7/4)*Log[Sqrt[b] + Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x]
+ Sqrt[c]*x])/(64*Sqrt[2]*b^(19/4))

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 290

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

Rule 325

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

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rubi steps

\begin {align*} \int \frac {x^{3/2}}{\left (b x^2+c x^4\right )^3} \, dx &=\int \frac {1}{x^{9/2} \left (b+c x^2\right )^3} \, dx\\ &=\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}+\frac {15 \int \frac {1}{x^{9/2} \left (b+c x^2\right )^2} \, dx}{8 b}\\ &=\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}+\frac {15}{16 b^2 x^{7/2} \left (b+c x^2\right )}+\frac {165 \int \frac {1}{x^{9/2} \left (b+c x^2\right )} \, dx}{32 b^2}\\ &=-\frac {165}{112 b^3 x^{7/2}}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}+\frac {15}{16 b^2 x^{7/2} \left (b+c x^2\right )}-\frac {(165 c) \int \frac {1}{x^{5/2} \left (b+c x^2\right )} \, dx}{32 b^3}\\ &=-\frac {165}{112 b^3 x^{7/2}}+\frac {55 c}{16 b^4 x^{3/2}}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}+\frac {15}{16 b^2 x^{7/2} \left (b+c x^2\right )}+\frac {\left (165 c^2\right ) \int \frac {1}{\sqrt {x} \left (b+c x^2\right )} \, dx}{32 b^4}\\ &=-\frac {165}{112 b^3 x^{7/2}}+\frac {55 c}{16 b^4 x^{3/2}}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}+\frac {15}{16 b^2 x^{7/2} \left (b+c x^2\right )}+\frac {\left (165 c^2\right ) \operatorname {Subst}\left (\int \frac {1}{b+c x^4} \, dx,x,\sqrt {x}\right )}{16 b^4}\\ &=-\frac {165}{112 b^3 x^{7/2}}+\frac {55 c}{16 b^4 x^{3/2}}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}+\frac {15}{16 b^2 x^{7/2} \left (b+c x^2\right )}+\frac {\left (165 c^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}-\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^{9/2}}+\frac {\left (165 c^2\right ) \operatorname {Subst}\left (\int \frac {\sqrt {b}+\sqrt {c} x^2}{b+c x^4} \, dx,x,\sqrt {x}\right )}{32 b^{9/2}}\\ &=-\frac {165}{112 b^3 x^{7/2}}+\frac {55 c}{16 b^4 x^{3/2}}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}+\frac {15}{16 b^2 x^{7/2} \left (b+c x^2\right )}+\frac {\left (165 c^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {b}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{64 b^{9/2}}+\frac {\left (165 c^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\frac {\sqrt {b}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}+x^2} \, dx,x,\sqrt {x}\right )}{64 b^{9/2}}-\frac {\left (165 c^{7/4}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}+2 x}{-\frac {\sqrt {b}}{\sqrt {c}}-\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} b^{19/4}}-\frac {\left (165 c^{7/4}\right ) \operatorname {Subst}\left (\int \frac {\frac {\sqrt {2} \sqrt [4]{b}}{\sqrt [4]{c}}-2 x}{-\frac {\sqrt {b}}{\sqrt {c}}+\frac {\sqrt {2} \sqrt [4]{b} x}{\sqrt [4]{c}}-x^2} \, dx,x,\sqrt {x}\right )}{64 \sqrt {2} b^{19/4}}\\ &=-\frac {165}{112 b^3 x^{7/2}}+\frac {55 c}{16 b^4 x^{3/2}}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}+\frac {15}{16 b^2 x^{7/2} \left (b+c x^2\right )}-\frac {165 c^{7/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}+\frac {165 c^{7/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}+\frac {\left (165 c^{7/4}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{19/4}}-\frac {\left (165 c^{7/4}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{19/4}}\\ &=-\frac {165}{112 b^3 x^{7/2}}+\frac {55 c}{16 b^4 x^{3/2}}+\frac {1}{4 b x^{7/2} \left (b+c x^2\right )^2}+\frac {15}{16 b^2 x^{7/2} \left (b+c x^2\right )}-\frac {165 c^{7/4} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{19/4}}+\frac {165 c^{7/4} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right )}{32 \sqrt {2} b^{19/4}}-\frac {165 c^{7/4} \log \left (\sqrt {b}-\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}+\frac {165 c^{7/4} \log \left (\sqrt {b}+\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}+\sqrt {c} x\right )}{64 \sqrt {2} b^{19/4}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 29, normalized size = 0.11 \begin {gather*} -\frac {2 \, _2F_1\left (-\frac {7}{4},3;-\frac {3}{4};-\frac {c x^2}{b}\right )}{7 b^3 x^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(b*x^2 + c*x^4)^3,x]

[Out]

(-2*Hypergeometric2F1[-7/4, 3, -3/4, -((c*x^2)/b)])/(7*b^3*x^(7/2))

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IntegrateAlgebraic [A]  time = 0.47, size = 171, normalized size = 0.65 \begin {gather*} -\frac {165 c^{7/4} \tan ^{-1}\left (\frac {\frac {\sqrt [4]{b}}{\sqrt {2} \sqrt [4]{c}}-\frac {\sqrt [4]{c} x}{\sqrt {2} \sqrt [4]{b}}}{\sqrt {x}}\right )}{32 \sqrt {2} b^{19/4}}+\frac {165 c^{7/4} \tanh ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt [4]{c} \sqrt {x}}{\sqrt {b}+\sqrt {c} x}\right )}{32 \sqrt {2} b^{19/4}}+\frac {-32 b^3+160 b^2 c x^2+605 b c^2 x^4+385 c^3 x^6}{112 b^4 x^{7/2} \left (b+c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(3/2)/(b*x^2 + c*x^4)^3,x]

[Out]

(-32*b^3 + 160*b^2*c*x^2 + 605*b*c^2*x^4 + 385*c^3*x^6)/(112*b^4*x^(7/2)*(b + c*x^2)^2) - (165*c^(7/4)*ArcTan[
(b^(1/4)/(Sqrt[2]*c^(1/4)) - (c^(1/4)*x)/(Sqrt[2]*b^(1/4)))/Sqrt[x]])/(32*Sqrt[2]*b^(19/4)) + (165*c^(7/4)*Arc
Tanh[(Sqrt[2]*b^(1/4)*c^(1/4)*Sqrt[x])/(Sqrt[b] + Sqrt[c]*x)])/(32*Sqrt[2]*b^(19/4))

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fricas [A]  time = 0.60, size = 300, normalized size = 1.14 \begin {gather*} \frac {4620 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {1}{4}} \arctan \left (-\frac {b^{14} c^{2} \sqrt {x} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {3}{4}} - \sqrt {b^{10} \sqrt {-\frac {c^{7}}{b^{19}}} + c^{4} x} b^{14} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {3}{4}}}{c^{7}}\right ) + 1155 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {1}{4}} \log \left (165 \, b^{5} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {1}{4}} + 165 \, c^{2} \sqrt {x}\right ) - 1155 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {1}{4}} \log \left (-165 \, b^{5} \left (-\frac {c^{7}}{b^{19}}\right )^{\frac {1}{4}} + 165 \, c^{2} \sqrt {x}\right ) + 4 \, {\left (385 \, c^{3} x^{6} + 605 \, b c^{2} x^{4} + 160 \, b^{2} c x^{2} - 32 \, b^{3}\right )} \sqrt {x}}{448 \, {\left (b^{4} c^{2} x^{8} + 2 \, b^{5} c x^{6} + b^{6} x^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

1/448*(4620*(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4)*(-c^7/b^19)^(1/4)*arctan(-(b^14*c^2*sqrt(x)*(-c^7/b^19)^(3/4
) - sqrt(b^10*sqrt(-c^7/b^19) + c^4*x)*b^14*(-c^7/b^19)^(3/4))/c^7) + 1155*(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^
4)*(-c^7/b^19)^(1/4)*log(165*b^5*(-c^7/b^19)^(1/4) + 165*c^2*sqrt(x)) - 1155*(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*
x^4)*(-c^7/b^19)^(1/4)*log(-165*b^5*(-c^7/b^19)^(1/4) + 165*c^2*sqrt(x)) + 4*(385*c^3*x^6 + 605*b*c^2*x^4 + 16
0*b^2*c*x^2 - 32*b^3)*sqrt(x))/(b^4*c^2*x^8 + 2*b^5*c*x^6 + b^6*x^4)

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giac [A]  time = 0.21, size = 224, normalized size = 0.85 \begin {gather*} \frac {165 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{5}} + \frac {165 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {b}{c}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {b}{c}\right )^{\frac {1}{4}}}\right )}{64 \, b^{5}} + \frac {165 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \log \left (\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{5}} - \frac {165 \, \sqrt {2} \left (b c^{3}\right )^{\frac {1}{4}} c \log \left (-\sqrt {2} \sqrt {x} \left (\frac {b}{c}\right )^{\frac {1}{4}} + x + \sqrt {\frac {b}{c}}\right )}{128 \, b^{5}} + \frac {23 \, c^{3} x^{\frac {5}{2}} + 27 \, b c^{2} \sqrt {x}}{16 \, {\left (c x^{2} + b\right )}^{2} b^{4}} + \frac {2 \, {\left (7 \, c x^{2} - b\right )}}{7 \, b^{4} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

165/64*sqrt(2)*(b*c^3)^(1/4)*c*arctan(1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) + 2*sqrt(x))/(b/c)^(1/4))/b^5 + 165/64*
sqrt(2)*(b*c^3)^(1/4)*c*arctan(-1/2*sqrt(2)*(sqrt(2)*(b/c)^(1/4) - 2*sqrt(x))/(b/c)^(1/4))/b^5 + 165/128*sqrt(
2)*(b*c^3)^(1/4)*c*log(sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^5 - 165/128*sqrt(2)*(b*c^3)^(1/4)*c*log(
-sqrt(2)*sqrt(x)*(b/c)^(1/4) + x + sqrt(b/c))/b^5 + 1/16*(23*c^3*x^(5/2) + 27*b*c^2*sqrt(x))/((c*x^2 + b)^2*b^
4) + 2/7*(7*c*x^2 - b)/(b^4*x^(7/2))

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maple [A]  time = 0.02, size = 198, normalized size = 0.75 \begin {gather*} \frac {23 c^{3} x^{\frac {5}{2}}}{16 \left (c \,x^{2}+b \right )^{2} b^{4}}+\frac {27 c^{2} \sqrt {x}}{16 \left (c \,x^{2}+b \right )^{2} b^{3}}+\frac {165 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}-1\right )}{64 b^{5}}+\frac {165 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, c^{2} \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {b}{c}\right )^{\frac {1}{4}}}+1\right )}{64 b^{5}}+\frac {165 \left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, c^{2} \ln \left (\frac {x +\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}{x -\left (\frac {b}{c}\right )^{\frac {1}{4}} \sqrt {2}\, \sqrt {x}+\sqrt {\frac {b}{c}}}\right )}{128 b^{5}}+\frac {2 c}{b^{4} x^{\frac {3}{2}}}-\frac {2}{7 b^{3} x^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(c*x^4+b*x^2)^3,x)

[Out]

23/16/b^4*c^3/(c*x^2+b)^2*x^(5/2)+27/16/b^3*c^2/(c*x^2+b)^2*x^(1/2)+165/128/b^5*c^2*(b/c)^(1/4)*2^(1/2)*ln((x+
(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2))/(x-(b/c)^(1/4)*2^(1/2)*x^(1/2)+(b/c)^(1/2)))+165/64/b^5*c^2*(b/c)^(1/
4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)*x^(1/2)+1)+165/64/b^5*c^2*(b/c)^(1/4)*2^(1/2)*arctan(2^(1/2)/(b/c)^(1/4)
*x^(1/2)-1)-2/7/b^3/x^(7/2)+2*c/b^4/x^(3/2)

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maxima [A]  time = 2.93, size = 246, normalized size = 0.93 \begin {gather*} \frac {385 \, c^{3} x^{6} + 605 \, b c^{2} x^{4} + 160 \, b^{2} c x^{2} - 32 \, b^{3}}{112 \, {\left (b^{4} c^{2} x^{\frac {15}{2}} + 2 \, b^{5} c x^{\frac {11}{2}} + b^{6} x^{\frac {7}{2}}\right )}} + \frac {165 \, {\left (\frac {2 \, \sqrt {2} c^{2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} + 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {2 \, \sqrt {2} c^{2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} - 2 \, \sqrt {c} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {b} \sqrt {c}}}\right )}{\sqrt {b} \sqrt {\sqrt {b} \sqrt {c}}} + \frac {\sqrt {2} c^{\frac {7}{4}} \log \left (\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}}} - \frac {\sqrt {2} c^{\frac {7}{4}} \log \left (-\sqrt {2} b^{\frac {1}{4}} c^{\frac {1}{4}} \sqrt {x} + \sqrt {c} x + \sqrt {b}\right )}{b^{\frac {3}{4}}}\right )}}{128 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)/(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

1/112*(385*c^3*x^6 + 605*b*c^2*x^4 + 160*b^2*c*x^2 - 32*b^3)/(b^4*c^2*x^(15/2) + 2*b^5*c*x^(11/2) + b^6*x^(7/2
)) + 165/128*(2*sqrt(2)*c^2*arctan(1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) + 2*sqrt(c)*sqrt(x))/sqrt(sqrt(b)*sqrt
(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + 2*sqrt(2)*c^2*arctan(-1/2*sqrt(2)*(sqrt(2)*b^(1/4)*c^(1/4) - 2*sqrt(c)
*sqrt(x))/sqrt(sqrt(b)*sqrt(c)))/(sqrt(b)*sqrt(sqrt(b)*sqrt(c))) + sqrt(2)*c^(7/4)*log(sqrt(2)*b^(1/4)*c^(1/4)
*sqrt(x) + sqrt(c)*x + sqrt(b))/b^(3/4) - sqrt(2)*c^(7/4)*log(-sqrt(2)*b^(1/4)*c^(1/4)*sqrt(x) + sqrt(c)*x + s
qrt(b))/b^(3/4))/b^4

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mupad [B]  time = 4.35, size = 109, normalized size = 0.41 \begin {gather*} \frac {\frac {10\,c\,x^2}{7\,b^2}-\frac {2}{7\,b}+\frac {605\,c^2\,x^4}{112\,b^3}+\frac {55\,c^3\,x^6}{16\,b^4}}{b^2\,x^{7/2}+c^2\,x^{15/2}+2\,b\,c\,x^{11/2}}+\frac {165\,{\left (-c\right )}^{7/4}\,\mathrm {atan}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{32\,b^{19/4}}+\frac {165\,{\left (-c\right )}^{7/4}\,\mathrm {atanh}\left (\frac {{\left (-c\right )}^{1/4}\,\sqrt {x}}{b^{1/4}}\right )}{32\,b^{19/4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(b*x^2 + c*x^4)^3,x)

[Out]

((10*c*x^2)/(7*b^2) - 2/(7*b) + (605*c^2*x^4)/(112*b^3) + (55*c^3*x^6)/(16*b^4))/(b^2*x^(7/2) + c^2*x^(15/2) +
2*b*c*x^(11/2)) + (165*(-c)^(7/4)*atan(((-c)^(1/4)*x^(1/2))/b^(1/4)))/(32*b^(19/4)) + (165*(-c)^(7/4)*atanh((
(-c)^(1/4)*x^(1/2))/b^(1/4)))/(32*b^(19/4))

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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)/(c*x**4+b*x**2)**3,x)

[Out]

Timed out

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