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

Optimal. Leaf size=308 \[ \frac{3 \log \left (-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{11/8} c^{5/8}}-\frac{3 \log \left (\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{11/8} c^{5/8}}+\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{11/8} c^{5/8}}-\frac{3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}+1\right )}{16 \sqrt{2} (-a)^{11/8} c^{5/8}}+\frac{3 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{11/8} c^{5/8}}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{11/8} c^{5/8}}+\frac{x^{5/2}}{4 a \left (a+c x^4\right )} \]

[Out]

x^(5/2)/(4*a*(a + c*x^4)) + (3*ArcTan[1 - (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*Sqrt[2]*(-a)^(11/8)*c^(5/
8)) - (3*ArcTan[1 + (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*Sqrt[2]*(-a)^(11/8)*c^(5/8)) + (3*ArcTan[(c^(1/
8)*Sqrt[x])/(-a)^(1/8)])/(16*(-a)^(11/8)*c^(5/8)) + (3*ArcTanh[(c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*(-a)^(11/8)*
c^(5/8)) + (3*Log[(-a)^(1/4) - Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt[x] + c^(1/4)*x])/(32*Sqrt[2]*(-a)^(11/8)*c^(5/8
)) - (3*Log[(-a)^(1/4) + Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt[x] + c^(1/4)*x])/(32*Sqrt[2]*(-a)^(11/8)*c^(5/8))

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Rubi [A]  time = 0.260625, antiderivative size = 308, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 12, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.8, Rules used = {290, 329, 301, 211, 1165, 628, 1162, 617, 204, 212, 208, 205} \[ \frac{3 \log \left (-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{11/8} c^{5/8}}-\frac{3 \log \left (\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{-a}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{11/8} c^{5/8}}+\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{11/8} c^{5/8}}-\frac{3 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}+1\right )}{16 \sqrt{2} (-a)^{11/8} c^{5/8}}+\frac{3 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{11/8} c^{5/8}}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{11/8} c^{5/8}}+\frac{x^{5/2}}{4 a \left (a+c x^4\right )} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(a + c*x^4)^2,x]

[Out]

x^(5/2)/(4*a*(a + c*x^4)) + (3*ArcTan[1 - (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*Sqrt[2]*(-a)^(11/8)*c^(5/
8)) - (3*ArcTan[1 + (Sqrt[2]*c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*Sqrt[2]*(-a)^(11/8)*c^(5/8)) + (3*ArcTan[(c^(1/
8)*Sqrt[x])/(-a)^(1/8)])/(16*(-a)^(11/8)*c^(5/8)) + (3*ArcTanh[(c^(1/8)*Sqrt[x])/(-a)^(1/8)])/(16*(-a)^(11/8)*
c^(5/8)) + (3*Log[(-a)^(1/4) - Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt[x] + c^(1/4)*x])/(32*Sqrt[2]*(-a)^(11/8)*c^(5/8
)) - (3*Log[(-a)^(1/4) + Sqrt[2]*(-a)^(1/8)*c^(1/8)*Sqrt[x] + c^(1/4)*x])/(32*Sqrt[2]*(-a)^(11/8)*c^(5/8))

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 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 301

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(
a/b), 2]]}, Dist[s/(2*b), Int[x^(m - n/2)/(r + s*x^(n/2)), x], x] - Dist[s/(2*b), Int[x^(m - n/2)/(r - s*x^(n/
2)), x], x]] /; FreeQ[{a, b}, x] && IGtQ[n/4, 0] && IGtQ[m, 0] && LeQ[n/2, m] && LtQ[m, n] &&  !GtQ[a/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 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 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 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 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 212

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

Rule 208

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

Rule 205

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

Rubi steps

\begin{align*} \int \frac{x^{3/2}}{\left (a+c x^4\right )^2} \, dx &=\frac{x^{5/2}}{4 a \left (a+c x^4\right )}+\frac{3 \int \frac{x^{3/2}}{a+c x^4} \, dx}{8 a}\\ &=\frac{x^{5/2}}{4 a \left (a+c x^4\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{x^4}{a+c x^8} \, dx,x,\sqrt{x}\right )}{4 a}\\ &=\frac{x^{5/2}}{4 a \left (a+c x^4\right )}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a}-\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{8 a \sqrt{c}}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a}+\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{8 a \sqrt{c}}\\ &=\frac{x^{5/2}}{4 a \left (a+c x^4\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{-a}-\sqrt [4]{c} x^2} \, dx,x,\sqrt{x}\right )}{16 (-a)^{5/4} \sqrt{c}}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{\sqrt [4]{-a}+\sqrt [4]{c} x^2} \, dx,x,\sqrt{x}\right )}{16 (-a)^{5/4} \sqrt{c}}-\frac{3 \operatorname{Subst}\left (\int \frac{\sqrt [4]{-a}-\sqrt [4]{c} x^2}{\sqrt{-a}+\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{16 (-a)^{5/4} \sqrt{c}}-\frac{3 \operatorname{Subst}\left (\int \frac{\sqrt [4]{-a}+\sqrt [4]{c} x^2}{\sqrt{-a}+\sqrt{c} x^4} \, dx,x,\sqrt{x}\right )}{16 (-a)^{5/4} \sqrt{c}}\\ &=\frac{x^{5/2}}{4 a \left (a+c x^4\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{11/8} c^{5/8}}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{11/8} c^{5/8}}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt [4]{-a}}{\sqrt [4]{c}}-\frac{\sqrt{2} \sqrt [8]{-a} x}{\sqrt [8]{c}}+x^2} \, dx,x,\sqrt{x}\right )}{32 (-a)^{5/4} c^{3/4}}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{\frac{\sqrt [4]{-a}}{\sqrt [4]{c}}+\frac{\sqrt{2} \sqrt [8]{-a} x}{\sqrt [8]{c}}+x^2} \, dx,x,\sqrt{x}\right )}{32 (-a)^{5/4} c^{3/4}}+\frac{3 \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [8]{-a}}{\sqrt [8]{c}}+2 x}{-\frac{\sqrt [4]{-a}}{\sqrt [4]{c}}-\frac{\sqrt{2} \sqrt [8]{-a} x}{\sqrt [8]{c}}-x^2} \, dx,x,\sqrt{x}\right )}{32 \sqrt{2} (-a)^{11/8} c^{5/8}}+\frac{3 \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt [8]{-a}}{\sqrt [8]{c}}-2 x}{-\frac{\sqrt [4]{-a}}{\sqrt [4]{c}}+\frac{\sqrt{2} \sqrt [8]{-a} x}{\sqrt [8]{c}}-x^2} \, dx,x,\sqrt{x}\right )}{32 \sqrt{2} (-a)^{11/8} c^{5/8}}\\ &=\frac{x^{5/2}}{4 a \left (a+c x^4\right )}+\frac{3 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{11/8} c^{5/8}}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{11/8} c^{5/8}}+\frac{3 \log \left (\sqrt [4]{-a}-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{11/8} c^{5/8}}-\frac{3 \log \left (\sqrt [4]{-a}+\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{11/8} c^{5/8}}-\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{11/8} c^{5/8}}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{11/8} c^{5/8}}\\ &=\frac{x^{5/2}}{4 a \left (a+c x^4\right )}+\frac{3 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{11/8} c^{5/8}}-\frac{3 \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 \sqrt{2} (-a)^{11/8} c^{5/8}}+\frac{3 \tan ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{11/8} c^{5/8}}+\frac{3 \tanh ^{-1}\left (\frac{\sqrt [8]{c} \sqrt{x}}{\sqrt [8]{-a}}\right )}{16 (-a)^{11/8} c^{5/8}}+\frac{3 \log \left (\sqrt [4]{-a}-\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{11/8} c^{5/8}}-\frac{3 \log \left (\sqrt [4]{-a}+\sqrt{2} \sqrt [8]{-a} \sqrt [8]{c} \sqrt{x}+\sqrt [4]{c} x\right )}{32 \sqrt{2} (-a)^{11/8} c^{5/8}}\\ \end{align*}

Mathematica [C]  time = 0.0061858, size = 29, normalized size = 0.09 \[ \frac{2 x^{5/2} \, _2F_1\left (\frac{5}{8},2;\frac{13}{8};-\frac{c x^4}{a}\right )}{5 a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(a + c*x^4)^2,x]

[Out]

(2*x^(5/2)*Hypergeometric2F1[5/8, 2, 13/8, -((c*x^4)/a)])/(5*a^2)

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Maple [C]  time = 0.011, size = 50, normalized size = 0.2 \begin{align*}{\frac{1}{4\,a \left ( c{x}^{4}+a \right ) }{x}^{{\frac{5}{2}}}}+{\frac{3}{32\,ac}\sum _{{\it \_R}={\it RootOf} \left ({{\it \_Z}}^{8}c+a \right ) }{\frac{1}{{{\it \_R}}^{3}}\ln \left ( \sqrt{x}-{\it \_R} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^(5/2)/a/(c*x^4+a)+3/32/a/c*sum(1/_R^3*ln(x^(1/2)-_R),_R=RootOf(_Z^8*c+a))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{x^{\frac{5}{2}}}{4 \,{\left (a c x^{4} + a^{2}\right )}} + 3 \, \int \frac{x^{\frac{3}{2}}}{8 \,{\left (a c x^{4} + a^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^(5/2)/(a*c*x^4 + a^2) + 3*integrate(1/8*x^(3/2)/(a*c*x^4 + a^2), x)

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Fricas [B]  time = 1.73104, size = 1501, normalized size = 4.87 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(12*sqrt(2)*(a*c*x^4 + a^2)*(-1/(a^11*c^5))^(1/8)*arctan(sqrt(2)*sqrt(sqrt(2)*a^7*c^3*sqrt(x)*(-1/(a^11*c
^5))^(5/8) - a^3*c*(-1/(a^11*c^5))^(1/4) + x)*a^4*c^2*(-1/(a^11*c^5))^(3/8) - sqrt(2)*a^4*c^2*sqrt(x)*(-1/(a^1
1*c^5))^(3/8) + 1) + 12*sqrt(2)*(a*c*x^4 + a^2)*(-1/(a^11*c^5))^(1/8)*arctan(sqrt(2)*sqrt(-sqrt(2)*a^7*c^3*sqr
t(x)*(-1/(a^11*c^5))^(5/8) - a^3*c*(-1/(a^11*c^5))^(1/4) + x)*a^4*c^2*(-1/(a^11*c^5))^(3/8) - sqrt(2)*a^4*c^2*
sqrt(x)*(-1/(a^11*c^5))^(3/8) - 1) + 3*sqrt(2)*(a*c*x^4 + a^2)*(-1/(a^11*c^5))^(1/8)*log(sqrt(2)*a^7*c^3*sqrt(
x)*(-1/(a^11*c^5))^(5/8) - a^3*c*(-1/(a^11*c^5))^(1/4) + x) - 3*sqrt(2)*(a*c*x^4 + a^2)*(-1/(a^11*c^5))^(1/8)*
log(-sqrt(2)*a^7*c^3*sqrt(x)*(-1/(a^11*c^5))^(5/8) - a^3*c*(-1/(a^11*c^5))^(1/4) + x) + 16*x^(5/2) - 24*(a*c*x
^4 + a^2)*(-1/(a^11*c^5))^(1/8)*arctan(sqrt(-a^3*c*(-1/(a^11*c^5))^(1/4) + x)*a^4*c^2*(-1/(a^11*c^5))^(3/8) -
a^4*c^2*sqrt(x)*(-1/(a^11*c^5))^(3/8)) - 6*(a*c*x^4 + a^2)*(-1/(a^11*c^5))^(1/8)*log(a^7*c^3*(-1/(a^11*c^5))^(
5/8) + sqrt(x)) + 6*(a*c*x^4 + a^2)*(-1/(a^11*c^5))^(1/8)*log(-a^7*c^3*(-1/(a^11*c^5))^(5/8) + sqrt(x)))/(a*c*
x^4 + a^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)/(c*x**4+a)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.54997, size = 613, normalized size = 1.99 \begin{align*} \frac{x^{\frac{5}{2}}}{4 \,{\left (c x^{4} + a\right )} a} - \frac{3 \, \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \arctan \left (\frac{\sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} + 2 \, \sqrt{x}}{\sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}}}\right )}{32 \, a^{2}} - \frac{3 \, \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \arctan \left (-\frac{\sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} - 2 \, \sqrt{x}}{\sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}}}\right )}{32 \, a^{2}} + \frac{3 \, \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \arctan \left (\frac{\sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} + 2 \, \sqrt{x}}{\sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}}}\right )}{32 \, a^{2}} + \frac{3 \, \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \arctan \left (-\frac{\sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} - 2 \, \sqrt{x}}{\sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}}}\right )}{32 \, a^{2}} - \frac{3 \, \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \log \left (\sqrt{x} \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} + x + \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}{64 \, a^{2}} + \frac{3 \, \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \log \left (-\sqrt{x} \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} + x + \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}{64 \, a^{2}} + \frac{3 \, \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \log \left (\sqrt{x} \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} + x + \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}{64 \, a^{2}} - \frac{3 \, \sqrt{\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{5}{8}} \log \left (-\sqrt{x} \sqrt{-\sqrt{2} + 2} \left (\frac{a}{c}\right )^{\frac{1}{8}} + x + \left (\frac{a}{c}\right )^{\frac{1}{4}}\right )}{64 \, a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x^(5/2)/((c*x^4 + a)*a) - 3/32*sqrt(-sqrt(2) + 2)*(a/c)^(5/8)*arctan((sqrt(-sqrt(2) + 2)*(a/c)^(1/8) + 2*s
qrt(x))/(sqrt(sqrt(2) + 2)*(a/c)^(1/8)))/a^2 - 3/32*sqrt(-sqrt(2) + 2)*(a/c)^(5/8)*arctan(-(sqrt(-sqrt(2) + 2)
*(a/c)^(1/8) - 2*sqrt(x))/(sqrt(sqrt(2) + 2)*(a/c)^(1/8)))/a^2 + 3/32*sqrt(sqrt(2) + 2)*(a/c)^(5/8)*arctan((sq
rt(sqrt(2) + 2)*(a/c)^(1/8) + 2*sqrt(x))/(sqrt(-sqrt(2) + 2)*(a/c)^(1/8)))/a^2 + 3/32*sqrt(sqrt(2) + 2)*(a/c)^
(5/8)*arctan(-(sqrt(sqrt(2) + 2)*(a/c)^(1/8) - 2*sqrt(x))/(sqrt(-sqrt(2) + 2)*(a/c)^(1/8)))/a^2 - 3/64*sqrt(-s
qrt(2) + 2)*(a/c)^(5/8)*log(sqrt(x)*sqrt(sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/a^2 + 3/64*sqrt(-sqrt(2)
+ 2)*(a/c)^(5/8)*log(-sqrt(x)*sqrt(sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/a^2 + 3/64*sqrt(sqrt(2) + 2)*(a
/c)^(5/8)*log(sqrt(x)*sqrt(-sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/a^2 - 3/64*sqrt(sqrt(2) + 2)*(a/c)^(5/
8)*log(-sqrt(x)*sqrt(-sqrt(2) + 2)*(a/c)^(1/8) + x + (a/c)^(1/4))/a^2