3.560 \(\int \frac{1}{x^2 (a c+b c x^3+d \sqrt{a+b x^3})} \, dx\)

Optimal. Leaf size=319 \[ \frac{d \sqrt{\frac{b x^3}{a}+1} F_1\left (-\frac{1}{3};\frac{1}{2},1;\frac{2}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )}{x \sqrt{a+b x^3} \left (a c^2-d^2\right )}-\frac{\sqrt [3]{b} c^{5/3} \log \left (-\sqrt [3]{b} c^{2/3} x \sqrt [3]{a c^2-d^2}+\left (a c^2-d^2\right )^{2/3}+b^{2/3} c^{4/3} x^2\right )}{6 \left (a c^2-d^2\right )^{4/3}}+\frac{\sqrt [3]{b} c^{5/3} \log \left (\sqrt [3]{a c^2-d^2}+\sqrt [3]{b} c^{2/3} x\right )}{3 \left (a c^2-d^2\right )^{4/3}}+\frac{\sqrt [3]{b} c^{5/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} c^{2/3} x}{\sqrt [3]{a c^2-d^2}}}{\sqrt{3}}\right )}{\sqrt{3} \left (a c^2-d^2\right )^{4/3}}-\frac{c}{x \left (a c^2-d^2\right )} \]

[Out]

-(c/((a*c^2 - d^2)*x)) + (d*Sqrt[1 + (b*x^3)/a]*AppellF1[-1/3, 1/2, 1, 2/3, -((b*x^3)/a), -((b*c^2*x^3)/(a*c^2
- d^2))])/((a*c^2 - d^2)*x*Sqrt[a + b*x^3]) + (b^(1/3)*c^(5/3)*ArcTan[(1 - (2*b^(1/3)*c^(2/3)*x)/(a*c^2 - d^2
)^(1/3))/Sqrt[3]])/(Sqrt[3]*(a*c^2 - d^2)^(4/3)) + (b^(1/3)*c^(5/3)*Log[(a*c^2 - d^2)^(1/3) + b^(1/3)*c^(2/3)*
x])/(3*(a*c^2 - d^2)^(4/3)) - (b^(1/3)*c^(5/3)*Log[(a*c^2 - d^2)^(2/3) - b^(1/3)*c^(2/3)*(a*c^2 - d^2)^(1/3)*x
+ b^(2/3)*c^(4/3)*x^2])/(6*(a*c^2 - d^2)^(4/3))

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Rubi [A]  time = 0.407952, antiderivative size = 319, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.345, Rules used = {2156, 325, 292, 31, 634, 617, 204, 628, 511, 510} \[ \frac{d \sqrt{\frac{b x^3}{a}+1} F_1\left (-\frac{1}{3};\frac{1}{2},1;\frac{2}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )}{x \sqrt{a+b x^3} \left (a c^2-d^2\right )}-\frac{\sqrt [3]{b} c^{5/3} \log \left (-\sqrt [3]{b} c^{2/3} x \sqrt [3]{a c^2-d^2}+\left (a c^2-d^2\right )^{2/3}+b^{2/3} c^{4/3} x^2\right )}{6 \left (a c^2-d^2\right )^{4/3}}+\frac{\sqrt [3]{b} c^{5/3} \log \left (\sqrt [3]{a c^2-d^2}+\sqrt [3]{b} c^{2/3} x\right )}{3 \left (a c^2-d^2\right )^{4/3}}+\frac{\sqrt [3]{b} c^{5/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} c^{2/3} x}{\sqrt [3]{a c^2-d^2}}}{\sqrt{3}}\right )}{\sqrt{3} \left (a c^2-d^2\right )^{4/3}}-\frac{c}{x \left (a c^2-d^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^2*(a*c + b*c*x^3 + d*Sqrt[a + b*x^3])),x]

[Out]

-(c/((a*c^2 - d^2)*x)) + (d*Sqrt[1 + (b*x^3)/a]*AppellF1[-1/3, 1/2, 1, 2/3, -((b*x^3)/a), -((b*c^2*x^3)/(a*c^2
- d^2))])/((a*c^2 - d^2)*x*Sqrt[a + b*x^3]) + (b^(1/3)*c^(5/3)*ArcTan[(1 - (2*b^(1/3)*c^(2/3)*x)/(a*c^2 - d^2
)^(1/3))/Sqrt[3]])/(Sqrt[3]*(a*c^2 - d^2)^(4/3)) + (b^(1/3)*c^(5/3)*Log[(a*c^2 - d^2)^(1/3) + b^(1/3)*c^(2/3)*
x])/(3*(a*c^2 - d^2)^(4/3)) - (b^(1/3)*c^(5/3)*Log[(a*c^2 - d^2)^(2/3) - b^(1/3)*c^(2/3)*(a*c^2 - d^2)^(1/3)*x
+ b^(2/3)*c^(4/3)*x^2])/(6*(a*c^2 - d^2)^(4/3))

Rule 2156

Int[(u_.)/((c_) + (d_.)*(x_)^(n_) + (e_.)*Sqrt[(a_) + (b_.)*(x_)^(n_)]), x_Symbol] :> Dist[c, Int[u/(c^2 - a*e
^2 + c*d*x^n), x], x] - Dist[a*e, Int[u/((c^2 - a*e^2 + c*d*x^n)*Sqrt[a + b*x^n]), x], x] /; FreeQ[{a, b, c, d
, e, n}, x] && EqQ[b*c - a*d, 0]

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 292

Int[(x_)/((a_) + (b_.)*(x_)^3), x_Symbol] :> -Dist[(3*Rt[a, 3]*Rt[b, 3])^(-1), Int[1/(Rt[a, 3] + Rt[b, 3]*x),
x], x] + Dist[1/(3*Rt[a, 3]*Rt[b, 3]), Int[(Rt[a, 3] + Rt[b, 3]*x)/(Rt[a, 3]^2 - Rt[a, 3]*Rt[b, 3]*x + Rt[b, 3
]^2*x^2), x], x] /; FreeQ[{a, b}, x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 634

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Dist[(2*c*d - b*e)/(2*c), Int[1/(a +
b*x + c*x^2), x], x] + Dist[e/(2*c), Int[(b + 2*c*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e}, x] &
& NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] &&  !NiceSqrtQ[b^2 - 4*a*c]

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 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 511

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

Rule 510

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(a^p*c^q
*(e*x)^(m + 1)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{1}{x^2 \left (a c+b c x^3+d \sqrt{a+b x^3}\right )} \, dx &=(a c) \int \frac{1}{x^2 \left (a^2 c^2-a d^2+a b c^2 x^3\right )} \, dx-(a d) \int \frac{1}{x^2 \sqrt{a+b x^3} \left (a^2 c^2-a d^2+a b c^2 x^3\right )} \, dx\\ &=-\frac{c}{\left (a c^2-d^2\right ) x}-\frac{\left (a b c^3\right ) \int \frac{x}{a^2 c^2-a d^2+a b c^2 x^3} \, dx}{a c^2-d^2}-\frac{\left (a d \sqrt{1+\frac{b x^3}{a}}\right ) \int \frac{1}{x^2 \sqrt{1+\frac{b x^3}{a}} \left (a^2 c^2-a d^2+a b c^2 x^3\right )} \, dx}{\sqrt{a+b x^3}}\\ &=-\frac{c}{\left (a c^2-d^2\right ) x}+\frac{d \sqrt{1+\frac{b x^3}{a}} F_1\left (-\frac{1}{3};\frac{1}{2},1;\frac{2}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) x \sqrt{a+b x^3}}+\frac{\left (\sqrt [3]{a} b^{2/3} c^{7/3}\right ) \int \frac{1}{\sqrt [3]{a} \sqrt [3]{a c^2-d^2}+\sqrt [3]{a} \sqrt [3]{b} c^{2/3} x} \, dx}{3 \left (a c^2-d^2\right )^{4/3}}-\frac{\left (\sqrt [3]{a} b^{2/3} c^{7/3}\right ) \int \frac{\sqrt [3]{a} \sqrt [3]{a c^2-d^2}+\sqrt [3]{a} \sqrt [3]{b} c^{2/3} x}{a^{2/3} \left (a c^2-d^2\right )^{2/3}-a^{2/3} \sqrt [3]{b} c^{2/3} \sqrt [3]{a c^2-d^2} x+a^{2/3} b^{2/3} c^{4/3} x^2} \, dx}{3 \left (a c^2-d^2\right )^{4/3}}\\ &=-\frac{c}{\left (a c^2-d^2\right ) x}+\frac{d \sqrt{1+\frac{b x^3}{a}} F_1\left (-\frac{1}{3};\frac{1}{2},1;\frac{2}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) x \sqrt{a+b x^3}}+\frac{\sqrt [3]{b} c^{5/3} \log \left (\sqrt [3]{a c^2-d^2}+\sqrt [3]{b} c^{2/3} x\right )}{3 \left (a c^2-d^2\right )^{4/3}}-\frac{\left (\sqrt [3]{b} c^{5/3}\right ) \int \frac{-a^{2/3} \sqrt [3]{b} c^{2/3} \sqrt [3]{a c^2-d^2}+2 a^{2/3} b^{2/3} c^{4/3} x}{a^{2/3} \left (a c^2-d^2\right )^{2/3}-a^{2/3} \sqrt [3]{b} c^{2/3} \sqrt [3]{a c^2-d^2} x+a^{2/3} b^{2/3} c^{4/3} x^2} \, dx}{6 \left (a c^2-d^2\right )^{4/3}}-\frac{\left (a^{2/3} b^{2/3} c^{7/3}\right ) \int \frac{1}{a^{2/3} \left (a c^2-d^2\right )^{2/3}-a^{2/3} \sqrt [3]{b} c^{2/3} \sqrt [3]{a c^2-d^2} x+a^{2/3} b^{2/3} c^{4/3} x^2} \, dx}{2 \left (a c^2-d^2\right )}\\ &=-\frac{c}{\left (a c^2-d^2\right ) x}+\frac{d \sqrt{1+\frac{b x^3}{a}} F_1\left (-\frac{1}{3};\frac{1}{2},1;\frac{2}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) x \sqrt{a+b x^3}}+\frac{\sqrt [3]{b} c^{5/3} \log \left (\sqrt [3]{a c^2-d^2}+\sqrt [3]{b} c^{2/3} x\right )}{3 \left (a c^2-d^2\right )^{4/3}}-\frac{\sqrt [3]{b} c^{5/3} \log \left (\left (a c^2-d^2\right )^{2/3}-\sqrt [3]{b} c^{2/3} \sqrt [3]{a c^2-d^2} x+b^{2/3} c^{4/3} x^2\right )}{6 \left (a c^2-d^2\right )^{4/3}}-\frac{\left (\sqrt [3]{b} c^{5/3}\right ) \operatorname{Subst}\left (\int \frac{1}{-3-x^2} \, dx,x,1-\frac{2 \sqrt [3]{b} c^{2/3} x}{\sqrt [3]{a c^2-d^2}}\right )}{\left (a c^2-d^2\right )^{4/3}}\\ &=-\frac{c}{\left (a c^2-d^2\right ) x}+\frac{d \sqrt{1+\frac{b x^3}{a}} F_1\left (-\frac{1}{3};\frac{1}{2},1;\frac{2}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )}{\left (a c^2-d^2\right ) x \sqrt{a+b x^3}}+\frac{\sqrt [3]{b} c^{5/3} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} c^{2/3} x}{\sqrt [3]{a c^2-d^2}}}{\sqrt{3}}\right )}{\sqrt{3} \left (a c^2-d^2\right )^{4/3}}+\frac{\sqrt [3]{b} c^{5/3} \log \left (\sqrt [3]{a c^2-d^2}+\sqrt [3]{b} c^{2/3} x\right )}{3 \left (a c^2-d^2\right )^{4/3}}-\frac{\sqrt [3]{b} c^{5/3} \log \left (\left (a c^2-d^2\right )^{2/3}-\sqrt [3]{b} c^{2/3} \sqrt [3]{a c^2-d^2} x+b^{2/3} c^{4/3} x^2\right )}{6 \left (a c^2-d^2\right )^{4/3}}\\ \end{align*}

Mathematica [A]  time = 0.701374, size = 496, normalized size = 1.55 \[ \frac{-6 b^2 c^2 d x^6 \sqrt{\frac{b x^3}{a}+1} \sqrt [3]{a c^2-d^2} F_1\left (\frac{5}{3};\frac{1}{2},1;\frac{8}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )+15 b d x^3 \sqrt{\frac{b x^3}{a}+1} \sqrt [3]{a c^2-d^2} \left (a c^2+d^2\right ) F_1\left (\frac{2}{3};\frac{1}{2},1;\frac{5}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )-10 \left (a c^2-d^2\right ) \left (a \sqrt [3]{b} c^{5/3} x \sqrt{a+b x^3} \log \left (-\sqrt [3]{b} c^{2/3} x \sqrt [3]{a c^2-d^2}+\left (a c^2-d^2\right )^{2/3}+b^{2/3} c^{4/3} x^2\right )-6 b d x^3 \sqrt [3]{a c^2-d^2}+6 a c \sqrt{a+b x^3} \sqrt [3]{a c^2-d^2}-2 a \sqrt [3]{b} c^{5/3} x \sqrt{a+b x^3} \log \left (\sqrt [3]{a c^2-d^2}+\sqrt [3]{b} c^{2/3} x\right )+2 \sqrt{3} a \sqrt [3]{b} c^{5/3} x \sqrt{a+b x^3} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} c^{2/3} x}{\sqrt [3]{a c^2-d^2}}-1}{\sqrt{3}}\right )-6 a d \sqrt [3]{a c^2-d^2}\right )}{60 a x \sqrt{a+b x^3} \left (a c^2-d^2\right )^{7/3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^2*(a*c + b*c*x^3 + d*Sqrt[a + b*x^3])),x]

[Out]

(15*b*d*(a*c^2 - d^2)^(1/3)*(a*c^2 + d^2)*x^3*Sqrt[1 + (b*x^3)/a]*AppellF1[2/3, 1/2, 1, 5/3, -((b*x^3)/a), -((
b*c^2*x^3)/(a*c^2 - d^2))] - 6*b^2*c^2*d*(a*c^2 - d^2)^(1/3)*x^6*Sqrt[1 + (b*x^3)/a]*AppellF1[5/3, 1/2, 1, 8/3
, -((b*x^3)/a), -((b*c^2*x^3)/(a*c^2 - d^2))] - 10*(a*c^2 - d^2)*(-6*a*d*(a*c^2 - d^2)^(1/3) - 6*b*d*(a*c^2 -
d^2)^(1/3)*x^3 + 6*a*c*(a*c^2 - d^2)^(1/3)*Sqrt[a + b*x^3] + 2*Sqrt[3]*a*b^(1/3)*c^(5/3)*x*Sqrt[a + b*x^3]*Arc
Tan[(-1 + (2*b^(1/3)*c^(2/3)*x)/(a*c^2 - d^2)^(1/3))/Sqrt[3]] - 2*a*b^(1/3)*c^(5/3)*x*Sqrt[a + b*x^3]*Log[(a*c
^2 - d^2)^(1/3) + b^(1/3)*c^(2/3)*x] + a*b^(1/3)*c^(5/3)*x*Sqrt[a + b*x^3]*Log[(a*c^2 - d^2)^(2/3) - b^(1/3)*c
^(2/3)*(a*c^2 - d^2)^(1/3)*x + b^(2/3)*c^(4/3)*x^2]))/(60*a*(a*c^2 - d^2)^(7/3)*x*Sqrt[a + b*x^3])

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Maple [C]  time = 0.034, size = 3560, normalized size = 11.2 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(a*c+b*c*x^3+d*(b*x^3+a)^(1/2)),x)

[Out]

1/3*a*c^3/(a*c^2-d^2)/d^2/(1/c^2/b*(a*c^2-d^2))^(1/3)*ln(x+(1/c^2/b*(a*c^2-d^2))^(1/3))-1/6*a*c^3/(a*c^2-d^2)/
d^2/(1/c^2/b*(a*c^2-d^2))^(1/3)*ln(x^2-(1/c^2/b*(a*c^2-d^2))^(1/3)*x+(1/c^2/b*(a*c^2-d^2))^(2/3))-1/3*a*c^3/(a
*c^2-d^2)/d^2*3^(1/2)/(1/c^2/b*(a*c^2-d^2))^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c^2/b*(a*c^2-d^2))^(1/3)*x-1))-c/(a
*c^2-d^2)/x-1/3*c/d^2/(1/c^2/b*(a*c^2-d^2))^(1/3)*ln(x+(1/c^2/b*(a*c^2-d^2))^(1/3))+1/6*c/d^2/(1/c^2/b*(a*c^2-
d^2))^(1/3)*ln(x^2-(1/c^2/b*(a*c^2-d^2))^(1/3)*x+(1/c^2/b*(a*c^2-d^2))^(2/3))+1/3*c/d^2*3^(1/2)/(1/c^2/b*(a*c^
2-d^2))^(1/3)*arctan(1/3*3^(1/2)*(2/(1/c^2/b*(a*c^2-d^2))^(1/3)*x-1))+2/3*I/a/d*3^(1/2)*(-a*b^2)^(2/3)*(I*(x+1
/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3))/(-
3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^
2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)/b*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1
/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1
/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))-1/a/d*(-a*b^2)^(2/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-
a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(
-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1
/2)/(b*x^3+a)^(1/2)*EllipticE(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b
/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/
2))/b-I/a/d*3^(1/2)*(-a*b^2)^(2/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^
2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1
/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticE(
1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1
/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))/b+I/b*c^2/(a*c^2-d^2)/d*3^
(1/2)*(-a*b^2)^(2/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2
)*((x-1/b*(-a*b^2)^(1/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^
(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticE(1/3*3^(1/2)*(I
*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)
^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))+1/b*c^2/(a*c^2-d^2)/d*(-a*b^2)^(2/3)*(I*
(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3)
)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-
a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticE(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)
-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^
(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))+1/3*I/b^2*c^2/(a*c^2-d^2)/d*2^(1/2)*sum(1/_alpha*(-a*b^2)^(1/3)*
(1/2*I*b*(2*x+1/b*((-a*b^2)^(1/3)-I*3^(1/2)*(-a*b^2)^(1/3)))/(-a*b^2)^(1/3))^(1/2)*(b*(x-1/b*(-a*b^2)^(1/3))/(
-3*(-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3)))^(1/2)*(-1/2*I*b*(2*x+1/b*((-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3)
))/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*(I*(-a*b^2)^(1/3)*3^(1/2)*_alpha*b-I*(-a*b^2)^(2/3)*3^(1/2)+2*_alpha^
2*b^2-(-a*b^2)^(1/3)*_alpha*b-(-a*b^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/
b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),-1/2*c^2/b*(2*I*(-a*b^2)^(1/3)*3^(1/2)*_alpha^2*b-I*(-a*b^2)
^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*a*b-3*(-a*b^2)^(2/3)*_alpha-3*a*b)/d^2,(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a
*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)),_alpha=RootOf(_Z^3*b*c^2+a*c^2-d^2))-3/2*I*d/a/(a*c^2-d^2)
*3^(1/2)*(-a*b^2)^(2/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(
1/2)*((x-1/b*(-a*b^2)^(1/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^
2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticE(1/3*3^(1/2)
*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b
^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))/b+d/a/(a*c^2-d^2)/x*(b*x^3+a)^(1/2)-2
/3*I/b*c^2/(a*c^2-d^2)/d*3^(1/2)*(-a*b^2)^(2/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(
1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(
1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/
2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^
(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))-3/2*d/a/(a*c^
2-d^2)*(-a*b^2)^(2/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/
2)*((x-1/b*(-a*b^2)^(1/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)
^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticE(1/3*3^(1/2)*(
I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2
)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))/b+I*d/a/(a*c^2-d^2)*3^(1/2)*(-a*b^2)^(2
/3)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2
)^(1/3))/(-3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/
2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)/b*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b
^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*
(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b c x^{3} + a c + \sqrt{b x^{3} + a} d\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a*c+b*c*x^3+d*(b*x^3+a)^(1/2)),x, algorithm="maxima")

[Out]

integrate(1/((b*c*x^3 + a*c + sqrt(b*x^3 + a)*d)*x^2), x)

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Fricas [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(1/x^2/(a*c+b*c*x^3+d*(b*x^3+a)^(1/2)),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{2} \left (a c + b c x^{3} + d \sqrt{a + b x^{3}}\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**2/(a*c+b*c*x**3+d*(b*x**3+a)**(1/2)),x)

[Out]

Integral(1/(x**2*(a*c + b*c*x**3 + d*sqrt(a + b*x**3))), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b c x^{3} + a c + \sqrt{b x^{3} + a} d\right )} x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^2/(a*c+b*c*x^3+d*(b*x^3+a)^(1/2)),x, algorithm="giac")

[Out]

integrate(1/((b*c*x^3 + a*c + sqrt(b*x^3 + a)*d)*x^2), x)