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

Optimal. Leaf size=311 \[ -\frac{d x^4 \sqrt{\frac{b x^3}{a}+1} F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )}{4 \sqrt{a+b x^3} \left (a c^2-d^2\right )}+\frac{\sqrt [3]{a c^2-d^2} \log \left (-\sqrt [3]{b} c^{2/3} x \sqrt [3]{a c^2-d^2}+\left (a c^2-d^2\right )^{2/3}+b^{2/3} c^{4/3} x^2\right )}{6 b^{4/3} c^{5/3}}-\frac{\sqrt [3]{a c^2-d^2} \log \left (\sqrt [3]{a c^2-d^2}+\sqrt [3]{b} c^{2/3} x\right )}{3 b^{4/3} c^{5/3}}+\frac{\sqrt [3]{a c^2-d^2} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} c^{2/3} x}{\sqrt [3]{a c^2-d^2}}}{\sqrt{3}}\right )}{\sqrt{3} b^{4/3} c^{5/3}}+\frac{x}{b c} \]

[Out]

x/(b*c) - (d*x^4*Sqrt[1 + (b*x^3)/a]*AppellF1[4/3, 1/2, 1, 7/3, -((b*x^3)/a), -((b*c^2*x^3)/(a*c^2 - d^2))])/(
4*(a*c^2 - d^2)*Sqrt[a + b*x^3]) + ((a*c^2 - d^2)^(1/3)*ArcTan[(1 - (2*b^(1/3)*c^(2/3)*x)/(a*c^2 - d^2)^(1/3))
/Sqrt[3]])/(Sqrt[3]*b^(4/3)*c^(5/3)) - ((a*c^2 - d^2)^(1/3)*Log[(a*c^2 - d^2)^(1/3) + b^(1/3)*c^(2/3)*x])/(3*b
^(4/3)*c^(5/3)) + ((a*c^2 - d^2)^(1/3)*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*b^(4/3)*c^(5/3))

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Rubi [A]  time = 0.503803, antiderivative size = 311, 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, 321, 200, 31, 634, 617, 204, 628, 511, 510} \[ -\frac{d x^4 \sqrt{\frac{b x^3}{a}+1} F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )}{4 \sqrt{a+b x^3} \left (a c^2-d^2\right )}+\frac{\sqrt [3]{a c^2-d^2} \log \left (-\sqrt [3]{b} c^{2/3} x \sqrt [3]{a c^2-d^2}+\left (a c^2-d^2\right )^{2/3}+b^{2/3} c^{4/3} x^2\right )}{6 b^{4/3} c^{5/3}}-\frac{\sqrt [3]{a c^2-d^2} \log \left (\sqrt [3]{a c^2-d^2}+\sqrt [3]{b} c^{2/3} x\right )}{3 b^{4/3} c^{5/3}}+\frac{\sqrt [3]{a c^2-d^2} \tan ^{-1}\left (\frac{1-\frac{2 \sqrt [3]{b} c^{2/3} x}{\sqrt [3]{a c^2-d^2}}}{\sqrt{3}}\right )}{\sqrt{3} b^{4/3} c^{5/3}}+\frac{x}{b c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

x/(b*c) - (d*x^4*Sqrt[1 + (b*x^3)/a]*AppellF1[4/3, 1/2, 1, 7/3, -((b*x^3)/a), -((b*c^2*x^3)/(a*c^2 - d^2))])/(
4*(a*c^2 - d^2)*Sqrt[a + b*x^3]) + ((a*c^2 - d^2)^(1/3)*ArcTan[(1 - (2*b^(1/3)*c^(2/3)*x)/(a*c^2 - d^2)^(1/3))
/Sqrt[3]])/(Sqrt[3]*b^(4/3)*c^(5/3)) - ((a*c^2 - d^2)^(1/3)*Log[(a*c^2 - d^2)^(1/3) + b^(1/3)*c^(2/3)*x])/(3*b
^(4/3)*c^(5/3)) + ((a*c^2 - d^2)^(1/3)*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*b^(4/3)*c^(5/3))

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 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 200

Int[((a_) + (b_.)*(x_)^3)^(-1), x_Symbol] :> Dist[1/(3*Rt[a, 3]^2), Int[1/(Rt[a, 3] + Rt[b, 3]*x), x], x] + Di
st[1/(3*Rt[a, 3]^2), Int[(2*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{x^3}{a c+b c x^3+d \sqrt{a+b x^3}} \, dx &=(a c) \int \frac{x^3}{a^2 c^2-a d^2+a b c^2 x^3} \, dx-(a d) \int \frac{x^3}{\sqrt{a+b x^3} \left (a^2 c^2-a d^2+a b c^2 x^3\right )} \, dx\\ &=\frac{x}{b c}-\frac{\left (a \left (a c^2-d^2\right )\right ) \int \frac{1}{a^2 c^2-a d^2+a b c^2 x^3} \, dx}{b c}-\frac{\left (a d \sqrt{1+\frac{b x^3}{a}}\right ) \int \frac{x^3}{\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{x}{b c}-\frac{d x^4 \sqrt{1+\frac{b x^3}{a}} F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )}{4 \left (a c^2-d^2\right ) \sqrt{a+b x^3}}-\frac{\left (\sqrt [3]{a} \sqrt [3]{a c^2-d^2}\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 b c}-\frac{\left (\sqrt [3]{a} \sqrt [3]{a c^2-d^2}\right ) \int \frac{2 \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 b c}\\ &=\frac{x}{b c}-\frac{d x^4 \sqrt{1+\frac{b x^3}{a}} F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )}{4 \left (a c^2-d^2\right ) \sqrt{a+b x^3}}-\frac{\sqrt [3]{a c^2-d^2} \log \left (\sqrt [3]{a c^2-d^2}+\sqrt [3]{b} c^{2/3} x\right )}{3 b^{4/3} c^{5/3}}+\frac{\sqrt [3]{a c^2-d^2} \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 b^{4/3} c^{5/3}}-\frac{\left (a^{2/3} \left (a c^2-d^2\right )^{2/3}\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 b c}\\ &=\frac{x}{b c}-\frac{d x^4 \sqrt{1+\frac{b x^3}{a}} F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )}{4 \left (a c^2-d^2\right ) \sqrt{a+b x^3}}-\frac{\sqrt [3]{a c^2-d^2} \log \left (\sqrt [3]{a c^2-d^2}+\sqrt [3]{b} c^{2/3} x\right )}{3 b^{4/3} c^{5/3}}+\frac{\sqrt [3]{a c^2-d^2} \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 b^{4/3} c^{5/3}}-\frac{\sqrt [3]{a c^2-d^2} \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 )}{b^{4/3} c^{5/3}}\\ &=\frac{x}{b c}-\frac{d x^4 \sqrt{1+\frac{b x^3}{a}} F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )}{4 \left (a c^2-d^2\right ) \sqrt{a+b x^3}}+\frac{\sqrt [3]{a c^2-d^2} \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} b^{4/3} c^{5/3}}-\frac{\sqrt [3]{a c^2-d^2} \log \left (\sqrt [3]{a c^2-d^2}+\sqrt [3]{b} c^{2/3} x\right )}{3 b^{4/3} c^{5/3}}+\frac{\sqrt [3]{a c^2-d^2} \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 b^{4/3} c^{5/3}}\\ \end{align*}

Mathematica [A]  time = 0.561473, size = 296, normalized size = 0.95 \[ \frac{\sqrt [3]{a c^2-d^2} \log \left (-\sqrt [3]{b} c^{2/3} x \sqrt [3]{a c^2-d^2}+\left (a c^2-d^2\right )^{2/3}+b^{2/3} c^{4/3} x^2\right )-2 \sqrt [3]{a c^2-d^2} \log \left (\sqrt [3]{a c^2-d^2}+\sqrt [3]{b} c^{2/3} x\right )-2 \sqrt{3} \sqrt [3]{a c^2-d^2} \tan ^{-1}\left (\frac{\frac{2 \sqrt [3]{b} c^{2/3} x}{\sqrt [3]{a c^2-d^2}}-1}{\sqrt{3}}\right )+6 \sqrt [3]{b} c^{2/3} x}{6 b^{4/3} c^{5/3}}-\frac{d x^4 \sqrt{\frac{b x^3}{a}+1} F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{b x^3}{a},-\frac{b c^2 x^3}{a c^2-d^2}\right )}{\sqrt{a+b x^3} \left (4 a c^2-4 d^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((d*x^4*Sqrt[1 + (b*x^3)/a]*AppellF1[4/3, 1/2, 1, 7/3, -((b*x^3)/a), -((b*c^2*x^3)/(a*c^2 - d^2))])/((4*a*c^2
- 4*d^2)*Sqrt[a + b*x^3])) + (6*b^(1/3)*c^(2/3)*x - 2*Sqrt[3]*(a*c^2 - d^2)^(1/3)*ArcTan[(-1 + (2*b^(1/3)*c^(
2/3)*x)/(a*c^2 - d^2)^(1/3))/Sqrt[3]] - 2*(a*c^2 - d^2)^(1/3)*Log[(a*c^2 - d^2)^(1/3) + b^(1/3)*c^(2/3)*x] + (
a*c^2 - d^2)^(1/3)*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*
b^(4/3)*c^(5/3))

________________________________________________________________________________________

Maple [C]  time = 0.045, size = 1544, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*I*d/b^2/c^2*3^(1/2)*(-a*b^2)^(1/3)*(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)*Ellipt
icF(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/d/b^4*2^(1/2)*su
m(1/_alpha^2*(-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))*a-1/3*I*d/b^4/c^2*2^(1/2)*sum(1/_alpha^2*(-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))*E
llipticPi(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))-1/3*a/c/b^2/(1/c^2/b*(a*c^2-d^2))^(2/3)*ln(x+(1/c^2/b*(a*c^2-d^2))^(
1/3))+1/6*a/c/b^2/(1/c^2/b*(a*c^2-d^2))^(2/3)*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/b^2/(1/c^2/b*(a*c^2-d^2))^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/c^2/b*(a*c^2-d^2))^(1/3)*x-1))+x/b/
c+1/3/b^2/c^3*d^2/(1/c^2/b*(a*c^2-d^2))^(2/3)*ln(x+(1/c^2/b*(a*c^2-d^2))^(1/3))-1/6/b^2/c^3*d^2/(1/c^2/b*(a*c^
2-d^2))^(2/3)*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/b^2/c^3*d^2/(1/c^2/b*(a*c^
2-d^2))^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/(1/c^2/b*(a*c^2-d^2))^(1/3)*x-1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3/(b*c*x^3 + a*c + sqrt(b*x^3 + a)*d), 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(x^3/(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{x^{3}}{a c + b c x^{3} + d \sqrt{a + b x^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3/(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{x^{3}}{b c x^{3} + a c + \sqrt{b x^{3} + a} d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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