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-1/3*(a*c^2-d^2)^(1/3)*ln((a*c^2-d^2)^(1/3)+b^(1/3)*c^(2/3)*x)/b^(4/3)/c^(5/3)+1/6*(a*c^2-d^2)^(1/3)*ln((
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)/b^(4/3)/c^(5/3)+1/3*(a*c^2-d^2)^(1/3
)*arctan(1/3*(1-2*b^(1/3)*c^(2/3)*x/(a*c^2-d^2)^(1/3))*3^(1/2))/b^(4/3)/c^(5/3)*3^(1/2)-1/4*d*x^4*AppellF1(4/3
,1/2,1,7/3,-b*x^3/a,-b*c^2*x^3/(a*c^2-d^2))*(1+b*x^3/a)^(1/2)/(a*c^2-d^2)/(b*x^3+a)^(1/2)

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Rubi [A]  time = 0.50, antiderivative size = 311, normalized size of antiderivative = 1.00, 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 31

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

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 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 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 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]

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*}

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Mathematica [A]  time = 0.55, 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))

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{b c x^{3} + a c + \sqrt {b x^{3} + a} d}\,{d x} \]

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)

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maple [C]  time = 0.08, size = 1544, normalized size = 4.96 \[ \text {result too large to display} \]

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*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*3^(1/2)/(-a*
b^2)^(1/3)*b)^(1/2)*((x-(-a*b^2)^(1/3)/b)/(-3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b))^(1/2)*(-I*(x
+1/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*Elliptic
F(1/3*3^(1/2)*(I*(x+1/2*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*3^(1/2)/(-a*b^2)^(1/3)*b)^(1/2),(I*3^
(1/2)*(-a*b^2)^(1/3)/(-3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/b)^(1/2))+1/3*I/b^4/d*2^(1/2)*sum(
1/_alpha^2*(-a*b^2)^(1/3)*(1/2*I*(2*x+((-a*b^2)^(1/3)-I*3^(1/2)*(-a*b^2)^(1/3))/b)/(-a*b^2)^(1/3)*b)^(1/2)*((x
-(-a*b^2)^(1/3)/b)/(-3*(-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3))*b)^(1/2)*(-1/2*I*(2*x+((-a*b^2)^(1/3)+I*3^(1/2
)*(-a*b^2)^(1/3))/b)/(-a*b^2)^(1/3)*b)^(1/2)/(b*x^3+a)^(1/2)*(2*_alpha^2*b^2+I*(-a*b^2)^(1/3)*3^(1/2)*_alpha*b
-(-a*b^2)^(1/3)*_alpha*b-I*(-a*b^2)^(2/3)*3^(1/2)-(-a*b^2)^(2/3))*EllipticPi(1/3*3^(1/2)*(I*(x+1/2*(-a*b^2)^(1
/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*3^(1/2)/(-a*b^2)^(1/3)*b)^(1/2),-1/2*(2*I*(-a*b^2)^(1/3)*3^(1/2)*_alpha^
2*b+I*3^(1/2)*a*b-3*a*b-I*(-a*b^2)^(2/3)*3^(1/2)*_alpha-3*(-a*b^2)^(2/3)*_alpha)/b*c^2/d^2,(I*3^(1/2)*(-a*b^2)
^(1/3)/(-3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/b)^(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*(2*x+((-a*b^2)^(1/3)-I*3^(1/2)*(-a*b^2)^(1/3))/b
)/(-a*b^2)^(1/3)*b)^(1/2)*((x-(-a*b^2)^(1/3)/b)/(-3*(-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3))*b)^(1/2)*(-1/2*I*
(2*x+((-a*b^2)^(1/3)+I*3^(1/2)*(-a*b^2)^(1/3))/b)/(-a*b^2)^(1/3)*b)^(1/2)/(b*x^3+a)^(1/2)*(2*_alpha^2*b^2+I*(-
a*b^2)^(1/3)*3^(1/2)*_alpha*b-(-a*b^2)^(1/3)*_alpha*b-I*(-a*b^2)^(2/3)*3^(1/2)-(-a*b^2)^(2/3))*EllipticPi(1/3*
3^(1/2)*(I*(x+1/2*(-a*b^2)^(1/3)/b-1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)*3^(1/2)/(-a*b^2)^(1/3)*b)^(1/2),-1/2*(2*I*(
-a*b^2)^(1/3)*3^(1/2)*_alpha^2*b+I*3^(1/2)*a*b-3*a*b-I*(-a*b^2)^(2/3)*3^(1/2)*_alpha-3*(-a*b^2)^(2/3)*_alpha)/
b*c^2/d^2,(I*3^(1/2)*(-a*b^2)^(1/3)/(-3/2*(-a*b^2)^(1/3)/b+1/2*I*3^(1/2)*(-a*b^2)^(1/3)/b)/b)^(1/2)),_alpha=Ro
otOf(_Z^3*b*c^2+a*c^2-d^2))-1/3*a/c/b^2/((a*c^2-d^2)/b/c^2)^(2/3)*ln(x+((a*c^2-d^2)/b/c^2)^(1/3))+1/6*a/c/b^2/
((a*c^2-d^2)/b/c^2)^(2/3)*ln(x^2-((a*c^2-d^2)/b/c^2)^(1/3)*x+((a*c^2-d^2)/b/c^2)^(2/3))-1/3*a/c/b^2/((a*c^2-d^
2)/b/c^2)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)*(2/((a*c^2-d^2)/b/c^2)^(1/3)*x-1))+x/b/c+1/3/b^2/c^3*d^2/((a*c^2-d^
2)/b/c^2)^(2/3)*ln(x+((a*c^2-d^2)/b/c^2)^(1/3))-1/6/b^2/c^3*d^2/((a*c^2-d^2)/b/c^2)^(2/3)*ln(x^2-((a*c^2-d^2)/
b/c^2)^(1/3)*x+((a*c^2-d^2)/b/c^2)^(2/3))+1/3/b^2/c^3*d^2/((a*c^2-d^2)/b/c^2)^(2/3)*3^(1/2)*arctan(1/3*3^(1/2)
*(2/((a*c^2-d^2)/b/c^2)^(1/3)*x-1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3}}{b c x^{3} + a c + \sqrt {b x^{3} + a} d}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^3}{a\,c+d\,\sqrt {b\,x^3+a}+b\,c\,x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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