3.3.14 \(\int \frac {x}{\sqrt {c+d x^3} (4 c+d x^3)} \, dx\)

Optimal. Leaf size=206 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3\ 2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}+\frac {\tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3\ 2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3\ 2^{2/3} c^{5/6} d^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{9\ 2^{2/3} c^{5/6} d^{2/3}} \]

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Rubi [A]  time = 0.03, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.042, Rules used = {484} \begin {gather*} -\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3\ 2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}+\frac {\tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3\ 2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3\ 2^{2/3} c^{5/6} d^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{9\ 2^{2/3} c^{5/6} d^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x/(Sqrt[c + d*x^3]*(4*c + d*x^3)),x]

[Out]

-ArcTan[(Sqrt[3]*c^(1/6)*(c^(1/3) + 2^(1/3)*d^(1/3)*x))/Sqrt[c + d*x^3]]/(3*2^(2/3)*Sqrt[3]*c^(5/6)*d^(2/3)) +
ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Sqrt[c])]/(3*2^(2/3)*Sqrt[3]*c^(5/6)*d^(2/3)) - ArcTanh[(c^(1/6)*(c^(1/3) - 2
^(1/3)*d^(1/3)*x))/Sqrt[c + d*x^3]]/(3*2^(2/3)*c^(5/6)*d^(2/3)) + ArcTanh[Sqrt[c + d*x^3]/Sqrt[c]]/(9*2^(2/3)*
c^(5/6)*d^(2/3))

Rule 484

Int[(x_)/(((a_) + (b_.)*(x_)^3)*Sqrt[(c_) + (d_.)*(x_)^3]), x_Symbol] :> With[{q = Rt[d/c, 3]}, Simp[(q*ArcTan
h[Sqrt[c + d*x^3]/Rt[c, 2]])/(9*2^(2/3)*b*Rt[c, 2]), x] + (-Simp[(q*ArcTanh[(Rt[c, 2]*(1 - 2^(1/3)*q*x))/Sqrt[
c + d*x^3]])/(3*2^(2/3)*b*Rt[c, 2]), x] + Simp[(q*ArcTan[Sqrt[c + d*x^3]/(Sqrt[3]*Rt[c, 2])])/(3*2^(2/3)*Sqrt[
3]*b*Rt[c, 2]), x] - Simp[(q*ArcTan[(Sqrt[3]*Rt[c, 2]*(1 + 2^(1/3)*q*x))/Sqrt[c + d*x^3]])/(3*2^(2/3)*Sqrt[3]*
b*Rt[c, 2]), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[4*b*c - a*d, 0] && PosQ[c]

Rubi steps

\begin {align*} \int \frac {x}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx &=-\frac {\tan ^{-1}\left (\frac {\sqrt {3} \sqrt [6]{c} \left (\sqrt [3]{c}+\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3\ 2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}+\frac {\tan ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {3} \sqrt {c}}\right )}{3\ 2^{2/3} \sqrt {3} c^{5/6} d^{2/3}}-\frac {\tanh ^{-1}\left (\frac {\sqrt [6]{c} \left (\sqrt [3]{c}-\sqrt [3]{2} \sqrt [3]{d} x\right )}{\sqrt {c+d x^3}}\right )}{3\ 2^{2/3} c^{5/6} d^{2/3}}+\frac {\tanh ^{-1}\left (\frac {\sqrt {c+d x^3}}{\sqrt {c}}\right )}{9\ 2^{2/3} c^{5/6} d^{2/3}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 67, normalized size = 0.33 \begin {gather*} \frac {x^2 \sqrt {\frac {c+d x^3}{c}} F_1\left (\frac {2}{3};\frac {1}{2},1;\frac {5}{3};-\frac {d x^3}{c},-\frac {d x^3}{4 c}\right )}{8 c \sqrt {c+d x^3}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[x/(Sqrt[c + d*x^3]*(4*c + d*x^3)),x]

[Out]

(x^2*Sqrt[(c + d*x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), -1/4*(d*x^3)/c])/(8*c*Sqrt[c + d*x^3])

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IntegrateAlgebraic [F]  time = 15.36, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {c+d x^3} \left (4 c+d x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[x/(Sqrt[c + d*x^3]*(4*c + d*x^3)),x]

[Out]

Defer[IntegrateAlgebraic][x/(Sqrt[c + d*x^3]*(4*c + d*x^3)), x]

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fricas [B]  time = 2.37, size = 2274, normalized size = 11.04

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="fricas")

[Out]

1/9*sqrt(3)*(1/432)^(1/6)*(-1/(c^5*d^4))^(1/6)*arctan(-1/3*(3*(sqrt(3)*sqrt(1/3)*c^3*d^2*x*sqrt(-1/(c^5*d^4))
+ 2*sqrt(3)*(1/432)^(1/6)*c*d*x^2*(-1/(c^5*d^4))^(1/6) + 24*sqrt(3)*(1/432)^(5/6)*(c^4*d^4*x^3 + 4*c^5*d^3)*(-
1/(c^5*d^4))^(5/6))*sqrt(d*x^3 + c) + (2*sqrt(3)*(1/2)^(1/3)*(c^2*d^2*x^3 + c^3*d)*(-1/(c^5*d^4))^(1/3) + sqrt
(3)*(d*x^4 + c*x) + 3*(sqrt(3)*sqrt(1/3)*c^3*d^2*x*sqrt(-1/(c^5*d^4)) + 2*sqrt(3)*(1/432)^(1/6)*c*d*x^2*(-1/(c
^5*d^4))^(1/6) - 24*sqrt(3)*(1/432)^(5/6)*(c^4*d^4*x^3 - 2*c^5*d^3)*(-1/(c^5*d^4))^(5/6))*sqrt(d*x^3 + c))*sqr
t((d^3*x^9 + 60*c*d^2*x^6 - 32*c^3 - 24*(1/2)^(2/3)*(c^4*d^5*x^7 + 5*c^5*d^4*x^4 + 4*c^6*d^3*x)*(-1/(c^5*d^4))
^(2/3) + 12*(1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^2*x^2)*(-1/(c^5*d^4))^(1/3) + 12*(648*(1/432)^(
5/6)*c^5*d^5*x^5*(-1/(c^5*d^4))^(5/6) - sqrt(1/3)*(c^3*d^4*x^6 - 16*c^4*d^3*x^3 - 8*c^5*d^2)*sqrt(-1/(c^5*d^4)
) - (1/432)^(1/6)*(c*d^3*x^7 + 2*c^2*d^2*x^4 - 8*c^3*d*x)*(-1/(c^5*d^4))^(1/6))*sqrt(d*x^3 + c))/(d^3*x^9 + 12
*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)))/(d*x^4 + c*x)) + 1/9*sqrt(3)*(1/432)^(1/6)*(-1/(c^5*d^4))^(1/6)*arctan(-
1/3*(3*(sqrt(3)*sqrt(1/3)*c^3*d^2*x*sqrt(-1/(c^5*d^4)) + 2*sqrt(3)*(1/432)^(1/6)*c*d*x^2*(-1/(c^5*d^4))^(1/6)
+ 24*sqrt(3)*(1/432)^(5/6)*(c^4*d^4*x^3 + 4*c^5*d^3)*(-1/(c^5*d^4))^(5/6))*sqrt(d*x^3 + c) - (2*sqrt(3)*(1/2)^
(1/3)*(c^2*d^2*x^3 + c^3*d)*(-1/(c^5*d^4))^(1/3) + sqrt(3)*(d*x^4 + c*x) - 3*(sqrt(3)*sqrt(1/3)*c^3*d^2*x*sqrt
(-1/(c^5*d^4)) + 2*sqrt(3)*(1/432)^(1/6)*c*d*x^2*(-1/(c^5*d^4))^(1/6) - 24*sqrt(3)*(1/432)^(5/6)*(c^4*d^4*x^3
- 2*c^5*d^3)*(-1/(c^5*d^4))^(5/6))*sqrt(d*x^3 + c))*sqrt((d^3*x^9 + 60*c*d^2*x^6 - 32*c^3 - 24*(1/2)^(2/3)*(c^
4*d^5*x^7 + 5*c^5*d^4*x^4 + 4*c^6*d^3*x)*(-1/(c^5*d^4))^(2/3) + 12*(1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 -
8*c^4*d^2*x^2)*(-1/(c^5*d^4))^(1/3) - 12*(648*(1/432)^(5/6)*c^5*d^5*x^5*(-1/(c^5*d^4))^(5/6) - sqrt(1/3)*(c^3*
d^4*x^6 - 16*c^4*d^3*x^3 - 8*c^5*d^2)*sqrt(-1/(c^5*d^4)) - (1/432)^(1/6)*(c*d^3*x^7 + 2*c^2*d^2*x^4 - 8*c^3*d*
x)*(-1/(c^5*d^4))^(1/6))*sqrt(d*x^3 + c))/(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)))/(d*x^4 + c*x)) -
1/36*(1/432)^(1/6)*(-1/(c^5*d^4))^(1/6)*log((d^3*x^9 + 60*c*d^2*x^6 - 32*c^3 - 24*(1/2)^(2/3)*(c^4*d^5*x^7 + 5
*c^5*d^4*x^4 + 4*c^6*d^3*x)*(-1/(c^5*d^4))^(2/3) + 12*(1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^2*x^2
)*(-1/(c^5*d^4))^(1/3) + 12*(648*(1/432)^(5/6)*c^5*d^5*x^5*(-1/(c^5*d^4))^(5/6) - sqrt(1/3)*(c^3*d^4*x^6 - 16*
c^4*d^3*x^3 - 8*c^5*d^2)*sqrt(-1/(c^5*d^4)) - (1/432)^(1/6)*(c*d^3*x^7 + 2*c^2*d^2*x^4 - 8*c^3*d*x)*(-1/(c^5*d
^4))^(1/6))*sqrt(d*x^3 + c))/(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) + 1/36*(1/432)^(1/6)*(-1/(c^5*d
^4))^(1/6)*log((d^3*x^9 + 60*c*d^2*x^6 - 32*c^3 - 24*(1/2)^(2/3)*(c^4*d^5*x^7 + 5*c^5*d^4*x^4 + 4*c^6*d^3*x)*(
-1/(c^5*d^4))^(2/3) + 12*(1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^2*x^2)*(-1/(c^5*d^4))^(1/3) - 12*(
648*(1/432)^(5/6)*c^5*d^5*x^5*(-1/(c^5*d^4))^(5/6) - sqrt(1/3)*(c^3*d^4*x^6 - 16*c^4*d^3*x^3 - 8*c^5*d^2)*sqrt
(-1/(c^5*d^4)) - (1/432)^(1/6)*(c*d^3*x^7 + 2*c^2*d^2*x^4 - 8*c^3*d*x)*(-1/(c^5*d^4))^(1/6))*sqrt(d*x^3 + c))/
(d^3*x^9 + 12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) + 1/18*(1/432)^(1/6)*(-1/(c^5*d^4))^(1/6)*log((d^3*x^9 - 66*
c*d^2*x^6 - 72*c^2*d*x^3 - 32*c^3 + 48*(1/2)^(2/3)*(c^4*d^5*x^7 - c^5*d^4*x^4 - 2*c^6*d^3*x)*(-1/(c^5*d^4))^(2
/3) + 12*(1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^2*x^2)*(-1/(c^5*d^4))^(1/3) + 6*(1296*(1/432)^(5/6
)*c^5*d^5*x^5*(-1/(c^5*d^4))^(5/6) + sqrt(1/3)*(5*c^3*d^4*x^6 - 20*c^4*d^3*x^3 - 16*c^5*d^2)*sqrt(-1/(c^5*d^4)
) + 2*(1/432)^(1/6)*(c*d^3*x^7 - 16*c^2*d^2*x^4 - 8*c^3*d*x)*(-1/(c^5*d^4))^(1/6))*sqrt(d*x^3 + c))/(d^3*x^9 +
12*c*d^2*x^6 + 48*c^2*d*x^3 + 64*c^3)) - 1/18*(1/432)^(1/6)*(-1/(c^5*d^4))^(1/6)*log((d^3*x^9 - 66*c*d^2*x^6
- 72*c^2*d*x^3 - 32*c^3 + 48*(1/2)^(2/3)*(c^4*d^5*x^7 - c^5*d^4*x^4 - 2*c^6*d^3*x)*(-1/(c^5*d^4))^(2/3) + 12*(
1/2)^(1/3)*(c^2*d^4*x^8 - 7*c^3*d^3*x^5 - 8*c^4*d^2*x^2)*(-1/(c^5*d^4))^(1/3) - 6*(1296*(1/432)^(5/6)*c^5*d^5*
x^5*(-1/(c^5*d^4))^(5/6) + sqrt(1/3)*(5*c^3*d^4*x^6 - 20*c^4*d^3*x^3 - 16*c^5*d^2)*sqrt(-1/(c^5*d^4)) + 2*(1/4
32)^(1/6)*(c*d^3*x^7 - 16*c^2*d^2*x^4 - 8*c^3*d*x)*(-1/(c^5*d^4))^(1/6))*sqrt(d*x^3 + c))/(d^3*x^9 + 12*c*d^2*
x^6 + 48*c^2*d*x^3 + 64*c^3))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{{\left (d x^{3} + 4 \, c\right )} \sqrt {d x^{3} + c}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="giac")

[Out]

integrate(x/((d*x^3 + 4*c)*sqrt(d*x^3 + c)), x)

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maple [C]  time = 0.18, size = 416, normalized size = 2.02 \begin {gather*} -\frac {i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {\frac {i \left (2 x +\frac {-i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {\frac {\left (x -\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{-3 \left (-c \,d^{2}\right )^{\frac {1}{3}}+i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \sqrt {-\frac {i \left (2 x +\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}+\left (-c \,d^{2}\right )^{\frac {1}{3}}}{d}\right ) d}{2 \left (-c \,d^{2}\right )^{\frac {1}{3}}}}\, \left (2 \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )^{2} d^{2}+i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right ) d -\left (-c \,d^{2}\right )^{\frac {1}{3}} \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right ) d -i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {2}{3}}-\left (-c \,d^{2}\right )^{\frac {2}{3}}\right ) \EllipticPi \left (\frac {\sqrt {3}\, \sqrt {\frac {i \left (x +\frac {\left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}-\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) \sqrt {3}\, d}{\left (-c \,d^{2}\right )^{\frac {1}{3}}}}}{3}, \frac {2 i \left (-c \,d^{2}\right )^{\frac {1}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )^{2} d +i \sqrt {3}\, c d -3 c d -i \left (-c \,d^{2}\right )^{\frac {2}{3}} \sqrt {3}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )-3 \left (-c \,d^{2}\right )^{\frac {2}{3}} \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )}{6 c d}, \sqrt {\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{\left (-\frac {3 \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}+\frac {i \sqrt {3}\, \left (-c \,d^{2}\right )^{\frac {1}{3}}}{2 d}\right ) d}}\right )}{9 c \,d^{3} \sqrt {d \,x^{3}+c}\, \RootOf \left (d \,\textit {\_Z}^{3}+4 c \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/(d*x^3+4*c)/(d*x^3+c)^(1/2),x)

[Out]

-1/9*I/d^3/c*2^(1/2)*sum(1/_alpha*(-c*d^2)^(1/3)*(1/2*I*(2*x+(-I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c
*d^2)^(1/3)*d)^(1/2)*((x-(-c*d^2)^(1/3)/d)/(-3*(-c*d^2)^(1/3)+I*3^(1/2)*(-c*d^2)^(1/3))*d)^(1/2)*(-1/2*I*(2*x+
(I*3^(1/2)*(-c*d^2)^(1/3)+(-c*d^2)^(1/3))/d)/(-c*d^2)^(1/3)*d)^(1/2)/(d*x^3+c)^(1/2)*(2*_alpha^2*d^2+I*(-c*d^2
)^(1/3)*3^(1/2)*_alpha*d-(-c*d^2)^(1/3)*_alpha*d-I*3^(1/2)*(-c*d^2)^(2/3)-(-c*d^2)^(2/3))*EllipticPi(1/3*3^(1/
2)*(I*(x+1/2*(-c*d^2)^(1/3)/d-1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*3^(1/2)/(-c*d^2)^(1/3)*d)^(1/2),1/6*(2*I*(-c*d^2
)^(1/3)*3^(1/2)*_alpha^2*d+I*3^(1/2)*c*d-3*c*d-I*(-c*d^2)^(2/3)*3^(1/2)*_alpha-3*(-c*d^2)^(2/3)*_alpha)/c/d,(I
*3^(1/2)*(-c*d^2)^(1/3)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)/d)^(1/2)),_alpha=RootOf(_Z^3*d+
4*c))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{{\left (d x^{3} + 4 \, c\right )} \sqrt {d x^{3} + c}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(d*x^3+4*c)/(d*x^3+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(x/((d*x^3 + 4*c)*sqrt(d*x^3 + c)), x)

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mupad [B]  time = 25.80, size = 453, normalized size = 2.20 \begin {gather*} \frac {\sqrt {3}\,{314928}^{1/3}\,\ln \left (\frac {{\left (\sqrt {d\,x^3+c}+\sqrt {3}\,\sqrt {-c}-2^{1/3}\,\sqrt {3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x\right )}^3\,\left (54\,\sqrt {d\,x^3+c}-54\,\sqrt {3}\,\sqrt {-c}+54\,2^{1/3}\,\sqrt {3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x\right )}{{\left (d^{1/3}\,x-2^{2/3}\,{\left (-c\right )}^{1/3}\right )}^6}\right )}{2916\,{\left (-c\right )}^{5/6}\,d^{2/3}}+\frac {\sqrt {3}\,{314928}^{1/3}\,\ln \left (\frac {{\left (2\,\sqrt {3}\,\sqrt {-c}-2\,\sqrt {d\,x^3+c}+2^{1/3}\,\sqrt {3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x+2^{1/3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x\,3{}\mathrm {i}\right )}^3\,\left (108\,\sqrt {d\,x^3+c}+108\,\sqrt {3}\,\sqrt {-c}+54\,2^{1/3}\,\sqrt {3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x+2^{1/3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x\,162{}\mathrm {i}\right )}{{\left (2\,d^{1/3}\,x+2^{2/3}\,{\left (-c\right )}^{1/3}-2^{2/3}\,\sqrt {3}\,{\left (-c\right )}^{1/3}\,1{}\mathrm {i}\right )}^6}\right )\,\sqrt {-\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}}{2916\,{\left (-c\right )}^{5/6}\,d^{2/3}}+\frac {\sqrt {3}\,{314928}^{1/3}\,\ln \left (\frac {{\left (2\,\sqrt {d\,x^3+c}+2\,\sqrt {3}\,\sqrt {-c}+2^{1/3}\,\sqrt {3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x-2^{1/3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x\,3{}\mathrm {i}\right )}^3\,\left (108\,\sqrt {d\,x^3+c}-108\,\sqrt {3}\,\sqrt {-c}-54\,2^{1/3}\,\sqrt {3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x+2^{1/3}\,{\left (-c\right )}^{1/6}\,d^{1/3}\,x\,162{}\mathrm {i}\right )}{{\left (2\,d^{1/3}\,x+2^{2/3}\,{\left (-c\right )}^{1/3}+2^{2/3}\,\sqrt {3}\,{\left (-c\right )}^{1/3}\,1{}\mathrm {i}\right )}^6}\right )\,\sqrt {\frac {1}{2}+\frac {\sqrt {3}\,1{}\mathrm {i}}{2}}\,1{}\mathrm {i}}{2916\,{\left (-c\right )}^{5/6}\,d^{2/3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x/((c + d*x^3)^(1/2)*(4*c + d*x^3)),x)

[Out]

(3^(1/2)*314928^(1/3)*log((((c + d*x^3)^(1/2) + 3^(1/2)*(-c)^(1/2) - 2^(1/3)*3^(1/2)*(-c)^(1/6)*d^(1/3)*x)^3*(
54*(c + d*x^3)^(1/2) - 54*3^(1/2)*(-c)^(1/2) + 54*2^(1/3)*3^(1/2)*(-c)^(1/6)*d^(1/3)*x))/(d^(1/3)*x - 2^(2/3)*
(-c)^(1/3))^6))/(2916*(-c)^(5/6)*d^(2/3)) + (3^(1/2)*314928^(1/3)*log(((2*3^(1/2)*(-c)^(1/2) - 2*(c + d*x^3)^(
1/2) + 2^(1/3)*(-c)^(1/6)*d^(1/3)*x*3i + 2^(1/3)*3^(1/2)*(-c)^(1/6)*d^(1/3)*x)^3*(108*(c + d*x^3)^(1/2) + 108*
3^(1/2)*(-c)^(1/2) + 2^(1/3)*(-c)^(1/6)*d^(1/3)*x*162i + 54*2^(1/3)*3^(1/2)*(-c)^(1/6)*d^(1/3)*x))/(2*d^(1/3)*
x + 2^(2/3)*(-c)^(1/3) - 2^(2/3)*3^(1/2)*(-c)^(1/3)*1i)^6)*((3^(1/2)*1i)/2 - 1/2)^(1/2))/(2916*(-c)^(5/6)*d^(2
/3)) + (3^(1/2)*314928^(1/3)*log(((2*(c + d*x^3)^(1/2) + 2*3^(1/2)*(-c)^(1/2) - 2^(1/3)*(-c)^(1/6)*d^(1/3)*x*3
i + 2^(1/3)*3^(1/2)*(-c)^(1/6)*d^(1/3)*x)^3*(108*(c + d*x^3)^(1/2) - 108*3^(1/2)*(-c)^(1/2) + 2^(1/3)*(-c)^(1/
6)*d^(1/3)*x*162i - 54*2^(1/3)*3^(1/2)*(-c)^(1/6)*d^(1/3)*x))/(2*d^(1/3)*x + 2^(2/3)*(-c)^(1/3) + 2^(2/3)*3^(1
/2)*(-c)^(1/3)*1i)^6)*((3^(1/2)*1i)/2 + 1/2)^(1/2)*1i)/(2916*(-c)^(5/6)*d^(2/3))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x}{\sqrt {c + d x^{3}} \left (4 c + d x^{3}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x/(d*x**3+4*c)/(d*x**3+c)**(1/2),x)

[Out]

Integral(x/(sqrt(c + d*x**3)*(4*c + d*x**3)), x)

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