3.338 \(\int \frac {x^3}{(8 c-d x^3) (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=66 \[ \frac {x^4 \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {4}{3};1,\frac {3}{2};\frac {7}{3};\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{32 c^2 \sqrt {c+d x^3}} \]

[Out]

1/32*x^4*AppellF1(4/3,3/2,1,7/3,-d*x^3/c,1/8*d*x^3/c)*(1+d*x^3/c)^(1/2)/c^2/(d*x^3+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {511, 510} \[ \frac {x^4 \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {4}{3};1,\frac {3}{2};\frac {7}{3};\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{32 c^2 \sqrt {c+d x^3}} \]

Antiderivative was successfully verified.

[In]

Int[x^3/((8*c - d*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

(x^4*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1, 3/2, 7/3, (d*x^3)/(8*c), -((d*x^3)/c)])/(32*c^2*Sqrt[c + d*x^3])

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])

Rubi steps

\begin {align*} \int \frac {x^3}{\left (8 c-d x^3\right ) \left (c+d x^3\right )^{3/2}} \, dx &=\frac {\sqrt {1+\frac {d x^3}{c}} \int \frac {x^3}{\left (8 c-d x^3\right ) \left (1+\frac {d x^3}{c}\right )^{3/2}} \, dx}{c \sqrt {c+d x^3}}\\ &=\frac {x^4 \sqrt {1+\frac {d x^3}{c}} F_1\left (\frac {4}{3};1,\frac {3}{2};\frac {7}{3};\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{32 c^2 \sqrt {c+d x^3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.21, size = 233, normalized size = 3.53 \[ \frac {x \left (\frac {64 c \left (\frac {256 c^2 F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{\left (8 c-d x^3\right ) \left (3 d x^3 \left (F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )-4 F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )+32 c F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )}-1\right )}{d}+x^3 \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {4}{3};\frac {1}{2},1;\frac {7}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )}{864 c^2 \sqrt {c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[x^3/((8*c - d*x^3)*(c + d*x^3)^(3/2)),x]

[Out]

(x*(x^3*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), (d*x^3)/(8*c)] + (64*c*(-1 + (256*c^2*App
ellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), (d*x^3)/(8*c)])/((8*c - d*x^3)*(32*c*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3
)/c), (d*x^3)/(8*c)] + 3*d*x^3*(AppellF1[4/3, 1/2, 2, 7/3, -((d*x^3)/c), (d*x^3)/(8*c)] - 4*AppellF1[4/3, 3/2,
1, 7/3, -((d*x^3)/c), (d*x^3)/(8*c)])))))/d))/(864*c^2*Sqrt[c + d*x^3])

________________________________________________________________________________________

fricas [F]  time = 2.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {d x^{3} + c} x^{3}}{d^{3} x^{9} - 6 \, c d^{2} x^{6} - 15 \, c^{2} d x^{3} - 8 \, c^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(d*x^3 + c)*x^3/(d^3*x^9 - 6*c*d^2*x^6 - 15*c^2*d*x^3 - 8*c^3), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {x^{3}}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="giac")

[Out]

integrate(-x^3/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)), x)

________________________________________________________________________________________

maple [C]  time = 0.24, size = 1038, normalized size = 15.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x)

[Out]

-1/d*(2/3/c*x/((x^3+c/d)*d)^(1/2)-2/9*I/c*3^(1/2)*(-c*d^2)^(1/3)/d*(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)*((x-(-c*d^2)^(1/3)/d)/(-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c
*d^2)^(1/3)/d))^(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)/(d*x^3+c)^(1/2)*EllipticF(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),(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)))-8*c/d*(-2/27/c^2*x/((x^3+c/d)*d)^(1/2)+2/81*I/c^2*3^(1/2)*(-c*d^2)^(1/3)/d*(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)*((x-(-c*d^2)^(1/3)/d)/(-3/2*(-c*d^2)^(1/3)/d+1/2*
I*3^(1/2)*(-c*d^2)^(1/3)/d))^(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)/(d*x^3+c)^(1/2)*EllipticF(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),(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))+1/243*I/c^2/d^3*2^(1/2)*sum(1/_alpha^2*(-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),-1/18*(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-8*c)))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {x^{3}}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3/(-d*x^3+8*c)/(d*x^3+c)^(3/2),x, algorithm="maxima")

[Out]

-integrate(x^3/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x^3}{{\left (d\,x^3+c\right )}^{3/2}\,\left (8\,c-d\,x^3\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3/((c + d*x^3)^(3/2)*(8*c - d*x^3)),x)

[Out]

int(x^3/((c + d*x^3)^(3/2)*(8*c - d*x^3)), x)

________________________________________________________________________________________

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/(-d*x**3+8*c)/(d*x**3+c)**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________