3.457 \(\int \frac {1}{(8 c-d x^3)^2 (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=64 \[ \frac {x \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {1}{3};2,\frac {3}{2};\frac {4}{3};\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{64 c^3 \sqrt {c+d x^3}} \]

[Out]

1/64*x*AppellF1(1/3,3/2,2,4/3,-d*x^3/c,1/8*d*x^3/c)*(1+d*x^3/c)^(1/2)/c^3/(d*x^3+c)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {430, 429} \[ \frac {x \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {1}{3};2,\frac {3}{2};\frac {4}{3};\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{64 c^3 \sqrt {c+d x^3}} \]

Antiderivative was successfully verified.

[In]

Int[1/((8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]

[Out]

(x*Sqrt[1 + (d*x^3)/c]*AppellF1[1/3, 2, 3/2, 4/3, (d*x^3)/(8*c), -((d*x^3)/c)])/(64*c^3*Sqrt[c + d*x^3])

Rule 429

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

Rule 430

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

Rubi steps

\begin {align*} \int \frac {1}{\left (8 c-d x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx &=\frac {\sqrt {1+\frac {d x^3}{c}} \int \frac {1}{\left (8 c-d x^3\right )^2 \left (1+\frac {d x^3}{c}\right )^{3/2}} \, dx}{c \sqrt {c+d x^3}}\\ &=\frac {x \sqrt {1+\frac {d x^3}{c}} F_1\left (\frac {1}{3};2,\frac {3}{2};\frac {4}{3};\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{64 c^3 \sqrt {c+d x^3}}\\ \end {align*}

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Mathematica [B]  time = 0.33, size = 253, normalized size = 3.95 \[ \frac {x \left (192 c \left (\frac {1216 c^2 F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{\left (8 c-d x^3\right ) \left (3 d x^3 \left (F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )-4 F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )+32 c F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )}+\frac {5 d x^3-43 c}{d x^3-8 c}\right )-15 d x^3 \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {4}{3};\frac {1}{2},1;\frac {7}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )}{124416 c^4 \sqrt {c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((8*c - d*x^3)^2*(c + d*x^3)^(3/2)),x]

[Out]

(x*(-15*d*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)] + 192*c*((-43*c + 5*
d*x^3)/(-8*c + d*x^3) + (1216*c^2*AppellF1[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)]))))))/(124416*c^4*Sqrt[c + d*x^3])

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fricas [F]  time = 3.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {d x^{3} + c}}{d^{4} x^{12} - 14 \, c d^{3} x^{9} + 33 \, c^{2} d^{2} x^{6} + 112 \, c^{3} d x^{3} + 64 \, c^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*x^3 + c)/(d^4*x^12 - 14*c*d^3*x^9 + 33*c^2*d^2*x^6 + 112*c^3*d*x^3 + 64*c^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)^2), x)

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maple [C]  time = 0.17, size = 747, normalized size = 11.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x)

[Out]

-1/1944*(d*x^3+c)^(1/2)/(d*x^3-8*c)/c^3*x+2/243/((x^3+c/d)*d)^(1/2)/c^3*x-5/1944*I/c^3*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))-1/972*I/c^3/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))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (d x^{3} + c\right )}^{\frac {3}{2}} {\left (d x^{3} - 8 \, c\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-d*x^3+8*c)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((d*x^3 + c)^(3/2)*(d*x^3 - 8*c)^2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c + d*x^3)^(3/2)*(8*c - d*x^3)^2),x)

[Out]

int(1/((c + d*x^3)^(3/2)*(8*c - d*x^3)^2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (- 8 c + d x^{3}\right )^{2} \left (c + d x^{3}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-d*x**3+8*c)**2/(d*x**3+c)**(3/2),x)

[Out]

Integral(1/((-8*c + d*x**3)**2*(c + d*x**3)**(3/2)), x)

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