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

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

[Out]

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

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Rubi [A]  time = 0.0298604, antiderivative size = 64, normalized size of antiderivative = 1., 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} \[ \frac{x \sqrt{\frac{d x^3}{c}+1} F_1\left (\frac{1}{3};1,\frac{3}{2};\frac{4}{3};\frac{d x^3}{8 c},-\frac{d x^3}{c}\right )}{8 c^2 \sqrt{c+d x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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])

Rubi steps

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

Mathematica [B]  time = 0.128031, size = 230, normalized size = 3.59 \[ \frac{x \left (64 \left (\frac{176 F_1\left (\frac{1}{3};\frac{1}{2},1;\frac{4}{3};-\frac{d x^3}{c},\frac{d x^3}{8 c}\right )}{\left (8 c-d x^3\right ) \left (3 d x^3 \left (F_1\left (\frac{4}{3};\frac{1}{2},2;\frac{7}{3};-\frac{d x^3}{c},\frac{d x^3}{8 c}\right )-4 F_1\left (\frac{4}{3};\frac{3}{2},1;\frac{7}{3};-\frac{d x^3}{c},\frac{d x^3}{8 c}\right )\right )+32 c F_1\left (\frac{1}{3};\frac{1}{2},1;\frac{4}{3};-\frac{d x^3}{c},\frac{d x^3}{8 c}\right )\right )}+\frac{1}{c^2}\right )-\frac{d x^3 \sqrt{\frac{d x^3}{c}+1} F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{d x^3}{c},\frac{d x^3}{8 c}\right )}{c^3}\right )}{864 \sqrt{c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x*(-((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)])/c^3) + 64*(c^(-2) +
(176*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)]))))))/(864*Sqrt[c + d*x^3])

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Maple [C]  time = 0.005, size = 721, normalized size = 11.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/27/c^2*x/((x^3+1/d*c)*d)^(1/2)-2/81*I/c^2*3^(1/2)/d*(-d^2*c)^(1/3)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/
d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/2)*((x-1/d*(-d^2*c)^(1/3))/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)
/d*(-d^2*c)^(1/3)))^(1/2)*(-I*(x+1/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3)
)^(1/2)/(d*x^3+c)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/
2)*d/(-d^2*c)^(1/3))^(1/2),(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))
^(1/2))-1/243*I/c^2/d^3*2^(1/2)*sum(1/_alpha^2*(-d^2*c)^(1/3)*(1/2*I*d*(2*x+1/d*(-I*3^(1/2)*(-d^2*c)^(1/3)+(-d
^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)*(d*(x-1/d*(-d^2*c)^(1/3))/(-3*(-d^2*c)^(1/3)+I*3^(1/2)*(-d^2*c)^(1/3)))^(1
/2)*(-1/2*I*d*(2*x+1/d*(I*3^(1/2)*(-d^2*c)^(1/3)+(-d^2*c)^(1/3)))/(-d^2*c)^(1/3))^(1/2)/(d*x^3+c)^(1/2)*(I*(-d
^2*c)^(1/3)*_alpha*3^(1/2)*d-I*3^(1/2)*(-d^2*c)^(2/3)+2*_alpha^2*d^2-(-d^2*c)^(1/3)*_alpha*d-(-d^2*c)^(2/3))*E
llipticPi(1/3*3^(1/2)*(I*(x+1/2/d*(-d^2*c)^(1/3)-1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*3^(1/2)*d/(-d^2*c)^(1/3))^(1/
2),-1/18/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_alpha^2*d-I*(-d^2*c)^(2/3)*3^(1/2)*_alpha+I*3^(1/2)*c*d-3*(-d^2*c)^(2/
3)*_alpha-3*c*d)/c,(I*3^(1/2)/d*(-d^2*c)^(1/3)/(-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3)))^(1/2)),
_alpha=RootOf(_Z^3*d-8*c))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{1}{{\left (d x^{3} + c\right )}^{\frac{3}{2}}{\left (d x^{3} - 8 \, c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{\sqrt{d x^{3} + c}}{d^{3} x^{9} - 6 \, c d^{2} x^{6} - 15 \, c^{2} d x^{3} - 8 \, c^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int -\frac{1}{{\left (d x^{3} + c\right )}^{\frac{3}{2}}{\left (d x^{3} - 8 \, c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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