3.324 \(\int \frac {1}{x^6 (8 c-d x^3) \sqrt {c+d x^3}} \, dx\)

Optimal. Leaf size=66 \[ -\frac {\sqrt {\frac {d x^3}{c}+1} F_1\left (-\frac {5}{3};1,\frac {1}{2};-\frac {2}{3};\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{40 c x^5 \sqrt {c+d x^3}} \]

[Out]

-1/40*AppellF1(-5/3,1/2,1,-2/3,-d*x^3/c,1/8*d*x^3/c)*(1+d*x^3/c)^(1/2)/c/x^5/(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} \[ -\frac {\sqrt {\frac {d x^3}{c}+1} F_1\left (-\frac {5}{3};1,\frac {1}{2};-\frac {2}{3};\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{40 c x^5 \sqrt {c+d x^3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^6*(8*c - d*x^3)*Sqrt[c + d*x^3]),x]

[Out]

-(Sqrt[1 + (d*x^3)/c]*AppellF1[-5/3, 1, 1/2, -2/3, (d*x^3)/(8*c), -((d*x^3)/c)])/(40*c*x^5*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 {1}{x^6 \left (8 c-d x^3\right ) \sqrt {c+d x^3}} \, dx &=\frac {\sqrt {1+\frac {d x^3}{c}} \int \frac {1}{x^6 \left (8 c-d x^3\right ) \sqrt {1+\frac {d x^3}{c}}} \, dx}{\sqrt {c+d x^3}}\\ &=-\frac {\sqrt {1+\frac {d x^3}{c}} F_1\left (-\frac {5}{3};1,\frac {1}{2};-\frac {2}{3};\frac {d x^3}{8 c},-\frac {d x^3}{c}\right )}{40 c x^5 \sqrt {c+d x^3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 0.25, size = 261, normalized size = 3.95 \[ \frac {64 c \left (\frac {3264 c^2 d^2 x^6 F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{\left (8 c-d x^3\right ) \left (3 d x^3 \left (F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )-4 F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )+32 c F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )\right )}-16 c^2+7 c d x^3+23 d^2 x^6\right )-23 d^3 x^9 \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {4}{3};\frac {1}{2},1;\frac {7}{3};-\frac {d x^3}{c},\frac {d x^3}{8 c}\right )}{40960 c^4 x^5 \sqrt {c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(x^6*(8*c - d*x^3)*Sqrt[c + d*x^3]),x]

[Out]

(-23*d^3*x^9*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*(-16*c^2 + 7*c
*d*x^3 + 23*d^2*x^6 + (3264*c^2*d^2*x^6*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)])))))/(40960*c^4*x^5*Sqrt[c +
d*x^3])

________________________________________________________________________________________

fricas [F]  time = 11.25, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {d x^{3} + c}}{d^{2} x^{12} - 7 \, c d x^{9} - 8 \, c^{2} x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="fricas")

[Out]

integral(-sqrt(d*x^3 + c)/(d^2*x^12 - 7*c*d*x^9 - 8*c^2*x^6), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {1}{\sqrt {d x^{3} + c} {\left (d x^{3} - 8 \, c\right )} x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="giac")

[Out]

integrate(-1/(sqrt(d*x^3 + c)*(d*x^3 - 8*c)*x^6), x)

________________________________________________________________________________________

maple [C]  time = 0.20, size = 1047, normalized size = 15.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^6/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x)

[Out]

1/8/c*(-1/5/c*(d*x^3+c)^(1/2)/x^5+7/20/c^2*d*(d*x^3+c)^(1/2)/x^2-7/60*I/c^2*d*3^(1/2)*(-c*d^2)^(1/3)*(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)))+1/64/c^2*d*(-1/2*(d*x^3+c)^(1/2)/c/x^2+1/6*I/c*3^(1/2)*(-c*d^2)^(1/3)*(
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)))-1/1728*I/c^3/d*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 \[ -\int \frac {1}{\sqrt {d x^{3} + c} {\left (d x^{3} - 8 \, c\right )} x^{6}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^6/(-d*x^3+8*c)/(d*x^3+c)^(1/2),x, algorithm="maxima")

[Out]

-integrate(1/(sqrt(d*x^3 + c)*(d*x^3 - 8*c)*x^6), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {1}{x^6\,\sqrt {d\,x^3+c}\,\left (8\,c-d\,x^3\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^6*(c + d*x^3)^(1/2)*(8*c - d*x^3)),x)

[Out]

int(1/(x^6*(c + d*x^3)^(1/2)*(8*c - d*x^3)), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {1}{- 8 c x^{6} \sqrt {c + d x^{3}} + d x^{9} \sqrt {c + d x^{3}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**6/(-d*x**3+8*c)/(d*x**3+c)**(1/2),x)

[Out]

-Integral(1/(-8*c*x**6*sqrt(c + d*x**3) + d*x**9*sqrt(c + d*x**3)), x)

________________________________________________________________________________________