3.397 \(\int \frac {1}{x^3 (a+b x^3) (c+d x^3)^{3/2}} \, dx\)

Optimal. Leaf size=67 \[ -\frac {\sqrt {\frac {d x^3}{c}+1} F_1\left (-\frac {2}{3};1,\frac {3}{2};\frac {1}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a c x^2 \sqrt {c+d x^3}} \]

[Out]

-1/2*AppellF1(-2/3,1,3/2,1/3,-b*x^3/a,-d*x^3/c)*(1+d*x^3/c)^(1/2)/a/c/x^2/(d*x^3+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 67, 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 = {511, 510} \[ -\frac {\sqrt {\frac {d x^3}{c}+1} F_1\left (-\frac {2}{3};1,\frac {3}{2};\frac {1}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a c x^2 \sqrt {c+d x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Mathematica [B]  time = 0.73, size = 425, normalized size = 6.34 \[ \frac {\frac {8 a \left (3 x^3 \left (a^2 d \left (3 c+7 d x^3\right )+a b \left (7 d^2 x^6-3 c^2\right )-3 b^2 c x^3 \left (c+d x^3\right )\right ) \left (2 b c F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )-4 a c \left (3 a^2 d \left (2 c+7 d x^3\right )+a b \left (-6 c^2-3 c d x^3+14 d^2 x^6\right )-6 b^2 c x^3 \left (3 c+d x^3\right )\right ) F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )}{\left (a+b x^3\right ) \left (8 a c F_1\left (\frac {1}{3};\frac {1}{2},1;\frac {4}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )-3 x^3 \left (2 b c F_1\left (\frac {4}{3};\frac {1}{2},2;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+a d F_1\left (\frac {4}{3};\frac {3}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )\right )\right )}+b d x^6 \sqrt {\frac {d x^3}{c}+1} (3 b c-7 a d) F_1\left (\frac {4}{3};\frac {1}{2},1;\frac {7}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )}{48 a^2 c^2 x^2 \sqrt {c+d x^3} (a d-b c)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(b*d*(3*b*c - 7*a*d)*x^6*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1/2, 1, 7/3, -((d*x^3)/c), -((b*x^3)/a)] + (8*a*(-4
*a*c*(-6*b^2*c*x^3*(3*c + d*x^3) + 3*a^2*d*(2*c + 7*d*x^3) + a*b*(-6*c^2 - 3*c*d*x^3 + 14*d^2*x^6))*AppellF1[1
/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)] + 3*x^3*(-3*b^2*c*x^3*(c + d*x^3) + a^2*d*(3*c + 7*d*x^3) + a*b*(
-3*c^2 + 7*d^2*x^6))*(2*b*c*AppellF1[4/3, 1/2, 2, 7/3, -((d*x^3)/c), -((b*x^3)/a)] + a*d*AppellF1[4/3, 3/2, 1,
7/3, -((d*x^3)/c), -((b*x^3)/a)])))/((a + b*x^3)*(8*a*c*AppellF1[1/3, 1/2, 1, 4/3, -((d*x^3)/c), -((b*x^3)/a)
] - 3*x^3*(2*b*c*AppellF1[4/3, 1/2, 2, 7/3, -((d*x^3)/c), -((b*x^3)/a)] + a*d*AppellF1[4/3, 3/2, 1, 7/3, -((d*
x^3)/c), -((b*x^3)/a)]))))/(48*a^2*c^2*(-(b*c) + a*d)*x^2*Sqrt[c + d*x^3])

________________________________________________________________________________________

fricas [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(1/x^3/(b*x^3+a)/(d*x^3+c)^(3/2),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*x^3 + a)*(d*x^3 + c)^(3/2)*x^3), x)

________________________________________________________________________________________

maple [C]  time = 0.23, size = 1084, normalized size = 16.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a*b*(2/3/(a*d-b*c)/((x^3+c/d)*d)^(1/2)/c*d*x-2/9*I/c/(a*d-b*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))+1/3*I*b/d^2*2^(1/2)*sum(1/(a*d-b*c)^2/_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),1/2*(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)/(a*d-b*c)*b/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*b+a)))+1/a*(-1/2/c^2*(d*x^3+c)^(1/2)/x^2-2/3*d/c^2*x/((x^3+c/d)
*d)^(1/2)+7/18*I/c^2*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)*Ell
ipticF(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)))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((b*x^3 + a)*(d*x^3 + c)^(3/2)*x^3), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^3\,\left (b\,x^3+a\right )\,{\left (d\,x^3+c\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/(x**3*(a + b*x**3)*(c + d*x**3)**(3/2)), x)

________________________________________________________________________________________