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]
-(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])
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Rubi [A] time = 0.0572465, antiderivative size = 67, 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 =
{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 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])
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])
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 = 0.578751, 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])
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Maple [C] time = 0.008, size = 1084, normalized size = 16.2 \begin{align*} \text{result too large to display} \end{align*}
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*(-2/3*d/c^2*x/((x^3+1/d*c)*d)^(1/2)-1/2/c^2*(d*x^3+c)^(1/2)/x^2+7/18*I/c^2*3^(1/2)*(-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)))-b/a*(2/3*d/c*x/(a*d-b*c)/((x^3+1/d*c)*d)^(1/2)-2/9*I/c/(a*d-b*c)*3^(
1/2)*(-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))+1/3*I/d^2*b*2^(1/2)*sum(1/(a*d-b*c)^2/_al
pha^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))*EllipticPi(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),1/2*b/d*(2*I*(-d^2*c)^(1/3)*3^(1/2)*_a
lpha^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)/(a*d-b*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*b+a)))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a\right )}{\left (d x^{3} + c\right )}^{\frac{3}{2}} x^{3}}\,{d x} \end{align*}
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)
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Fricas [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}
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]
Exception raised: UnboundLocalError
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{x^{3} \left (a + b x^{3}\right ) \left (c + d x^{3}\right )^{\frac{3}{2}}}\, dx \end{align*}
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)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b x^{3} + a\right )}{\left (d x^{3} + c\right )}^{\frac{3}{2}} x^{3}}\,{d x} \end{align*}
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)