3.496 \(\int \frac{x}{(a+b x^3)^2 (c+d x^3)^{3/2}} \, dx\)
Optimal. Leaf size=67 \[ \frac{x^2 \sqrt{\frac{d x^3}{c}+1} F_1\left (\frac{2}{3};2,\frac{3}{2};\frac{5}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{2 a^2 c \sqrt{c+d x^3}} \]
[Out]
(x^2*Sqrt[1 + (d*x^3)/c]*AppellF1[2/3, 2, 3/2, 5/3, -((b*x^3)/a), -((d*x^3)/c)])/(2*a^2*c*Sqrt[c + d*x^3])
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Rubi [A] time = 0.0425051, antiderivative size = 67, normalized size of antiderivative = 1.,
number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used =
{511, 510} \[ \frac{x^2 \sqrt{\frac{d x^3}{c}+1} F_1\left (\frac{2}{3};2,\frac{3}{2};\frac{5}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{2 a^2 c \sqrt{c+d x^3}} \]
Antiderivative was successfully verified.
[In]
Int[x/((a + b*x^3)^2*(c + d*x^3)^(3/2)),x]
[Out]
(x^2*Sqrt[1 + (d*x^3)/c]*AppellF1[2/3, 2, 3/2, 5/3, -((b*x^3)/a), -((d*x^3)/c)])/(2*a^2*c*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{x}{\left (a+b x^3\right )^2 \left (c+d x^3\right )^{3/2}} \, dx &=\frac{\sqrt{1+\frac{d x^3}{c}} \int \frac{x}{\left (a+b x^3\right )^2 \left (1+\frac{d x^3}{c}\right )^{3/2}} \, dx}{c \sqrt{c+d x^3}}\\ &=\frac{x^2 \sqrt{1+\frac{d x^3}{c}} F_1\left (\frac{2}{3};2,\frac{3}{2};\frac{5}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{2 a^2 c \sqrt{c+d x^3}}\\ \end{align*}
Mathematica [B] time = 0.257256, size = 216, normalized size = 3.22 \[ -\frac{x^2 \left (5 \left (a+b x^3\right ) \sqrt{\frac{d x^3}{c}+1} \left (a^2 d^2+6 a b c d-b^2 c^2\right ) F_1\left (\frac{2}{3};\frac{1}{2},1;\frac{5}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )-10 a \left (2 a^2 d^2+2 a b d^2 x^3+b^2 c \left (c+d x^3\right )\right )+b d x^3 \left (a+b x^3\right ) \sqrt{\frac{d x^3}{c}+1} (2 a d+b c) F_1\left (\frac{5}{3};\frac{1}{2},1;\frac{8}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )\right )}{30 a^2 c \left (a+b x^3\right ) \sqrt{c+d x^3} (b c-a d)^2} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[x/((a + b*x^3)^2*(c + d*x^3)^(3/2)),x]
[Out]
-(x^2*(-10*a*(2*a^2*d^2 + 2*a*b*d^2*x^3 + b^2*c*(c + d*x^3)) + 5*(-(b^2*c^2) + 6*a*b*c*d + a^2*d^2)*(a + b*x^3
)*Sqrt[1 + (d*x^3)/c]*AppellF1[2/3, 1/2, 1, 5/3, -((d*x^3)/c), -((b*x^3)/a)] + b*d*(b*c + 2*a*d)*x^3*(a + b*x^
3)*Sqrt[1 + (d*x^3)/c]*AppellF1[5/3, 1/2, 1, 8/3, -((d*x^3)/c), -((b*x^3)/a)]))/(30*a^2*c*(b*c - a*d)^2*(a + b
*x^3)*Sqrt[c + d*x^3])
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Maple [C] time = 0.039, size = 986, normalized size = 14.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x/(b*x^3+a)^2/(d*x^3+c)^(3/2),x)
[Out]
2/3*d^2/c*x^2/(a*d-b*c)^2/((x^3+1/d*c)*d)^(1/2)+1/3*b^2/(a*d-b*c)^2/a*x^2*(d*x^3+c)^(1/2)/(b*x^3+a)-2/3*I*(-1/
3*d^2/c/(a*d-b*c)^2-1/6*b*d/(a*d-b*c)^2/a)*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)*((-3/2/d*(-d^2*c)^(1/3)+1/2*I*3^(1/2)/d*(-d^2*c)^(1/3))*EllipticE(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/d*(-d^2*c)^(1/3)*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)))+1/18*I/a/d^2*b*2^(1/2)*sum((11*a*d-2*b*c)/(
a*d-b*c)^3/_alpha*(-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)*_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)/(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{x}{{\left (b x^{3} + a\right )}^{2}{\left (d x^{3} + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="maxima")
[Out]
integrate(x/((b*x^3 + a)^2*(d*x^3 + c)^(3/2)), x)
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Fricas [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(x/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="fricas")
[Out]
Timed out
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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(x/(b*x**3+a)**2/(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{x}{{\left (b x^{3} + a\right )}^{2}{\left (d x^{3} + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x/(b*x^3+a)^2/(d*x^3+c)^(3/2),x, algorithm="giac")
[Out]
integrate(x/((b*x^3 + a)^2*(d*x^3 + c)^(3/2)), x)