3.486 \(\int \frac {x}{(a+b x^3)^2 \sqrt {c+d x^3}} \, dx\)

Optimal. Leaf size=64 \[ \frac {x^2 \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {2}{3};2,\frac {1}{2};\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a^2 \sqrt {c+d x^3}} \]

[Out]

1/2*x^2*AppellF1(2/3,2,1/2,5/3,-b*x^3/a,-d*x^3/c)*(1+d*x^3/c)^(1/2)/a^2/(d*x^3+c)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 64, normalized size of antiderivative = 1.00, 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} \[ \frac {x^2 \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {2}{3};2,\frac {1}{2};\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a^2 \sqrt {c+d x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [B]  time = 0.16, size = 172, normalized size = 2.69 \[ \frac {-b d x^5 \left (a+b x^3\right ) \sqrt {\frac {d x^3}{c}+1} F_1\left (\frac {5}{3};\frac {1}{2},1;\frac {8}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+5 x^2 \left (a+b x^3\right ) \sqrt {\frac {d x^3}{c}+1} (b c-3 a d) F_1\left (\frac {2}{3};\frac {1}{2},1;\frac {5}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+10 a b x^2 \left (c+d x^3\right )}{30 a^2 \left (a+b x^3\right ) \sqrt {c+d x^3} (b c-a d)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(10*a*b*x^2*(c + d*x^3) + 5*(b*c - 3*a*d)*x^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*x^5*(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*(b*c - a*d)*(a + b*x^3)*Sqrt[c + d*x^3])

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.40, size = 923, normalized size = 14.42 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/a*b/(a*d-b*c)*x^2*(d*x^3+c)^(1/2)/(b*x^3+a)-1/9*I/(a*d-b*c)/a*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)*((-3/2*(-c*d^2)^(1/3)/d+1/2*I*3^(1/2)*(-c*d^2)^(1/3)/d)*EllipticE(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))+(-c*d^2)^(1/3)/d*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)))+1/18*I/a/d^2*2^(1/2)*sum(
(-5*a*d+2*b*c)/(a*d-b*c)^2/_alpha*(-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=Root
Of(_Z^3*b+a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {x}{{\left (b\,x^3+a\right )}^2\,\sqrt {d\,x^3+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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