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

Optimal. Leaf size=65 \[ \frac {c x^2 \sqrt {c+d x^3} F_1\left (\frac {2}{3};2,-\frac {3}{2};\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a^2 \sqrt {\frac {d x^3}{c}+1}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 65, 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} \[ \frac {c x^2 \sqrt {c+d x^3} F_1\left (\frac {2}{3};2,-\frac {3}{2};\frac {5}{3};-\frac {b x^3}{a},-\frac {d x^3}{c}\right )}{2 a^2 \sqrt {\frac {d x^3}{c}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

________________________________________________________________________________________

Mathematica [B]  time = 0.26, size = 177, normalized size = 2.72 \[ \frac {x^2 \left (-d x^3 \left (a+b x^3\right ) \sqrt {\frac {d x^3}{c}+1} (b c-7 a d) F_1\left (\frac {5}{3};\frac {1}{2},1;\frac {8}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )+5 c \left (a+b x^3\right ) \sqrt {\frac {d x^3}{c}+1} (2 a d+b c) F_1\left (\frac {2}{3};\frac {1}{2},1;\frac {5}{3};-\frac {d x^3}{c},-\frac {b x^3}{a}\right )-10 a \left (c+d x^3\right ) (a d-b c)\right )}{30 a^2 b \left (a+b x^3\right ) \sqrt {c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(x^2*(-10*a*(-(b*c) + a*d)*(c + d*x^3) + 5*c*(b*c + 2*a*d)*(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)] - d*(b*c - 7*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*b*(a + b*x^3)*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(x*(d*x^3+c)^(3/2)/(b*x^3+a)^2,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.36, size = 955, normalized size = 14.69 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(a*d-b*c)/a/b*x^2*(d*x^3+c)^(1/2)/(b*x^3+a)-2/3*I*(1/b^2*d^2+1/6*(a*d-b*c)/a/b^2*d)*3^(1/2)*(-c*d^2)^(1/3
)/d*(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/b^2/d^2*2^(1/2)*sum((7*a^2*d^2-5*a*b*c*d-2*b^2*c^2)/_alpha/(a*d-b*c)*(-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))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________