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

Optimal. Leaf size=64 \[ \frac{x^4 \sqrt{\frac{d x^3}{c}+1} F_1\left (\frac{4}{3};1,\frac{1}{2};\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{4 a \sqrt{c+d x^3}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0512265, antiderivative size = 64, 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} \[ \frac{x^4 \sqrt{\frac{d x^3}{c}+1} F_1\left (\frac{4}{3};1,\frac{1}{2};\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{4 a \sqrt{c+d x^3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^4*Sqrt[1 + (d*x^3)/c]*AppellF1[4/3, 1, 1/2, 7/3, -((b*x^3)/a), -((d*x^3)/c)])/(4*a*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^3}{\left (a+b x^3\right ) \sqrt{c+d x^3}} \, dx &=\frac{\sqrt{1+\frac{d x^3}{c}} \int \frac{x^3}{\left (a+b x^3\right ) \sqrt{1+\frac{d x^3}{c}}} \, dx}{\sqrt{c+d x^3}}\\ &=\frac{x^4 \sqrt{1+\frac{d x^3}{c}} F_1\left (\frac{4}{3};1,\frac{1}{2};\frac{7}{3};-\frac{b x^3}{a},-\frac{d x^3}{c}\right )}{4 a \sqrt{c+d x^3}}\\ \end{align*}

Mathematica [A]  time = 0.0292805, size = 65, normalized size = 1.02 \[ \frac{x^4 \sqrt{\frac{c+d x^3}{c}} F_1\left (\frac{4}{3};\frac{1}{2},1;\frac{7}{3};-\frac{d x^3}{c},-\frac{b x^3}{a}\right )}{4 a \sqrt{c+d x^3}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.03, size = 719, normalized size = 11.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*I/b*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)*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*a/b/d^2*2^(1/2)*sum(1
/_alpha^2/(a*d-b*c)*(-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))

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\left (b x^{3} + a\right )} \sqrt{d x^{3} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{\left (a + b x^{3}\right ) \sqrt{c + d x^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{3}}{{\left (b x^{3} + a\right )} \sqrt{d x^{3} + c}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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