3.3.35 \(\int \frac {x^3 (A+B x^3)}{(a+b x^3)^{3/2}} \, dx\) [235]

Optimal. Leaf size=269 \[ -\frac {2 (5 A b-8 a B) x}{15 b^2 \sqrt {a+b x^3}}+\frac {2 B x^4}{5 b \sqrt {a+b x^3}}+\frac {4 \sqrt {2+\sqrt {3}} (5 A b-8 a B) \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} b^{7/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}} \]

[Out]

-2/15*(5*A*b-8*B*a)*x/b^2/(b*x^3+a)^(1/2)+2/5*B*x^4/b/(b*x^3+a)^(1/2)+4/45*(5*A*b-8*B*a)*(a^(1/3)+b^(1/3)*x)*E
llipticF((b^(1/3)*x+a^(1/3)*(1-3^(1/2)))/(b^(1/3)*x+a^(1/3)*(1+3^(1/2))),I*3^(1/2)+2*I)*(1/2*6^(1/2)+1/2*2^(1/
2))*((a^(2/3)-a^(1/3)*b^(1/3)*x+b^(2/3)*x^2)/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)*3^(3/4)/b^(7/3)/(b*x^3+a
)^(1/2)/(a^(1/3)*(a^(1/3)+b^(1/3)*x)/(b^(1/3)*x+a^(1/3)*(1+3^(1/2)))^2)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 269, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {470, 294, 224} \begin {gather*} \frac {4 \sqrt {2+\sqrt {3}} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (5 A b-8 a B) F\left (\text {ArcSin}\left (\frac {\sqrt [3]{b} x+\left (1-\sqrt {3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt {3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} b^{7/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}-\frac {2 x (5 A b-8 a B)}{15 b^2 \sqrt {a+b x^3}}+\frac {2 B x^4}{5 b \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(A + B*x^3))/(a + b*x^3)^(3/2),x]

[Out]

(-2*(5*A*b - 8*a*B)*x)/(15*b^2*Sqrt[a + b*x^3]) + (2*B*x^4)/(5*b*Sqrt[a + b*x^3]) + (4*Sqrt[2 + Sqrt[3]]*(5*A*
b - 8*a*B)*(a^(1/3) + b^(1/3)*x)*Sqrt[(a^(2/3) - a^(1/3)*b^(1/3)*x + b^(2/3)*x^2)/((1 + Sqrt[3])*a^(1/3) + b^(
1/3)*x)^2]*EllipticF[ArcSin[((1 - Sqrt[3])*a^(1/3) + b^(1/3)*x)/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)], -7 - 4*S
qrt[3]])/(15*3^(1/4)*b^(7/3)*Sqrt[(a^(1/3)*(a^(1/3) + b^(1/3)*x))/((1 + Sqrt[3])*a^(1/3) + b^(1/3)*x)^2]*Sqrt[
a + b*x^3])

Rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt
[2 + Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s*x + r^2*x^2)/((1 + Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sq
rt[s*((s + r*x)/((1 + Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 - Sqrt[3])*s + r*x)/((1 + Sqrt[3])*s + r*x)
], -7 - 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x] && PosQ[a]

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[d*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p + 1) + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m +
n*(p + 1) + 1)), Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0]
&& NeQ[m + n*(p + 1) + 1, 0]

Rubi steps

\begin {align*} \int \frac {x^3 \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx &=\frac {2 B x^4}{5 b \sqrt {a+b x^3}}-\frac {\left (2 \left (-\frac {5 A b}{2}+4 a B\right )\right ) \int \frac {x^3}{\left (a+b x^3\right )^{3/2}} \, dx}{5 b}\\ &=-\frac {2 (5 A b-8 a B) x}{15 b^2 \sqrt {a+b x^3}}+\frac {2 B x^4}{5 b \sqrt {a+b x^3}}+\frac {(2 (5 A b-8 a B)) \int \frac {1}{\sqrt {a+b x^3}} \, dx}{15 b^2}\\ &=-\frac {2 (5 A b-8 a B) x}{15 b^2 \sqrt {a+b x^3}}+\frac {2 B x^4}{5 b \sqrt {a+b x^3}}+\frac {4 \sqrt {2+\sqrt {3}} (5 A b-8 a B) \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt {\frac {a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1-\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right )|-7-4 \sqrt {3}\right )}{15 \sqrt [4]{3} b^{7/3} \sqrt {\frac {\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt {3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt {a+b x^3}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.05, size = 78, normalized size = 0.29 \begin {gather*} \frac {2 x \left (-5 A b+8 a B+3 b B x^3+(5 A b-8 a B) \sqrt {1+\frac {b x^3}{a}} \, _2F_1\left (\frac {1}{3},\frac {1}{2};\frac {4}{3};-\frac {b x^3}{a}\right )\right )}{15 b^2 \sqrt {a+b x^3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(A + B*x^3))/(a + b*x^3)^(3/2),x]

[Out]

(2*x*(-5*A*b + 8*a*B + 3*b*B*x^3 + (5*A*b - 8*a*B)*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/3, 1/2, 4/3, -((b*x
^3)/a)]))/(15*b^2*Sqrt[a + b*x^3])

________________________________________________________________________________________

Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 626 vs. \(2 (206 ) = 412\).
time = 0.35, size = 627, normalized size = 2.33 Too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(B*x^3+A)/(b*x^3+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

B*(2/3/b^2*a*x/((x^3+a/b)*b)^(1/2)+2/5*x*(b*x^3+a)^(1/2)/b^2+32/45*I*a/b^3*3^(1/2)*(-a*b^2)^(1/3)*(I*(x+1/2/b*
(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3))/(-3/2/b
*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1
/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2*I*3^
(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3)+1/2
*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)))+A*(-2/3/b*x/((x^3+a/b)*b)^(1/2)-4/9*I/b^2*3^(1/2)*(-a*b^2)^(1/3)*(I*(x+1
/2/b*(-a*b^2)^(1/3)-1/2*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)*((x-1/b*(-a*b^2)^(1/3))/(-
3/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)*(-I*(x+1/2/b*(-a*b^2)^(1/3)+1/2*I*3^(1/2)/b*(-a*b^
2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2)/(b*x^3+a)^(1/2)*EllipticF(1/3*3^(1/2)*(I*(x+1/2/b*(-a*b^2)^(1/3)-1/2
*I*3^(1/2)/b*(-a*b^2)^(1/3))*3^(1/2)*b/(-a*b^2)^(1/3))^(1/2),(I*3^(1/2)/b*(-a*b^2)^(1/3)/(-3/2/b*(-a*b^2)^(1/3
)+1/2*I*3^(1/2)/b*(-a*b^2)^(1/3)))^(1/2)))

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^3 + A)*x^3/(b*x^3 + a)^(3/2), x)

________________________________________________________________________________________

Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.53, size = 95, normalized size = 0.35 \begin {gather*} -\frac {2 \, {\left (2 \, {\left ({\left (8 \, B a b - 5 \, A b^{2}\right )} x^{3} + 8 \, B a^{2} - 5 \, A a b\right )} \sqrt {b} {\rm weierstrassPInverse}\left (0, -\frac {4 \, a}{b}, x\right ) - {\left (3 \, B b^{2} x^{4} + {\left (8 \, B a b - 5 \, A b^{2}\right )} x\right )} \sqrt {b x^{3} + a}\right )}}{15 \, {\left (b^{4} x^{3} + a b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(B*x^3+A)/(b*x^3+a)^(3/2),x, algorithm="fricas")

[Out]

-2/15*(2*((8*B*a*b - 5*A*b^2)*x^3 + 8*B*a^2 - 5*A*a*b)*sqrt(b)*weierstrassPInverse(0, -4*a/b, x) - (3*B*b^2*x^
4 + (8*B*a*b - 5*A*b^2)*x)*sqrt(b*x^3 + a))/(b^4*x^3 + a*b^3)

________________________________________________________________________________________

Sympy [A]
time = 5.17, size = 80, normalized size = 0.30 \begin {gather*} \frac {A x^{4} \Gamma \left (\frac {4}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {4}{3}, \frac {3}{2} \\ \frac {7}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {3}{2}} \Gamma \left (\frac {7}{3}\right )} + \frac {B x^{7} \Gamma \left (\frac {7}{3}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {7}{3} \\ \frac {10}{3} \end {matrix}\middle | {\frac {b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac {3}{2}} \Gamma \left (\frac {10}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(B*x**3+A)/(b*x**3+a)**(3/2),x)

[Out]

A*x**4*gamma(4/3)*hyper((4/3, 3/2), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(3/2)*gamma(7/3)) + B*x**7*gamma(7
/3)*hyper((3/2, 7/3), (10/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(3/2)*gamma(10/3))

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*x^3 + A)*x^3/(b*x^3 + a)^(3/2), x)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^3\,\left (B\,x^3+A\right )}{{\left (b\,x^3+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(A + B*x^3))/(a + b*x^3)^(3/2),x)

[Out]

int((x^3*(A + B*x^3))/(a + b*x^3)^(3/2), x)

________________________________________________________________________________________