3.251 \(\int \frac{A+B x^3}{(a+b x^3)^{5/2}} \, dx\)

Optimal. Leaf size=283 \[ \frac{2 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (2 a B+7 A b) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt{3}\right )}{27 \sqrt [4]{3} a^2 b^{4/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{2 x (2 a B+7 A b)}{27 a^2 b \sqrt{a+b x^3}}+\frac{2 x (A b-a B)}{9 a b \left (a+b x^3\right )^{3/2}} \]

[Out]

(2*(A*b - a*B)*x)/(9*a*b*(a + b*x^3)^(3/2)) + (2*(7*A*b + 2*a*B)*x)/(27*a^2*b*Sqrt[a + b*x^3]) + (2*Sqrt[2 + S
qrt[3]]*(7*A*b + 2*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*Sqrt[3]])/(27*3^(1/4)*a^2*b^(4/3)*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])

________________________________________________________________________________________

Rubi [A]  time = 0.09851, antiderivative size = 283, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {385, 199, 218} \[ \frac{2 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (2 a B+7 A b) F\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{27 \sqrt [4]{3} a^2 b^{4/3} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{2 x (2 a B+7 A b)}{27 a^2 b \sqrt{a+b x^3}}+\frac{2 x (A b-a B)}{9 a b \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(a + b*x^3)^(5/2),x]

[Out]

(2*(A*b - a*B)*x)/(9*a*b*(a + b*x^3)^(3/2)) + (2*(7*A*b + 2*a*B)*x)/(27*a^2*b*Sqrt[a + b*x^3]) + (2*Sqrt[2 + S
qrt[3]]*(7*A*b + 2*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*Sqrt[3]])/(27*3^(1/4)*a^2*b^(4/3)*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 385

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

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

Rule 218

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

Rubi steps

\begin{align*} \int \frac{A+B x^3}{\left (a+b x^3\right )^{5/2}} \, dx &=\frac{2 (A b-a B) x}{9 a b \left (a+b x^3\right )^{3/2}}+\frac{\left (2 \left (\frac{7 A b}{2}+a B\right )\right ) \int \frac{1}{\left (a+b x^3\right )^{3/2}} \, dx}{9 a b}\\ &=\frac{2 (A b-a B) x}{9 a b \left (a+b x^3\right )^{3/2}}+\frac{2 (7 A b+2 a B) x}{27 a^2 b \sqrt{a+b x^3}}+\frac{(7 A b+2 a B) \int \frac{1}{\sqrt{a+b x^3}} \, dx}{27 a^2 b}\\ &=\frac{2 (A b-a B) x}{9 a b \left (a+b x^3\right )^{3/2}}+\frac{2 (7 A b+2 a B) x}{27 a^2 b \sqrt{a+b x^3}}+\frac{2 \sqrt{2+\sqrt{3}} (7 A b+2 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 )}{27 \sqrt [4]{3} a^2 b^{4/3} \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]  time = 0.0545086, size = 103, normalized size = 0.36 \[ \frac{-2 a^2 B x+x \left (a+b x^3\right ) \sqrt{\frac{b x^3}{a}+1} (2 a B+7 A b) \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{b x^3}{a}\right )+4 a b x \left (5 A+B x^3\right )+14 A b^2 x^4}{27 a^2 b \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(a + b*x^3)^(5/2),x]

[Out]

(-2*a^2*B*x + 14*A*b^2*x^4 + 4*a*b*x*(5*A + B*x^3) + (7*A*b + 2*a*B)*x*(a + b*x^3)*Sqrt[1 + (b*x^3)/a]*Hyperge
ometric2F1[1/3, 1/2, 4/3, -((b*x^3)/a)])/(27*a^2*b*(a + b*x^3)^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.018, size = 674, normalized size = 2.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^3+A)/(b*x^3+a)^(5/2),x)

[Out]

B*(-2/9*x/b^3*(b*x^3+a)^(1/2)/(x^3+a/b)^2+4/27/b/a*x/((x^3+a/b)*b)^(1/2)-4/81*I/b^2/a*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*(2/9/a*x/b^2*(b*x^3+a)^(1/2)/(x^3+a/b)^2+14/27/a^2*x/((x^3+
a/b)*b)^(1/2)-14/81*I/a^2*3^(1/2)/b*(-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., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)/(b*x^3 + a)^(5/2), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B x^{3} + A\right )} \sqrt{b x^{3} + a}}{b^{3} x^{9} + 3 \, a b^{2} x^{6} + 3 \, a^{2} b x^{3} + a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="fricas")

[Out]

integral((B*x^3 + A)*sqrt(b*x^3 + a)/(b^3*x^9 + 3*a*b^2*x^6 + 3*a^2*b*x^3 + a^3), x)

________________________________________________________________________________________

Sympy [A]  time = 170.077, size = 78, normalized size = 0.28 \begin{align*} \frac{A x \Gamma \left (\frac{1}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{1}{3}, \frac{5}{2} \\ \frac{4}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac{5}{2}} \Gamma \left (\frac{4}{3}\right )} + \frac{B x^{4} \Gamma \left (\frac{4}{3}\right ){{}_{2}F_{1}\left (\begin{matrix} \frac{4}{3}, \frac{5}{2} \\ \frac{7}{3} \end{matrix}\middle |{\frac{b x^{3} e^{i \pi }}{a}} \right )}}{3 a^{\frac{5}{2}} \Gamma \left (\frac{7}{3}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**3+A)/(b*x**3+a)**(5/2),x)

[Out]

A*x*gamma(1/3)*hyper((1/3, 5/2), (4/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(5/2)*gamma(4/3)) + B*x**4*gamma(4/3)
*hyper((4/3, 5/2), (7/3,), b*x**3*exp_polar(I*pi)/a)/(3*a**(5/2)*gamma(7/3))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x^{3} + A}{{\left (b x^{3} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="giac")

[Out]

integrate((B*x^3 + A)/(b*x^3 + a)^(5/2), x)