3.253 \(\int \frac{A+B x^3}{x^6 (a+b x^3)^{5/2}} \, dx\)

Optimal. Leaf size=334 \[ \frac{91 \sqrt{2+\sqrt{3}} b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (19 A b-10 a B) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt{3}\right )}{540 \sqrt [4]{3} a^4 \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{91 \sqrt{a+b x^3} (19 A b-10 a B)}{540 a^4 x^2}-\frac{13 (19 A b-10 a B)}{135 a^3 x^2 \sqrt{a+b x^3}}-\frac{19 A b-10 a B}{45 a^2 x^2 \left (a+b x^3\right )^{3/2}}-\frac{A}{5 a x^5 \left (a+b x^3\right )^{3/2}} \]

[Out]

-A/(5*a*x^5*(a + b*x^3)^(3/2)) - (19*A*b - 10*a*B)/(45*a^2*x^2*(a + b*x^3)^(3/2)) - (13*(19*A*b - 10*a*B))/(13
5*a^3*x^2*Sqrt[a + b*x^3]) + (91*(19*A*b - 10*a*B)*Sqrt[a + b*x^3])/(540*a^4*x^2) + (91*Sqrt[2 + Sqrt[3]]*b^(2
/3)*(19*A*b - 10*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]])/(540*3^(1/4)*a^4*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 = 0.166023, antiderivative size = 334, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.182, Rules used = {453, 290, 325, 218} \[ \frac{91 \sqrt{2+\sqrt{3}} b^{2/3} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (19 A b-10 a B) F\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{540 \sqrt [4]{3} a^4 \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}+\frac{91 \sqrt{a+b x^3} (19 A b-10 a B)}{540 a^4 x^2}-\frac{13 (19 A b-10 a B)}{135 a^3 x^2 \sqrt{a+b x^3}}-\frac{19 A b-10 a B}{45 a^2 x^2 \left (a+b x^3\right )^{3/2}}-\frac{A}{5 a x^5 \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^3)/(x^6*(a + b*x^3)^(5/2)),x]

[Out]

-A/(5*a*x^5*(a + b*x^3)^(3/2)) - (19*A*b - 10*a*B)/(45*a^2*x^2*(a + b*x^3)^(3/2)) - (13*(19*A*b - 10*a*B))/(13
5*a^3*x^2*Sqrt[a + b*x^3]) + (91*(19*A*b - 10*a*B)*Sqrt[a + b*x^3])/(540*a^4*x^2) + (91*Sqrt[2 + Sqrt[3]]*b^(2
/3)*(19*A*b - 10*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]])/(540*3^(1/4)*a^4*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 453

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

Rule 290

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

Rule 325

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

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}{x^6 \left (a+b x^3\right )^{5/2}} \, dx &=-\frac{A}{5 a x^5 \left (a+b x^3\right )^{3/2}}-\frac{\left (\frac{19 A b}{2}-5 a B\right ) \int \frac{1}{x^3 \left (a+b x^3\right )^{5/2}} \, dx}{5 a}\\ &=-\frac{A}{5 a x^5 \left (a+b x^3\right )^{3/2}}-\frac{19 A b-10 a B}{45 a^2 x^2 \left (a+b x^3\right )^{3/2}}-\frac{(13 (19 A b-10 a B)) \int \frac{1}{x^3 \left (a+b x^3\right )^{3/2}} \, dx}{90 a^2}\\ &=-\frac{A}{5 a x^5 \left (a+b x^3\right )^{3/2}}-\frac{19 A b-10 a B}{45 a^2 x^2 \left (a+b x^3\right )^{3/2}}-\frac{13 (19 A b-10 a B)}{135 a^3 x^2 \sqrt{a+b x^3}}-\frac{(91 (19 A b-10 a B)) \int \frac{1}{x^3 \sqrt{a+b x^3}} \, dx}{270 a^3}\\ &=-\frac{A}{5 a x^5 \left (a+b x^3\right )^{3/2}}-\frac{19 A b-10 a B}{45 a^2 x^2 \left (a+b x^3\right )^{3/2}}-\frac{13 (19 A b-10 a B)}{135 a^3 x^2 \sqrt{a+b x^3}}+\frac{91 (19 A b-10 a B) \sqrt{a+b x^3}}{540 a^4 x^2}+\frac{(91 b (19 A b-10 a B)) \int \frac{1}{\sqrt{a+b x^3}} \, dx}{1080 a^4}\\ &=-\frac{A}{5 a x^5 \left (a+b x^3\right )^{3/2}}-\frac{19 A b-10 a B}{45 a^2 x^2 \left (a+b x^3\right )^{3/2}}-\frac{13 (19 A b-10 a B)}{135 a^3 x^2 \sqrt{a+b x^3}}+\frac{91 (19 A b-10 a B) \sqrt{a+b x^3}}{540 a^4 x^2}+\frac{91 \sqrt{2+\sqrt{3}} b^{2/3} (19 A b-10 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 )}{540 \sqrt [4]{3} a^4 \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 = 0.0412599, size = 83, normalized size = 0.25 \[ \frac{x^3 \left (a+b x^3\right ) \sqrt{\frac{b x^3}{a}+1} \left (\frac{19 A b}{2}-5 a B\right ) \, _2F_1\left (-\frac{2}{3},\frac{5}{2};\frac{1}{3};-\frac{b x^3}{a}\right )-2 a^2 A}{10 a^3 x^5 \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^3)/(x^6*(a + b*x^3)^(5/2)),x]

[Out]

(-2*a^2*A + ((19*A*b)/2 - 5*a*B)*x^3*(a + b*x^3)*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[-2/3, 5/2, 1/3, -((b*x^
3)/a)])/(10*a^3*x^5*(a + b*x^3)^(3/2))

________________________________________________________________________________________

Maple [B]  time = 0.029, size = 722, normalized size = 2.2 \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)/x^6/(b*x^3+a)^(5/2),x)

[Out]

B*(-1/2/a^3*(b*x^3+a)^(1/2)/x^2-2/9/a^2*x/b*(b*x^3+a)^(1/2)/(x^3+a/b)^2-32/27*b/a^3*x/((x^3+a/b)*b)^(1/2)+91/1
62*I/a^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*(-1/5/a^3*(b*x^3+a)^(1/2)/
x^5+27/20*b/a^4*(b*x^3+a)^(1/2)/x^2+2/9/a^3*x*(b*x^3+a)^(1/2)/(x^3+a/b)^2+50/27*b^2/a^4*x/((x^3+a/b)*b)^(1/2)-
1729/1620*I*b/a^4*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)*Elli
pticF(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}} x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^6/(b*x^3+a)^(5/2),x, algorithm="maxima")

[Out]

integrate((B*x^3 + A)/((b*x^3 + a)^(5/2)*x^6), 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^{15} + 3 \, a b^{2} x^{12} + 3 \, a^{2} b x^{9} + a^{3} x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^6/(b*x^3+a)^(5/2),x, algorithm="fricas")

[Out]

integral((B*x^3 + A)*sqrt(b*x^3 + a)/(b^3*x^15 + 3*a*b^2*x^12 + 3*a^2*b*x^9 + a^3*x^6), x)

________________________________________________________________________________________

Sympy [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((B*x**3+A)/x**6/(b*x**3+a)**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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}} x^{6}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^3+A)/x^6/(b*x^3+a)^(5/2),x, algorithm="giac")

[Out]

integrate((B*x^3 + A)/((b*x^3 + a)^(5/2)*x^6), x)