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

Optimal. Leaf size=300 \[ -\frac{7 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (13 A b-4 a B) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\left (1-\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}{\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x}\right ),-7-4 \sqrt{3}\right )}{54 \sqrt [4]{3} a^3 \sqrt [3]{b} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{7 x (13 A b-4 a B)}{54 a^3 \sqrt{a+b x^3}}-\frac{x (13 A b-4 a B)}{18 a^2 \left (a+b x^3\right )^{3/2}}-\frac{A}{2 a x^2 \left (a+b x^3\right )^{3/2}} \]

[Out]

-A/(2*a*x^2*(a + b*x^3)^(3/2)) - ((13*A*b - 4*a*B)*x)/(18*a^2*(a + b*x^3)^(3/2)) - (7*(13*A*b - 4*a*B)*x)/(54*
a^3*Sqrt[a + b*x^3]) - (7*Sqrt[2 + Sqrt[3]]*(13*A*b - 4*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]])/(54*3^(1/4)*a^3*b^(1/3)*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])

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Rubi [A]  time = 0.129629, antiderivative size = 300, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.136, Rules used = {453, 199, 218} \[ -\frac{7 \sqrt{2+\sqrt{3}} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right ) \sqrt{\frac{a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} x+b^{2/3} x^2}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} (13 A b-4 a B) F\left (\sin ^{-1}\left (\frac{\sqrt [3]{b} x+\left (1-\sqrt{3}\right ) \sqrt [3]{a}}{\sqrt [3]{b} x+\left (1+\sqrt{3}\right ) \sqrt [3]{a}}\right )|-7-4 \sqrt{3}\right )}{54 \sqrt [4]{3} a^3 \sqrt [3]{b} \sqrt{\frac{\sqrt [3]{a} \left (\sqrt [3]{a}+\sqrt [3]{b} x\right )}{\left (\left (1+\sqrt{3}\right ) \sqrt [3]{a}+\sqrt [3]{b} x\right )^2}} \sqrt{a+b x^3}}-\frac{7 x (13 A b-4 a B)}{54 a^3 \sqrt{a+b x^3}}-\frac{x (13 A b-4 a B)}{18 a^2 \left (a+b x^3\right )^{3/2}}-\frac{A}{2 a x^2 \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-A/(2*a*x^2*(a + b*x^3)^(3/2)) - ((13*A*b - 4*a*B)*x)/(18*a^2*(a + b*x^3)^(3/2)) - (7*(13*A*b - 4*a*B)*x)/(54*
a^3*Sqrt[a + b*x^3]) - (7*Sqrt[2 + Sqrt[3]]*(13*A*b - 4*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]])/(54*3^(1/4)*a^3*b^(1/3)*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 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 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}{x^3 \left (a+b x^3\right )^{5/2}} \, dx &=-\frac{A}{2 a x^2 \left (a+b x^3\right )^{3/2}}-\frac{\left (\frac{13 A b}{2}-2 a B\right ) \int \frac{1}{\left (a+b x^3\right )^{5/2}} \, dx}{2 a}\\ &=-\frac{A}{2 a x^2 \left (a+b x^3\right )^{3/2}}-\frac{(13 A b-4 a B) x}{18 a^2 \left (a+b x^3\right )^{3/2}}-\frac{(7 (13 A b-4 a B)) \int \frac{1}{\left (a+b x^3\right )^{3/2}} \, dx}{36 a^2}\\ &=-\frac{A}{2 a x^2 \left (a+b x^3\right )^{3/2}}-\frac{(13 A b-4 a B) x}{18 a^2 \left (a+b x^3\right )^{3/2}}-\frac{7 (13 A b-4 a B) x}{54 a^3 \sqrt{a+b x^3}}-\frac{(7 (13 A b-4 a B)) \int \frac{1}{\sqrt{a+b x^3}} \, dx}{108 a^3}\\ &=-\frac{A}{2 a x^2 \left (a+b x^3\right )^{3/2}}-\frac{(13 A b-4 a B) x}{18 a^2 \left (a+b x^3\right )^{3/2}}-\frac{7 (13 A b-4 a B) x}{54 a^3 \sqrt{a+b x^3}}-\frac{7 \sqrt{2+\sqrt{3}} (13 A b-4 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 )}{54 \sqrt [4]{3} a^3 \sqrt [3]{b} \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 = 0.0582123, size = 116, normalized size = 0.39 \[ \frac{a^2 \left (80 B x^3-54 A\right )+7 x^3 \left (a+b x^3\right ) \sqrt{\frac{b x^3}{a}+1} (4 a B-13 A b) \, _2F_1\left (\frac{1}{3},\frac{1}{2};\frac{4}{3};-\frac{b x^3}{a}\right )+a \left (56 b B x^6-260 A b x^3\right )-182 A b^2 x^6}{108 a^3 x^2 \left (a+b x^3\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-182*A*b^2*x^6 + a^2*(-54*A + 80*B*x^3) + a*(-260*A*b*x^3 + 56*b*B*x^6) + 7*(-13*A*b + 4*a*B)*x^3*(a + b*x^3)
*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[1/3, 1/2, 4/3, -((b*x^3)/a)])/(108*a^3*x^2*(a + b*x^3)^(3/2))

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Maple [B]  time = 0.026, size = 689, normalized size = 2.3 \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^3/(b*x^3+a)^(5/2),x)

[Out]

B*(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)))+A*(-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)+91/162*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)))

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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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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^{12} + 3 \, a b^{2} x^{9} + 3 \, a^{2} b x^{6} + a^{3} x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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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^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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