3.427 \(\int \frac{1}{x^7 (a+b x^3)^{3/2}} \, dx\)

Optimal. Leaf size=95 \[ \frac{5 b^2}{4 a^3 \sqrt{a+b x^3}}-\frac{5 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{4 a^{7/2}}+\frac{5 b}{12 a^2 x^3 \sqrt{a+b x^3}}-\frac{1}{6 a x^6 \sqrt{a+b x^3}} \]

[Out]

(5*b^2)/(4*a^3*Sqrt[a + b*x^3]) - 1/(6*a*x^6*Sqrt[a + b*x^3]) + (5*b)/(12*a^2*x^3*Sqrt[a + b*x^3]) - (5*b^2*Ar
cTanh[Sqrt[a + b*x^3]/Sqrt[a]])/(4*a^(7/2))

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Rubi [A]  time = 0.050526, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ -\frac{5 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{4 a^{7/2}}+\frac{5 b \sqrt{a+b x^3}}{4 a^3 x^3}-\frac{5 \sqrt{a+b x^3}}{6 a^2 x^6}+\frac{2}{3 a x^6 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^7*(a + b*x^3)^(3/2)),x]

[Out]

2/(3*a*x^6*Sqrt[a + b*x^3]) - (5*Sqrt[a + b*x^3])/(6*a^2*x^6) + (5*b*Sqrt[a + b*x^3])/(4*a^3*x^3) - (5*b^2*Arc
Tanh[Sqrt[a + b*x^3]/Sqrt[a]])/(4*a^(7/2))

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{1}{x^7 \left (a+b x^3\right )^{3/2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{x^3 (a+b x)^{3/2}} \, dx,x,x^3\right )\\ &=\frac{2}{3 a x^6 \sqrt{a+b x^3}}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{a+b x}} \, dx,x,x^3\right )}{3 a}\\ &=\frac{2}{3 a x^6 \sqrt{a+b x^3}}-\frac{5 \sqrt{a+b x^3}}{6 a^2 x^6}-\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{a+b x}} \, dx,x,x^3\right )}{4 a^2}\\ &=\frac{2}{3 a x^6 \sqrt{a+b x^3}}-\frac{5 \sqrt{a+b x^3}}{6 a^2 x^6}+\frac{5 b \sqrt{a+b x^3}}{4 a^3 x^3}+\frac{\left (5 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+b x}} \, dx,x,x^3\right )}{8 a^3}\\ &=\frac{2}{3 a x^6 \sqrt{a+b x^3}}-\frac{5 \sqrt{a+b x^3}}{6 a^2 x^6}+\frac{5 b \sqrt{a+b x^3}}{4 a^3 x^3}+\frac{(5 b) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^3}\right )}{4 a^3}\\ &=\frac{2}{3 a x^6 \sqrt{a+b x^3}}-\frac{5 \sqrt{a+b x^3}}{6 a^2 x^6}+\frac{5 b \sqrt{a+b x^3}}{4 a^3 x^3}-\frac{5 b^2 \tanh ^{-1}\left (\frac{\sqrt{a+b x^3}}{\sqrt{a}}\right )}{4 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0080245, size = 39, normalized size = 0.41 \[ \frac{2 b^2 \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{b x^3}{a}+1\right )}{3 a^3 \sqrt{a+b x^3}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^7*(a + b*x^3)^(3/2)),x]

[Out]

(2*b^2*Hypergeometric2F1[-1/2, 3, 1/2, 1 + (b*x^3)/a])/(3*a^3*Sqrt[a + b*x^3])

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Maple [A]  time = 0.023, size = 80, normalized size = 0.8 \begin{align*}{\frac{2\,{b}^{2}}{3\,{a}^{3}}{\frac{1}{\sqrt{ \left ({x}^{3}+{\frac{a}{b}} \right ) b}}}}-{\frac{1}{6\,{a}^{2}{x}^{6}}\sqrt{b{x}^{3}+a}}+{\frac{7\,b}{12\,{a}^{3}{x}^{3}}\sqrt{b{x}^{3}+a}}-{\frac{5\,{b}^{2}}{4}{\it Artanh} \left ({\sqrt{b{x}^{3}+a}{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^7/(b*x^3+a)^(3/2),x)

[Out]

2/3*b^2/a^3/((x^3+1/b*a)*b)^(1/2)-1/6/a^2*(b*x^3+a)^(1/2)/x^6+7/12*b/a^3*(b*x^3+a)^(1/2)/x^3-5/4*b^2*arctanh((
b*x^3+a)^(1/2)/a^(1/2))/a^(7/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^3+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55799, size = 444, normalized size = 4.67 \begin{align*} \left [\frac{15 \,{\left (b^{3} x^{9} + a b^{2} x^{6}\right )} \sqrt{a} \log \left (\frac{b x^{3} - 2 \, \sqrt{b x^{3} + a} \sqrt{a} + 2 \, a}{x^{3}}\right ) + 2 \,{\left (15 \, a b^{2} x^{6} + 5 \, a^{2} b x^{3} - 2 \, a^{3}\right )} \sqrt{b x^{3} + a}}{24 \,{\left (a^{4} b x^{9} + a^{5} x^{6}\right )}}, \frac{15 \,{\left (b^{3} x^{9} + a b^{2} x^{6}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x^{3} + a} \sqrt{-a}}{a}\right ) +{\left (15 \, a b^{2} x^{6} + 5 \, a^{2} b x^{3} - 2 \, a^{3}\right )} \sqrt{b x^{3} + a}}{12 \,{\left (a^{4} b x^{9} + a^{5} x^{6}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^3+a)^(3/2),x, algorithm="fricas")

[Out]

[1/24*(15*(b^3*x^9 + a*b^2*x^6)*sqrt(a)*log((b*x^3 - 2*sqrt(b*x^3 + a)*sqrt(a) + 2*a)/x^3) + 2*(15*a*b^2*x^6 +
5*a^2*b*x^3 - 2*a^3)*sqrt(b*x^3 + a))/(a^4*b*x^9 + a^5*x^6), 1/12*(15*(b^3*x^9 + a*b^2*x^6)*sqrt(-a)*arctan(s
qrt(b*x^3 + a)*sqrt(-a)/a) + (15*a*b^2*x^6 + 5*a^2*b*x^3 - 2*a^3)*sqrt(b*x^3 + a))/(a^4*b*x^9 + a^5*x^6)]

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Sympy [A]  time = 6.27099, size = 112, normalized size = 1.18 \begin{align*} - \frac{1}{6 a \sqrt{b} x^{\frac{15}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} + \frac{5 \sqrt{b}}{12 a^{2} x^{\frac{9}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} + \frac{5 b^{\frac{3}{2}}}{4 a^{3} x^{\frac{3}{2}} \sqrt{\frac{a}{b x^{3}} + 1}} - \frac{5 b^{2} \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{\frac{3}{2}}} \right )}}{4 a^{\frac{7}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**7/(b*x**3+a)**(3/2),x)

[Out]

-1/(6*a*sqrt(b)*x**(15/2)*sqrt(a/(b*x**3) + 1)) + 5*sqrt(b)/(12*a**2*x**(9/2)*sqrt(a/(b*x**3) + 1)) + 5*b**(3/
2)/(4*a**3*x**(3/2)*sqrt(a/(b*x**3) + 1)) - 5*b**2*asinh(sqrt(a)/(sqrt(b)*x**(3/2)))/(4*a**(7/2))

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Giac [A]  time = 1.11811, size = 108, normalized size = 1.14 \begin{align*} \frac{1}{12} \, b^{2}{\left (\frac{15 \, \arctan \left (\frac{\sqrt{b x^{3} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{3}} + \frac{8}{\sqrt{b x^{3} + a} a^{3}} + \frac{7 \,{\left (b x^{3} + a\right )}^{\frac{3}{2}} - 9 \, \sqrt{b x^{3} + a} a}{a^{3} b^{2} x^{6}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^7/(b*x^3+a)^(3/2),x, algorithm="giac")

[Out]

1/12*b^2*(15*arctan(sqrt(b*x^3 + a)/sqrt(-a))/(sqrt(-a)*a^3) + 8/(sqrt(b*x^3 + a)*a^3) + (7*(b*x^3 + a)^(3/2)
- 9*sqrt(b*x^3 + a)*a)/(a^3*b^2*x^6))