3.855 \(\int \frac{1}{x^5 (a+b x^4)^{3/2}} \, dx\)

Optimal. Leaf size=69 \[ -\frac{3 b}{4 a^2 \sqrt{a+b x^4}}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{4 a^{5/2}}-\frac{1}{4 a x^4 \sqrt{a+b x^4}} \]

[Out]

(-3*b)/(4*a^2*Sqrt[a + b*x^4]) - 1/(4*a*x^4*Sqrt[a + b*x^4]) + (3*b*ArcTanh[Sqrt[a + b*x^4]/Sqrt[a]])/(4*a^(5/
2))

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Rubi [A]  time = 0.0376518, antiderivative size = 71, normalized size of antiderivative = 1.03, number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {266, 51, 63, 208} \[ -\frac{3 \sqrt{a+b x^4}}{4 a^2 x^4}+\frac{3 b \tanh ^{-1}\left (\frac{\sqrt{a+b x^4}}{\sqrt{a}}\right )}{4 a^{5/2}}+\frac{1}{2 a x^4 \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^5*(a + b*x^4)^(3/2)),x]

[Out]

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

Mathematica [C]  time = 0.0073428, size = 37, normalized size = 0.54 \[ -\frac{b \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{b x^4}{a}+1\right )}{2 a^2 \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^5*(a + b*x^4)^(3/2)),x]

[Out]

-(b*Hypergeometric2F1[-1/2, 2, 1/2, 1 + (b*x^4)/a])/(2*a^2*Sqrt[a + b*x^4])

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Maple [A]  time = 0.012, size = 63, normalized size = 0.9 \begin{align*} -{\frac{1}{4\,a{x}^{4}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}-{\frac{3\,b}{4\,{a}^{2}}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{3\,b}{4}\ln \left ({\frac{1}{{x}^{2}} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{4}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^5/(b*x^4+a)^(3/2),x)

[Out]

-1/4/a/x^4/(b*x^4+a)^(1/2)-3/4*b/a^2/(b*x^4+a)^(1/2)+3/4*b/a^(5/2)*ln((2*a+2*a^(1/2)*(b*x^4+a)^(1/2))/x^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^5/(b*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.55663, size = 383, normalized size = 5.55 \begin{align*} \left [\frac{3 \,{\left (b^{2} x^{8} + a b x^{4}\right )} \sqrt{a} \log \left (\frac{b x^{4} + 2 \, \sqrt{b x^{4} + a} \sqrt{a} + 2 \, a}{x^{4}}\right ) - 2 \,{\left (3 \, a b x^{4} + a^{2}\right )} \sqrt{b x^{4} + a}}{8 \,{\left (a^{3} b x^{8} + a^{4} x^{4}\right )}}, -\frac{3 \,{\left (b^{2} x^{8} + a b x^{4}\right )} \sqrt{-a} \arctan \left (\frac{\sqrt{b x^{4} + a} \sqrt{-a}}{a}\right ) +{\left (3 \, a b x^{4} + a^{2}\right )} \sqrt{b x^{4} + a}}{4 \,{\left (a^{3} b x^{8} + a^{4} x^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(3*(b^2*x^8 + a*b*x^4)*sqrt(a)*log((b*x^4 + 2*sqrt(b*x^4 + a)*sqrt(a) + 2*a)/x^4) - 2*(3*a*b*x^4 + a^2)*s
qrt(b*x^4 + a))/(a^3*b*x^8 + a^4*x^4), -1/4*(3*(b^2*x^8 + a*b*x^4)*sqrt(-a)*arctan(sqrt(b*x^4 + a)*sqrt(-a)/a)
+ (3*a*b*x^4 + a^2)*sqrt(b*x^4 + a))/(a^3*b*x^8 + a^4*x^4)]

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Sympy [A]  time = 3.59818, size = 76, normalized size = 1.1 \begin{align*} - \frac{1}{4 a \sqrt{b} x^{6} \sqrt{\frac{a}{b x^{4}} + 1}} - \frac{3 \sqrt{b}}{4 a^{2} x^{2} \sqrt{\frac{a}{b x^{4}} + 1}} + \frac{3 b \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x^{2}} \right )}}{4 a^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**5/(b*x**4+a)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.09851, size = 89, normalized size = 1.29 \begin{align*} -\frac{1}{4} \, b{\left (\frac{3 \, \arctan \left (\frac{\sqrt{b x^{4} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{3 \, b x^{4} + a}{{\left ({\left (b x^{4} + a\right )}^{\frac{3}{2}} - \sqrt{b x^{4} + a} a\right )} a^{2}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^5/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

-1/4*b*(3*arctan(sqrt(b*x^4 + a)/sqrt(-a))/(sqrt(-a)*a^2) + (3*b*x^4 + a)/(((b*x^4 + a)^(3/2) - sqrt(b*x^4 + a
)*a)*a^2))