3.2075 \(\int (a+\frac{b}{x^4})^{5/2} x \, dx\)

Optimal. Leaf size=91 \[ -\frac{15}{16} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b}}{x^2 \sqrt{a+\frac{b}{x^4}}}\right )+\frac{1}{2} x^2 \left (a+\frac{b}{x^4}\right )^{5/2}-\frac{5 b \left (a+\frac{b}{x^4}\right )^{3/2}}{8 x^2}-\frac{15 a b \sqrt{a+\frac{b}{x^4}}}{16 x^2} \]

[Out]

(-15*a*b*Sqrt[a + b/x^4])/(16*x^2) - (5*b*(a + b/x^4)^(3/2))/(8*x^2) + ((a + b/x^4)^(5/2)*x^2)/2 - (15*a^2*Sqr
t[b]*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^4]*x^2)])/16

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Rubi [A]  time = 0.0733903, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.462, Rules used = {335, 275, 277, 195, 217, 206} \[ -\frac{15}{16} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b}}{x^2 \sqrt{a+\frac{b}{x^4}}}\right )+\frac{1}{2} x^2 \left (a+\frac{b}{x^4}\right )^{5/2}-\frac{5 b \left (a+\frac{b}{x^4}\right )^{3/2}}{8 x^2}-\frac{15 a b \sqrt{a+\frac{b}{x^4}}}{16 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^4)^(5/2)*x,x]

[Out]

(-15*a*b*Sqrt[a + b/x^4])/(16*x^2) - (5*b*(a + b/x^4)^(3/2))/(8*x^2) + ((a + b/x^4)^(5/2)*x^2)/2 - (15*a^2*Sqr
t[b]*ArcTanh[Sqrt[b]/(Sqrt[a + b/x^4]*x^2)])/16

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 277

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

Rule 195

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

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \left (a+\frac{b}{x^4}\right )^{5/2} x \, dx &=-\operatorname{Subst}\left (\int \frac{\left (a+b x^4\right )^{5/2}}{x^3} \, dx,x,\frac{1}{x}\right )\\ &=-\left (\frac{1}{2} \operatorname{Subst}\left (\int \frac{\left (a+b x^2\right )^{5/2}}{x^2} \, dx,x,\frac{1}{x^2}\right )\right )\\ &=\frac{1}{2} \left (a+\frac{b}{x^4}\right )^{5/2} x^2-\frac{1}{2} (5 b) \operatorname{Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{5 b \left (a+\frac{b}{x^4}\right )^{3/2}}{8 x^2}+\frac{1}{2} \left (a+\frac{b}{x^4}\right )^{5/2} x^2-\frac{1}{8} (15 a b) \operatorname{Subst}\left (\int \sqrt{a+b x^2} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{15 a b \sqrt{a+\frac{b}{x^4}}}{16 x^2}-\frac{5 b \left (a+\frac{b}{x^4}\right )^{3/2}}{8 x^2}+\frac{1}{2} \left (a+\frac{b}{x^4}\right )^{5/2} x^2-\frac{1}{16} \left (15 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\frac{1}{x^2}\right )\\ &=-\frac{15 a b \sqrt{a+\frac{b}{x^4}}}{16 x^2}-\frac{5 b \left (a+\frac{b}{x^4}\right )^{3/2}}{8 x^2}+\frac{1}{2} \left (a+\frac{b}{x^4}\right )^{5/2} x^2-\frac{1}{16} \left (15 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{1}{\sqrt{a+\frac{b}{x^4}} x^2}\right )\\ &=-\frac{15 a b \sqrt{a+\frac{b}{x^4}}}{16 x^2}-\frac{5 b \left (a+\frac{b}{x^4}\right )^{3/2}}{8 x^2}+\frac{1}{2} \left (a+\frac{b}{x^4}\right )^{5/2} x^2-\frac{15}{16} a^2 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b}}{\sqrt{a+\frac{b}{x^4}} x^2}\right )\\ \end{align*}

Mathematica [C]  time = 0.0152759, size = 49, normalized size = 0.54 \[ -\frac{a^2 x^{10} \left (a+\frac{b}{x^4}\right )^{5/2} \left (a x^4+b\right ) \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};\frac{a x^4}{b}+1\right )}{14 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^4)^(5/2)*x,x]

[Out]

-(a^2*(a + b/x^4)^(5/2)*x^10*(b + a*x^4)*Hypergeometric2F1[3, 7/2, 9/2, 1 + (a*x^4)/b])/(14*b^3)

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Maple [A]  time = 0.016, size = 108, normalized size = 1.2 \begin{align*} -{\frac{{x}^{2}}{16} \left ({\frac{a{x}^{4}+b}{{x}^{4}}} \right ) ^{{\frac{5}{2}}} \left ( 15\,\sqrt{b}{a}^{2}\ln \left ( 2\,{\frac{\sqrt{b}\sqrt{a{x}^{4}+b}+b}{{x}^{2}}} \right ){x}^{8}-8\,{a}^{2}{x}^{8}\sqrt{a{x}^{4}+b}+9\,ba\sqrt{a{x}^{4}+b}{x}^{4}+2\,{b}^{2}\sqrt{a{x}^{4}+b} \right ) \left ( a{x}^{4}+b \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((a*x^4+b)/x^4)^(5/2)*x^2*(15*b^(1/2)*a^2*ln(2*(b^(1/2)*(a*x^4+b)^(1/2)+b)/x^2)*x^8-8*a^2*x^8*(a*x^4+b)^
(1/2)+9*b*a*(a*x^4+b)^(1/2)*x^4+2*b^2*(a*x^4+b)^(1/2))/(a*x^4+b)^(5/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52616, size = 387, normalized size = 4.25 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{b} x^{6} \log \left (\frac{a x^{4} - 2 \, \sqrt{b} x^{2} \sqrt{\frac{a x^{4} + b}{x^{4}}} + 2 \, b}{x^{4}}\right ) + 2 \,{\left (8 \, a^{2} x^{8} - 9 \, a b x^{4} - 2 \, b^{2}\right )} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{32 \, x^{6}}, \frac{15 \, a^{2} \sqrt{-b} x^{6} \arctan \left (\frac{\sqrt{-b} x^{2} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{b}\right ) +{\left (8 \, a^{2} x^{8} - 9 \, a b x^{4} - 2 \, b^{2}\right )} \sqrt{\frac{a x^{4} + b}{x^{4}}}}{16 \, x^{6}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*(15*a^2*sqrt(b)*x^6*log((a*x^4 - 2*sqrt(b)*x^2*sqrt((a*x^4 + b)/x^4) + 2*b)/x^4) + 2*(8*a^2*x^8 - 9*a*b*
x^4 - 2*b^2)*sqrt((a*x^4 + b)/x^4))/x^6, 1/16*(15*a^2*sqrt(-b)*x^6*arctan(sqrt(-b)*x^2*sqrt((a*x^4 + b)/x^4)/b
) + (8*a^2*x^8 - 9*a*b*x^4 - 2*b^2)*sqrt((a*x^4 + b)/x^4))/x^6]

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Sympy [A]  time = 5.07993, size = 124, normalized size = 1.36 \begin{align*} \frac{a^{\frac{5}{2}} x^{2}}{2 \sqrt{1 + \frac{b}{a x^{4}}}} - \frac{a^{\frac{3}{2}} b}{16 x^{2} \sqrt{1 + \frac{b}{a x^{4}}}} - \frac{11 \sqrt{a} b^{2}}{16 x^{6} \sqrt{1 + \frac{b}{a x^{4}}}} - \frac{15 a^{2} \sqrt{b} \operatorname{asinh}{\left (\frac{\sqrt{b}}{\sqrt{a} x^{2}} \right )}}{16} - \frac{b^{3}}{8 \sqrt{a} x^{10} \sqrt{1 + \frac{b}{a x^{4}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**(5/2)*x**2/(2*sqrt(1 + b/(a*x**4))) - a**(3/2)*b/(16*x**2*sqrt(1 + b/(a*x**4))) - 11*sqrt(a)*b**2/(16*x**6*
sqrt(1 + b/(a*x**4))) - 15*a**2*sqrt(b)*asinh(sqrt(b)/(sqrt(a)*x**2))/16 - b**3/(8*sqrt(a)*x**10*sqrt(1 + b/(a
*x**4)))

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Giac [A]  time = 1.10505, size = 103, normalized size = 1.13 \begin{align*} \frac{1}{16} \,{\left (\frac{15 \, b \arctan \left (\frac{\sqrt{a x^{4} + b}}{\sqrt{-b}}\right )}{\sqrt{-b}} + 8 \, \sqrt{a x^{4} + b} - \frac{9 \,{\left (a x^{4} + b\right )}^{\frac{3}{2}} b - 7 \, \sqrt{a x^{4} + b} b^{2}}{a^{2} x^{8}}\right )} a^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(15*b*arctan(sqrt(a*x^4 + b)/sqrt(-b))/sqrt(-b) + 8*sqrt(a*x^4 + b) - (9*(a*x^4 + b)^(3/2)*b - 7*sqrt(a*x
^4 + b)*b^2)/(a^2*x^8))*a^2