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

Optimal. Leaf size=52 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 b^{3/2}}-\frac{x^2}{2 b \sqrt{a+b x^4}} \]

[Out]

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

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Rubi [A]  time = 0.0286138, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {275, 288, 217, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a+b x^4}}\right )}{2 b^{3/2}}-\frac{x^2}{2 b \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Int[x^5/(a + b*x^4)^(3/2),x]

[Out]

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

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 288

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

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

Mathematica [A]  time = 0.0560084, size = 66, normalized size = 1.27 \[ \frac{\sqrt{a} \sqrt{\frac{b x^4}{a}+1} \sinh ^{-1}\left (\frac{\sqrt{b} x^2}{\sqrt{a}}\right )-\sqrt{b} x^2}{2 b^{3/2} \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/(a + b*x^4)^(3/2),x]

[Out]

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

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Maple [A]  time = 0.01, size = 42, normalized size = 0.8 \begin{align*} -{\frac{{x}^{2}}{2\,b}{\frac{1}{\sqrt{b{x}^{4}+a}}}}+{\frac{1}{2}\ln \left ({x}^{2}\sqrt{b}+\sqrt{b{x}^{4}+a} \right ){b}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.58783, size = 316, normalized size = 6.08 \begin{align*} \left [-\frac{2 \, \sqrt{b x^{4} + a} b x^{2} -{\left (b x^{4} + a\right )} \sqrt{b} \log \left (-2 \, b x^{4} - 2 \, \sqrt{b x^{4} + a} \sqrt{b} x^{2} - a\right )}{4 \,{\left (b^{3} x^{4} + a b^{2}\right )}}, -\frac{\sqrt{b x^{4} + a} b x^{2} +{\left (b x^{4} + a\right )} \sqrt{-b} \arctan \left (\frac{\sqrt{-b} x^{2}}{\sqrt{b x^{4} + a}}\right )}{2 \,{\left (b^{3} x^{4} + a b^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 2.11745, size = 44, normalized size = 0.85 \begin{align*} \frac{\operatorname{asinh}{\left (\frac{\sqrt{b} x^{2}}{\sqrt{a}} \right )}}{2 b^{\frac{3}{2}}} - \frac{x^{2}}{2 \sqrt{a} b \sqrt{1 + \frac{b x^{4}}{a}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.13141, size = 58, normalized size = 1.12 \begin{align*} -\frac{x^{2}}{2 \, \sqrt{b x^{4} + a} b} - \frac{\log \left ({\left | -\sqrt{b} x^{2} + \sqrt{b x^{4} + a} \right |}\right )}{2 \, b^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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