3.852 \(\int \frac{x^7}{(a+b x^4)^{3/2}} \, dx\)

Optimal. Leaf size=38 \[ \frac{a}{2 b^2 \sqrt{a+b x^4}}+\frac{\sqrt{a+b x^4}}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0221735, antiderivative size = 38, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {266, 43} \[ \frac{a}{2 b^2 \sqrt{a+b x^4}}+\frac{\sqrt{a+b x^4}}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^7/(a + b*x^4)^(3/2),x]

[Out]

a/(2*b^2*Sqrt[a + b*x^4]) + Sqrt[a + b*x^4]/(2*b^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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0111201, size = 27, normalized size = 0.71 \[ \frac{2 a+b x^4}{2 b^2 \sqrt{a+b x^4}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7/(a + b*x^4)^(3/2),x]

[Out]

(2*a + b*x^4)/(2*b^2*Sqrt[a + b*x^4])

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Maple [A]  time = 0.006, size = 24, normalized size = 0.6 \begin{align*}{\frac{b{x}^{4}+2\,a}{2\,{b}^{2}}{\frac{1}{\sqrt{b{x}^{4}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(b*x^4+a)^(3/2),x)

[Out]

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

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Maxima [A]  time = 0.954405, size = 41, normalized size = 1.08 \begin{align*} \frac{\sqrt{b x^{4} + a}}{2 \, b^{2}} + \frac{a}{2 \, \sqrt{b x^{4} + a} b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.47133, size = 72, normalized size = 1.89 \begin{align*} \frac{{\left (b x^{4} + 2 \, a\right )} \sqrt{b x^{4} + a}}{2 \,{\left (b^{3} x^{4} + a b^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.51455, size = 41, normalized size = 1.08 \begin{align*} \begin{cases} \frac{a}{b^{2} \sqrt{a + b x^{4}}} + \frac{x^{4}}{2 b \sqrt{a + b x^{4}}} & \text{for}\: b \neq 0 \\\frac{x^{8}}{8 a^{\frac{3}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(b*x**4+a)**(3/2),x)

[Out]

Piecewise((a/(b**2*sqrt(a + b*x**4)) + x**4/(2*b*sqrt(a + b*x**4)), Ne(b, 0)), (x**8/(8*a**(3/2)), True))

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Giac [A]  time = 1.10081, size = 35, normalized size = 0.92 \begin{align*} \frac{\sqrt{b x^{4} + a} + \frac{a}{\sqrt{b x^{4} + a}}}{2 \, b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^4+a)^(3/2),x, algorithm="giac")

[Out]

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