3.2752 \(\int \frac{x^m}{(a+b x^{2+2 m})^{7/2}} \, dx\)

Optimal. Leaf size=102 \[ \frac{8 b^2 x^{5 (m+1)}}{15 a^3 (m+1) \left (a+b x^{2 (m+1)}\right )^{5/2}}+\frac{4 b x^{3 (m+1)}}{3 a^2 (m+1) \left (a+b x^{2 (m+1)}\right )^{5/2}}+\frac{x^{m+1}}{a (m+1) \left (a+b x^{2 (m+1)}\right )^{5/2}} \]

[Out]

x^(1 + m)/(a*(1 + m)*(a + b*x^(2*(1 + m)))^(5/2)) + (4*b*x^(3*(1 + m)))/(3*a^2*(1 + m)*(a + b*x^(2*(1 + m)))^(
5/2)) + (8*b^2*x^(5*(1 + m)))/(15*a^3*(1 + m)*(a + b*x^(2*(1 + m)))^(5/2))

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Rubi [A]  time = 0.0537406, antiderivative size = 102, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {271, 264} \[ \frac{8 b^2 x^{5 (m+1)}}{15 a^3 (m+1) \left (a+b x^{2 (m+1)}\right )^{5/2}}+\frac{4 b x^{3 (m+1)}}{3 a^2 (m+1) \left (a+b x^{2 (m+1)}\right )^{5/2}}+\frac{x^{m+1}}{a (m+1) \left (a+b x^{2 (m+1)}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[x^m/(a + b*x^(2 + 2*m))^(7/2),x]

[Out]

x^(1 + m)/(a*(1 + m)*(a + b*x^(2*(1 + m)))^(5/2)) + (4*b*x^(3*(1 + m)))/(3*a^2*(1 + m)*(a + b*x^(2*(1 + m)))^(
5/2)) + (8*b^2*x^(5*(1 + m)))/(15*a^3*(1 + m)*(a + b*x^(2*(1 + m)))^(5/2))

Rule 271

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

Rule 264

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

Rubi steps

\begin{align*} \int \frac{x^m}{\left (a+b x^{2+2 m}\right )^{7/2}} \, dx &=\frac{x^{1+m}}{a (1+m) \left (a+b x^{2 (1+m)}\right )^{5/2}}+\frac{(4 b) \int \frac{x^{2+3 m}}{\left (a+b x^{2+2 m}\right )^{7/2}} \, dx}{a}\\ &=\frac{x^{1+m}}{a (1+m) \left (a+b x^{2 (1+m)}\right )^{5/2}}+\frac{4 b x^{3 (1+m)}}{3 a^2 (1+m) \left (a+b x^{2 (1+m)}\right )^{5/2}}+\frac{\left (8 b^2\right ) \int \frac{x^{4+5 m}}{\left (a+b x^{2+2 m}\right )^{7/2}} \, dx}{3 a^2}\\ &=\frac{x^{1+m}}{a (1+m) \left (a+b x^{2 (1+m)}\right )^{5/2}}+\frac{4 b x^{3 (1+m)}}{3 a^2 (1+m) \left (a+b x^{2 (1+m)}\right )^{5/2}}+\frac{8 b^2 x^{5 (1+m)}}{15 a^3 (1+m) \left (a+b x^{2 (1+m)}\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0488818, size = 61, normalized size = 0.6 \[ \frac{x^{m+1} \left (15 a^2+20 a b x^{2 m+2}+8 b^2 x^{4 m+4}\right )}{15 a^3 (m+1) \left (a+b x^{2 m+2}\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(1 + m)*(15*a^2 + 20*a*b*x^(2 + 2*m) + 8*b^2*x^(4 + 4*m)))/(15*a^3*(1 + m)*(a + b*x^(2 + 2*m))^(5/2))

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Maple [F]  time = 0.047, size = 0, normalized size = 0. \begin{align*} \int{{x}^{m} \left ( a+b{x}^{2+2\,m} \right ) ^{-{\frac{7}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{{\left (b x^{2 \, m + 2} + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m/(b*x^(2*m + 2) + a)^(7/2), x)

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Fricas [A]  time = 1.21762, size = 281, normalized size = 2.75 \begin{align*} \frac{{\left (8 \, b^{2} x^{5} x^{5 \, m} + 20 \, a b x^{3} x^{3 \, m} + 15 \, a^{2} x x^{m}\right )} \sqrt{b x^{2} x^{2 \, m} + a}}{15 \,{\left ({\left (a^{3} b^{3} m + a^{3} b^{3}\right )} x^{6} x^{6 \, m} + a^{6} m + a^{6} + 3 \,{\left (a^{4} b^{2} m + a^{4} b^{2}\right )} x^{4} x^{4 \, m} + 3 \,{\left (a^{5} b m + a^{5} b\right )} x^{2} x^{2 \, m}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(8*b^2*x^5*x^(5*m) + 20*a*b*x^3*x^(3*m) + 15*a^2*x*x^m)*sqrt(b*x^2*x^(2*m) + a)/((a^3*b^3*m + a^3*b^3)*x^
6*x^(6*m) + a^6*m + a^6 + 3*(a^4*b^2*m + a^4*b^2)*x^4*x^(4*m) + 3*(a^5*b*m + a^5*b)*x^2*x^(2*m))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{{\left (b x^{2 \, m + 2} + a\right )}^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^m/(b*x^(2*m + 2) + a)^(7/2), x)