\(\int \frac {x^m}{(a+b x^{2+2 m})^{7/2}} \, dx\) [2752]

Optimal result

Integrand size = 19, antiderivative size = 102 \[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^{7/2}} \, dx=\frac {x^{1+m}}{a (1+m) \left (a+b x^{2 (1+m)}\right )^{5/2}}+\frac {4 b x^{3 (1+m)}}{3 a^2 (1+m) \left (a+b x^{2 (1+m)}\right )^{5/2}}+\frac {8 b^2 x^{5 (1+m)}}{15 a^3 (1+m) \left (a+b x^{2 (1+m)}\right )^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {277, 270} \[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^{7/2}} \, dx=\frac {8 b^2 x^{5 (m+1)}}{15 a^3 (m+1) \left (a+b x^{2 (m+1)}\right )^{5/2}}+\frac {4 b x^{3 (m+1)}}{3 a^2 (m+1) \left (a+b x^{2 (m+1)}\right )^{5/2}}+\frac {x^{m+1}}{a (m+1) \left (a+b x^{2 (m+1)}\right )^{5/2}} \]

[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 270

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]

Rule 277

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]

Rubi steps \begin{align*} \text {integral}& = \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] (verified)

Time = 0.13 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.60 \[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^{7/2}} \, dx=\frac {x^{1+m} \left (15 a^2+20 a b x^{2+2 m}+8 b^2 x^{4+4 m}\right )}{15 a^3 (1+m) \left (a+b x^{2+2 m}\right )^{5/2}} \]

[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))

Maple [F]

\[\int \frac {x^{m}}{\left (a +b \,x^{2+2 m}\right )^{\frac {7}{2}}}d x\]

[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)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.32 \[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^{7/2}} \, dx=\frac {{\left (8 \, b^{2} x^{5} x^{5 \, m} + 20 \, a b x^{3} x^{3 \, m} + 15 \, a^{2} x x^{m}\right )} \sqrt {b x^{2} x^{2 \, m} + a}}{15 \, {\left ({\left (a^{3} b^{3} m + a^{3} b^{3}\right )} x^{6} x^{6 \, m} + a^{6} m + a^{6} + 3 \, {\left (a^{4} b^{2} m + a^{4} b^{2}\right )} x^{4} x^{4 \, m} + 3 \, {\left (a^{5} b m + a^{5} b\right )} x^{2} x^{2 \, m}\right )}} \]

[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))

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 50.74 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.05 \[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^{7/2}} \, dx=\frac {\sqrt {\pi } \sqrt {a} a^{- \frac {m}{2 m + 2} - \frac {7}{2} - \frac {1}{2 m + 2}} x^{m + 1} {{}_{2}F_{1}\left (\begin {matrix} \frac {7}{2}, \frac {1}{2} \\ \frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2} \end {matrix}\middle | {\frac {b x^{2 m + 2} e^{i \pi }}{a}} \right )}}{2 m \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right ) + 2 \Gamma \left (\frac {m}{2 m + 2} + 1 + \frac {1}{2 m + 2}\right )} \]

[In]

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

[Out]

sqrt(pi)*sqrt(a)*a**(-m/(2*m + 2) - 7/2 - 1/(2*m + 2))*x**(m + 1)*hyper((7/2, 1/2), (m/(2*m + 2) + 1 + 1/(2*m
+ 2),), b*x**(2*m + 2)*exp_polar(I*pi)/a)/(2*m*gamma(m/(2*m + 2) + 1 + 1/(2*m + 2)) + 2*gamma(m/(2*m + 2) + 1
+ 1/(2*m + 2)))

Maxima [F]

\[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^{7/2}} \, dx=\int { \frac {x^{m}}{{\left (b x^{2 \, m + 2} + a\right )}^{\frac {7}{2}}} \,d x } \]

[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)

Giac [F]

\[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^{7/2}} \, dx=\int { \frac {x^{m}}{{\left (b x^{2 \, m + 2} + a\right )}^{\frac {7}{2}}} \,d x } \]

[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)

Mupad [B] (verification not implemented)

Time = 5.79 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.53 \[ \int \frac {x^m}{\left (a+b x^{2+2 m}\right )^{7/2}} \, dx=\frac {x^{m+1}\,\left (\frac {8\,b^2\,x^{4\,m+4}}{15}+a^2+\frac {4\,a\,b\,x^{2\,m+2}}{3}\right )}{a^3\,{\left (a+b\,x^{2\,m+2}\right )}^{5/2}\,\left (m+1\right )} \]

[In]

int(x^m/(a + b*x^(2*m + 2))^(7/2),x)

[Out]

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