\(\int \frac {x^{-1+m} (2 a m+b (2 m-n) x^n)}{2 (a+b x^n)^{3/2}} \, dx\) [2698]

Optimal result

Integrand size = 37, antiderivative size = 15 \[ \int \frac {x^{-1+m} \left (2 a m+b (2 m-n) x^n\right )}{2 \left (a+b x^n\right )^{3/2}} \, dx=\frac {x^m}{\sqrt {a+b x^n}} \]

[Out]

x^m/(a+b*x^n)^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 15, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {12, 460} \[ \int \frac {x^{-1+m} \left (2 a m+b (2 m-n) x^n\right )}{2 \left (a+b x^n\right )^{3/2}} \, dx=\frac {x^m}{\sqrt {a+b x^n}} \]

[In]

Int[(x^(-1 + m)*(2*a*m + b*(2*m - n)*x^n))/(2*(a + b*x^n)^(3/2)),x]

[Out]

x^m/Sqrt[a + b*x^n]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 460

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \int \frac {x^{-1+m} \left (2 a m+b (2 m-n) x^n\right )}{\left (a+b x^n\right )^{3/2}} \, dx \\ & = \frac {x^m}{\sqrt {a+b x^n}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.05 (sec) , antiderivative size = 111, normalized size of antiderivative = 7.40 \[ \int \frac {x^{-1+m} \left (2 a m+b (2 m-n) x^n\right )}{2 \left (a+b x^n\right )^{3/2}} \, dx=\frac {x^m \sqrt {1+\frac {b x^n}{a}} \left (2 a (m+n) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m}{n},\frac {m+n}{n},-\frac {b x^n}{a}\right )+b (2 m-n) x^n \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+n}{n},2+\frac {m}{n},-\frac {b x^n}{a}\right )\right )}{2 a (m+n) \sqrt {a+b x^n}} \]

[In]

Integrate[(x^(-1 + m)*(2*a*m + b*(2*m - n)*x^n))/(2*(a + b*x^n)^(3/2)),x]

[Out]

(x^m*Sqrt[1 + (b*x^n)/a]*(2*a*(m + n)*Hypergeometric2F1[3/2, m/n, (m + n)/n, -((b*x^n)/a)] + b*(2*m - n)*x^n*H
ypergeometric2F1[3/2, (m + n)/n, 2 + m/n, -((b*x^n)/a)]))/(2*a*(m + n)*Sqrt[a + b*x^n])

Maple [F]

\[\int \frac {x^{-1+m} \left (2 a m +b \left (2 m -n \right ) x^{n}\right )}{2 \left (a +b \,x^{n}\right )^{\frac {3}{2}}}d x\]

[In]

int(1/2*x^(-1+m)*(2*a*m+b*(2*m-n)*x^n)/(a+b*x^n)^(3/2),x)

[Out]

int(1/2*x^(-1+m)*(2*a*m+b*(2*m-n)*x^n)/(a+b*x^n)^(3/2),x)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 16, normalized size of antiderivative = 1.07 \[ \int \frac {x^{-1+m} \left (2 a m+b (2 m-n) x^n\right )}{2 \left (a+b x^n\right )^{3/2}} \, dx=\frac {x x^{m - 1}}{\sqrt {b x^{n} + a}} \]

[In]

integrate(1/2*x^(-1+m)*(2*a*m+b*(2*m-n)*x^n)/(a+b*x^n)^(3/2),x, algorithm="fricas")

[Out]

x*x^(m - 1)/sqrt(b*x^n + a)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.29 (sec) , antiderivative size = 122, normalized size of antiderivative = 8.13 \[ \int \frac {x^{-1+m} \left (2 a m+b (2 m-n) x^n\right )}{2 \left (a+b x^n\right )^{3/2}} \, dx=\frac {a a^{\frac {m}{n}} a^{- \frac {m}{n} - \frac {3}{2}} m x^{m} \Gamma \left (\frac {m}{n}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{n} \\ \frac {m}{n} + 1 \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{n \Gamma \left (\frac {m}{n} + 1\right )} + \frac {a^{- \frac {m}{n} - \frac {5}{2}} a^{\frac {m}{n} + 1} b x^{m + n} \left (2 m - n\right ) \Gamma \left (\frac {m}{n} + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{n} + 1 \\ \frac {m}{n} + 2 \end {matrix}\middle | {\frac {b x^{n} e^{i \pi }}{a}} \right )}}{2 n \Gamma \left (\frac {m}{n} + 2\right )} \]

[In]

integrate(1/2*x**(-1+m)*(2*a*m+b*(2*m-n)*x**n)/(a+b*x**n)**(3/2),x)

[Out]

a*a**(m/n)*a**(-m/n - 3/2)*m*x**m*gamma(m/n)*hyper((3/2, m/n), (m/n + 1,), b*x**n*exp_polar(I*pi)/a)/(n*gamma(
m/n + 1)) + a**(-m/n - 5/2)*a**(m/n + 1)*b*x**(m + n)*(2*m - n)*gamma(m/n + 1)*hyper((3/2, m/n + 1), (m/n + 2,
), b*x**n*exp_polar(I*pi)/a)/(2*n*gamma(m/n + 2))

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {x^{-1+m} \left (2 a m+b (2 m-n) x^n\right )}{2 \left (a+b x^n\right )^{3/2}} \, dx=\frac {x^{m}}{\sqrt {b x^{n} + a}} \]

[In]

integrate(1/2*x^(-1+m)*(2*a*m+b*(2*m-n)*x^n)/(a+b*x^n)^(3/2),x, algorithm="maxima")

[Out]

x^m/sqrt(b*x^n + a)

Giac [F]

\[ \int \frac {x^{-1+m} \left (2 a m+b (2 m-n) x^n\right )}{2 \left (a+b x^n\right )^{3/2}} \, dx=\int { \frac {{\left (b {\left (2 \, m - n\right )} x^{n} + 2 \, a m\right )} x^{m - 1}}{2 \, {\left (b x^{n} + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(1/2*x^(-1+m)*(2*a*m+b*(2*m-n)*x^n)/(a+b*x^n)^(3/2),x, algorithm="giac")

[Out]

integrate(1/2*(b*(2*m - n)*x^n + 2*a*m)*x^(m - 1)/(b*x^n + a)^(3/2), x)

Mupad [B] (verification not implemented)

Time = 6.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 0.87 \[ \int \frac {x^{-1+m} \left (2 a m+b (2 m-n) x^n\right )}{2 \left (a+b x^n\right )^{3/2}} \, dx=\frac {x^m}{\sqrt {a+b\,x^n}} \]

[In]

int((x^(m - 1)*(2*a*m + b*x^n*(2*m - n)))/(2*(a + b*x^n)^(3/2)),x)

[Out]

x^m/(a + b*x^n)^(1/2)