\(\int \frac {x^{-1+m} (2 a m+b (-1+2 m) x)}{2 (a+b x)^{3/2}} \, dx\) [368]

Optimal result

Integrand size = 31, antiderivative size = 13 \[ \int \frac {x^{-1+m} (2 a m+b (-1+2 m) x)}{2 (a+b x)^{3/2}} \, dx=\frac {x^m}{\sqrt {a+b x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {12, 75} \[ \int \frac {x^{-1+m} (2 a m+b (-1+2 m) x)}{2 (a+b x)^{3/2}} \, dx=\frac {x^m}{\sqrt {a+b x}} \]

[In]

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

[Out]

x^m/Sqrt[a + b*x]

Rule 12

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

Rule 75

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[b*(c + d*x)^
(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2, 0] &
& EqQ[a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)), 0]

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

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {x^{-1+m} (2 a m+b (-1+2 m) x)}{2 (a+b x)^{3/2}} \, dx=\frac {x^m}{\sqrt {a+b x}} \]

[In]

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

[Out]

x^m/Sqrt[a + b*x]

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 12, normalized size of antiderivative = 0.92

method	result	size
gosper	\(\frac {x^{m}}{\sqrt {b x +a}}\)	\(12\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 14, normalized size of antiderivative = 1.08 \[ \int \frac {x^{-1+m} (2 a m+b (-1+2 m) x)}{2 (a+b x)^{3/2}} \, dx=\frac {x x^{m - 1}}{\sqrt {b x + a}} \]

[In]

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

[Out]

x*x^(m - 1)/sqrt(b*x + a)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 10.51 (sec) , antiderivative size = 87, normalized size of antiderivative = 6.69 \[ \int \frac {x^{-1+m} (2 a m+b (-1+2 m) x)}{2 (a+b x)^{3/2}} \, dx=\frac {a a^{m} a^{- m - \frac {3}{2}} m x^{m} \Gamma \left (m\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, m \\ m + 1 \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{\Gamma \left (m + 1\right )} + \frac {b x^{m + 1} \cdot \left (2 m - 1\right ) \Gamma \left (m + 1\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, m + 1 \\ m + 2 \end {matrix}\middle | {\frac {b x e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} \Gamma \left (m + 2\right )} \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {x^{-1+m} (2 a m+b (-1+2 m) x)}{2 (a+b x)^{3/2}} \, dx=\frac {x^{m}}{\sqrt {b x + a}} \]

[In]

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

[Out]

x^m/sqrt(b*x + a)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.46 (sec) , antiderivative size = 11, normalized size of antiderivative = 0.85 \[ \int \frac {x^{-1+m} (2 a m+b (-1+2 m) x)}{2 (a+b x)^{3/2}} \, dx=\frac {x^m}{\sqrt {a+b\,x}} \]

[In]

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

[Out]

x^m/(a + b*x)^(1/2)