\(\int x^{-2+m} \sinh ^2(a+b x) \, dx\) [90]

Optimal result

Integrand size = 14, antiderivative size = 83 \[ \int x^{-2+m} \sinh ^2(a+b x) \, dx=\frac {x^{-1+m}}{2 (1-m)}+2^{-1-m} b e^{2 a} x^m (-b x)^{-m} \Gamma (-1+m,-2 b x)-2^{-1-m} b e^{-2 a} x^m (b x)^{-m} \Gamma (-1+m,2 b x) \]

[Out]

1/2*x^(-1+m)/(1-m)+2^(-1-m)*b*exp(2*a)*x^m*GAMMA(-1+m,-2*b*x)/((-b*x)^m)-2^(-1-m)*b*x^m*GAMMA(-1+m,2*b*x)/exp(
2*a)/((b*x)^m)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3393, 3388, 2212} \[ \int x^{-2+m} \sinh ^2(a+b x) \, dx=e^{2 a} b 2^{-m-1} x^m (-b x)^{-m} \Gamma (m-1,-2 b x)-e^{-2 a} b 2^{-m-1} x^m (b x)^{-m} \Gamma (m-1,2 b x)+\frac {x^{m-1}}{2 (1-m)} \]

[In]

Int[x^(-2 + m)*Sinh[a + b*x]^2,x]

[Out]

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

Rule 2212

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[(-F^(g*(e - c*(f/d))))*((c
+ d*x)^FracPart[m]/(d*((-f)*g*(Log[F]/d))^(IntPart[m] + 1)*((-f)*g*Log[F]*((c + d*x)/d))^FracPart[m]))*Gamma[m
+ 1, ((-f)*g*(Log[F]/d))*(c + d*x)], x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rule 3388

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Dist[I/2, Int[(c + d*x)^m/(E^(
I*k*Pi)*E^(I*(e + f*x))), x], x] - Dist[I/2, Int[(c + d*x)^m*E^(I*k*Pi)*E^(I*(e + f*x)), x], x] /; FreeQ[{c, d
, e, f, m}, x] && IntegerQ[2*k]

Rule 3393

Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> Int[ExpandTrigReduce[(c + d*x)^m, Sin
[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f, m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1])
)

Rubi steps \begin{align*} \text {integral}& = -\int \left (\frac {x^{-2+m}}{2}-\frac {1}{2} x^{-2+m} \cosh (2 a+2 b x)\right ) \, dx \\ & = \frac {x^{-1+m}}{2 (1-m)}+\frac {1}{2} \int x^{-2+m} \cosh (2 a+2 b x) \, dx \\ & = \frac {x^{-1+m}}{2 (1-m)}+\frac {1}{4} \int e^{-i (2 i a+2 i b x)} x^{-2+m} \, dx+\frac {1}{4} \int e^{i (2 i a+2 i b x)} x^{-2+m} \, dx \\ & = \frac {x^{-1+m}}{2 (1-m)}+2^{-1-m} b e^{2 a} x^m (-b x)^{-m} \Gamma (-1+m,-2 b x)-2^{-1-m} b e^{-2 a} x^m (b x)^{-m} \Gamma (-1+m,2 b x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.87 \[ \int x^{-2+m} \sinh ^2(a+b x) \, dx=\frac {1}{2} x^m \left (\frac {1}{x-m x}+2^{-m} b e^{2 a} (-b x)^{-m} \Gamma (-1+m,-2 b x)-2^{-m} b e^{-2 a} (b x)^{-m} \Gamma (-1+m,2 b x)\right ) \]

[In]

Integrate[x^(-2 + m)*Sinh[a + b*x]^2,x]

[Out]

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

Maple [F]

\[\int x^{m -2} \sinh \left (b x +a \right )^{2}d x\]

[In]

int(x^(m-2)*sinh(b*x+a)^2,x)

[Out]

int(x^(m-2)*sinh(b*x+a)^2,x)

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.64 \[ \int x^{-2+m} \sinh ^2(a+b x) \, dx=-\frac {4 \, b x \cosh \left ({\left (m - 2\right )} \log \left (x\right )\right ) + {\left (m - 1\right )} \cosh \left ({\left (m - 2\right )} \log \left (2 \, b\right ) + 2 \, a\right ) \Gamma \left (m - 1, 2 \, b x\right ) - {\left (m - 1\right )} \cosh \left ({\left (m - 2\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) \Gamma \left (m - 1, -2 \, b x\right ) - {\left (m - 1\right )} \Gamma \left (m - 1, 2 \, b x\right ) \sinh \left ({\left (m - 2\right )} \log \left (2 \, b\right ) + 2 \, a\right ) + {\left (m - 1\right )} \Gamma \left (m - 1, -2 \, b x\right ) \sinh \left ({\left (m - 2\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) + 4 \, b x \sinh \left ({\left (m - 2\right )} \log \left (x\right )\right )}{8 \, {\left (b m - b\right )}} \]

[In]

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

[Out]

-1/8*(4*b*x*cosh((m - 2)*log(x)) + (m - 1)*cosh((m - 2)*log(2*b) + 2*a)*gamma(m - 1, 2*b*x) - (m - 1)*cosh((m
- 2)*log(-2*b) - 2*a)*gamma(m - 1, -2*b*x) - (m - 1)*gamma(m - 1, 2*b*x)*sinh((m - 2)*log(2*b) + 2*a) + (m - 1
)*gamma(m - 1, -2*b*x)*sinh((m - 2)*log(-2*b) - 2*a) + 4*b*x*sinh((m - 2)*log(x)))/(b*m - b)

Sympy [F]

\[ \int x^{-2+m} \sinh ^2(a+b x) \, dx=\int x^{m - 2} \sinh ^{2}{\left (a + b x \right )}\, dx \]

[In]

integrate(x**(-2+m)*sinh(b*x+a)**2,x)

[Out]

Integral(x**(m - 2)*sinh(a + b*x)**2, x)

Maxima [F(-2)]

Exception generated. \[ \int x^{-2+m} \sinh ^2(a+b x) \, dx=\text {Exception raised: ValueError} \]

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(m-2>0)', see `assume?` for mor
e details)Is

Giac [F]

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

[In]

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

[Out]

integrate(x^(m - 2)*sinh(b*x + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int x^{-2+m} \sinh ^2(a+b x) \, dx=\int x^{m-2}\,{\mathrm {sinh}\left (a+b\,x\right )}^2 \,d x \]

[In]

int(x^(m - 2)*sinh(a + b*x)^2,x)

[Out]

int(x^(m - 2)*sinh(a + b*x)^2, x)