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

Optimal. Leaf size=83 \[ e^{2 a} b 2^{-m-1} x^m (-b x)^{-m} \text{Gamma}(m-1,-2 b x)-e^{-2 a} b 2^{-m-1} x^m (b x)^{-m} \text{Gamma}(m-1,2 b x)+\frac{x^{m-1}}{2 (1-m)} \]

[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)

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Rubi [A]  time = 0.136826, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.214, Rules used = {3312, 3307, 2181} \[ e^{2 a} b 2^{-m-1} x^m (-b x)^{-m} \text{Gamma}(m-1,-2 b x)-e^{-2 a} b 2^{-m-1} x^m (b x)^{-m} \text{Gamma}(m-1,2 b x)+\frac{x^{m-1}}{2 (1-m)} \]

Antiderivative was successfully verified.

[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 3312

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])
)

Rule 3307

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 2181

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

Rubi steps

\begin{align*} \int x^{-2+m} \sinh ^2(a+b x) \, dx &=-\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]  time = 0.100091, size = 72, normalized size = 0.87 \[ \frac{1}{2} x^m \left (e^{2 a} b 2^{-m} (-b x)^{-m} \text{Gamma}(m-1,-2 b x)-e^{-2 a} b 2^{-m} (b x)^{-m} \text{Gamma}(m-1,2 b x)+\frac{1}{x-m x}\right ) \]

Antiderivative was successfully verified.

[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

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Maple [F]  time = 0.046, size = 0, normalized size = 0. \begin{align*} \int{x}^{-2+m} \left ( \sinh \left ( bx+a \right ) \right ) ^{2}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.79577, size = 424, normalized size = 5.11 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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