3.89 \(\int x^{-1+m} \sinh ^2(a+b x) \, dx\)

Optimal. Leaf size=72 \[ e^{2 a} \left (-2^{-m-2}\right ) x^m (-b x)^{-m} \text{Gamma}(m,-2 b x)-e^{-2 a} 2^{-m-2} x^m (b x)^{-m} \text{Gamma}(m,2 b x)-\frac{x^m}{2 m} \]

[Out]

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

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Rubi [A]  time = 0.126916, antiderivative size = 72, 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} \[ e^{2 a} \left (-2^{-m-2}\right ) x^m (-b x)^{-m} \text{Gamma}(m,-2 b x)-e^{-2 a} 2^{-m-2} x^m (b x)^{-m} \text{Gamma}(m,2 b x)-\frac{x^m}{2 m} \]

Antiderivative was successfully verified.

[In]

Int[x^(-1 + m)*Sinh[a + b*x]^2,x]

[Out]

-x^m/(2*m) - (2^(-2 - m)*E^(2*a)*x^m*Gamma[m, -2*b*x])/(-(b*x))^m - (2^(-2 - m)*x^m*Gamma[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^{-1+m} \sinh ^2(a+b x) \, dx &=-\int \left (\frac{x^{-1+m}}{2}-\frac{1}{2} x^{-1+m} \cosh (2 a+2 b x)\right ) \, dx\\ &=-\frac{x^m}{2 m}+\frac{1}{2} \int x^{-1+m} \cosh (2 a+2 b x) \, dx\\ &=-\frac{x^m}{2 m}+\frac{1}{4} \int e^{-i (2 i a+2 i b x)} x^{-1+m} \, dx+\frac{1}{4} \int e^{i (2 i a+2 i b x)} x^{-1+m} \, dx\\ &=-\frac{x^m}{2 m}-2^{-2-m} e^{2 a} x^m (-b x)^{-m} \Gamma (m,-2 b x)-2^{-2-m} e^{-2 a} x^m (b x)^{-m} \Gamma (m,2 b x)\\ \end{align*}

Mathematica [A]  time = 0.075868, size = 63, normalized size = 0.88 \[ -\frac{x^m \left (e^{2 a} 2^{-m} m (-b x)^{-m} \text{Gamma}(m,-2 b x)+e^{-2 a} 2^{-m} m (b x)^{-m} \text{Gamma}(m,2 b x)+2\right )}{4 m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x^m*(2 + (E^(2*a)*m*Gamma[m, -2*b*x])/(2^m*(-(b*x))^m) + (m*Gamma[m, 2*b*x])/(2^m*E^(2*a)*(b*x)^m)))/(4*m)

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int{x}^{-1+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^(-1+m)*sinh(b*x+a)^2,x)

[Out]

int(x^(-1+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^(-1+m)*sinh(b*x+a)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.75443, size = 365, normalized size = 5.07 \begin{align*} -\frac{4 \, b x \cosh \left ({\left (m - 1\right )} \log \left (x\right )\right ) + m \cosh \left ({\left (m - 1\right )} \log \left (2 \, b\right ) + 2 \, a\right ) \Gamma \left (m, 2 \, b x\right ) - m \cosh \left ({\left (m - 1\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) \Gamma \left (m, -2 \, b x\right ) - m \Gamma \left (m, 2 \, b x\right ) \sinh \left ({\left (m - 1\right )} \log \left (2 \, b\right ) + 2 \, a\right ) + m \Gamma \left (m, -2 \, b x\right ) \sinh \left ({\left (m - 1\right )} \log \left (-2 \, b\right ) - 2 \, a\right ) + 4 \, b x \sinh \left ({\left (m - 1\right )} \log \left (x\right )\right )}{8 \, b m} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m - 1} \sinh ^{2}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**(m - 1)*sinh(a + b*x)**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m - 1} \sinh \left (b x + a\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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