3.1.89 \(\int x^{-1+m} \sinh ^2(a+b x) \, dx\) [89]

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

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 72, normalized size of antiderivative = 1.00, 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 = {3393, 3388, 2212} \begin {gather*} 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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(-1 + m)*Sinh[a + b*x]^2,x]

[Out]

-1/2*x^m/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 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*} \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*}

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Mathematica [A]
time = 0.07, size = 63, normalized size = 0.88 \begin {gather*} -\frac {x^m \left (2+2^{-m} e^{2 a} m (-b x)^{-m} \Gamma (m,-2 b x)+2^{-m} e^{-2 a} m (b x)^{-m} \Gamma (m,2 b x)\right )}{4 m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(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)))/m

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Maple [F]
time = 0.28, size = 0, normalized size = 0.00 \[\int x^{-1+m} \left (\sinh ^{2}\left (b x +a \right )\right )\, dx\]

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 [A]
time = 0.09, size = 55, normalized size = 0.76 \begin {gather*} -\frac {x^{m} e^{\left (-2 \, a\right )} \Gamma \left (m, 2 \, b x\right )}{4 \, \left (2 \, b x\right )^{m}} - \frac {x^{m} e^{\left (2 \, a\right )} \Gamma \left (m, -2 \, b x\right )}{4 \, \left (-2 \, b x\right )^{m}} - \frac {x^{m}}{2 \, m} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*x^m*e^(-2*a)*gamma(m, 2*b*x)/(2*b*x)^m - 1/4*x^m*e^(2*a)*gamma(m, -2*b*x)/(-2*b*x)^m - 1/2*x^m/m

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Fricas [A]
time = 0.09, size = 117, normalized size = 1.62 \begin {gather*} -\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 {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{m - 1} \sinh ^{2}{\left (a + b x \right )}\, dx \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{m-1}\,{\mathrm {sinh}\left (a+b\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(m - 1)*sinh(a + b*x)^2,x)

[Out]

int(x^(m - 1)*sinh(a + b*x)^2, x)

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