\(\int x^m \sinh ^2(a+b x) \, dx\) [88]

Optimal result

Integrand size = 12, antiderivative size = 85 \[ \int x^m \sinh ^2(a+b x) \, dx=-\frac {x^{1+m}}{2 (1+m)}+\frac {2^{-3-m} e^{2 a} x^m (-b x)^{-m} \Gamma (1+m,-2 b x)}{b}-\frac {2^{-3-m} e^{-2 a} x^m (b x)^{-m} \Gamma (1+m,2 b x)}{b} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3393, 3388, 2212} \[ \int x^m \sinh ^2(a+b x) \, dx=\frac {e^{2 a} 2^{-m-3} x^m (-b x)^{-m} \Gamma (m+1,-2 b x)}{b}-\frac {e^{-2 a} 2^{-m-3} x^m (b x)^{-m} \Gamma (m+1,2 b x)}{b}-\frac {x^{m+1}}{2 (m+1)} \]

[In]

Int[x^m*Sinh[a + b*x]^2,x]

[Out]

-1/2*x^(1 + m)/(1 + m) + (2^(-3 - m)*E^(2*a)*x^m*Gamma[1 + m, -2*b*x])/(b*(-(b*x))^m) - (2^(-3 - m)*x^m*Gamma[
1 + m, 2*b*x])/(b*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^m}{2}-\frac {1}{2} x^m \cosh (2 a+2 b x)\right ) \, dx \\ & = -\frac {x^{1+m}}{2 (1+m)}+\frac {1}{2} \int x^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^m \, dx+\frac {1}{4} \int e^{i (2 i a+2 i b x)} x^m \, dx \\ & = -\frac {x^{1+m}}{2 (1+m)}+\frac {2^{-3-m} e^{2 a} x^m (-b x)^{-m} \Gamma (1+m,-2 b x)}{b}-\frac {2^{-3-m} e^{-2 a} x^m (b x)^{-m} \Gamma (1+m,2 b x)}{b} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.89 \[ \int x^m \sinh ^2(a+b x) \, dx=\frac {1}{8} x^m \left (-\frac {4 x}{1+m}+\frac {2^{-m} e^{2 a} (-b x)^{-m} \Gamma (1+m,-2 b x)}{b}-\frac {2^{-m} e^{-2 a} (b x)^{-m} \Gamma (1+m,2 b x)}{b}\right ) \]

[In]

Integrate[x^m*Sinh[a + b*x]^2,x]

[Out]

(x^m*((-4*x)/(1 + m) + (E^(2*a)*Gamma[1 + m, -2*b*x])/(2^m*b*(-(b*x))^m) - Gamma[1 + m, 2*b*x]/(2^m*b*E^(2*a)*
(b*x)^m)))/8

Maple [F]

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

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.44 \[ \int x^m \sinh ^2(a+b x) \, dx=-\frac {4 \, b x \cosh \left (m \log \left (x\right )\right ) + {\left (m + 1\right )} \cosh \left (m \log \left (2 \, b\right ) + 2 \, a\right ) \Gamma \left (m + 1, 2 \, b x\right ) - {\left (m + 1\right )} \cosh \left (m \log \left (-2 \, b\right ) - 2 \, a\right ) \Gamma \left (m + 1, -2 \, b x\right ) - {\left (m + 1\right )} \Gamma \left (m + 1, 2 \, b x\right ) \sinh \left (m \log \left (2 \, b\right ) + 2 \, a\right ) + {\left (m + 1\right )} \Gamma \left (m + 1, -2 \, b x\right ) \sinh \left (m \log \left (-2 \, b\right ) - 2 \, a\right ) + 4 \, b x \sinh \left (m \log \left (x\right )\right )}{8 \, {\left (b m + b\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

Integral(x**m*sinh(a + b*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.08 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.84 \[ \int x^m \sinh ^2(a+b x) \, dx=-\frac {1}{4} \, \left (2 \, b x\right )^{-m - 1} x^{m + 1} e^{\left (-2 \, a\right )} \Gamma \left (m + 1, 2 \, b x\right ) - \frac {1}{4} \, \left (-2 \, b x\right )^{-m - 1} x^{m + 1} e^{\left (2 \, a\right )} \Gamma \left (m + 1, -2 \, b x\right ) - \frac {x^{m + 1}}{2 \, {\left (m + 1\right )}} \]

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

integrate(x^m*sinh(b*x + a)^2, x)

Mupad [F(-1)]

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

[In]

int(x^m*sinh(a + b*x)^2,x)

[Out]

int(x^m*sinh(a + b*x)^2, x)