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

Optimal. Leaf size=59 \[ \frac{e^{-a} x^m (b x)^{-m} \text{Gamma}(m+2,b x)}{2 b^2}-\frac{e^a x^m (-b x)^{-m} \text{Gamma}(m+2,-b x)}{2 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0736857, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {3308, 2181} \[ \frac{e^{-a} x^m (b x)^{-m} \text{Gamma}(m+2,b x)}{2 b^2}-\frac{e^a x^m (-b x)^{-m} \text{Gamma}(m+2,-b x)}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Int[x^(1 + m)*Sinh[a + b*x],x]

[Out]

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

Rule 3308

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

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 (a+b x) \, dx &=\frac{1}{2} \int e^{-i (i a+i b x)} x^{1+m} \, dx-\frac{1}{2} \int e^{i (i a+i b x)} x^{1+m} \, dx\\ &=-\frac{e^a x^m (-b x)^{-m} \Gamma (2+m,-b x)}{2 b^2}+\frac{e^{-a} x^m (b x)^{-m} \Gamma (2+m,b x)}{2 b^2}\\ \end{align*}

Mathematica [A]  time = 0.021923, size = 54, normalized size = 0.92 \[ \frac{e^{-a} x^m \left ((b x)^{-m} \text{Gamma}(m+2,b x)-e^{2 a} (-b x)^{-m} \text{Gamma}(m+2,-b x)\right )}{2 b^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(1 + m)*Sinh[a + b*x],x]

[Out]

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

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Maple [C]  time = 0.04, size = 73, normalized size = 1.2 \begin{align*}{\frac{{x}^{2+m}\sinh \left ( a \right ) }{2+m}{\mbox{$_1$F$_2$}(1+{\frac{m}{2}};\,{\frac{1}{2}},2+{\frac{m}{2}};\,{\frac{{x}^{2}{b}^{2}}{4}})}}+{\frac{b{x}^{3+m}\cosh \left ( a \right ) }{3+m}{\mbox{$_1$F$_2$}({\frac{3}{2}}+{\frac{m}{2}};\,{\frac{3}{2}},{\frac{5}{2}}+{\frac{m}{2}};\,{\frac{{x}^{2}{b}^{2}}{4}})}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1+m)*sinh(b*x+a),x)

[Out]

1/(2+m)*x^(2+m)*hypergeom([1+1/2*m],[1/2,2+1/2*m],1/4*x^2*b^2)*sinh(a)+b/(3+m)*x^(3+m)*hypergeom([3/2+1/2*m],[
3/2,5/2+1/2*m],1/4*x^2*b^2)*cosh(a)

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Maxima [A]  time = 1.29879, size = 74, normalized size = 1.25 \begin{align*} \frac{1}{2} \, \left (b x\right )^{-m - 2} x^{m + 2} e^{\left (-a\right )} \Gamma \left (m + 2, b x\right ) - \frac{1}{2} \, \left (-b x\right )^{-m - 2} x^{m + 2} e^{a} \Gamma \left (m + 2, -b x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.74414, size = 258, normalized size = 4.37 \begin{align*} \frac{\cosh \left ({\left (m + 1\right )} \log \left (b\right ) + a\right ) \Gamma \left (m + 2, b x\right ) + \cosh \left ({\left (m + 1\right )} \log \left (-b\right ) - a\right ) \Gamma \left (m + 2, -b x\right ) - \Gamma \left (m + 2, -b x\right ) \sinh \left ({\left (m + 1\right )} \log \left (-b\right ) - a\right ) - \Gamma \left (m + 2, b x\right ) \sinh \left ({\left (m + 1\right )} \log \left (b\right ) + a\right )}{2 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError

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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 )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^(m + 1)*sinh(b*x + a), x)