### 3.259 $$\int x^m \cosh ^2(a+b x) \sinh (a+b x) \, dx$$

Optimal. Leaf size=134 $\frac{e^{3 a} 3^{-m-1} x^m (-b x)^{-m} \text{Gamma}(m+1,-3 b x)}{8 b}+\frac{e^a x^m (-b x)^{-m} \text{Gamma}(m+1,-b x)}{8 b}+\frac{e^{-a} x^m (b x)^{-m} \text{Gamma}(m+1,b x)}{8 b}+\frac{e^{-3 a} 3^{-m-1} x^m (b x)^{-m} \text{Gamma}(m+1,3 b x)}{8 b}$

[Out]

(3^(-1 - m)*E^(3*a)*x^m*Gamma[1 + m, -3*b*x])/(8*b*(-(b*x))^m) + (E^a*x^m*Gamma[1 + m, -(b*x)])/(8*b*(-(b*x))^
m) + (x^m*Gamma[1 + m, b*x])/(8*b*E^a*(b*x)^m) + (3^(-1 - m)*x^m*Gamma[1 + m, 3*b*x])/(8*b*E^(3*a)*(b*x)^m)

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Rubi [A]  time = 0.196015, antiderivative size = 134, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 3, integrand size = 18, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.167, Rules used = {5448, 3308, 2181} $\frac{e^{3 a} 3^{-m-1} x^m (-b x)^{-m} \text{Gamma}(m+1,-3 b x)}{8 b}+\frac{e^a x^m (-b x)^{-m} \text{Gamma}(m+1,-b x)}{8 b}+\frac{e^{-a} x^m (b x)^{-m} \text{Gamma}(m+1,b x)}{8 b}+\frac{e^{-3 a} 3^{-m-1} x^m (b x)^{-m} \text{Gamma}(m+1,3 b x)}{8 b}$

Antiderivative was successfully veriﬁed.

[In]

Int[x^m*Cosh[a + b*x]^2*Sinh[a + b*x],x]

[Out]

(3^(-1 - m)*E^(3*a)*x^m*Gamma[1 + m, -3*b*x])/(8*b*(-(b*x))^m) + (E^a*x^m*Gamma[1 + m, -(b*x)])/(8*b*(-(b*x))^
m) + (x^m*Gamma[1 + m, b*x])/(8*b*E^a*(b*x)^m) + (3^(-1 - m)*x^m*Gamma[1 + m, 3*b*x])/(8*b*E^(3*a)*(b*x)^m)

Rule 5448

Int[Cosh[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sinh[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int
[ExpandTrigReduce[(c + d*x)^m, Sinh[a + b*x]^n*Cosh[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n,
0] && IGtQ[p, 0]

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^m \cosh ^2(a+b x) \sinh (a+b x) \, dx &=\int \left (\frac{1}{4} x^m \sinh (a+b x)+\frac{1}{4} x^m \sinh (3 a+3 b x)\right ) \, dx\\ &=\frac{1}{4} \int x^m \sinh (a+b x) \, dx+\frac{1}{4} \int x^m \sinh (3 a+3 b x) \, dx\\ &=\frac{1}{8} \int e^{-i (i a+i b x)} x^m \, dx-\frac{1}{8} \int e^{i (i a+i b x)} x^m \, dx+\frac{1}{8} \int e^{-i (3 i a+3 i b x)} x^m \, dx-\frac{1}{8} \int e^{i (3 i a+3 i b x)} x^m \, dx\\ &=\frac{3^{-1-m} e^{3 a} x^m (-b x)^{-m} \Gamma (1+m,-3 b x)}{8 b}+\frac{e^a x^m (-b x)^{-m} \Gamma (1+m,-b x)}{8 b}+\frac{e^{-a} x^m (b x)^{-m} \Gamma (1+m,b x)}{8 b}+\frac{3^{-1-m} e^{-3 a} x^m (b x)^{-m} \Gamma (1+m,3 b x)}{8 b}\\ \end{align*}

Mathematica [A]  time = 0.173345, size = 114, normalized size = 0.85 $\frac{e^{-3 a} x^m \left (3^{-m} \left (-b^2 x^2\right )^{-m} \left (e^{6 a} (b x)^m \text{Gamma}(m+1,-3 b x)+(-b x)^m \text{Gamma}(m+1,3 b x)\right )+3 e^{2 a} \left (e^{2 a} (-b x)^{-m} \text{Gamma}(m+1,-b x)+(b x)^{-m} \text{Gamma}(m+1,b x)\right )\right )}{24 b}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[x^m*Cosh[a + b*x]^2*Sinh[a + b*x],x]

[Out]

(x^m*(3*E^(2*a)*((E^(2*a)*Gamma[1 + m, -(b*x)])/(-(b*x))^m + Gamma[1 + m, b*x]/(b*x)^m) + (E^(6*a)*(b*x)^m*Gam
ma[1 + m, -3*b*x] + (-(b*x))^m*Gamma[1 + m, 3*b*x])/(3^m*(-(b^2*x^2))^m)))/(24*b*E^(3*a))

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Maple [F]  time = 0.061, size = 0, normalized size = 0. \begin{align*} \int{x}^{m} \left ( \cosh \left ( bx+a \right ) \right ) ^{2}\sinh \left ( bx+a \right ) \, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.3533, size = 153, normalized size = 1.14 \begin{align*} \frac{1}{8} \, \left (3 \, b x\right )^{-m - 1} x^{m + 1} e^{\left (-3 \, a\right )} \Gamma \left (m + 1, 3 \, b x\right ) + \frac{1}{8} \, \left (b x\right )^{-m - 1} x^{m + 1} e^{\left (-a\right )} \Gamma \left (m + 1, b x\right ) - \frac{1}{8} \, \left (-b x\right )^{-m - 1} x^{m + 1} e^{a} \Gamma \left (m + 1, -b x\right ) - \frac{1}{8} \, \left (-3 \, b x\right )^{-m - 1} x^{m + 1} e^{\left (3 \, a\right )} \Gamma \left (m + 1, -3 \, b x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(3*b*x)^(-m - 1)*x^(m + 1)*e^(-3*a)*gamma(m + 1, 3*b*x) + 1/8*(b*x)^(-m - 1)*x^(m + 1)*e^(-a)*gamma(m + 1,
b*x) - 1/8*(-b*x)^(-m - 1)*x^(m + 1)*e^a*gamma(m + 1, -b*x) - 1/8*(-3*b*x)^(-m - 1)*x^(m + 1)*e^(3*a)*gamma(m
+ 1, -3*b*x)

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Fricas [A]  time = 2.2133, size = 486, normalized size = 3.63 \begin{align*} \frac{\cosh \left (m \log \left (3 \, b\right ) + 3 \, a\right ) \Gamma \left (m + 1, 3 \, b x\right ) + 3 \, \cosh \left (m \log \left (b\right ) + a\right ) \Gamma \left (m + 1, b x\right ) + 3 \, \cosh \left (m \log \left (-b\right ) - a\right ) \Gamma \left (m + 1, -b x\right ) + \cosh \left (m \log \left (-3 \, b\right ) - 3 \, a\right ) \Gamma \left (m + 1, -3 \, b x\right ) - \Gamma \left (m + 1, 3 \, b x\right ) \sinh \left (m \log \left (3 \, b\right ) + 3 \, a\right ) - 3 \, \Gamma \left (m + 1, -b x\right ) \sinh \left (m \log \left (-b\right ) - a\right ) - \Gamma \left (m + 1, -3 \, b x\right ) \sinh \left (m \log \left (-3 \, b\right ) - 3 \, a\right ) - 3 \, \Gamma \left (m + 1, b x\right ) \sinh \left (m \log \left (b\right ) + a\right )}{24 \, b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(cosh(m*log(3*b) + 3*a)*gamma(m + 1, 3*b*x) + 3*cosh(m*log(b) + a)*gamma(m + 1, b*x) + 3*cosh(m*log(-b) -
a)*gamma(m + 1, -b*x) + cosh(m*log(-3*b) - 3*a)*gamma(m + 1, -3*b*x) - gamma(m + 1, 3*b*x)*sinh(m*log(3*b) +
3*a) - 3*gamma(m + 1, -b*x)*sinh(m*log(-b) - a) - gamma(m + 1, -3*b*x)*sinh(m*log(-3*b) - 3*a) - 3*gamma(m + 1
, b*x)*sinh(m*log(b) + a))/b

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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