3.307 \(\int x^m \cosh (a+b x) \sinh ^3(a+b x) \, dx\)

Optimal. Leaf size=141 \[ \frac{e^{4 a} 2^{-2 (m+3)} x^m (-b x)^{-m} \text{Gamma}(m+1,-4 b x)}{b}-\frac{e^{2 a} 2^{-m-4} x^m (-b x)^{-m} \text{Gamma}(m+1,-2 b x)}{b}-\frac{e^{-2 a} 2^{-m-4} x^m (b x)^{-m} \text{Gamma}(m+1,2 b x)}{b}+\frac{e^{-4 a} 2^{-2 (m+3)} x^m (b x)^{-m} \text{Gamma}(m+1,4 b x)}{b} \]

[Out]

(E^(4*a)*x^m*Gamma[1 + m, -4*b*x])/(2^(2*(3 + m))*b*(-(b*x))^m) - (2^(-4 - m)*E^(2*a)*x^m*Gamma[1 + m, -2*b*x]
)/(b*(-(b*x))^m) - (2^(-4 - m)*x^m*Gamma[1 + m, 2*b*x])/(b*E^(2*a)*(b*x)^m) + (x^m*Gamma[1 + m, 4*b*x])/(2^(2*
(3 + m))*b*E^(4*a)*(b*x)^m)

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Rubi [A]  time = 0.210249, antiderivative size = 141, 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} \[ \frac{e^{4 a} 2^{-2 (m+3)} x^m (-b x)^{-m} \text{Gamma}(m+1,-4 b x)}{b}-\frac{e^{2 a} 2^{-m-4} x^m (-b x)^{-m} \text{Gamma}(m+1,-2 b x)}{b}-\frac{e^{-2 a} 2^{-m-4} x^m (b x)^{-m} \text{Gamma}(m+1,2 b x)}{b}+\frac{e^{-4 a} 2^{-2 (m+3)} x^m (b x)^{-m} \text{Gamma}(m+1,4 b x)}{b} \]

Antiderivative was successfully verified.

[In]

Int[x^m*Cosh[a + b*x]*Sinh[a + b*x]^3,x]

[Out]

(E^(4*a)*x^m*Gamma[1 + m, -4*b*x])/(2^(2*(3 + m))*b*(-(b*x))^m) - (2^(-4 - m)*E^(2*a)*x^m*Gamma[1 + m, -2*b*x]
)/(b*(-(b*x))^m) - (2^(-4 - m)*x^m*Gamma[1 + m, 2*b*x])/(b*E^(2*a)*(b*x)^m) + (x^m*Gamma[1 + m, 4*b*x])/(2^(2*
(3 + m))*b*E^(4*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 (a+b x) \sinh ^3(a+b x) \, dx &=\int \left (-\frac{1}{4} x^m \sinh (2 a+2 b x)+\frac{1}{8} x^m \sinh (4 a+4 b x)\right ) \, dx\\ &=\frac{1}{8} \int x^m \sinh (4 a+4 b x) \, dx-\frac{1}{4} \int x^m \sinh (2 a+2 b x) \, dx\\ &=\frac{1}{16} \int e^{-i (4 i a+4 i b x)} x^m \, dx-\frac{1}{16} \int e^{i (4 i a+4 i b x)} x^m \, dx-\frac{1}{8} \int e^{-i (2 i a+2 i b x)} x^m \, dx+\frac{1}{8} \int e^{i (2 i a+2 i b x)} x^m \, dx\\ &=\frac{4^{-3-m} e^{4 a} x^m (-b x)^{-m} \Gamma (1+m,-4 b x)}{b}-\frac{2^{-4-m} e^{2 a} x^m (-b x)^{-m} \Gamma (1+m,-2 b x)}{b}-\frac{2^{-4-m} e^{-2 a} x^m (b x)^{-m} \Gamma (1+m,2 b x)}{b}+\frac{4^{-3-m} e^{-4 a} x^m (b x)^{-m} \Gamma (1+m,4 b x)}{b}\\ \end{align*}

Mathematica [A]  time = 0.136832, size = 112, normalized size = 0.79 \[ \frac{e^{-4 a} 4^{-m-3} x^m \left (-b^2 x^2\right )^{-m} \left ((-b x)^m \left (\text{Gamma}(m+1,4 b x)-e^{2 a} 2^{m+2} \text{Gamma}(m+1,2 b x)\right )+e^{8 a} (b x)^m \text{Gamma}(m+1,-4 b x)-e^{6 a} 2^{m+2} (b x)^m \text{Gamma}(m+1,-2 b x)\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m*Cosh[a + b*x]*Sinh[a + b*x]^3,x]

[Out]

(4^(-3 - m)*x^m*(E^(8*a)*(b*x)^m*Gamma[1 + m, -4*b*x] - 2^(2 + m)*E^(6*a)*(b*x)^m*Gamma[1 + m, -2*b*x] + (-(b*
x))^m*(-(2^(2 + m)*E^(2*a)*Gamma[1 + m, 2*b*x]) + Gamma[1 + m, 4*b*x])))/(b*E^(4*a)*(-(b^2*x^2))^m)

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Maple [F]  time = 0.053, size = 0, normalized size = 0. \begin{align*} \int{x}^{m}\cosh \left ( bx+a \right ) \left ( \sinh \left ( bx+a \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*cosh(b*x+a)*sinh(b*x+a)^3,x)

[Out]

int(x^m*cosh(b*x+a)*sinh(b*x+a)^3,x)

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Maxima [A]  time = 1.22765, size = 158, normalized size = 1.12 \begin{align*} \frac{1}{16} \, \left (4 \, b x\right )^{-m - 1} x^{m + 1} e^{\left (-4 \, a\right )} \Gamma \left (m + 1, 4 \, b x\right ) - \frac{1}{8} \, \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}{8} \, \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}{16} \, \left (-4 \, b x\right )^{-m - 1} x^{m + 1} e^{\left (4 \, a\right )} \Gamma \left (m + 1, -4 \, b x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*cosh(b*x+a)*sinh(b*x+a)^3,x, algorithm="maxima")

[Out]

1/16*(4*b*x)^(-m - 1)*x^(m + 1)*e^(-4*a)*gamma(m + 1, 4*b*x) - 1/8*(2*b*x)^(-m - 1)*x^(m + 1)*e^(-2*a)*gamma(m
+ 1, 2*b*x) + 1/8*(-2*b*x)^(-m - 1)*x^(m + 1)*e^(2*a)*gamma(m + 1, -2*b*x) - 1/16*(-4*b*x)^(-m - 1)*x^(m + 1)
*e^(4*a)*gamma(m + 1, -4*b*x)

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Fricas [A]  time = 1.86706, size = 518, normalized size = 3.67 \begin{align*} \frac{\cosh \left (m \log \left (4 \, b\right ) + 4 \, a\right ) \Gamma \left (m + 1, 4 \, b x\right ) - 4 \, \cosh \left (m \log \left (2 \, b\right ) + 2 \, a\right ) \Gamma \left (m + 1, 2 \, b x\right ) - 4 \, \cosh \left (m \log \left (-2 \, b\right ) - 2 \, a\right ) \Gamma \left (m + 1, -2 \, b x\right ) + \cosh \left (m \log \left (-4 \, b\right ) - 4 \, a\right ) \Gamma \left (m + 1, -4 \, b x\right ) - \Gamma \left (m + 1, 4 \, b x\right ) \sinh \left (m \log \left (4 \, b\right ) + 4 \, a\right ) + 4 \, \Gamma \left (m + 1, 2 \, b x\right ) \sinh \left (m \log \left (2 \, b\right ) + 2 \, a\right ) + 4 \, \Gamma \left (m + 1, -2 \, b x\right ) \sinh \left (m \log \left (-2 \, b\right ) - 2 \, a\right ) - \Gamma \left (m + 1, -4 \, b x\right ) \sinh \left (m \log \left (-4 \, b\right ) - 4 \, a\right )}{64 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*cosh(b*x+a)*sinh(b*x+a)^3,x, algorithm="fricas")

[Out]

1/64*(cosh(m*log(4*b) + 4*a)*gamma(m + 1, 4*b*x) - 4*cosh(m*log(2*b) + 2*a)*gamma(m + 1, 2*b*x) - 4*cosh(m*log
(-2*b) - 2*a)*gamma(m + 1, -2*b*x) + cosh(m*log(-4*b) - 4*a)*gamma(m + 1, -4*b*x) - gamma(m + 1, 4*b*x)*sinh(m
*log(4*b) + 4*a) + 4*gamma(m + 1, 2*b*x)*sinh(m*log(2*b) + 2*a) + 4*gamma(m + 1, -2*b*x)*sinh(m*log(-2*b) - 2*
a) - gamma(m + 1, -4*b*x)*sinh(m*log(-4*b) - 4*a))/b

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{m} \sinh ^{3}{\left (a + b x \right )} \cosh{\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)*sinh(b*x+a)**3,x)

[Out]

Integral(x**m*sinh(a + b*x)**3*cosh(a + b*x), 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 ) \sinh \left (b x + a\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m*cosh(b*x+a)*sinh(b*x+a)^3,x, algorithm="giac")

[Out]

integrate(x^m*cosh(b*x + a)*sinh(b*x + a)^3, x)