3.3.89 \(\int x^m \cosh ^2(a+b x) \sinh ^2(a+b x) \, dx\) [289]

Optimal. Leaf size=85 \[ -\frac {x^{1+m}}{8 (1+m)}+\frac {2^{-2 (3+m)} e^{4 a} x^m (-b x)^{-m} \Gamma (1+m,-4 b x)}{b}-\frac {2^{-2 (3+m)} e^{-4 a} x^m (b x)^{-m} \Gamma (1+m,4 b x)}{b} \]

[Out]

-1/8*x^(1+m)/(1+m)+exp(4*a)*x^m*GAMMA(1+m,-4*b*x)/(2^(6+2*m))/b/((-b*x)^m)-x^m*GAMMA(1+m,4*b*x)/(2^(6+2*m))/b/
exp(4*a)/((b*x)^m)

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Rubi [A]
time = 0.10, antiderivative size = 85, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {5556, 3388, 2212} \begin {gather*} \frac {e^{4 a} 2^{-2 (m+3)} x^m (-b x)^{-m} \text {Gamma}(m+1,-4 b x)}{b}-\frac {e^{-4 a} 2^{-2 (m+3)} x^m (b x)^{-m} \text {Gamma}(m+1,4 b x)}{b}-\frac {x^{m+1}}{8 (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*x^(1 + m)/(1 + m) + (E^(4*a)*x^m*Gamma[1 + m, -4*b*x])/(2^(2*(3 + m))*b*(-(b*x))^m) - (x^m*Gamma[1 + m, 4
*b*x])/(2^(2*(3 + m))*b*E^(4*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 5556

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]

Rubi steps

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

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Mathematica [A]
time = 0.13, size = 106, normalized size = 1.25 \begin {gather*} -\frac {2^{-2-2 (2+m)} e^{-4 a} x^m \left (-b^2 x^2\right )^{-m} \left (2^{3+2 m} b e^{4 a} x \left (-b^2 x^2\right )^m-e^{8 a} (1+m) (b x)^m \Gamma (1+m,-4 b x)+(1+m) (-b x)^m \Gamma (1+m,4 b x)\right )}{b (1+m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 1.42, size = 0, normalized size = 0.00 \[\int x^{m} \left (\cosh ^{2}\left (b x +a \right )\right ) \left (\sinh ^{2}\left (b x +a \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.08, size = 71, normalized size = 0.84 \begin {gather*} -\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}{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 {x^{m + 1}}{8 \, {\left (m + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.10, size = 122, normalized size = 1.44 \begin {gather*} -\frac {8 \, b x \cosh \left (m \log \left (x\right )\right ) + {\left (m + 1\right )} \cosh \left (m \log \left (4 \, b\right ) + 4 \, a\right ) \Gamma \left (m + 1, 4 \, b x\right ) - {\left (m + 1\right )} \cosh \left (m \log \left (-4 \, b\right ) - 4 \, a\right ) \Gamma \left (m + 1, -4 \, b x\right ) - {\left (m + 1\right )} \Gamma \left (m + 1, 4 \, b x\right ) \sinh \left (m \log \left (4 \, b\right ) + 4 \, a\right ) + {\left (m + 1\right )} \Gamma \left (m + 1, -4 \, b x\right ) \sinh \left (m \log \left (-4 \, b\right ) - 4 \, a\right ) + 8 \, b x \sinh \left (m \log \left (x\right )\right )}{64 \, {\left (b m + b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/64*(8*b*x*cosh(m*log(x)) + (m + 1)*cosh(m*log(4*b) + 4*a)*gamma(m + 1, 4*b*x) - (m + 1)*cosh(m*log(-4*b) -
4*a)*gamma(m + 1, -4*b*x) - (m + 1)*gamma(m + 1, 4*b*x)*sinh(m*log(4*b) + 4*a) + (m + 1)*gamma(m + 1, -4*b*x)*
sinh(m*log(-4*b) - 4*a) + 8*b*x*sinh(m*log(x)))/(b*m + b)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{m} \sinh ^{2}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

integrate(x^m*cosh(b*x + a)^2*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\,{\mathrm {cosh}\left (a+b\,x\right )}^2\,{\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*cosh(a + b*x)^2*sinh(a + b*x)^2,x)

[Out]

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

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