3.13 $$\int \cosh ^m(a+b x) \sinh ^3(a+b x) \, dx$$

Optimal. Leaf size=40 $\frac{\cosh ^{m+3}(a+b x)}{b (m+3)}-\frac{\cosh ^{m+1}(a+b x)}{b (m+1)}$

[Out]

-(Cosh[a + b*x]^(1 + m)/(b*(1 + m))) + Cosh[a + b*x]^(3 + m)/(b*(3 + m))

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Rubi [A]  time = 0.0502823, antiderivative size = 40, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 17, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.118, Rules used = {2565, 14} $\frac{\cosh ^{m+3}(a+b x)}{b (m+3)}-\frac{\cosh ^{m+1}(a+b x)}{b (m+1)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-(Cosh[a + b*x]^(1 + m)/(b*(1 + m))) + Cosh[a + b*x]^(3 + m)/(b*(3 + m))

Rule 2565

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

Mathematica [A]  time = 0.125609, size = 44, normalized size = 1.1 $\frac{\cosh ^{m+1}(a+b x) ((m+1) \cosh (2 (a+b x))-m-5)}{2 b (m+1) (m+3)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

(Cosh[a + b*x]^(1 + m)*(-5 - m + (1 + m)*Cosh[2*(a + b*x)]))/(2*b*(1 + m)*(3 + m))

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Maple [C]  time = 0.372, size = 923, normalized size = 23.1 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8/b/(3+m)*(1+exp(2*b*x+2*a))^m*(1/2)^m*exp(b*x+a)^(-m)*exp(-3*b*x-3*a)*exp(-1/2*I*Pi*m*csgn(I*(1+exp(2*b*x+2
*a)))*csgn(I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a))))*exp(1/2*I*Pi*m*csgn(I*(1+exp(2*b*x+2*a)))*cs
gn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(1/2*I*Pi*m*csgn(I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a
)))^2)*exp(-1/2*I*Pi*m*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^3)+1/8/b/(3+m)*(1+exp(2*b*x+2*a))^m*(1/2)^m*exp(
b*x+a)^(-m)*exp(3*b*x+3*a)*exp(-1/2*I*Pi*m*csgn(I*(1+exp(2*b*x+2*a)))*csgn(I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(
1+exp(2*b*x+2*a))))*exp(1/2*I*Pi*m*csgn(I*(1+exp(2*b*x+2*a)))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(1/
2*I*Pi*m*csgn(I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(-1/2*I*Pi*m*csgn(I*exp(-b*x-a)*(1+e
xp(2*b*x+2*a)))^3)-1/8*(m+9)/b/(m^2+4*m+3)*(1+exp(2*b*x+2*a))^m*(1/2)^m*exp(b*x+a)^(-m)*exp(-b*x-a)*exp(-1/2*I
*Pi*m*csgn(I*(1+exp(2*b*x+2*a)))*csgn(I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a))))*exp(1/2*I*Pi*m*cs
gn(I*(1+exp(2*b*x+2*a)))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(1/2*I*Pi*m*csgn(I*exp(-b*x-a))*csgn(I*e
xp(-b*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(-1/2*I*Pi*m*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^3)-1/8*(m+9)/b/(m^2+4
*m+3)*(1+exp(2*b*x+2*a))^m*(1/2)^m*exp(b*x+a)^(-m)*exp(b*x+a)*exp(-1/2*I*Pi*m*csgn(I*(1+exp(2*b*x+2*a)))*csgn(
I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a))))*exp(1/2*I*Pi*m*csgn(I*(1+exp(2*b*x+2*a)))*csgn(I*exp(-b
*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(1/2*I*Pi*m*csgn(I*exp(-b*x-a))*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^2)*exp(
-1/2*I*Pi*m*csgn(I*exp(-b*x-a)*(1+exp(2*b*x+2*a)))^3)

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Maxima [B]  time = 1.66797, size = 396, normalized size = 9.9 \begin{align*} \frac{m e^{\left ({\left (b x + a\right )} m + 3 \, b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) + 3 \, a\right )}}{8 \,{\left (2^{m} m^{2} + 2^{m + 2} m + 3 \cdot 2^{m}\right )} b} - \frac{{\left (m + 9\right )} e^{\left ({\left (b x + a\right )} m + b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) + a\right )}}{8 \,{\left (2^{m} m^{2} + 2^{m + 2} m + 3 \cdot 2^{m}\right )} b} - \frac{{\left (m + 9\right )} e^{\left ({\left (b x + a\right )} m - b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) - a\right )}}{8 \,{\left (2^{m} m^{2} + 2^{m + 2} m + 3 \cdot 2^{m}\right )} b} + \frac{{\left (m + 1\right )} e^{\left ({\left (b x + a\right )} m - 3 \, b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) - 3 \, a\right )}}{8 \,{\left (2^{m} m^{2} + 2^{m + 2} m + 3 \cdot 2^{m}\right )} b} + \frac{e^{\left ({\left (b x + a\right )} m + 3 \, b x + m \log \left (e^{\left (-2 \, b x - 2 \, a\right )} + 1\right ) + 3 \, a\right )}}{8 \,{\left (2^{m} m^{2} + 2^{m + 2} m + 3 \cdot 2^{m}\right )} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*m*e^((b*x + a)*m + 3*b*x + m*log(e^(-2*b*x - 2*a) + 1) + 3*a)/((2^m*m^2 + 2^(m + 2)*m + 3*2^m)*b) - 1/8*(m
+ 9)*e^((b*x + a)*m + b*x + m*log(e^(-2*b*x - 2*a) + 1) + a)/((2^m*m^2 + 2^(m + 2)*m + 3*2^m)*b) - 1/8*(m + 9
)*e^((b*x + a)*m - b*x + m*log(e^(-2*b*x - 2*a) + 1) - a)/((2^m*m^2 + 2^(m + 2)*m + 3*2^m)*b) + 1/8*(m + 1)*e^
((b*x + a)*m - 3*b*x + m*log(e^(-2*b*x - 2*a) + 1) - 3*a)/((2^m*m^2 + 2^(m + 2)*m + 3*2^m)*b) + 1/8*e^((b*x +
a)*m + 3*b*x + m*log(e^(-2*b*x - 2*a) + 1) + 3*a)/((2^m*m^2 + 2^(m + 2)*m + 3*2^m)*b)

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Fricas [B]  time = 2.18545, size = 520, normalized size = 13. \begin{align*} \frac{{\left ({\left (m + 1\right )} \cosh \left (b x + a\right )^{3} + 3 \,{\left (m + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} -{\left (m + 9\right )} \cosh \left (b x + a\right )\right )} \cosh \left (m \log \left (\cosh \left (b x + a\right )\right )\right ) +{\left ({\left (m + 1\right )} \cosh \left (b x + a\right )^{3} + 3 \,{\left (m + 1\right )} \cosh \left (b x + a\right ) \sinh \left (b x + a\right )^{2} -{\left (m + 9\right )} \cosh \left (b x + a\right )\right )} \sinh \left (m \log \left (\cosh \left (b x + a\right )\right )\right )}{4 \,{\left ({\left (b m^{2} + 4 \, b m + 3 \, b\right )} \cosh \left (b x + a\right )^{4} - 2 \,{\left (b m^{2} + 4 \, b m + 3 \, b\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} +{\left (b m^{2} + 4 \, b m + 3 \, b\right )} \sinh \left (b x + a\right )^{4}\right )}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(((m + 1)*cosh(b*x + a)^3 + 3*(m + 1)*cosh(b*x + a)*sinh(b*x + a)^2 - (m + 9)*cosh(b*x + a))*cosh(m*log(co
sh(b*x + a))) + ((m + 1)*cosh(b*x + a)^3 + 3*(m + 1)*cosh(b*x + a)*sinh(b*x + a)^2 - (m + 9)*cosh(b*x + a))*si
nh(m*log(cosh(b*x + a))))/((b*m^2 + 4*b*m + 3*b)*cosh(b*x + a)^4 - 2*(b*m^2 + 4*b*m + 3*b)*cosh(b*x + a)^2*sin
h(b*x + a)^2 + (b*m^2 + 4*b*m + 3*b)*sinh(b*x + a)^4)

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Sympy [A]  time = 14.7158, size = 678, normalized size = 16.95 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*sinh(a)**3*cosh(a)**m, Eq(b, 0)), (log(cosh(a + b*x))/b - sinh(a + b*x)**2/(2*b*cosh(a + b*x)**2)
, Eq(m, -3)), (-b*x*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) + 2*b*x*tanh(
a/2 + b*x/2)**2/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) - b*x/(b*tanh(a/2 + b*x/2)**4 - 2*b*ta
nh(a/2 + b*x/2)**2 + b) + 2*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh
(a/2 + b*x/2)**2 + b) - 4*log(tanh(a/2 + b*x/2) + 1)*tanh(a/2 + b*x/2)**2/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a
/2 + b*x/2)**2 + b) + 2*log(tanh(a/2 + b*x/2) + 1)/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) - l
og(tanh(a/2 + b*x/2)**2 + 1)*tanh(a/2 + b*x/2)**4/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) + 2*
log(tanh(a/2 + b*x/2)**2 + 1)*tanh(a/2 + b*x/2)**2/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) - l
og(tanh(a/2 + b*x/2)**2 + 1)/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) + tanh(a/2 + b*x/2)**4/(b
*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 + b) + 1/(b*tanh(a/2 + b*x/2)**4 - 2*b*tanh(a/2 + b*x/2)**2 +
b), Eq(m, -1)), (m*sinh(a + b*x)**2*cosh(a + b*x)*cosh(a + b*x)**m/(b*m**2 + 4*b*m + 3*b) + 3*sinh(a + b*x)**
2*cosh(a + b*x)*cosh(a + b*x)**m/(b*m**2 + 4*b*m + 3*b) - 2*cosh(a + b*x)**3*cosh(a + b*x)**m/(b*m**2 + 4*b*m
+ 3*b), True))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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