### 3.10 $$\int \cosh ^3(a+b x) \sinh ^n(a+b x) \, dx$$

Optimal. Leaf size=39 $\frac{\sinh ^{n+1}(a+b x)}{b (n+1)}+\frac{\sinh ^{n+3}(a+b x)}{b (n+3)}$

[Out]

Sinh[a + b*x]^(1 + n)/(b*(1 + n)) + Sinh[a + b*x]^(3 + n)/(b*(3 + n))

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Rubi [A]  time = 0.0448459, antiderivative size = 39, 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 = {2564, 14} $\frac{\sinh ^{n+1}(a+b x)}{b (n+1)}+\frac{\sinh ^{n+3}(a+b x)}{b (n+3)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Sinh[a + b*x]^(1 + n)/(b*(1 + n)) + Sinh[a + b*x]^(3 + n)/(b*(3 + n))

Rule 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, 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 ^3(a+b x) \sinh ^n(a+b x) \, dx &=\frac{\operatorname{Subst}\left (\int x^n \left (1+x^2\right ) \, dx,x,\sinh (a+b x)\right )}{b}\\ &=\frac{\operatorname{Subst}\left (\int \left (x^n+x^{2+n}\right ) \, dx,x,\sinh (a+b x)\right )}{b}\\ &=\frac{\sinh ^{1+n}(a+b x)}{b (1+n)}+\frac{\sinh ^{3+n}(a+b x)}{b (3+n)}\\ \end{align*}

Mathematica [A]  time = 0.0608808, size = 39, normalized size = 1. $\frac{\sinh ^{n+1}(a+b x)}{b (n+1)}+\frac{\sinh ^{n+3}(a+b x)}{b (n+3)}$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

Sinh[a + b*x]^(1 + n)/(b*(1 + n)) + Sinh[a + b*x]^(3 + n)/(b*(3 + n))

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Maple [F]  time = 0.5, size = 0, normalized size = 0. \begin{align*} \int \left ( \cosh \left ( bx+a \right ) \right ) ^{3} \left ( \sinh \left ( bx+a \right ) \right ) ^{n}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.83658, size = 504, normalized size = 12.92 \begin{align*} \frac{n e^{\left ({\left (b x + a\right )} n + 3 \, b x + n \log \left (e^{\left (-b x - a\right )} + 1\right ) + n \log \left (-e^{\left (-b x - a\right )} + 1\right ) + 3 \, a\right )}}{8 \,{\left (2^{n} n^{2} + 2^{n + 2} n + 3 \cdot 2^{n}\right )} b} + \frac{{\left (n + 9\right )} e^{\left ({\left (b x + a\right )} n + b x + n \log \left (e^{\left (-b x - a\right )} + 1\right ) + n \log \left (-e^{\left (-b x - a\right )} + 1\right ) + a\right )}}{8 \,{\left (2^{n} n^{2} + 2^{n + 2} n + 3 \cdot 2^{n}\right )} b} - \frac{{\left (n + 9\right )} e^{\left ({\left (b x + a\right )} n - b x + n \log \left (e^{\left (-b x - a\right )} + 1\right ) + n \log \left (-e^{\left (-b x - a\right )} + 1\right ) - a\right )}}{8 \,{\left (2^{n} n^{2} + 2^{n + 2} n + 3 \cdot 2^{n}\right )} b} - \frac{{\left (n + 1\right )} e^{\left ({\left (b x + a\right )} n - 3 \, b x + n \log \left (e^{\left (-b x - a\right )} + 1\right ) + n \log \left (-e^{\left (-b x - a\right )} + 1\right ) - 3 \, a\right )}}{8 \,{\left (2^{n} n^{2} + 2^{n + 2} n + 3 \cdot 2^{n}\right )} b} + \frac{e^{\left ({\left (b x + a\right )} n + 3 \, b x + n \log \left (e^{\left (-b x - a\right )} + 1\right ) + n \log \left (-e^{\left (-b x - a\right )} + 1\right ) + 3 \, a\right )}}{8 \,{\left (2^{n} n^{2} + 2^{n + 2} n + 3 \cdot 2^{n}\right )} b} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.26868, size = 482, normalized size = 12.36 \begin{align*} \frac{{\left ({\left (n + 1\right )} \sinh \left (b x + a\right )^{3} +{\left (3 \,{\left (n + 1\right )} \cosh \left (b x + a\right )^{2} + n + 9\right )} \sinh \left (b x + a\right )\right )} \cosh \left (n \log \left (\sinh \left (b x + a\right )\right )\right ) +{\left ({\left (n + 1\right )} \sinh \left (b x + a\right )^{3} +{\left (3 \,{\left (n + 1\right )} \cosh \left (b x + a\right )^{2} + n + 9\right )} \sinh \left (b x + a\right )\right )} \sinh \left (n \log \left (\sinh \left (b x + a\right )\right )\right )}{4 \,{\left ({\left (b n^{2} + 4 \, b n + 3 \, b\right )} \cosh \left (b x + a\right )^{4} - 2 \,{\left (b n^{2} + 4 \, b n + 3 \, b\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} +{\left (b n^{2} + 4 \, b n + 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)^3*sinh(b*x+a)^n,x, algorithm="fricas")

[Out]

1/4*(((n + 1)*sinh(b*x + a)^3 + (3*(n + 1)*cosh(b*x + a)^2 + n + 9)*sinh(b*x + a))*cosh(n*log(sinh(b*x + a)))
+ ((n + 1)*sinh(b*x + a)^3 + (3*(n + 1)*cosh(b*x + a)^2 + n + 9)*sinh(b*x + a))*sinh(n*log(sinh(b*x + a))))/((
b*n^2 + 4*b*n + 3*b)*cosh(b*x + a)^4 - 2*(b*n^2 + 4*b*n + 3*b)*cosh(b*x + a)^2*sinh(b*x + a)^2 + (b*n^2 + 4*b*
n + 3*b)*sinh(b*x + a)^4)

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Sympy [A]  time = 14.3459, size = 668, normalized size = 17.13 \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)**3*sinh(b*x+a)**n,x)

[Out]

Piecewise((x*sinh(a)**n*cosh(a)**3, Eq(b, 0)), (log(sinh(a + b*x))/b - cosh(a + b*x)**2/(2*b*sinh(a + b*x)**2)
, Eq(n, -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*tan
h(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) + lo
g(tanh(a/2 + b*x/2))*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))*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) + log(tanh(a/2 + b
*x/2))/(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(n, -1)), (n*si
nh(a + b*x)*sinh(a + b*x)**n*cosh(a + b*x)**2/(b*n**2 + 4*b*n + 3*b) - 2*sinh(a + b*x)**3*sinh(a + b*x)**n/(b*
n**2 + 4*b*n + 3*b) + 3*sinh(a + b*x)*sinh(a + b*x)**n*cosh(a + b*x)**2/(b*n**2 + 4*b*n + 3*b), True))

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

[Out]

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