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(b*x+a)^(1+n)/b/(1+n)+sinh(b*x+a)^(3+n)/b/(3+n)

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Rubi [A]  time = 0.04, antiderivative size = 39, normalized size of antiderivative = 1.00, 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 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]]

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])

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*}

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Mathematica [A]  time = 0.06, size = 39, normalized size = 1.00 \[ \frac {\sinh ^{n+1}(a+b x)}{b (n+1)}+\frac {\sinh ^{n+3}(a+b x)}{b (n+3)} \]

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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fricas [B]  time = 0.54, size = 175, normalized size = 4.49 \[ \frac {{\left ({\left (n + 1\right )} \sinh \left (b x + a\right )^{3} + {\left (3 \, {\left (n + 1\right )} \cosh \left (b x + a\right )^{2} + n + 9\right )} \sinh \left (b x + a\right )\right )} \cosh \left (n \log \left (\sinh \left (b x + a\right )\right )\right ) + {\left ({\left (n + 1\right )} \sinh \left (b x + a\right )^{3} + {\left (3 \, {\left (n + 1\right )} \cosh \left (b x + a\right )^{2} + n + 9\right )} \sinh \left (b x + a\right )\right )} \sinh \left (n \log \left (\sinh \left (b x + a\right )\right )\right )}{4 \, {\left ({\left (b n^{2} + 4 \, b n + 3 \, b\right )} \cosh \left (b x + a\right )^{4} - 2 \, {\left (b n^{2} + 4 \, b n + 3 \, b\right )} \cosh \left (b x + a\right )^{2} \sinh \left (b x + a\right )^{2} + {\left (b n^{2} + 4 \, b n + 3 \, b\right )} \sinh \left (b x + a\right )^{4}\right )}} \]

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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giac [B]  time = 0.32, size = 327, normalized size = 8.38 \[ \frac {n e^{\left (7 \, b x + n \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right ) + 7 \, a\right )} + n e^{\left (5 \, b x + n \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right ) + 5 \, a\right )} - n e^{\left (3 \, b x + n \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right ) + 3 \, a\right )} - n e^{\left (b x + n \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right ) + a\right )} + e^{\left (7 \, b x + n \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right ) + 7 \, a\right )} + 9 \, e^{\left (5 \, b x + n \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right ) + 5 \, a\right )} - 9 \, e^{\left (3 \, b x + n \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right ) + 3 \, a\right )} - e^{\left (b x + n \log \left (\frac {1}{2} \, {\left (e^{\left (2 \, b x + 2 \, a\right )} - 1\right )} e^{\left (-b x - a\right )}\right ) + a\right )}}{8 \, {\left (b n^{2} e^{\left (4 \, b x + 4 \, a\right )} + 4 \, b n e^{\left (4 \, b x + 4 \, a\right )} + 3 \, b e^{\left (4 \, b x + 4 \, a\right )}\right )}} \]

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]

1/8*(n*e^(7*b*x + n*log(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)) + 7*a) + n*e^(5*b*x + n*log(1/2*(e^(2*b*x + 2*
a) - 1)*e^(-b*x - a)) + 5*a) - n*e^(3*b*x + n*log(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)) + 3*a) - n*e^(b*x +
n*log(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)) + a) + e^(7*b*x + n*log(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a))
+ 7*a) + 9*e^(5*b*x + n*log(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)) + 5*a) - 9*e^(3*b*x + n*log(1/2*(e^(2*b*x
+ 2*a) - 1)*e^(-b*x - a)) + 3*a) - e^(b*x + n*log(1/2*(e^(2*b*x + 2*a) - 1)*e^(-b*x - a)) + a))/(b*n^2*e^(4*b*
x + 4*a) + 4*b*n*e^(4*b*x + 4*a) + 3*b*e^(4*b*x + 4*a))

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maple [F]  time = 1.28, size = 0, normalized size = 0.00 \[ \int \left (\cosh ^{3}\left (b x +a \right )\right ) \left (\sinh ^{n}\left (b x +a \right )\right )\, dx \]

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 = 0.82, size = 373, normalized size = 9.56 \[ \frac {n e^{\left ({\left (b x + a\right )} n + 3 \, b x + n \log \left (e^{\left (-b x - a\right )} + 1\right ) + n \log \left (-e^{\left (-b x - a\right )} + 1\right ) + 3 \, a\right )}}{8 \, {\left (2^{n} n^{2} + 2^{n + 2} n + 3 \cdot 2^{n}\right )} b} + \frac {{\left (n + 9\right )} e^{\left ({\left (b x + a\right )} n + b x + n \log \left (e^{\left (-b x - a\right )} + 1\right ) + n \log \left (-e^{\left (-b x - a\right )} + 1\right ) + a\right )}}{8 \, {\left (2^{n} n^{2} + 2^{n + 2} n + 3 \cdot 2^{n}\right )} b} - \frac {{\left (n + 9\right )} e^{\left ({\left (b x + a\right )} n - b x + n \log \left (e^{\left (-b x - a\right )} + 1\right ) + n \log \left (-e^{\left (-b x - a\right )} + 1\right ) - a\right )}}{8 \, {\left (2^{n} n^{2} + 2^{n + 2} n + 3 \cdot 2^{n}\right )} b} - \frac {{\left (n + 1\right )} e^{\left ({\left (b x + a\right )} n - 3 \, b x + n \log \left (e^{\left (-b x - a\right )} + 1\right ) + n \log \left (-e^{\left (-b x - a\right )} + 1\right ) - 3 \, a\right )}}{8 \, {\left (2^{n} n^{2} + 2^{n + 2} n + 3 \cdot 2^{n}\right )} b} + \frac {e^{\left ({\left (b x + a\right )} n + 3 \, b x + n \log \left (e^{\left (-b x - a\right )} + 1\right ) + n \log \left (-e^{\left (-b x - a\right )} + 1\right ) + 3 \, a\right )}}{8 \, {\left (2^{n} n^{2} + 2^{n + 2} n + 3 \cdot 2^{n}\right )} b} \]

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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mupad [B]  time = 1.64, size = 135, normalized size = 3.46 \[ -{\left (\frac {1}{2}\right )}^n\,{\mathrm {e}}^{-3\,a-3\,b\,x}\,{\left ({\mathrm {e}}^{a+b\,x}-{\mathrm {e}}^{-a-b\,x}\right )}^n\,\left (\frac {\frac {n}{8}+\frac {1}{8}}{b\,\left (n^2+4\,n+3\right )}+\frac {{\mathrm {e}}^{2\,a+2\,b\,x}\,\left (n+9\right )}{8\,b\,\left (n^2+4\,n+3\right )}-\frac {{\mathrm {e}}^{6\,a+6\,b\,x}\,\left (n+1\right )}{8\,b\,\left (n^2+4\,n+3\right )}-\frac {{\mathrm {e}}^{4\,a+4\,b\,x}\,\left (n+9\right )}{8\,b\,\left (n^2+4\,n+3\right )}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(1/2)^n*exp(- 3*a - 3*b*x)*(exp(a + b*x) - exp(- a - b*x))^n*((n/8 + 1/8)/(b*(4*n + n^2 + 3)) + (exp(2*a + 2*
b*x)*(n + 9))/(8*b*(4*n + n^2 + 3)) - (exp(6*a + 6*b*x)*(n + 1))/(8*b*(4*n + n^2 + 3)) - (exp(4*a + 4*b*x)*(n
+ 9))/(8*b*(4*n + n^2 + 3)))

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sympy [A]  time = 8.42, size = 638, normalized size = 16.36 \[ \begin {cases} x \sinh ^{n}{\relax (a )} \cosh ^{3}{\relax (a )} & \text {for}\: b = 0 \\\frac {\log {\left (\sinh {\left (a + b x \right )} \right )}}{b} - \frac {\cosh ^{2}{\left (a + b x \right )}}{2 b \sinh ^{2}{\left (a + b x \right )}} & \text {for}\: n = -3 \\\frac {b x \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} - \frac {2 b x \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} + \frac {b x}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} - \frac {2 \log {\left (\tanh {\left (\frac {a}{2} + \frac {b x}{2} \right )} + 1 \right )} \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} + \frac {4 \log {\left (\tanh {\left (\frac {a}{2} + \frac {b x}{2} \right )} + 1 \right )} \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} - \frac {2 \log {\left (\tanh {\left (\frac {a}{2} + \frac {b x}{2} \right )} + 1 \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} + \frac {\log {\left (\tanh {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )} \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} - \frac {2 \log {\left (\tanh {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )} \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} + \frac {\log {\left (\tanh {\left (\frac {a}{2} + \frac {b x}{2} \right )} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} + \frac {2 \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )}}{b \tanh ^{4}{\left (\frac {a}{2} + \frac {b x}{2} \right )} - 2 b \tanh ^{2}{\left (\frac {a}{2} + \frac {b x}{2} \right )} + b} & \text {for}\: n = -1 \\\frac {n \sinh {\left (a + b x \right )} \sinh ^{n}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{b n^{2} + 4 b n + 3 b} - \frac {2 \sinh ^{3}{\left (a + b x \right )} \sinh ^{n}{\left (a + b x \right )}}{b n^{2} + 4 b n + 3 b} + \frac {3 \sinh {\left (a + b x \right )} \sinh ^{n}{\left (a + b x \right )} \cosh ^{2}{\left (a + b x \right )}}{b n^{2} + 4 b n + 3 b} & \text {otherwise} \end {cases} \]

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) + 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), Eq(n, -1)), (n*sinh(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*cos
h(a + b*x)**2/(b*n**2 + 4*b*n + 3*b), True))

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