### 3.584 $$\int (a \cosh (x)+b \sinh (x))^5 \, dx$$

Optimal. Leaf size=61 $\frac{2}{3} \left (a^2-b^2\right ) (a \sinh (x)+b \cosh (x))^3+\left (a^2-b^2\right )^2 (a \sinh (x)+b \cosh (x))+\frac{1}{5} (a \sinh (x)+b \cosh (x))^5$

[Out]

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

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Rubi [A]  time = 0.0450399, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 11, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.182, Rules used = {3072, 194} $\frac{2}{3} \left (a^2-b^2\right ) (a \sinh (x)+b \cosh (x))^3+\left (a^2-b^2\right )^2 (a \sinh (x)+b \cosh (x))+\frac{1}{5} (a \sinh (x)+b \cosh (x))^5$

Antiderivative was successfully veriﬁed.

[In]

Int[(a*Cosh[x] + b*Sinh[x])^5,x]

[Out]

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

Rule 3072

Int[(cos[(c_.) + (d_.)*(x_)]*(a_.) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int
[(a^2 + b^2 - x^2)^((n - 1)/2), x], x, b*Cos[c + d*x] - a*Sin[c + d*x]], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 + b^2, 0] && IGtQ[(n - 1)/2, 0]

Rule 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int (a \cosh (x)+b \sinh (x))^5 \, dx &=i \operatorname{Subst}\left (\int \left (a^2-b^2-x^2\right )^2 \, dx,x,-i b \cosh (x)-i a \sinh (x)\right )\\ &=i \operatorname{Subst}\left (\int \left (a^4 \left (1+\frac{-2 a^2 b^2+b^4}{a^4}\right )-2 a^2 \left (1-\frac{b^2}{a^2}\right ) x^2+x^4\right ) \, dx,x,-i b \cosh (x)-i a \sinh (x)\right )\\ &=\left (a^2-b^2\right )^2 (b \cosh (x)+a \sinh (x))+\frac{2}{3} \left (a^2-b^2\right ) (b \cosh (x)+a \sinh (x))^3+\frac{1}{5} (b \cosh (x)+a \sinh (x))^5\\ \end{align*}

Mathematica [B]  time = 0.225237, size = 133, normalized size = 2.18 $\frac{1}{240} \left (150 a \left (a^2-b^2\right )^2 \sinh (x)+25 a \left (2 a^2 b^2+a^4-3 b^4\right ) \sinh (3 x)+3 a \left (10 a^2 b^2+a^4+5 b^4\right ) \sinh (5 x)+150 b \left (a^2-b^2\right )^2 \cosh (x)-25 b \left (2 a^2 b^2-3 a^4+b^4\right ) \cosh (3 x)+3 b \left (10 a^2 b^2+5 a^4+b^4\right ) \cosh (5 x)\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(a*Cosh[x] + b*Sinh[x])^5,x]

[Out]

(150*b*(a^2 - b^2)^2*Cosh[x] - 25*b*(-3*a^4 + 2*a^2*b^2 + b^4)*Cosh[3*x] + 3*b*(5*a^4 + 10*a^2*b^2 + b^4)*Cosh
[5*x] + 150*a*(a^2 - b^2)^2*Sinh[x] + 25*a*(a^4 + 2*a^2*b^2 - 3*b^4)*Sinh[3*x] + 3*a*(a^4 + 10*a^2*b^2 + 5*b^4
)*Sinh[5*x])/240

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Maple [B]  time = 0.056, size = 160, normalized size = 2.6 \begin{align*}{b}^{5} \left ({\frac{8}{15}}+{\frac{ \left ( \sinh \left ( x \right ) \right ) ^{4}}{5}}-{\frac{4\, \left ( \sinh \left ( x \right ) \right ) ^{2}}{15}} \right ) \cosh \left ( x \right ) +5\,a{b}^{4} \left ( 1/5\, \left ( \sinh \left ( x \right ) \right ) ^{3} \left ( \cosh \left ( x \right ) \right ) ^{2}-1/5\,\sinh \left ( x \right ) \left ( \cosh \left ( x \right ) \right ) ^{2}+1/5\,\sinh \left ( x \right ) \right ) +10\,{a}^{2}{b}^{3} \left ( 1/5\, \left ( \sinh \left ( x \right ) \right ) ^{2} \left ( \cosh \left ( x \right ) \right ) ^{3}-2/15\,\cosh \left ( x \right ) \left ( \sinh \left ( x \right ) \right ) ^{2}-2/15\,\cosh \left ( x \right ) \right ) +10\,{a}^{3}{b}^{2} \left ( 1/5\, \left ( \cosh \left ( x \right ) \right ) ^{4}\sinh \left ( x \right ) -1/5\, \left ( 2/3+1/3\, \left ( \cosh \left ( x \right ) \right ) ^{2} \right ) \sinh \left ( x \right ) \right ) +5\,{a}^{4}b \left ( 1/5\, \left ( \sinh \left ( x \right ) \right ) ^{2} \left ( \cosh \left ( x \right ) \right ) ^{3}+1/5\,\cosh \left ( x \right ) \left ( \sinh \left ( x \right ) \right ) ^{2}+1/5\,\cosh \left ( x \right ) \right ) +{a}^{5} \left ({\frac{8}{15}}+{\frac{ \left ( \cosh \left ( x \right ) \right ) ^{4}}{5}}+{\frac{4\, \left ( \cosh \left ( x \right ) \right ) ^{2}}{15}} \right ) \sinh \left ( x \right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int((a*cosh(x)+b*sinh(x))^5,x)

[Out]

b^5*(8/15+1/5*sinh(x)^4-4/15*sinh(x)^2)*cosh(x)+5*a*b^4*(1/5*sinh(x)^3*cosh(x)^2-1/5*sinh(x)*cosh(x)^2+1/5*sin
h(x))+10*a^2*b^3*(1/5*sinh(x)^2*cosh(x)^3-2/15*cosh(x)*sinh(x)^2-2/15*cosh(x))+10*a^3*b^2*(1/5*cosh(x)^4*sinh(
x)-1/5*(2/3+1/3*cosh(x)^2)*sinh(x))+5*a^4*b*(1/5*sinh(x)^2*cosh(x)^3+1/5*cosh(x)*sinh(x)^2+1/5*cosh(x))+a^5*(8
/15+1/5*cosh(x)^4+4/15*cosh(x)^2)*sinh(x)

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Maxima [B]  time = 1.16099, size = 258, normalized size = 4.23 \begin{align*} a^{4} b \cosh \left (x\right )^{5} + a b^{4} \sinh \left (x\right )^{5} + \frac{1}{48} \,{\left ({\left (5 \, e^{\left (-2 \, x\right )} - 30 \, e^{\left (-4 \, x\right )} + 3\right )} e^{\left (5 \, x\right )} + 30 \, e^{\left (-x\right )} - 5 \, e^{\left (-3 \, x\right )} - 3 \, e^{\left (-5 \, x\right )}\right )} a^{3} b^{2} - \frac{1}{48} \,{\left ({\left (5 \, e^{\left (-2 \, x\right )} + 30 \, e^{\left (-4 \, x\right )} - 3\right )} e^{\left (5 \, x\right )} + 30 \, e^{\left (-x\right )} + 5 \, e^{\left (-3 \, x\right )} - 3 \, e^{\left (-5 \, x\right )}\right )} a^{2} b^{3} + \frac{1}{480} \, a^{5}{\left (3 \, e^{\left (5 \, x\right )} + 25 \, e^{\left (3 \, x\right )} - 150 \, e^{\left (-x\right )} - 25 \, e^{\left (-3 \, x\right )} - 3 \, e^{\left (-5 \, x\right )} + 150 \, e^{x}\right )} + \frac{1}{480} \, b^{5}{\left (3 \, e^{\left (5 \, x\right )} - 25 \, e^{\left (3 \, x\right )} + 150 \, e^{\left (-x\right )} - 25 \, e^{\left (-3 \, x\right )} + 3 \, e^{\left (-5 \, x\right )} + 150 \, e^{x}\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)+b*sinh(x))^5,x, algorithm="maxima")

[Out]

a^4*b*cosh(x)^5 + a*b^4*sinh(x)^5 + 1/48*((5*e^(-2*x) - 30*e^(-4*x) + 3)*e^(5*x) + 30*e^(-x) - 5*e^(-3*x) - 3*
e^(-5*x))*a^3*b^2 - 1/48*((5*e^(-2*x) + 30*e^(-4*x) - 3)*e^(5*x) + 30*e^(-x) + 5*e^(-3*x) - 3*e^(-5*x))*a^2*b^
3 + 1/480*a^5*(3*e^(5*x) + 25*e^(3*x) - 150*e^(-x) - 25*e^(-3*x) - 3*e^(-5*x) + 150*e^x) + 1/480*b^5*(3*e^(5*x
) - 25*e^(3*x) + 150*e^(-x) - 25*e^(-3*x) + 3*e^(-5*x) + 150*e^x)

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Fricas [B]  time = 2.46245, size = 747, normalized size = 12.25 \begin{align*} \frac{1}{80} \,{\left (5 \, a^{4} b + 10 \, a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right )^{5} + \frac{1}{16} \,{\left (5 \, a^{4} b + 10 \, a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right ) \sinh \left (x\right )^{4} + \frac{1}{80} \,{\left (a^{5} + 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \sinh \left (x\right )^{5} + \frac{5}{48} \,{\left (3 \, a^{4} b - 2 \, a^{2} b^{3} - b^{5}\right )} \cosh \left (x\right )^{3} + \frac{1}{48} \,{\left (5 \, a^{5} + 10 \, a^{3} b^{2} - 15 \, a b^{4} + 6 \,{\left (a^{5} + 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right )^{3} + \frac{1}{16} \,{\left (2 \,{\left (5 \, a^{4} b + 10 \, a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right )^{3} + 5 \,{\left (3 \, a^{4} b - 2 \, a^{2} b^{3} - b^{5}\right )} \cosh \left (x\right )\right )} \sinh \left (x\right )^{2} + \frac{5}{8} \,{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cosh \left (x\right ) + \frac{1}{16} \,{\left (10 \, a^{5} - 20 \, a^{3} b^{2} + 10 \, a b^{4} +{\left (a^{5} + 10 \, a^{3} b^{2} + 5 \, a b^{4}\right )} \cosh \left (x\right )^{4} + 5 \,{\left (a^{5} + 2 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cosh \left (x\right )^{2}\right )} \sinh \left (x\right ) \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)+b*sinh(x))^5,x, algorithm="fricas")

[Out]

1/80*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(x)^5 + 1/16*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(x)*sinh(x)^4 + 1/80*(a^5
+ 10*a^3*b^2 + 5*a*b^4)*sinh(x)^5 + 5/48*(3*a^4*b - 2*a^2*b^3 - b^5)*cosh(x)^3 + 1/48*(5*a^5 + 10*a^3*b^2 - 15
*a*b^4 + 6*(a^5 + 10*a^3*b^2 + 5*a*b^4)*cosh(x)^2)*sinh(x)^3 + 1/16*(2*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(x)^3
+ 5*(3*a^4*b - 2*a^2*b^3 - b^5)*cosh(x))*sinh(x)^2 + 5/8*(a^4*b - 2*a^2*b^3 + b^5)*cosh(x) + 1/16*(10*a^5 - 20
*a^3*b^2 + 10*a*b^4 + (a^5 + 10*a^3*b^2 + 5*a*b^4)*cosh(x)^4 + 5*(a^5 + 2*a^3*b^2 - 3*a*b^4)*cosh(x)^2)*sinh(x
)

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Sympy [B]  time = 1.65456, size = 172, normalized size = 2.82 \begin{align*} \frac{8 a^{5} \sinh ^{5}{\left (x \right )}}{15} - \frac{4 a^{5} \sinh ^{3}{\left (x \right )} \cosh ^{2}{\left (x \right )}}{3} + a^{5} \sinh{\left (x \right )} \cosh ^{4}{\left (x \right )} + a^{4} b \cosh ^{5}{\left (x \right )} - \frac{4 a^{3} b^{2} \sinh ^{5}{\left (x \right )}}{3} + \frac{10 a^{3} b^{2} \sinh ^{3}{\left (x \right )} \cosh ^{2}{\left (x \right )}}{3} + \frac{10 a^{2} b^{3} \sinh ^{2}{\left (x \right )} \cosh ^{3}{\left (x \right )}}{3} - \frac{4 a^{2} b^{3} \cosh ^{5}{\left (x \right )}}{3} + a b^{4} \sinh ^{5}{\left (x \right )} + b^{5} \sinh ^{4}{\left (x \right )} \cosh{\left (x \right )} - \frac{4 b^{5} \sinh ^{2}{\left (x \right )} \cosh ^{3}{\left (x \right )}}{3} + \frac{8 b^{5} \cosh ^{5}{\left (x \right )}}{15} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)+b*sinh(x))**5,x)

[Out]

8*a**5*sinh(x)**5/15 - 4*a**5*sinh(x)**3*cosh(x)**2/3 + a**5*sinh(x)*cosh(x)**4 + a**4*b*cosh(x)**5 - 4*a**3*b
**2*sinh(x)**5/3 + 10*a**3*b**2*sinh(x)**3*cosh(x)**2/3 + 10*a**2*b**3*sinh(x)**2*cosh(x)**3/3 - 4*a**2*b**3*c
osh(x)**5/3 + a*b**4*sinh(x)**5 + b**5*sinh(x)**4*cosh(x) - 4*b**5*sinh(x)**2*cosh(x)**3/3 + 8*b**5*cosh(x)**5
/15

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Giac [B]  time = 1.16068, size = 464, normalized size = 7.61 \begin{align*} \frac{1}{160} \, a^{5} e^{\left (5 \, x\right )} + \frac{1}{32} \, a^{4} b e^{\left (5 \, x\right )} + \frac{1}{16} \, a^{3} b^{2} e^{\left (5 \, x\right )} + \frac{1}{16} \, a^{2} b^{3} e^{\left (5 \, x\right )} + \frac{1}{32} \, a b^{4} e^{\left (5 \, x\right )} + \frac{1}{160} \, b^{5} e^{\left (5 \, x\right )} + \frac{5}{96} \, a^{5} e^{\left (3 \, x\right )} + \frac{5}{32} \, a^{4} b e^{\left (3 \, x\right )} + \frac{5}{48} \, a^{3} b^{2} e^{\left (3 \, x\right )} - \frac{5}{48} \, a^{2} b^{3} e^{\left (3 \, x\right )} - \frac{5}{32} \, a b^{4} e^{\left (3 \, x\right )} - \frac{5}{96} \, b^{5} e^{\left (3 \, x\right )} + \frac{5}{16} \, a^{5} e^{x} + \frac{5}{16} \, a^{4} b e^{x} - \frac{5}{8} \, a^{3} b^{2} e^{x} - \frac{5}{8} \, a^{2} b^{3} e^{x} + \frac{5}{16} \, a b^{4} e^{x} + \frac{5}{16} \, b^{5} e^{x} - \frac{1}{480} \,{\left (150 \, a^{5} e^{\left (4 \, x\right )} - 150 \, a^{4} b e^{\left (4 \, x\right )} - 300 \, a^{3} b^{2} e^{\left (4 \, x\right )} + 300 \, a^{2} b^{3} e^{\left (4 \, x\right )} + 150 \, a b^{4} e^{\left (4 \, x\right )} - 150 \, b^{5} e^{\left (4 \, x\right )} + 25 \, a^{5} e^{\left (2 \, x\right )} - 75 \, a^{4} b e^{\left (2 \, x\right )} + 50 \, a^{3} b^{2} e^{\left (2 \, x\right )} + 50 \, a^{2} b^{3} e^{\left (2 \, x\right )} - 75 \, a b^{4} e^{\left (2 \, x\right )} + 25 \, b^{5} e^{\left (2 \, x\right )} + 3 \, a^{5} - 15 \, a^{4} b + 30 \, a^{3} b^{2} - 30 \, a^{2} b^{3} + 15 \, a b^{4} - 3 \, b^{5}\right )} e^{\left (-5 \, x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate((a*cosh(x)+b*sinh(x))^5,x, algorithm="giac")

[Out]

1/160*a^5*e^(5*x) + 1/32*a^4*b*e^(5*x) + 1/16*a^3*b^2*e^(5*x) + 1/16*a^2*b^3*e^(5*x) + 1/32*a*b^4*e^(5*x) + 1/
160*b^5*e^(5*x) + 5/96*a^5*e^(3*x) + 5/32*a^4*b*e^(3*x) + 5/48*a^3*b^2*e^(3*x) - 5/48*a^2*b^3*e^(3*x) - 5/32*a
*b^4*e^(3*x) - 5/96*b^5*e^(3*x) + 5/16*a^5*e^x + 5/16*a^4*b*e^x - 5/8*a^3*b^2*e^x - 5/8*a^2*b^3*e^x + 5/16*a*b
^4*e^x + 5/16*b^5*e^x - 1/480*(150*a^5*e^(4*x) - 150*a^4*b*e^(4*x) - 300*a^3*b^2*e^(4*x) + 300*a^2*b^3*e^(4*x)
+ 150*a*b^4*e^(4*x) - 150*b^5*e^(4*x) + 25*a^5*e^(2*x) - 75*a^4*b*e^(2*x) + 50*a^3*b^2*e^(2*x) + 50*a^2*b^3*e
^(2*x) - 75*a*b^4*e^(2*x) + 25*b^5*e^(2*x) + 3*a^5 - 15*a^4*b + 30*a^3*b^2 - 30*a^2*b^3 + 15*a*b^4 - 3*b^5)*e^
(-5*x)