3.62 \(\int (a+b \cosh (c+d x))^5 \, dx\)

Optimal. Leaf size=183 \[ \frac{b \left (192 a^2 b^2+107 a^4+16 b^4\right ) \sinh (c+d x)}{30 d}+\frac{b \left (47 a^2+16 b^2\right ) \sinh (c+d x) (a+b \cosh (c+d x))^2}{60 d}+\frac{7 a b^2 \left (22 a^2+23 b^2\right ) \sinh (c+d x) \cosh (c+d x)}{120 d}+\frac{1}{8} a x \left (40 a^2 b^2+8 a^4+15 b^4\right )+\frac{b \sinh (c+d x) (a+b \cosh (c+d x))^4}{5 d}+\frac{9 a b \sinh (c+d x) (a+b \cosh (c+d x))^3}{20 d} \]

[Out]

(a*(8*a^4 + 40*a^2*b^2 + 15*b^4)*x)/8 + (b*(107*a^4 + 192*a^2*b^2 + 16*b^4)*Sinh[c + d*x])/(30*d) + (7*a*b^2*(
22*a^2 + 23*b^2)*Cosh[c + d*x]*Sinh[c + d*x])/(120*d) + (b*(47*a^2 + 16*b^2)*(a + b*Cosh[c + d*x])^2*Sinh[c +
d*x])/(60*d) + (9*a*b*(a + b*Cosh[c + d*x])^3*Sinh[c + d*x])/(20*d) + (b*(a + b*Cosh[c + d*x])^4*Sinh[c + d*x]
)/(5*d)

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Rubi [A]  time = 0.260091, antiderivative size = 183, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2656, 2753, 2734} \[ \frac{b \left (192 a^2 b^2+107 a^4+16 b^4\right ) \sinh (c+d x)}{30 d}+\frac{b \left (47 a^2+16 b^2\right ) \sinh (c+d x) (a+b \cosh (c+d x))^2}{60 d}+\frac{7 a b^2 \left (22 a^2+23 b^2\right ) \sinh (c+d x) \cosh (c+d x)}{120 d}+\frac{1}{8} a x \left (40 a^2 b^2+8 a^4+15 b^4\right )+\frac{b \sinh (c+d x) (a+b \cosh (c+d x))^4}{5 d}+\frac{9 a b \sinh (c+d x) (a+b \cosh (c+d x))^3}{20 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cosh[c + d*x])^5,x]

[Out]

(a*(8*a^4 + 40*a^2*b^2 + 15*b^4)*x)/8 + (b*(107*a^4 + 192*a^2*b^2 + 16*b^4)*Sinh[c + d*x])/(30*d) + (7*a*b^2*(
22*a^2 + 23*b^2)*Cosh[c + d*x]*Sinh[c + d*x])/(120*d) + (b*(47*a^2 + 16*b^2)*(a + b*Cosh[c + d*x])^2*Sinh[c +
d*x])/(60*d) + (9*a*b*(a + b*Cosh[c + d*x])^3*Sinh[c + d*x])/(20*d) + (b*(a + b*Cosh[c + d*x])^4*Sinh[c + d*x]
)/(5*d)

Rule 2656

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

Rule 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rule 2734

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((2*a*c
+ b*d)*x)/2, x] + (-Simp[((b*c + a*d)*Cos[e + f*x])/f, x] - Simp[(b*d*Cos[e + f*x]*Sin[e + f*x])/(2*f), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]

Rubi steps

\begin{align*} \int (a+b \cosh (c+d x))^5 \, dx &=\frac{b (a+b \cosh (c+d x))^4 \sinh (c+d x)}{5 d}+\frac{1}{5} \int (a+b \cosh (c+d x))^3 \left (5 a^2+4 b^2+9 a b \cosh (c+d x)\right ) \, dx\\ &=\frac{9 a b (a+b \cosh (c+d x))^3 \sinh (c+d x)}{20 d}+\frac{b (a+b \cosh (c+d x))^4 \sinh (c+d x)}{5 d}+\frac{1}{20} \int (a+b \cosh (c+d x))^2 \left (a \left (20 a^2+43 b^2\right )+b \left (47 a^2+16 b^2\right ) \cosh (c+d x)\right ) \, dx\\ &=\frac{b \left (47 a^2+16 b^2\right ) (a+b \cosh (c+d x))^2 \sinh (c+d x)}{60 d}+\frac{9 a b (a+b \cosh (c+d x))^3 \sinh (c+d x)}{20 d}+\frac{b (a+b \cosh (c+d x))^4 \sinh (c+d x)}{5 d}+\frac{1}{60} \int (a+b \cosh (c+d x)) \left (60 a^4+223 a^2 b^2+32 b^4+7 a b \left (22 a^2+23 b^2\right ) \cosh (c+d x)\right ) \, dx\\ &=\frac{1}{8} a \left (8 a^4+40 a^2 b^2+15 b^4\right ) x+\frac{b \left (107 a^4+192 a^2 b^2+16 b^4\right ) \sinh (c+d x)}{30 d}+\frac{7 a b^2 \left (22 a^2+23 b^2\right ) \cosh (c+d x) \sinh (c+d x)}{120 d}+\frac{b \left (47 a^2+16 b^2\right ) (a+b \cosh (c+d x))^2 \sinh (c+d x)}{60 d}+\frac{9 a b (a+b \cosh (c+d x))^3 \sinh (c+d x)}{20 d}+\frac{b (a+b \cosh (c+d x))^4 \sinh (c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 0.349365, size = 133, normalized size = 0.73 \[ \frac{60 a \left (40 a^2 b^2+8 a^4+15 b^4\right ) (c+d x)+50 b^3 \left (8 a^2+b^2\right ) \sinh (3 (c+d x))+600 a b^2 \left (2 a^2+b^2\right ) \sinh (2 (c+d x))+300 b \left (12 a^2 b^2+8 a^4+b^4\right ) \sinh (c+d x)+75 a b^4 \sinh (4 (c+d x))+6 b^5 \sinh (5 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cosh[c + d*x])^5,x]

[Out]

(60*a*(8*a^4 + 40*a^2*b^2 + 15*b^4)*(c + d*x) + 300*b*(8*a^4 + 12*a^2*b^2 + b^4)*Sinh[c + d*x] + 600*a*b^2*(2*
a^2 + b^2)*Sinh[2*(c + d*x)] + 50*b^3*(8*a^2 + b^2)*Sinh[3*(c + d*x)] + 75*a*b^4*Sinh[4*(c + d*x)] + 6*b^5*Sin
h[5*(c + d*x)])/(480*d)

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Maple [A]  time = 0.014, size = 155, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({b}^{5} \left ({\frac{8}{15}}+{\frac{ \left ( \cosh \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{4\, \left ( \cosh \left ( dx+c \right ) \right ) ^{2}}{15}} \right ) \sinh \left ( dx+c \right ) +5\,a{b}^{4} \left ( \left ( 1/4\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}+3/8\,\cosh \left ( dx+c \right ) \right ) \sinh \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +10\,{a}^{2}{b}^{3} \left ( 2/3+1/3\, \left ( \cosh \left ( dx+c \right ) \right ) ^{2} \right ) \sinh \left ( dx+c \right ) +10\,{a}^{3}{b}^{2} \left ( 1/2\,\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +5\,{a}^{4}b\sinh \left ( dx+c \right ) +{a}^{5} \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(b^5*(8/15+1/5*cosh(d*x+c)^4+4/15*cosh(d*x+c)^2)*sinh(d*x+c)+5*a*b^4*((1/4*cosh(d*x+c)^3+3/8*cosh(d*x+c))*
sinh(d*x+c)+3/8*d*x+3/8*c)+10*a^2*b^3*(2/3+1/3*cosh(d*x+c)^2)*sinh(d*x+c)+10*a^3*b^2*(1/2*cosh(d*x+c)*sinh(d*x
+c)+1/2*d*x+1/2*c)+5*a^4*b*sinh(d*x+c)+a^5*(d*x+c))

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Maxima [A]  time = 1.12051, size = 369, normalized size = 2.02 \begin{align*} \frac{5}{64} \, a b^{4}{\left (24 \, x + \frac{e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac{8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac{8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac{5}{4} \, a^{3} b^{2}{\left (4 \, x + \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + a^{5} x + \frac{1}{480} \, b^{5}{\left (\frac{3 \, e^{\left (5 \, d x + 5 \, c\right )}}{d} + \frac{25 \, e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac{150 \, e^{\left (d x + c\right )}}{d} - \frac{150 \, e^{\left (-d x - c\right )}}{d} - \frac{25 \, e^{\left (-3 \, d x - 3 \, c\right )}}{d} - \frac{3 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d}\right )} + \frac{5}{12} \, a^{2} b^{3}{\left (\frac{e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac{9 \, e^{\left (d x + c\right )}}{d} - \frac{9 \, e^{\left (-d x - c\right )}}{d} - \frac{e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} + \frac{5 \, a^{4} b \sinh \left (d x + c\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/64*a*b^4*(24*x + e^(4*d*x + 4*c)/d + 8*e^(2*d*x + 2*c)/d - 8*e^(-2*d*x - 2*c)/d - e^(-4*d*x - 4*c)/d) + 5/4*
a^3*b^2*(4*x + e^(2*d*x + 2*c)/d - e^(-2*d*x - 2*c)/d) + a^5*x + 1/480*b^5*(3*e^(5*d*x + 5*c)/d + 25*e^(3*d*x
+ 3*c)/d + 150*e^(d*x + c)/d - 150*e^(-d*x - c)/d - 25*e^(-3*d*x - 3*c)/d - 3*e^(-5*d*x - 5*c)/d) + 5/12*a^2*b
^3*(e^(3*d*x + 3*c)/d + 9*e^(d*x + c)/d - 9*e^(-d*x - c)/d - e^(-3*d*x - 3*c)/d) + 5*a^4*b*sinh(d*x + c)/d

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Fricas [A]  time = 2.18402, size = 463, normalized size = 2.53 \begin{align*} \frac{3 \, b^{5} \sinh \left (d x + c\right )^{5} + 5 \,{\left (6 \, b^{5} \cosh \left (d x + c\right )^{2} + 30 \, a b^{4} \cosh \left (d x + c\right ) + 40 \, a^{2} b^{3} + 5 \, b^{5}\right )} \sinh \left (d x + c\right )^{3} + 30 \,{\left (8 \, a^{5} + 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} d x + 15 \,{\left (b^{5} \cosh \left (d x + c\right )^{4} + 10 \, a b^{4} \cosh \left (d x + c\right )^{3} + 80 \, a^{4} b + 120 \, a^{2} b^{3} + 10 \, b^{5} + 5 \,{\left (8 \, a^{2} b^{3} + b^{5}\right )} \cosh \left (d x + c\right )^{2} + 40 \,{\left (2 \, a^{3} b^{2} + a b^{4}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(3*b^5*sinh(d*x + c)^5 + 5*(6*b^5*cosh(d*x + c)^2 + 30*a*b^4*cosh(d*x + c) + 40*a^2*b^3 + 5*b^5)*sinh(d*
x + c)^3 + 30*(8*a^5 + 40*a^3*b^2 + 15*a*b^4)*d*x + 15*(b^5*cosh(d*x + c)^4 + 10*a*b^4*cosh(d*x + c)^3 + 80*a^
4*b + 120*a^2*b^3 + 10*b^5 + 5*(8*a^2*b^3 + b^5)*cosh(d*x + c)^2 + 40*(2*a^3*b^2 + a*b^4)*cosh(d*x + c))*sinh(
d*x + c))/d

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Sympy [A]  time = 2.66193, size = 314, normalized size = 1.72 \begin{align*} \begin{cases} a^{5} x + \frac{5 a^{4} b \sinh{\left (c + d x \right )}}{d} - 5 a^{3} b^{2} x \sinh ^{2}{\left (c + d x \right )} + 5 a^{3} b^{2} x \cosh ^{2}{\left (c + d x \right )} + \frac{5 a^{3} b^{2} \sinh{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{d} - \frac{20 a^{2} b^{3} \sinh ^{3}{\left (c + d x \right )}}{3 d} + \frac{10 a^{2} b^{3} \sinh{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{d} + \frac{15 a b^{4} x \sinh ^{4}{\left (c + d x \right )}}{8} - \frac{15 a b^{4} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{4} + \frac{15 a b^{4} x \cosh ^{4}{\left (c + d x \right )}}{8} - \frac{15 a b^{4} \sinh ^{3}{\left (c + d x \right )} \cosh{\left (c + d x \right )}}{8 d} + \frac{25 a b^{4} \sinh{\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{8 d} + \frac{8 b^{5} \sinh ^{5}{\left (c + d x \right )}}{15 d} - \frac{4 b^{5} \sinh ^{3}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{3 d} + \frac{b^{5} \sinh{\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \cosh{\left (c \right )}\right )^{5} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cosh(d*x+c))**5,x)

[Out]

Piecewise((a**5*x + 5*a**4*b*sinh(c + d*x)/d - 5*a**3*b**2*x*sinh(c + d*x)**2 + 5*a**3*b**2*x*cosh(c + d*x)**2
+ 5*a**3*b**2*sinh(c + d*x)*cosh(c + d*x)/d - 20*a**2*b**3*sinh(c + d*x)**3/(3*d) + 10*a**2*b**3*sinh(c + d*x
)*cosh(c + d*x)**2/d + 15*a*b**4*x*sinh(c + d*x)**4/8 - 15*a*b**4*x*sinh(c + d*x)**2*cosh(c + d*x)**2/4 + 15*a
*b**4*x*cosh(c + d*x)**4/8 - 15*a*b**4*sinh(c + d*x)**3*cosh(c + d*x)/(8*d) + 25*a*b**4*sinh(c + d*x)*cosh(c +
d*x)**3/(8*d) + 8*b**5*sinh(c + d*x)**5/(15*d) - 4*b**5*sinh(c + d*x)**3*cosh(c + d*x)**2/(3*d) + b**5*sinh(c
+ d*x)*cosh(c + d*x)**4/d, Ne(d, 0)), (x*(a + b*cosh(c))**5, True))

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Giac [A]  time = 1.18354, size = 362, normalized size = 1.98 \begin{align*} \frac{6 \, b^{5} e^{\left (5 \, d x + 5 \, c\right )} + 75 \, a b^{4} e^{\left (4 \, d x + 4 \, c\right )} + 400 \, a^{2} b^{3} e^{\left (3 \, d x + 3 \, c\right )} + 50 \, b^{5} e^{\left (3 \, d x + 3 \, c\right )} + 1200 \, a^{3} b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 600 \, a b^{4} e^{\left (2 \, d x + 2 \, c\right )} + 2400 \, a^{4} b e^{\left (d x + c\right )} + 3600 \, a^{2} b^{3} e^{\left (d x + c\right )} + 300 \, b^{5} e^{\left (d x + c\right )} + 120 \,{\left (8 \, a^{5} + 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )}{\left (d x + c\right )} -{\left (75 \, a b^{4} e^{\left (d x + c\right )} + 6 \, b^{5} + 300 \,{\left (8 \, a^{4} b + 12 \, a^{2} b^{3} + b^{5}\right )} e^{\left (4 \, d x + 4 \, c\right )} + 600 \,{\left (2 \, a^{3} b^{2} + a b^{4}\right )} e^{\left (3 \, d x + 3 \, c\right )} + 50 \,{\left (8 \, a^{2} b^{3} + b^{5}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )} e^{\left (-5 \, d x - 5 \, c\right )}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/960*(6*b^5*e^(5*d*x + 5*c) + 75*a*b^4*e^(4*d*x + 4*c) + 400*a^2*b^3*e^(3*d*x + 3*c) + 50*b^5*e^(3*d*x + 3*c)
+ 1200*a^3*b^2*e^(2*d*x + 2*c) + 600*a*b^4*e^(2*d*x + 2*c) + 2400*a^4*b*e^(d*x + c) + 3600*a^2*b^3*e^(d*x + c
) + 300*b^5*e^(d*x + c) + 120*(8*a^5 + 40*a^3*b^2 + 15*a*b^4)*(d*x + c) - (75*a*b^4*e^(d*x + c) + 6*b^5 + 300*
(8*a^4*b + 12*a^2*b^3 + b^5)*e^(4*d*x + 4*c) + 600*(2*a^3*b^2 + a*b^4)*e^(3*d*x + 3*c) + 50*(8*a^2*b^3 + b^5)*
e^(2*d*x + 2*c))*e^(-5*d*x - 5*c))/d