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

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

[Out]

1/8*a*(8*a^4+40*a^2*b^2+15*b^4)*x+1/30*b*(107*a^4+192*a^2*b^2+16*b^4)*sinh(d*x+c)/d+7/120*a*b^2*(22*a^2+23*b^2
)*cosh(d*x+c)*sinh(d*x+c)/d+1/60*b*(47*a^2+16*b^2)*(a+b*cosh(d*x+c))^2*sinh(d*x+c)/d+9/20*a*b*(a+b*cosh(d*x+c)
)^3*sinh(d*x+c)/d+1/5*b*(a+b*cosh(d*x+c))^4*sinh(d*x+c)/d

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Rubi [A]  time = 0.26, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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]

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]

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*}

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Mathematica [A]  time = 0.37, size = 133, normalized size = 0.73 \[ \frac {600 a b^2 \left (2 a^2+b^2\right ) \sinh (2 (c+d x))+50 b^3 \left (8 a^2+b^2\right ) \sinh (3 (c+d x))+60 a \left (8 a^4+40 a^2 b^2+15 b^4\right ) (c+d x)+300 b \left (8 a^4+12 a^2 b^2+b^4\right ) \sinh (c+d x)+75 a b^4 \sinh (4 (c+d x))+6 b^5 \sinh (5 (c+d x))}{480 d} \]

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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fricas [A]  time = 0.51, size = 190, normalized size = 1.04 \[ \frac {3 \, b^{5} \sinh \left (d x + c\right )^{5} + 5 \, {\left (6 \, b^{5} \cosh \left (d x + c\right )^{2} + 30 \, a b^{4} \cosh \left (d x + c\right ) + 40 \, a^{2} b^{3} + 5 \, b^{5}\right )} \sinh \left (d x + c\right )^{3} + 30 \, {\left (8 \, a^{5} + 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} d x + 15 \, {\left (b^{5} \cosh \left (d x + c\right )^{4} + 10 \, a b^{4} \cosh \left (d x + c\right )^{3} + 80 \, a^{4} b + 120 \, a^{2} b^{3} + 10 \, b^{5} + 5 \, {\left (8 \, a^{2} b^{3} + b^{5}\right )} \cosh \left (d x + c\right )^{2} + 40 \, {\left (2 \, a^{3} b^{2} + a b^{4}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{240 \, d} \]

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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giac [A]  time = 0.16, size = 263, normalized size = 1.44 \[ \frac {b^{5} e^{\left (5 \, d x + 5 \, c\right )}}{160 \, d} + \frac {5 \, a b^{4} e^{\left (4 \, d x + 4 \, c\right )}}{64 \, d} - \frac {5 \, a b^{4} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} - \frac {b^{5} e^{\left (-5 \, d x - 5 \, c\right )}}{160 \, d} + \frac {1}{8} \, {\left (8 \, a^{5} + 40 \, a^{3} b^{2} + 15 \, a b^{4}\right )} x + \frac {5 \, {\left (8 \, a^{2} b^{3} + b^{5}\right )} e^{\left (3 \, d x + 3 \, c\right )}}{96 \, d} + \frac {5 \, {\left (2 \, a^{3} b^{2} + a b^{4}\right )} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} + \frac {5 \, {\left (8 \, a^{4} b + 12 \, a^{2} b^{3} + b^{5}\right )} e^{\left (d x + c\right )}}{16 \, d} - \frac {5 \, {\left (8 \, a^{4} b + 12 \, a^{2} b^{3} + b^{5}\right )} e^{\left (-d x - c\right )}}{16 \, d} - \frac {5 \, {\left (2 \, a^{3} b^{2} + a b^{4}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, d} - \frac {5 \, {\left (8 \, a^{2} b^{3} + b^{5}\right )} e^{\left (-3 \, d x - 3 \, c\right )}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/160*b^5*e^(5*d*x + 5*c)/d + 5/64*a*b^4*e^(4*d*x + 4*c)/d - 5/64*a*b^4*e^(-4*d*x - 4*c)/d - 1/160*b^5*e^(-5*d
*x - 5*c)/d + 1/8*(8*a^5 + 40*a^3*b^2 + 15*a*b^4)*x + 5/96*(8*a^2*b^3 + b^5)*e^(3*d*x + 3*c)/d + 5/8*(2*a^3*b^
2 + a*b^4)*e^(2*d*x + 2*c)/d + 5/16*(8*a^4*b + 12*a^2*b^3 + b^5)*e^(d*x + c)/d - 5/16*(8*a^4*b + 12*a^2*b^3 +
b^5)*e^(-d*x - c)/d - 5/8*(2*a^3*b^2 + a*b^4)*e^(-2*d*x - 2*c)/d - 5/96*(8*a^2*b^3 + b^5)*e^(-3*d*x - 3*c)/d

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maple [A]  time = 0.30, size = 155, normalized size = 0.85 \[ \frac {b^{5} \left (\frac {8}{15}+\frac {\left (\cosh ^{4}\left (d x +c \right )\right )}{5}+\frac {4 \left (\cosh ^{2}\left (d x +c \right )\right )}{15}\right ) \sinh \left (d x +c \right )+5 a \,b^{4} \left (\left (\frac {\left (\cosh ^{3}\left (d x +c \right )\right )}{4}+\frac {3 \cosh \left (d x +c \right )}{8}\right ) \sinh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )+10 a^{2} b^{3} \left (\frac {2}{3}+\frac {\left (\cosh ^{2}\left (d x +c \right )\right )}{3}\right ) \sinh \left (d x +c \right )+10 a^{3} b^{2} \left (\frac {\cosh \left (d x +c \right ) \sinh \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+5 a^{4} b \sinh \left (d x +c \right )+a^{5} \left (d x +c \right )}{d} \]

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 = 0.36, size = 273, normalized size = 1.49 \[ \frac {5}{64} \, a b^{4} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} + \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + \frac {5}{4} \, a^{3} b^{2} {\left (4 \, x + \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} - \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + a^{5} x + \frac {1}{480} \, b^{5} {\left (\frac {3 \, e^{\left (5 \, d x + 5 \, c\right )}}{d} + \frac {25 \, e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac {150 \, e^{\left (d x + c\right )}}{d} - \frac {150 \, e^{\left (-d x - c\right )}}{d} - \frac {25 \, e^{\left (-3 \, d x - 3 \, c\right )}}{d} - \frac {3 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d}\right )} + \frac {5}{12} \, a^{2} b^{3} {\left (\frac {e^{\left (3 \, d x + 3 \, c\right )}}{d} + \frac {9 \, e^{\left (d x + c\right )}}{d} - \frac {9 \, e^{\left (-d x - c\right )}}{d} - \frac {e^{\left (-3 \, d x - 3 \, c\right )}}{d}\right )} + \frac {5 \, a^{4} b \sinh \left (d x + c\right )}{d} \]

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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mupad [B]  time = 1.14, size = 160, normalized size = 0.87 \[ \frac {75\,b^5\,\mathrm {sinh}\left (c+d\,x\right )+\frac {25\,b^5\,\mathrm {sinh}\left (3\,c+3\,d\,x\right )}{2}+\frac {3\,b^5\,\mathrm {sinh}\left (5\,c+5\,d\,x\right )}{2}+150\,a\,b^4\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+\frac {75\,a\,b^4\,\mathrm {sinh}\left (4\,c+4\,d\,x\right )}{4}+900\,a^2\,b^3\,\mathrm {sinh}\left (c+d\,x\right )+300\,a^3\,b^2\,\mathrm {sinh}\left (2\,c+2\,d\,x\right )+100\,a^2\,b^3\,\mathrm {sinh}\left (3\,c+3\,d\,x\right )+600\,a^4\,b\,\mathrm {sinh}\left (c+d\,x\right )+120\,a^5\,d\,x+225\,a\,b^4\,d\,x+600\,a^3\,b^2\,d\,x}{120\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cosh(c + d*x))^5,x)

[Out]

(75*b^5*sinh(c + d*x) + (25*b^5*sinh(3*c + 3*d*x))/2 + (3*b^5*sinh(5*c + 5*d*x))/2 + 150*a*b^4*sinh(2*c + 2*d*
x) + (75*a*b^4*sinh(4*c + 4*d*x))/4 + 900*a^2*b^3*sinh(c + d*x) + 300*a^3*b^2*sinh(2*c + 2*d*x) + 100*a^2*b^3*
sinh(3*c + 3*d*x) + 600*a^4*b*sinh(c + d*x) + 120*a^5*d*x + 225*a*b^4*d*x + 600*a^3*b^2*d*x)/(120*d)

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sympy [A]  time = 2.25, size = 314, normalized size = 1.72 \[ \begin {cases} a^{5} x + \frac {5 a^{4} b \sinh {\left (c + d x \right )}}{d} - 5 a^{3} b^{2} x \sinh ^{2}{\left (c + d x \right )} + 5 a^{3} b^{2} x \cosh ^{2}{\left (c + d x \right )} + \frac {5 a^{3} b^{2} \sinh {\left (c + d x \right )} \cosh {\left (c + d x \right )}}{d} - \frac {20 a^{2} b^{3} \sinh ^{3}{\left (c + d x \right )}}{3 d} + \frac {10 a^{2} b^{3} \sinh {\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{d} + \frac {15 a b^{4} x \sinh ^{4}{\left (c + d x \right )}}{8} - \frac {15 a b^{4} x \sinh ^{2}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{4} + \frac {15 a b^{4} x \cosh ^{4}{\left (c + d x \right )}}{8} - \frac {15 a b^{4} \sinh ^{3}{\left (c + d x \right )} \cosh {\left (c + d x \right )}}{8 d} + \frac {25 a b^{4} \sinh {\left (c + d x \right )} \cosh ^{3}{\left (c + d x \right )}}{8 d} + \frac {8 b^{5} \sinh ^{5}{\left (c + d x \right )}}{15 d} - \frac {4 b^{5} \sinh ^{3}{\left (c + d x \right )} \cosh ^{2}{\left (c + d x \right )}}{3 d} + \frac {b^{5} \sinh {\left (c + d x \right )} \cosh ^{4}{\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \cosh {\relax (c )}\right )^{5} & \text {otherwise} \end {cases} \]

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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