3.41 \(\int (a+a \tanh (c+d x))^5 \, dx\)

Optimal. Leaf size=100 \[ -\frac {8 a^5 \tanh (c+d x)}{d}+\frac {16 a^5 \log (\cosh (c+d x))}{d}+16 a^5 x-\frac {2 a^2 (a \tanh (c+d x)+a)^3}{3 d}-\frac {2 a \left (a^2 \tanh (c+d x)+a^2\right )^2}{d}-\frac {a (a \tanh (c+d x)+a)^4}{4 d} \]

[Out]

16*a^5*x+16*a^5*ln(cosh(d*x+c))/d-8*a^5*tanh(d*x+c)/d-2/3*a^2*(a+a*tanh(d*x+c))^3/d-1/4*a*(a+a*tanh(d*x+c))^4/
d-2*a*(a^2+a^2*tanh(d*x+c))^2/d

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Rubi [A]  time = 0.07, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3478, 3477, 3475} \[ -\frac {8 a^5 \tanh (c+d x)}{d}-\frac {2 a^2 (a \tanh (c+d x)+a)^3}{3 d}-\frac {2 a \left (a^2 \tanh (c+d x)+a^2\right )^2}{d}+\frac {16 a^5 \log (\cosh (c+d x))}{d}+16 a^5 x-\frac {a (a \tanh (c+d x)+a)^4}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Tanh[c + d*x])^5,x]

[Out]

16*a^5*x + (16*a^5*Log[Cosh[c + d*x]])/d - (8*a^5*Tanh[c + d*x])/d - (2*a^2*(a + a*Tanh[c + d*x])^3)/(3*d) - (
a*(a + a*Tanh[c + d*x])^4)/(4*d) - (2*a*(a^2 + a^2*Tanh[c + d*x])^2)/d

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3477

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[(a^2 - b^2)*x, x] + (Dist[2*a*b, Int[Tan[c + d
*x], x], x] + Simp[(b^2*Tan[c + d*x])/d, x]) /; FreeQ[{a, b, c, d}, x]

Rule 3478

Int[((a_) + (b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a + b*Tan[c + d*x])^(n - 1))/(d*(n - 1)
), x] + Dist[2*a, Int[(a + b*Tan[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0] && G
tQ[n, 1]

Rubi steps

\begin {align*} \int (a+a \tanh (c+d x))^5 \, dx &=-\frac {a (a+a \tanh (c+d x))^4}{4 d}+(2 a) \int (a+a \tanh (c+d x))^4 \, dx\\ &=-\frac {2 a^2 (a+a \tanh (c+d x))^3}{3 d}-\frac {a (a+a \tanh (c+d x))^4}{4 d}+\left (4 a^2\right ) \int (a+a \tanh (c+d x))^3 \, dx\\ &=-\frac {2 a^3 (a+a \tanh (c+d x))^2}{d}-\frac {2 a^2 (a+a \tanh (c+d x))^3}{3 d}-\frac {a (a+a \tanh (c+d x))^4}{4 d}+\left (8 a^3\right ) \int (a+a \tanh (c+d x))^2 \, dx\\ &=16 a^5 x-\frac {8 a^5 \tanh (c+d x)}{d}-\frac {2 a^3 (a+a \tanh (c+d x))^2}{d}-\frac {2 a^2 (a+a \tanh (c+d x))^3}{3 d}-\frac {a (a+a \tanh (c+d x))^4}{4 d}+\left (16 a^5\right ) \int \tanh (c+d x) \, dx\\ &=16 a^5 x+\frac {16 a^5 \log (\cosh (c+d x))}{d}-\frac {8 a^5 \tanh (c+d x)}{d}-\frac {2 a^3 (a+a \tanh (c+d x))^2}{d}-\frac {2 a^2 (a+a \tanh (c+d x))^3}{3 d}-\frac {a (a+a \tanh (c+d x))^4}{4 d}\\ \end {align*}

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Mathematica [B]  time = 1.67, size = 202, normalized size = 2.02 \[ \frac {a^5 \text {sech}(c) \text {sech}^4(c+d x) (-70 \sinh (c+2 d x)+30 \sinh (3 c+2 d x)-25 \sinh (3 c+4 d x)+48 d x \cosh (3 c+2 d x)+18 \cosh (3 c+2 d x)+12 d x \cosh (3 c+4 d x)+12 d x \cosh (5 c+4 d x)+48 \cosh (3 c+2 d x) \log (\cosh (c+d x))+12 \cosh (3 c+4 d x) \log (\cosh (c+d x))+12 \cosh (5 c+4 d x) \log (\cosh (c+d x))+6 \cosh (c+2 d x) (8 \log (\cosh (c+d x))+8 d x+3)+\cosh (c) (72 \log (\cosh (c+d x))+72 d x+33)+75 \sinh (c))}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Tanh[c + d*x])^5,x]

[Out]

(a^5*Sech[c]*Sech[c + d*x]^4*(18*Cosh[3*c + 2*d*x] + 48*d*x*Cosh[3*c + 2*d*x] + 12*d*x*Cosh[3*c + 4*d*x] + 12*
d*x*Cosh[5*c + 4*d*x] + 48*Cosh[3*c + 2*d*x]*Log[Cosh[c + d*x]] + 12*Cosh[3*c + 4*d*x]*Log[Cosh[c + d*x]] + 12
*Cosh[5*c + 4*d*x]*Log[Cosh[c + d*x]] + 6*Cosh[c + 2*d*x]*(3 + 8*d*x + 8*Log[Cosh[c + d*x]]) + Cosh[c]*(33 + 7
2*d*x + 72*Log[Cosh[c + d*x]]) + 75*Sinh[c] - 70*Sinh[c + 2*d*x] + 30*Sinh[3*c + 2*d*x] - 25*Sinh[3*c + 4*d*x]
))/(12*d)

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fricas [B]  time = 0.72, size = 907, normalized size = 9.07 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*(48*a^5*cosh(d*x + c)^6 + 288*a^5*cosh(d*x + c)*sinh(d*x + c)^5 + 48*a^5*sinh(d*x + c)^6 + 108*a^5*cosh(d*
x + c)^4 + 88*a^5*cosh(d*x + c)^2 + 25*a^5 + 36*(20*a^5*cosh(d*x + c)^2 + 3*a^5)*sinh(d*x + c)^4 + 48*(20*a^5*
cosh(d*x + c)^3 + 9*a^5*cosh(d*x + c))*sinh(d*x + c)^3 + 8*(90*a^5*cosh(d*x + c)^4 + 81*a^5*cosh(d*x + c)^2 +
11*a^5)*sinh(d*x + c)^2 + 12*(a^5*cosh(d*x + c)^8 + 8*a^5*cosh(d*x + c)*sinh(d*x + c)^7 + a^5*sinh(d*x + c)^8
+ 4*a^5*cosh(d*x + c)^6 + 6*a^5*cosh(d*x + c)^4 + 4*a^5*cosh(d*x + c)^2 + 4*(7*a^5*cosh(d*x + c)^2 + a^5)*sinh
(d*x + c)^6 + 8*(7*a^5*cosh(d*x + c)^3 + 3*a^5*cosh(d*x + c))*sinh(d*x + c)^5 + a^5 + 2*(35*a^5*cosh(d*x + c)^
4 + 30*a^5*cosh(d*x + c)^2 + 3*a^5)*sinh(d*x + c)^4 + 8*(7*a^5*cosh(d*x + c)^5 + 10*a^5*cosh(d*x + c)^3 + 3*a^
5*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(7*a^5*cosh(d*x + c)^6 + 15*a^5*cosh(d*x + c)^4 + 9*a^5*cosh(d*x + c)^2 +
a^5)*sinh(d*x + c)^2 + 8*(a^5*cosh(d*x + c)^7 + 3*a^5*cosh(d*x + c)^5 + 3*a^5*cosh(d*x + c)^3 + a^5*cosh(d*x
+ c))*sinh(d*x + c))*log(2*cosh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 16*(18*a^5*cosh(d*x + c)^5 + 27*a^
5*cosh(d*x + c)^3 + 11*a^5*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^8 + 8*d*cosh(d*x + c)*sinh(d*x + c)^
7 + d*sinh(d*x + c)^8 + 4*d*cosh(d*x + c)^6 + 4*(7*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^6 + 8*(7*d*cosh(d*x +
c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^5 + 6*d*cosh(d*x + c)^4 + 2*(35*d*cosh(d*x + c)^4 + 30*d*cosh(d*x + c)
^2 + 3*d)*sinh(d*x + c)^4 + 8*(7*d*cosh(d*x + c)^5 + 10*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^3
+ 4*d*cosh(d*x + c)^2 + 4*(7*d*cosh(d*x + c)^6 + 15*d*cosh(d*x + c)^4 + 9*d*cosh(d*x + c)^2 + d)*sinh(d*x + c
)^2 + 8*(d*cosh(d*x + c)^7 + 3*d*cosh(d*x + c)^5 + 3*d*cosh(d*x + c)^3 + d*cosh(d*x + c))*sinh(d*x + c) + d)

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giac [A]  time = 0.14, size = 85, normalized size = 0.85 \[ \frac {4 \, {\left (12 \, a^{5} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) + \frac {48 \, a^{5} e^{\left (6 \, d x + 6 \, c\right )} + 108 \, a^{5} e^{\left (4 \, d x + 4 \, c\right )} + 88 \, a^{5} e^{\left (2 \, d x + 2 \, c\right )} + 25 \, a^{5}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{4}}\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4/3*(12*a^5*log(e^(2*d*x + 2*c) + 1) + (48*a^5*e^(6*d*x + 6*c) + 108*a^5*e^(4*d*x + 4*c) + 88*a^5*e^(2*d*x + 2
*c) + 25*a^5)/(e^(2*d*x + 2*c) + 1)^4)/d

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maple [A]  time = 0.01, size = 81, normalized size = 0.81 \[ -\frac {a^{5} \left (\tanh ^{4}\left (d x +c \right )\right )}{4 d}-\frac {5 a^{5} \left (\tanh ^{3}\left (d x +c \right )\right )}{3 d}-\frac {11 a^{5} \left (\tanh ^{2}\left (d x +c \right )\right )}{2 d}-\frac {15 a^{5} \tanh \left (d x +c \right )}{d}-\frac {16 a^{5} \ln \left (\tanh \left (d x +c \right )-1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/d*a^5*tanh(d*x+c)^4-5/3/d*a^5*tanh(d*x+c)^3-11/2/d*a^5*tanh(d*x+c)^2-15*a^5*tanh(d*x+c)/d-16/d*a^5*ln(tan
h(d*x+c)-1)

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maxima [B]  time = 0.42, size = 302, normalized size = 3.02 \[ \frac {5}{3} \, a^{5} {\left (3 \, x + \frac {3 \, c}{d} - \frac {4 \, {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + 2\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} + 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} + 1\right )}}\right )} + a^{5} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {4 \, {\left (e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{d {\left (4 \, e^{\left (-2 \, d x - 2 \, c\right )} + 6 \, e^{\left (-4 \, d x - 4 \, c\right )} + 4 \, e^{\left (-6 \, d x - 6 \, c\right )} + e^{\left (-8 \, d x - 8 \, c\right )} + 1\right )}}\right )} + 10 \, a^{5} {\left (x + \frac {c}{d} + \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + 10 \, a^{5} {\left (x + \frac {c}{d} - \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + a^{5} x + \frac {5 \, a^{5} \log \left (\cosh \left (d x + c\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/3*a^5*(3*x + 3*c/d - 4*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x - 4*c) + 2)/(d*(3*e^(-2*d*x - 2*c) + 3*e^(-4*d*x -
4*c) + e^(-6*d*x - 6*c) + 1))) + a^5*(x + c/d + log(e^(-2*d*x - 2*c) + 1)/d + 4*(e^(-2*d*x - 2*c) + e^(-4*d*x
- 4*c) + e^(-6*d*x - 6*c))/(d*(4*e^(-2*d*x - 2*c) + 6*e^(-4*d*x - 4*c) + 4*e^(-6*d*x - 6*c) + e^(-8*d*x - 8*c)
+ 1))) + 10*a^5*(x + c/d + log(e^(-2*d*x - 2*c) + 1)/d + 2*e^(-2*d*x - 2*c)/(d*(2*e^(-2*d*x - 2*c) + e^(-4*d*
x - 4*c) + 1))) + 10*a^5*(x + c/d - 2/(d*(e^(-2*d*x - 2*c) + 1))) + a^5*x + 5*a^5*log(cosh(d*x + c))/d

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mupad [B]  time = 0.16, size = 65, normalized size = 0.65 \[ 32\,a^5\,x-\frac {a^5\,\left (192\,\ln \left (\mathrm {tanh}\left (c+d\,x\right )+1\right )+180\,\mathrm {tanh}\left (c+d\,x\right )+66\,{\mathrm {tanh}\left (c+d\,x\right )}^2+20\,{\mathrm {tanh}\left (c+d\,x\right )}^3+3\,{\mathrm {tanh}\left (c+d\,x\right )}^4\right )}{12\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tanh(c + d*x))^5,x)

[Out]

32*a^5*x - (a^5*(192*log(tanh(c + d*x) + 1) + 180*tanh(c + d*x) + 66*tanh(c + d*x)^2 + 20*tanh(c + d*x)^3 + 3*
tanh(c + d*x)^4))/(12*d)

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sympy [A]  time = 0.47, size = 95, normalized size = 0.95 \[ \begin {cases} 32 a^{5} x - \frac {16 a^{5} \log {\left (\tanh {\left (c + d x \right )} + 1 \right )}}{d} - \frac {a^{5} \tanh ^{4}{\left (c + d x \right )}}{4 d} - \frac {5 a^{5} \tanh ^{3}{\left (c + d x \right )}}{3 d} - \frac {11 a^{5} \tanh ^{2}{\left (c + d x \right )}}{2 d} - \frac {15 a^{5} \tanh {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a \tanh {\relax (c )} + a\right )^{5} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*tanh(d*x+c))**5,x)

[Out]

Piecewise((32*a**5*x - 16*a**5*log(tanh(c + d*x) + 1)/d - a**5*tanh(c + d*x)**4/(4*d) - 5*a**5*tanh(c + d*x)**
3/(3*d) - 11*a**5*tanh(c + d*x)**2/(2*d) - 15*a**5*tanh(c + d*x)/d, Ne(d, 0)), (x*(a*tanh(c) + a)**5, True))

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