3.77 \(\int (a+b \coth (c+d x))^5 \, dx\)

Optimal. Leaf size=142 \[ -\frac{b \left (3 a^2+b^2\right ) (a+b \coth (c+d x))^2}{2 d}-\frac{4 a b^2 \left (a^2+b^2\right ) \coth (c+d x)}{d}+\frac{b \left (10 a^2 b^2+5 a^4+b^4\right ) \log (\sinh (c+d x))}{d}+a x \left (10 a^2 b^2+a^4+5 b^4\right )-\frac{b (a+b \coth (c+d x))^4}{4 d}-\frac{2 a b (a+b \coth (c+d x))^3}{3 d} \]

[Out]

a*(a^4 + 10*a^2*b^2 + 5*b^4)*x - (4*a*b^2*(a^2 + b^2)*Coth[c + d*x])/d - (b*(3*a^2 + b^2)*(a + b*Coth[c + d*x]
)^2)/(2*d) - (2*a*b*(a + b*Coth[c + d*x])^3)/(3*d) - (b*(a + b*Coth[c + d*x])^4)/(4*d) + (b*(5*a^4 + 10*a^2*b^
2 + b^4)*Log[Sinh[c + d*x]])/d

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Rubi [A]  time = 0.208768, antiderivative size = 142, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3482, 3528, 3525, 3475} \[ -\frac{b \left (3 a^2+b^2\right ) (a+b \coth (c+d x))^2}{2 d}-\frac{4 a b^2 \left (a^2+b^2\right ) \coth (c+d x)}{d}+\frac{b \left (10 a^2 b^2+5 a^4+b^4\right ) \log (\sinh (c+d x))}{d}+a x \left (10 a^2 b^2+a^4+5 b^4\right )-\frac{b (a+b \coth (c+d x))^4}{4 d}-\frac{2 a b (a+b \coth (c+d x))^3}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Coth[c + d*x])^5,x]

[Out]

a*(a^4 + 10*a^2*b^2 + 5*b^4)*x - (4*a*b^2*(a^2 + b^2)*Coth[c + d*x])/d - (b*(3*a^2 + b^2)*(a + b*Coth[c + d*x]
)^2)/(2*d) - (2*a*b*(a + b*Coth[c + d*x])^3)/(3*d) - (b*(a + b*Coth[c + d*x])^4)/(4*d) + (b*(5*a^4 + 10*a^2*b^
2 + b^4)*Log[Sinh[c + d*x]])/d

Rule 3482

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

Rule 3528

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*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]

Rule 3525

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

Rule 3475

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

Rubi steps

\begin{align*} \int (a+b \coth (c+d x))^5 \, dx &=-\frac{b (a+b \coth (c+d x))^4}{4 d}+\int (a+b \coth (c+d x))^3 \left (a^2+b^2+2 a b \coth (c+d x)\right ) \, dx\\ &=-\frac{2 a b (a+b \coth (c+d x))^3}{3 d}-\frac{b (a+b \coth (c+d x))^4}{4 d}+\int (a+b \coth (c+d x))^2 \left (a \left (a^2+3 b^2\right )+b \left (3 a^2+b^2\right ) \coth (c+d x)\right ) \, dx\\ &=-\frac{b \left (3 a^2+b^2\right ) (a+b \coth (c+d x))^2}{2 d}-\frac{2 a b (a+b \coth (c+d x))^3}{3 d}-\frac{b (a+b \coth (c+d x))^4}{4 d}+\int (a+b \coth (c+d x)) \left (a^4+6 a^2 b^2+b^4+4 a b \left (a^2+b^2\right ) \coth (c+d x)\right ) \, dx\\ &=a \left (a^4+10 a^2 b^2+5 b^4\right ) x-\frac{4 a b^2 \left (a^2+b^2\right ) \coth (c+d x)}{d}-\frac{b \left (3 a^2+b^2\right ) (a+b \coth (c+d x))^2}{2 d}-\frac{2 a b (a+b \coth (c+d x))^3}{3 d}-\frac{b (a+b \coth (c+d x))^4}{4 d}+\left (b \left (5 a^4+10 a^2 b^2+b^4\right )\right ) \int \coth (c+d x) \, dx\\ &=a \left (a^4+10 a^2 b^2+5 b^4\right ) x-\frac{4 a b^2 \left (a^2+b^2\right ) \coth (c+d x)}{d}-\frac{b \left (3 a^2+b^2\right ) (a+b \coth (c+d x))^2}{2 d}-\frac{2 a b (a+b \coth (c+d x))^3}{3 d}-\frac{b (a+b \coth (c+d x))^4}{4 d}+\frac{b \left (5 a^4+10 a^2 b^2+b^4\right ) \log (\sinh (c+d x))}{d}\\ \end{align*}

Mathematica [A]  time = 0.798875, size = 141, normalized size = 0.99 \[ -\frac{6 b^3 \left (10 a^2+b^2\right ) \coth ^2(c+d x)+60 a b^2 \left (2 a^2+b^2\right ) \coth (c+d x)-12 b \left (10 a^2 b^2+5 a^4+b^4\right ) \log (\tanh (c+d x))+20 a b^4 \coth ^3(c+d x)-6 (a-b)^5 \log (\tanh (c+d x)+1)+6 (a+b)^5 \log (1-\tanh (c+d x))+3 b^5 \coth ^4(c+d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Coth[c + d*x])^5,x]

[Out]

-(60*a*b^2*(2*a^2 + b^2)*Coth[c + d*x] + 6*b^3*(10*a^2 + b^2)*Coth[c + d*x]^2 + 20*a*b^4*Coth[c + d*x]^3 + 3*b
^5*Coth[c + d*x]^4 + 6*(a + b)^5*Log[1 - Tanh[c + d*x]] - 12*b*(5*a^4 + 10*a^2*b^2 + b^4)*Log[Tanh[c + d*x]] -
6*(a - b)^5*Log[1 + Tanh[c + d*x]])/(12*d)

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Maple [B]  time = 0.006, size = 322, normalized size = 2.3 \begin{align*} -5\,{\frac{ \left ({\rm coth} \left (dx+c\right ) \right ) ^{2}{a}^{2}{b}^{3}}{d}}+{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ){a}^{5}}{2\,d}}-{\frac{5\,\ln \left ({\rm coth} \left (dx+c\right )+1 \right ){a}^{4}b}{2\,d}}+5\,{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ){a}^{3}{b}^{2}}{d}}-5\,{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ){a}^{2}{b}^{3}}{d}}+{\frac{5\,\ln \left ({\rm coth} \left (dx+c\right )+1 \right ) a{b}^{4}}{2\,d}}-{\frac{\ln \left ({\rm coth} \left (dx+c\right )+1 \right ){b}^{5}}{2\,d}}-{\frac{\ln \left ({\rm coth} \left (dx+c\right )-1 \right ){a}^{5}}{2\,d}}-{\frac{5\,\ln \left ({\rm coth} \left (dx+c\right )-1 \right ){a}^{4}b}{2\,d}}-5\,{\frac{\ln \left ({\rm coth} \left (dx+c\right )-1 \right ){a}^{3}{b}^{2}}{d}}-5\,{\frac{\ln \left ({\rm coth} \left (dx+c\right )-1 \right ){a}^{2}{b}^{3}}{d}}-{\frac{5\,\ln \left ({\rm coth} \left (dx+c\right )-1 \right ) a{b}^{4}}{2\,d}}-{\frac{\ln \left ({\rm coth} \left (dx+c\right )-1 \right ){b}^{5}}{2\,d}}-{\frac{ \left ({\rm coth} \left (dx+c\right ) \right ) ^{2}{b}^{5}}{2\,d}}-{\frac{{b}^{5} \left ({\rm coth} \left (dx+c\right ) \right ) ^{4}}{4\,d}}-{\frac{5\, \left ({\rm coth} \left (dx+c\right ) \right ) ^{3}a{b}^{4}}{3\,d}}-5\,{\frac{a{b}^{4}{\rm coth} \left (dx+c\right )}{d}}-10\,{\frac{{a}^{3}{b}^{2}{\rm coth} \left (dx+c\right )}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/d*coth(d*x+c)^2*a^2*b^3+1/2/d*ln(coth(d*x+c)+1)*a^5-5/2/d*ln(coth(d*x+c)+1)*a^4*b+5/d*ln(coth(d*x+c)+1)*a^3
*b^2-5/d*ln(coth(d*x+c)+1)*a^2*b^3+5/2/d*ln(coth(d*x+c)+1)*a*b^4-1/2/d*ln(coth(d*x+c)+1)*b^5-1/2/d*ln(coth(d*x
+c)-1)*a^5-5/2/d*ln(coth(d*x+c)-1)*a^4*b-5/d*ln(coth(d*x+c)-1)*a^3*b^2-5/d*ln(coth(d*x+c)-1)*a^2*b^3-5/2/d*ln(
coth(d*x+c)-1)*a*b^4-1/2/d*ln(coth(d*x+c)-1)*b^5-1/2/d*coth(d*x+c)^2*b^5-1/4/d*b^5*coth(d*x+c)^4-5/3/d*coth(d*
x+c)^3*a*b^4-5/d*a*b^4*coth(d*x+c)-10/d*a^3*b^2*coth(d*x+c)

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Maxima [B]  time = 1.08076, size = 470, normalized size = 3.31 \begin{align*} \frac{5}{3} \, a b^{4}{\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 )} + b^{5}{\left (x + \frac{c}{d} + \frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac{\log \left (e^{\left (-d x - 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^{2} b^{3}{\left (x + \frac{c}{d} + \frac{\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac{\log \left (e^{\left (-d x - 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^{3} b^{2}{\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^{4} b \log \left (\sinh \left (d x + c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5/3*a*b^4*(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))) + b^5*(x + c/d + log(e^(-d*x - c) + 1)/d + log(e^(-d*x - 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^2*b^3*(x + c/d + log(e^(-d*x - c) + 1)/d + log(e^(-d*x - 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^3*b^2*(x + c/d + 2/(d*(e^(-2*d*x - 2*
c) - 1))) + a^5*x + 5*a^4*b*log(sinh(d*x + c))/d

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Fricas [B]  time = 2.76279, size = 6413, normalized size = 45.16 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*x + c)^8 + 24*(a^5 - 5*a^4*b + 10*
a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*x + c)*sinh(d*x + c)^7 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a
^2*b^3 + 5*a*b^4 - b^5)*d*x*sinh(d*x + c)^8 - 12*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4 + b^5 + (a^5 - 5*a^4*b + 10*
a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x + c)^6 - 12*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4 + b^5 - 7*(a^
5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*x + c)^2 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*
a^2*b^3 + 5*a*b^4 - b^5)*d*x)*sinh(d*x + c)^6 + 24*(7*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5
)*d*x*cosh(d*x + c)^3 - 3*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4 + b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 +
5*a*b^4 - b^5)*d*x)*cosh(d*x + c))*sinh(d*x + c)^5 + 60*a^3*b^2 + 40*a*b^4 + 6*(30*a^3*b^2 + 20*a^2*b^3 + 20*a
*b^4 + 2*b^5 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x + c)^4 + 6*(35*(a^5 -
5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*x + c)^4 + 30*a^3*b^2 + 20*a^2*b^3 + 20*a*b^4 +
2*b^5 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x - 30*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4
+ b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + 24*
(7*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*x + c)^5 - 10*(5*a^3*b^2 + 5*a^2*b^3 +
5*a*b^4 + b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x + c)^3 + (30*a^3*b^2
+ 20*a^2*b^3 + 20*a*b^4 + 2*b^5 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x +
c))*sinh(d*x + c)^3 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x - 4*(45*a^3*b^2 + 15*a^2
*b^3 + 25*a*b^4 + 3*b^5 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x + c)^2 + 4
*(21*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*x + c)^6 - 45*a^3*b^2 - 15*a^2*b^3 -
25*a*b^4 - 3*b^5 - 45*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4 + b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a
*b^4 - b^5)*d*x)*cosh(d*x + c)^4 - 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x + 9*(30*a^3
*b^2 + 20*a^2*b^3 + 20*a*b^4 + 2*b^5 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d
*x + c)^2)*sinh(d*x + c)^2 + 3*((5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^8 + 8*(5*a^4*b + 10*a^2*b^3 + b^5)*
cosh(d*x + c)*sinh(d*x + c)^7 + (5*a^4*b + 10*a^2*b^3 + b^5)*sinh(d*x + c)^8 - 4*(5*a^4*b + 10*a^2*b^3 + b^5)*
cosh(d*x + c)^6 - 4*(5*a^4*b + 10*a^2*b^3 + b^5 - 7*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^2)*sinh(d*x + c
)^6 + 8*(7*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^3 - 3*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c))*sinh(d
*x + c)^5 + 5*a^4*b + 10*a^2*b^3 + b^5 + 6*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^4 + 2*(15*a^4*b + 30*a^2
*b^3 + 3*b^5 + 35*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^4 - 30*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)
^2)*sinh(d*x + c)^4 + 8*(7*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^5 - 10*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh
(d*x + c)^3 + 3*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c))*sinh(d*x + c)^3 - 4*(5*a^4*b + 10*a^2*b^3 + b^5)*c
osh(d*x + c)^2 + 4*(7*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^6 - 5*a^4*b - 10*a^2*b^3 - b^5 - 15*(5*a^4*b
+ 10*a^2*b^3 + b^5)*cosh(d*x + c)^4 + 9*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 8*((5*
a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^7 - 3*(5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c)^5 + 3*(5*a^4*b + 10*a
^2*b^3 + b^5)*cosh(d*x + c)^3 - (5*a^4*b + 10*a^2*b^3 + b^5)*cosh(d*x + c))*sinh(d*x + c))*log(2*sinh(d*x + c)
/(cosh(d*x + c) - sinh(d*x + c))) + 8*(3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x*cosh(d*
x + c)^7 - 9*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4 + b^5 + (a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5
)*d*x)*cosh(d*x + c)^5 + 3*(30*a^3*b^2 + 20*a^2*b^3 + 20*a*b^4 + 2*b^5 + 3*(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^
2*b^3 + 5*a*b^4 - b^5)*d*x)*cosh(d*x + c)^3 - (45*a^3*b^2 + 15*a^2*b^3 + 25*a*b^4 + 3*b^5 + 3*(a^5 - 5*a^4*b +
10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*d*x)*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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Sympy [A]  time = 22.9375, size = 325, normalized size = 2.29 \begin{align*} \begin{cases} x \left (a + b \coth{\left (c \right )}\right )^{5} & \text{for}\: d = 0 \\a^{5} x + \tilde{\infty } a^{4} b x + \tilde{\infty } a^{3} b^{2} x + \tilde{\infty } a^{2} b^{3} x + \tilde{\infty } a b^{4} x + \tilde{\infty } b^{5} x & \text{for}\: c = \log{\left (- e^{- d x} \right )} \vee c = \log{\left (e^{- d x} \right )} \\a^{5} x + 5 a^{4} b x - \frac{5 a^{4} b \log{\left (\tanh{\left (c + d x \right )} + 1 \right )}}{d} + \frac{5 a^{4} b \log{\left (\tanh{\left (c + d x \right )} \right )}}{d} + 10 a^{3} b^{2} x - \frac{10 a^{3} b^{2}}{d \tanh{\left (c + d x \right )}} + 10 a^{2} b^{3} x - \frac{10 a^{2} b^{3} \log{\left (\tanh{\left (c + d x \right )} + 1 \right )}}{d} + \frac{10 a^{2} b^{3} \log{\left (\tanh{\left (c + d x \right )} \right )}}{d} - \frac{5 a^{2} b^{3}}{d \tanh ^{2}{\left (c + d x \right )}} + 5 a b^{4} x - \frac{5 a b^{4}}{d \tanh{\left (c + d x \right )}} - \frac{5 a b^{4}}{3 d \tanh ^{3}{\left (c + d x \right )}} + b^{5} x - \frac{b^{5} \log{\left (\tanh{\left (c + d x \right )} + 1 \right )}}{d} + \frac{b^{5} \log{\left (\tanh{\left (c + d x \right )} \right )}}{d} - \frac{b^{5}}{2 d \tanh ^{2}{\left (c + d x \right )}} - \frac{b^{5}}{4 d \tanh ^{4}{\left (c + d x \right )}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((x*(a + b*coth(c))**5, Eq(d, 0)), (a**5*x + zoo*a**4*b*x + zoo*a**3*b**2*x + zoo*a**2*b**3*x + zoo*a
*b**4*x + zoo*b**5*x, Eq(c, log(exp(-d*x))) | Eq(c, log(-exp(-d*x)))), (a**5*x + 5*a**4*b*x - 5*a**4*b*log(tan
h(c + d*x) + 1)/d + 5*a**4*b*log(tanh(c + d*x))/d + 10*a**3*b**2*x - 10*a**3*b**2/(d*tanh(c + d*x)) + 10*a**2*
b**3*x - 10*a**2*b**3*log(tanh(c + d*x) + 1)/d + 10*a**2*b**3*log(tanh(c + d*x))/d - 5*a**2*b**3/(d*tanh(c + d
*x)**2) + 5*a*b**4*x - 5*a*b**4/(d*tanh(c + d*x)) - 5*a*b**4/(3*d*tanh(c + d*x)**3) + b**5*x - b**5*log(tanh(c
+ d*x) + 1)/d + b**5*log(tanh(c + d*x))/d - b**5/(2*d*tanh(c + d*x)**2) - b**5/(4*d*tanh(c + d*x)**4), True))

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Giac [A]  time = 1.15487, size = 308, normalized size = 2.17 \begin{align*} \frac{{\left (a^{5} - 5 \, a^{4} b + 10 \, a^{3} b^{2} - 10 \, a^{2} b^{3} + 5 \, a b^{4} - b^{5}\right )}{\left (d x + c\right )}}{d} + \frac{{\left (5 \, a^{4} b + 10 \, a^{2} b^{3} + b^{5}\right )} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right )}{d} + \frac{4 \,{\left (15 \, a^{3} b^{2} + 10 \, a b^{4} - 3 \,{\left (5 \, a^{3} b^{2} + 5 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} e^{\left (6 \, d x + 6 \, c\right )} + 3 \,{\left (15 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 10 \, a b^{4} + b^{5}\right )} e^{\left (4 \, d x + 4 \, c\right )} -{\left (45 \, a^{3} b^{2} + 15 \, a^{2} b^{3} + 25 \, a b^{4} + 3 \, b^{5}\right )} e^{\left (2 \, d x + 2 \, c\right )}\right )}}{3 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a^5 - 5*a^4*b + 10*a^3*b^2 - 10*a^2*b^3 + 5*a*b^4 - b^5)*(d*x + c)/d + (5*a^4*b + 10*a^2*b^3 + b^5)*log(abs(e
^(2*d*x + 2*c) - 1))/d + 4/3*(15*a^3*b^2 + 10*a*b^4 - 3*(5*a^3*b^2 + 5*a^2*b^3 + 5*a*b^4 + b^5)*e^(6*d*x + 6*c
) + 3*(15*a^3*b^2 + 10*a^2*b^3 + 10*a*b^4 + b^5)*e^(4*d*x + 4*c) - (45*a^3*b^2 + 15*a^2*b^3 + 25*a*b^4 + 3*b^5
)*e^(2*d*x + 2*c))/(d*(e^(2*d*x + 2*c) - 1)^4)