3.156 \(\int \frac{\coth ^3(c+d x)}{(a+b \text{sech}^2(c+d x))^2} \, dx\)
Optimal. Leaf size=110 \[ \frac{b^3}{2 a^2 d (a+b)^2 \left (a \cosh ^2(c+d x)+b\right )}+\frac{b^2 (3 a+b) \log \left (a \cosh ^2(c+d x)+b\right )}{2 a^2 d (a+b)^3}-\frac{\text{csch}^2(c+d x)}{2 d (a+b)^2}+\frac{(a+3 b) \log (\sinh (c+d x))}{d (a+b)^3} \]
[Out]
b^3/(2*a^2*(a + b)^2*d*(b + a*Cosh[c + d*x]^2)) - Csch[c + d*x]^2/(2*(a + b)^2*d) + (b^2*(3*a + b)*Log[b + a*C
osh[c + d*x]^2])/(2*a^2*(a + b)^3*d) + ((a + 3*b)*Log[Sinh[c + d*x]])/((a + b)^3*d)
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Rubi [A] time = 0.174502, antiderivative size = 110, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used =
{4138, 446, 88} \[ \frac{b^3}{2 a^2 d (a+b)^2 \left (a \cosh ^2(c+d x)+b\right )}+\frac{b^2 (3 a+b) \log \left (a \cosh ^2(c+d x)+b\right )}{2 a^2 d (a+b)^3}-\frac{\text{csch}^2(c+d x)}{2 d (a+b)^2}+\frac{(a+3 b) \log (\sinh (c+d x))}{d (a+b)^3} \]
Antiderivative was successfully verified.
[In]
Int[Coth[c + d*x]^3/(a + b*Sech[c + d*x]^2)^2,x]
[Out]
b^3/(2*a^2*(a + b)^2*d*(b + a*Cosh[c + d*x]^2)) - Csch[c + d*x]^2/(2*(a + b)^2*d) + (b^2*(3*a + b)*Log[b + a*C
osh[c + d*x]^2])/(2*a^2*(a + b)^3*d) + ((a + 3*b)*Log[Sinh[c + d*x]])/((a + b)^3*d)
Rule 4138
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]
Rule 446
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]
Rule 88
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))
Rubi steps
\begin{align*} \int \frac{\coth ^3(c+d x)}{\left (a+b \text{sech}^2(c+d x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^7}{\left (1-x^2\right )^2 \left (b+a x^2\right )^2} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x^3}{(1-x)^2 (b+a x)^2} \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=\frac{\operatorname{Subst}\left (\int \left (\frac{1}{(a+b)^2 (-1+x)^2}+\frac{a+3 b}{(a+b)^3 (-1+x)}-\frac{b^3}{a (a+b)^2 (b+a x)^2}+\frac{b^2 (3 a+b)}{a (a+b)^3 (b+a x)}\right ) \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=\frac{b^3}{2 a^2 (a+b)^2 d \left (b+a \cosh ^2(c+d x)\right )}-\frac{\text{csch}^2(c+d x)}{2 (a+b)^2 d}+\frac{b^2 (3 a+b) \log \left (b+a \cosh ^2(c+d x)\right )}{2 a^2 (a+b)^3 d}+\frac{(a+3 b) \log (\sinh (c+d x))}{(a+b)^3 d}\\ \end{align*}
Mathematica [A] time = 1.29486, size = 130, normalized size = 1.18 \[ \frac{\text{sech}^4(c+d x) (a \cosh (2 (c+d x))+a+2 b)^2 \left (\frac{b^3 (a+b)}{a^2 \left (a \sinh ^2(c+d x)+a+b\right )}+\frac{b^2 (3 a+b) \log \left (a \sinh ^2(c+d x)+a+b\right )}{a^2}-(a+b) \text{csch}^2(c+d x)+2 (a+3 b) \log (\sinh (c+d x))\right )}{8 d (a+b)^3 \left (a+b \text{sech}^2(c+d x)\right )^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Coth[c + d*x]^3/(a + b*Sech[c + d*x]^2)^2,x]
[Out]
((a + 2*b + a*Cosh[2*(c + d*x)])^2*Sech[c + d*x]^4*(-((a + b)*Csch[c + d*x]^2) + 2*(a + 3*b)*Log[Sinh[c + d*x]
] + (b^2*(3*a + b)*Log[a + b + a*Sinh[c + d*x]^2])/a^2 + (b^3*(a + b))/(a^2*(a + b + a*Sinh[c + d*x]^2))))/(8*
(a + b)^3*d*(a + b*Sech[c + d*x]^2)^2)
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Maple [B] time = 0.096, size = 367, normalized size = 3.3 \begin{align*} -{\frac{1}{8\,d \left ({a}^{2}+2\,ab+{b}^{2} \right ) } \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{1}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-2\,{\frac{{b}^{3} \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{da \left ( a+b \right ) ^{3} \left ( \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}a+b \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }}+{\frac{3\,{b}^{2}}{2\,da \left ( a+b \right ) ^{3}}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+b \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }+{\frac{{b}^{3}}{2\,d{a}^{2} \left ( a+b \right ) ^{3}}\ln \left ( \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}a+b \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{4}+2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a-2\, \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b \right ) }-{\frac{1}{8\,d \left ( a+b \right ) ^{2}} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{a}{d \left ( a+b \right ) ^{3}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+3\,{\frac{\ln \left ( \tanh \left ( 1/2\,dx+c/2 \right ) \right ) b}{d \left ( a+b \right ) ^{3}}}-{\frac{1}{d{a}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(coth(d*x+c)^3/(a+b*sech(d*x+c)^2)^2,x)
[Out]
-1/8/d*tanh(1/2*d*x+1/2*c)^2/(a^2+2*a*b+b^2)-1/d/a^2*ln(tanh(1/2*d*x+1/2*c)+1)-2/d*b^3/a/(a+b)^3*tanh(1/2*d*x+
1/2*c)^2/(tanh(1/2*d*x+1/2*c)^4*a+b*tanh(1/2*d*x+1/2*c)^4+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+
a+b)+3/2/d*b^2/a/(a+b)^3*ln(tanh(1/2*d*x+1/2*c)^4*a+b*tanh(1/2*d*x+1/2*c)^4+2*tanh(1/2*d*x+1/2*c)^2*a-2*tanh(1
/2*d*x+1/2*c)^2*b+a+b)+1/2/d*b^3/a^2/(a+b)^3*ln(tanh(1/2*d*x+1/2*c)^4*a+b*tanh(1/2*d*x+1/2*c)^4+2*tanh(1/2*d*x
+1/2*c)^2*a-2*tanh(1/2*d*x+1/2*c)^2*b+a+b)-1/8/d/(a+b)^2/tanh(1/2*d*x+1/2*c)^2+1/d/(a+b)^3*ln(tanh(1/2*d*x+1/2
*c))*a+3/d/(a+b)^3*ln(tanh(1/2*d*x+1/2*c))*b-1/d/a^2*ln(tanh(1/2*d*x+1/2*c)-1)
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Maxima [B] time = 1.26792, size = 518, normalized size = 4.71 \begin{align*} \frac{{\left (3 \, a b^{2} + b^{3}\right )} \log \left (2 \,{\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{2 \,{\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} d} + \frac{{\left (a + 3 \, b\right )} \log \left (e^{\left (-d x - c\right )} + 1\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} + \frac{{\left (a + 3 \, b\right )} \log \left (e^{\left (-d x - c\right )} - 1\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} d} - \frac{2 \,{\left ({\left (a^{3} - b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + 2 \,{\left (a^{3} + 2 \, a^{2} b + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} +{\left (a^{3} - b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )}\right )}}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2} + 4 \,{\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )} - 2 \,{\left (a^{5} + 6 \, a^{4} b + 9 \, a^{3} b^{2} + 4 \, a^{2} b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )} + 4 \,{\left (a^{4} b + 2 \, a^{3} b^{2} + a^{2} b^{3}\right )} e^{\left (-6 \, d x - 6 \, c\right )} +{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} e^{\left (-8 \, d x - 8 \, c\right )}\right )} d} + \frac{d x + c}{a^{2} d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(d*x+c)^3/(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")
[Out]
1/2*(3*a*b^2 + b^3)*log(2*(a + 2*b)*e^(-2*d*x - 2*c) + a*e^(-4*d*x - 4*c) + a)/((a^5 + 3*a^4*b + 3*a^3*b^2 + a
^2*b^3)*d) + (a + 3*b)*log(e^(-d*x - c) + 1)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d) + (a + 3*b)*log(e^(-d*x - c)
- 1)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*d) - 2*((a^3 - b^3)*e^(-2*d*x - 2*c) + 2*(a^3 + 2*a^2*b + b^3)*e^(-4*d*x
- 4*c) + (a^3 - b^3)*e^(-6*d*x - 6*c))/((a^5 + 2*a^4*b + a^3*b^2 + 4*(a^4*b + 2*a^3*b^2 + a^2*b^3)*e^(-2*d*x
- 2*c) - 2*(a^5 + 6*a^4*b + 9*a^3*b^2 + 4*a^2*b^3)*e^(-4*d*x - 4*c) + 4*(a^4*b + 2*a^3*b^2 + a^2*b^3)*e^(-6*d*
x - 6*c) + (a^5 + 2*a^4*b + a^3*b^2)*e^(-8*d*x - 8*c))*d) + (d*x + c)/(a^2*d)
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Fricas [B] time = 4.7792, size = 8284, normalized size = 75.31 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(d*x+c)^3/(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")
[Out]
-1/2*(2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*x*cosh(d*x + c)^8 + 16*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*x*c
osh(d*x + c)*sinh(d*x + c)^7 + 2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*x*sinh(d*x + c)^8 + 4*(a^4 + a^3*b - a*
b^3 - b^4 + 2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*x)*cosh(d*x + c)^6 + 4*(14*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*
b^3)*d*x*cosh(d*x + c)^2 + a^4 + a^3*b - a*b^3 - b^4 + 2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*x)*sinh(d*x + c
)^6 + 8*(14*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*x*cosh(d*x + c)^3 + 3*(a^4 + a^3*b - a*b^3 - b^4 + 2*(a^3*b
+ 3*a^2*b^2 + 3*a*b^3 + b^4)*d*x)*cosh(d*x + c))*sinh(d*x + c)^5 + 4*(2*a^4 + 6*a^3*b + 4*a^2*b^2 + 2*a*b^3 +
2*b^4 - (a^4 + 7*a^3*b + 15*a^2*b^2 + 13*a*b^3 + 4*b^4)*d*x)*cosh(d*x + c)^4 + 4*(35*(a^4 + 3*a^3*b + 3*a^2*b^
2 + a*b^3)*d*x*cosh(d*x + c)^4 + 2*a^4 + 6*a^3*b + 4*a^2*b^2 + 2*a*b^3 + 2*b^4 - (a^4 + 7*a^3*b + 15*a^2*b^2 +
13*a*b^3 + 4*b^4)*d*x + 15*(a^4 + a^3*b - a*b^3 - b^4 + 2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*x)*cosh(d*x +
c)^2)*sinh(d*x + c)^4 + 16*(7*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*x*cosh(d*x + c)^5 + 5*(a^4 + a^3*b - a*b^
3 - b^4 + 2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*x)*cosh(d*x + c)^3 + (2*a^4 + 6*a^3*b + 4*a^2*b^2 + 2*a*b^3
+ 2*b^4 - (a^4 + 7*a^3*b + 15*a^2*b^2 + 13*a*b^3 + 4*b^4)*d*x)*cosh(d*x + c))*sinh(d*x + c)^3 + 2*(a^4 + 3*a^3
*b + 3*a^2*b^2 + a*b^3)*d*x + 4*(a^4 + a^3*b - a*b^3 - b^4 + 2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*x)*cosh(d
*x + c)^2 + 4*(14*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*x*cosh(d*x + c)^6 + 15*(a^4 + a^3*b - a*b^3 - b^4 + 2*
(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*x)*cosh(d*x + c)^4 + a^4 + a^3*b - a*b^3 - b^4 + 2*(a^3*b + 3*a^2*b^2 +
3*a*b^3 + b^4)*d*x + 6*(2*a^4 + 6*a^3*b + 4*a^2*b^2 + 2*a*b^3 + 2*b^4 - (a^4 + 7*a^3*b + 15*a^2*b^2 + 13*a*b^3
+ 4*b^4)*d*x)*cosh(d*x + c)^2)*sinh(d*x + c)^2 - ((3*a^2*b^2 + a*b^3)*cosh(d*x + c)^8 + 8*(3*a^2*b^2 + a*b^3)
*cosh(d*x + c)*sinh(d*x + c)^7 + (3*a^2*b^2 + a*b^3)*sinh(d*x + c)^8 + 4*(3*a*b^3 + b^4)*cosh(d*x + c)^6 + 4*(
3*a*b^3 + b^4 + 7*(3*a^2*b^2 + a*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^6 + 8*(7*(3*a^2*b^2 + a*b^3)*cosh(d*x + c
)^3 + 3*(3*a*b^3 + b^4)*cosh(d*x + c))*sinh(d*x + c)^5 - 2*(3*a^2*b^2 + 13*a*b^3 + 4*b^4)*cosh(d*x + c)^4 + 2*
(35*(3*a^2*b^2 + a*b^3)*cosh(d*x + c)^4 - 3*a^2*b^2 - 13*a*b^3 - 4*b^4 + 30*(3*a*b^3 + b^4)*cosh(d*x + c)^2)*s
inh(d*x + c)^4 + 3*a^2*b^2 + a*b^3 + 8*(7*(3*a^2*b^2 + a*b^3)*cosh(d*x + c)^5 + 10*(3*a*b^3 + b^4)*cosh(d*x +
c)^3 - (3*a^2*b^2 + 13*a*b^3 + 4*b^4)*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(3*a*b^3 + b^4)*cosh(d*x + c)^2 + 4*(
7*(3*a^2*b^2 + a*b^3)*cosh(d*x + c)^6 + 15*(3*a*b^3 + b^4)*cosh(d*x + c)^4 + 3*a*b^3 + b^4 - 3*(3*a^2*b^2 + 13
*a*b^3 + 4*b^4)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 8*((3*a^2*b^2 + a*b^3)*cosh(d*x + c)^7 + 3*(3*a*b^3 + b^4)*
cosh(d*x + c)^5 - (3*a^2*b^2 + 13*a*b^3 + 4*b^4)*cosh(d*x + c)^3 + (3*a*b^3 + b^4)*cosh(d*x + c))*sinh(d*x + c
))*log(2*(a*cosh(d*x + c)^2 + a*sinh(d*x + c)^2 + a + 2*b)/(cosh(d*x + c)^2 - 2*cosh(d*x + c)*sinh(d*x + c) +
sinh(d*x + c)^2)) - 2*((a^4 + 3*a^3*b)*cosh(d*x + c)^8 + 8*(a^4 + 3*a^3*b)*cosh(d*x + c)*sinh(d*x + c)^7 + (a^
4 + 3*a^3*b)*sinh(d*x + c)^8 + 4*(a^3*b + 3*a^2*b^2)*cosh(d*x + c)^6 + 4*(a^3*b + 3*a^2*b^2 + 7*(a^4 + 3*a^3*b
)*cosh(d*x + c)^2)*sinh(d*x + c)^6 + 8*(7*(a^4 + 3*a^3*b)*cosh(d*x + c)^3 + 3*(a^3*b + 3*a^2*b^2)*cosh(d*x + c
))*sinh(d*x + c)^5 - 2*(a^4 + 7*a^3*b + 12*a^2*b^2)*cosh(d*x + c)^4 + 2*(35*(a^4 + 3*a^3*b)*cosh(d*x + c)^4 -
a^4 - 7*a^3*b - 12*a^2*b^2 + 30*(a^3*b + 3*a^2*b^2)*cosh(d*x + c)^2)*sinh(d*x + c)^4 + a^4 + 3*a^3*b + 8*(7*(a
^4 + 3*a^3*b)*cosh(d*x + c)^5 + 10*(a^3*b + 3*a^2*b^2)*cosh(d*x + c)^3 - (a^4 + 7*a^3*b + 12*a^2*b^2)*cosh(d*x
+ c))*sinh(d*x + c)^3 + 4*(a^3*b + 3*a^2*b^2)*cosh(d*x + c)^2 + 4*(7*(a^4 + 3*a^3*b)*cosh(d*x + c)^6 + 15*(a^
3*b + 3*a^2*b^2)*cosh(d*x + c)^4 + a^3*b + 3*a^2*b^2 - 3*(a^4 + 7*a^3*b + 12*a^2*b^2)*cosh(d*x + c)^2)*sinh(d*
x + c)^2 + 8*((a^4 + 3*a^3*b)*cosh(d*x + c)^7 + 3*(a^3*b + 3*a^2*b^2)*cosh(d*x + c)^5 - (a^4 + 7*a^3*b + 12*a^
2*b^2)*cosh(d*x + c)^3 + (a^3*b + 3*a^2*b^2)*cosh(d*x + c))*sinh(d*x + c))*log(2*sinh(d*x + c)/(cosh(d*x + c)
- sinh(d*x + c))) + 8*(2*(a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*d*x*cosh(d*x + c)^7 + 3*(a^4 + a^3*b - a*b^3 - b^
4 + 2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*x)*cosh(d*x + c)^5 + 2*(2*a^4 + 6*a^3*b + 4*a^2*b^2 + 2*a*b^3 + 2*
b^4 - (a^4 + 7*a^3*b + 15*a^2*b^2 + 13*a*b^3 + 4*b^4)*d*x)*cosh(d*x + c)^3 + (a^4 + a^3*b - a*b^3 - b^4 + 2*(a
^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*d*x)*cosh(d*x + c))*sinh(d*x + c))/((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*d*
cosh(d*x + c)^8 + 8*(a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*d*cosh(d*x + c)*sinh(d*x + c)^7 + (a^6 + 3*a^5*b + 3
*a^4*b^2 + a^3*b^3)*d*sinh(d*x + c)^8 + 4*(a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*d*cosh(d*x + c)^6 + 4*(7*(
a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*d*cosh(d*x + c)^2 + (a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*d)*sinh(d*x
+ c)^6 - 2*(a^6 + 7*a^5*b + 15*a^4*b^2 + 13*a^3*b^3 + 4*a^2*b^4)*d*cosh(d*x + c)^4 + 8*(7*(a^6 + 3*a^5*b + 3*
a^4*b^2 + a^3*b^3)*d*cosh(d*x + c)^3 + 3*(a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*d*cosh(d*x + c))*sinh(d*x +
c)^5 + 2*(35*(a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*d*cosh(d*x + c)^4 + 30*(a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^
2*b^4)*d*cosh(d*x + c)^2 - (a^6 + 7*a^5*b + 15*a^4*b^2 + 13*a^3*b^3 + 4*a^2*b^4)*d)*sinh(d*x + c)^4 + 4*(a^5*b
+ 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*d*cosh(d*x + c)^2 + 8*(7*(a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*d*cosh(d*x
+ c)^5 + 10*(a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*d*cosh(d*x + c)^3 - (a^6 + 7*a^5*b + 15*a^4*b^2 + 13*a^3
*b^3 + 4*a^2*b^4)*d*cosh(d*x + c))*sinh(d*x + c)^3 + 4*(7*(a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*d*cosh(d*x + c
)^6 + 15*(a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*d*cosh(d*x + c)^4 - 3*(a^6 + 7*a^5*b + 15*a^4*b^2 + 13*a^3*
b^3 + 4*a^2*b^4)*d*cosh(d*x + c)^2 + (a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*d)*sinh(d*x + c)^2 + (a^6 + 3*a
^5*b + 3*a^4*b^2 + a^3*b^3)*d + 8*((a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*d*cosh(d*x + c)^7 + 3*(a^5*b + 3*a^4*
b^2 + 3*a^3*b^3 + a^2*b^4)*d*cosh(d*x + c)^5 - (a^6 + 7*a^5*b + 15*a^4*b^2 + 13*a^3*b^3 + 4*a^2*b^4)*d*cosh(d*
x + c)^3 + (a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*d*cosh(d*x + c))*sinh(d*x + c))
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\coth ^{3}{\left (c + d x \right )}}{\left (a + b \operatorname{sech}^{2}{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(d*x+c)**3/(a+b*sech(d*x+c)**2)**2,x)
[Out]
Integral(coth(c + d*x)**3/(a + b*sech(c + d*x)**2)**2, x)
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Giac [B] time = 2.69152, size = 517, normalized size = 4.7 \begin{align*} \frac{\frac{{\left (3 \, a b^{2} + b^{3}\right )} \log \left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}{a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}} + \frac{2 \,{\left (a e^{\left (2 \, c\right )} + 3 \, b e^{\left (2 \, c\right )}\right )} \log \left ({\left | -e^{\left (2 \, d x + 2 \, c\right )} + 1 \right |}\right )}{a^{3} e^{\left (2 \, c\right )} + 3 \, a^{2} b e^{\left (2 \, c\right )} + 3 \, a b^{2} e^{\left (2 \, c\right )} + b^{3} e^{\left (2 \, c\right )}} - \frac{2 \, d x}{a^{2}} - \frac{3 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 10 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a b^{2} + b^{3}}{{\left (a^{4} + 3 \, a^{3} b + 3 \, a^{2} b^{2} + a b^{3}\right )}{\left (a e^{\left (4 \, d x + 4 \, c\right )} + 2 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, b e^{\left (2 \, d x + 2 \, c\right )} + a\right )}} - \frac{3 \, a e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b e^{\left (4 \, d x + 4 \, c\right )} - 2 \, a e^{\left (2 \, d x + 2 \, c\right )} - 14 \, b e^{\left (2 \, d x + 2 \, c\right )} + 3 \, a + 9 \, b}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )}{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{2}}}{2 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(coth(d*x+c)^3/(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")
[Out]
1/2*((3*a*b^2 + b^3)*log(a*e^(4*d*x + 4*c) + 2*a*e^(2*d*x + 2*c) + 4*b*e^(2*d*x + 2*c) + a)/(a^5 + 3*a^4*b + 3
*a^3*b^2 + a^2*b^3) + 2*(a*e^(2*c) + 3*b*e^(2*c))*log(abs(-e^(2*d*x + 2*c) + 1))/(a^3*e^(2*c) + 3*a^2*b*e^(2*c
) + 3*a*b^2*e^(2*c) + b^3*e^(2*c)) - 2*d*x/a^2 - (3*a*b^2*e^(4*d*x + 4*c) + b^3*e^(4*d*x + 4*c) + 6*a*b^2*e^(2
*d*x + 2*c) + 10*b^3*e^(2*d*x + 2*c) + 3*a*b^2 + b^3)/((a^4 + 3*a^3*b + 3*a^2*b^2 + a*b^3)*(a*e^(4*d*x + 4*c)
+ 2*a*e^(2*d*x + 2*c) + 4*b*e^(2*d*x + 2*c) + a)) - (3*a*e^(4*d*x + 4*c) + 9*b*e^(4*d*x + 4*c) - 2*a*e^(2*d*x
+ 2*c) - 14*b*e^(2*d*x + 2*c) + 3*a + 9*b)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(e^(2*d*x + 2*c) - 1)^2))/d