3.146 \(\int \frac {\coth ^3(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx\)

Optimal. Leaf size=73 \[ \frac {b^2 \log \left (a \cosh ^2(c+d x)+b\right )}{2 a d (a+b)^2}-\frac {\text {csch}^2(c+d x)}{2 d (a+b)}+\frac {(a+2 b) \log (\sinh (c+d x))}{d (a+b)^2} \]

[Out]

-1/2*csch(d*x+c)^2/(a+b)/d+1/2*b^2*ln(b+a*cosh(d*x+c)^2)/a/(a+b)^2/d+(a+2*b)*ln(sinh(d*x+c))/(a+b)^2/d

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4138, 446, 88} \[ \frac {b^2 \log \left (a \cosh ^2(c+d x)+b\right )}{2 a d (a+b)^2}-\frac {\text {csch}^2(c+d x)}{2 d (a+b)}+\frac {(a+2 b) \log (\sinh (c+d x))}{d (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^3/(a + b*Sech[c + d*x]^2),x]

[Out]

-Csch[c + d*x]^2/(2*(a + b)*d) + (b^2*Log[b + a*Cosh[c + d*x]^2])/(2*a*(a + b)^2*d) + ((a + 2*b)*Log[Sinh[c +
d*x]])/((a + b)^2*d)

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]))

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 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]

Rubi steps

\begin {align*} \int \frac {\coth ^3(c+d x)}{a+b \text {sech}^2(c+d x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5}{\left (1-x^2\right )^2 \left (b+a x^2\right )} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^2}{(1-x)^2 (b+a x)} \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (\frac {1}{(a+b) (-1+x)^2}+\frac {a+2 b}{(a+b)^2 (-1+x)}+\frac {b^2}{(a+b)^2 (b+a x)}\right ) \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=-\frac {\text {csch}^2(c+d x)}{2 (a+b) d}+\frac {b^2 \log \left (b+a \cosh ^2(c+d x)\right )}{2 a (a+b)^2 d}+\frac {(a+2 b) \log (\sinh (c+d x))}{(a+b)^2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.24, size = 100, normalized size = 1.37 \[ -\frac {\text {sech}^2(c+d x) (a \cosh (2 (c+d x))+a+2 b) \left (b^2 \left (-\log \left (a \sinh ^2(c+d x)+a+b\right )\right )+a (a+b) \text {csch}^2(c+d x)-2 a (a+2 b) \log (\sinh (c+d x))\right )}{4 a d (a+b)^2 \left (a+b \text {sech}^2(c+d x)\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^3/(a + b*Sech[c + d*x]^2),x]

[Out]

-1/4*((a + 2*b + a*Cosh[2*(c + d*x)])*(a*(a + b)*Csch[c + d*x]^2 - 2*a*(a + 2*b)*Log[Sinh[c + d*x]] - b^2*Log[
a + b + a*Sinh[c + d*x]^2])*Sech[c + d*x]^2)/(a*(a + b)^2*d*(a + b*Sech[c + d*x]^2))

________________________________________________________________________________________

fricas [B]  time = 0.52, size = 862, normalized size = 11.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sech(d*x+c)^2),x, algorithm="fricas")

[Out]

-1/2*(2*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^4 + 8*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + 2*
(a^2 + 2*a*b + b^2)*d*x*sinh(d*x + c)^4 + 2*(a^2 + 2*a*b + b^2)*d*x - 4*((a^2 + 2*a*b + b^2)*d*x - a^2 - a*b)*
cosh(d*x + c)^2 + 4*(3*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^2 - (a^2 + 2*a*b + b^2)*d*x + a^2 + a*b)*sinh(d*x
+ c)^2 - (b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^3 + b^2*sinh(d*x + c)^4 - 2*b^2*cosh(d*x +
c)^2 + 2*(3*b^2*cosh(d*x + c)^2 - b^2)*sinh(d*x + c)^2 + b^2 + 4*(b^2*cosh(d*x + c)^3 - b^2*cosh(d*x + c))*sin
h(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^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b)*cosh(d*x + c)*sinh(d*x + c)^3
+ (a^2 + 2*a*b)*sinh(d*x + c)^4 - 2*(a^2 + 2*a*b)*cosh(d*x + c)^2 + 2*(3*(a^2 + 2*a*b)*cosh(d*x + c)^2 - a^2 -
2*a*b)*sinh(d*x + c)^2 + a^2 + 2*a*b + 4*((a^2 + 2*a*b)*cosh(d*x + c)^3 - (a^2 + 2*a*b)*cosh(d*x + c))*sinh(d
*x + c))*log(2*sinh(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) + 8*((a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^3 - (
(a^2 + 2*a*b + b^2)*d*x - a^2 - a*b)*cosh(d*x + c))*sinh(d*x + c))/((a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^4
+ 4*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)*sinh(d*x + c)^3 + (a^3 + 2*a^2*b + a*b^2)*d*sinh(d*x + c)^4 - 2*(a
^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^2 + 2*(3*(a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^2 - (a^3 + 2*a^2*b + a*
b^2)*d)*sinh(d*x + c)^2 + (a^3 + 2*a^2*b + a*b^2)*d + 4*((a^3 + 2*a^2*b + a*b^2)*d*cosh(d*x + c)^3 - (a^3 + 2*
a^2*b + a*b^2)*d*cosh(d*x + c))*sinh(d*x + c))

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sech(d*x+c)^2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Erro
r index.cc index_gcd Error: Bad Argument ValueError index.cc index_gcd Error: Bad Argument ValueError index.cc
index_gcd Error: Bad Argument ValueError index.cc index_gcd Error: Bad Argument ValueEvaluation time: 0.58Don
e

________________________________________________________________________________________

maple [B]  time = 0.48, size = 199, normalized size = 2.73 \[ -\frac {\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \left (a +b \right )}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}-\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d a}+\frac {b^{2} \ln \left (\left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +b \left (\tanh ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a -2 \left (\tanh ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b \right )}{2 d a \left (a +b \right )^{2}}-\frac {1}{8 d \left (a +b \right ) \tanh \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {\ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a}{d \left (a +b \right )^{2}}+\frac {2 \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{d \left (a +b \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^3/(a+b*sech(d*x+c)^2),x)

[Out]

-1/8/d*tanh(1/2*d*x+1/2*c)^2/(a+b)-1/d/a*ln(tanh(1/2*d*x+1/2*c)-1)-1/d/a*ln(tanh(1/2*d*x+1/2*c)+1)+1/2/d*b^2/a
/(a+b)^2*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)/tanh(1/2*d*x+1/2*c)^2+1/d/(a+b)^2*ln(tanh(1/2*d*x+1/2*c))*a+2/d/(a+b)^2*ln(tanh(1/2*d*x+1/2
*c))*b

________________________________________________________________________________________

maxima [B]  time = 0.44, size = 187, normalized size = 2.56 \[ \frac {b^{2} \log \left (2 \, {\left (a + 2 \, b\right )} e^{\left (-2 \, d x - 2 \, c\right )} + a e^{\left (-4 \, d x - 4 \, c\right )} + a\right )}{2 \, {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} d} + \frac {{\left (a + 2 \, b\right )} \log \left (e^{\left (-d x - c\right )} + 1\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {{\left (a + 2 \, b\right )} \log \left (e^{\left (-d x - c\right )} - 1\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} d} + \frac {d x + c}{a d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{{\left (2 \, {\left (a + b\right )} e^{\left (-2 \, d x - 2 \, c\right )} - {\left (a + b\right )} e^{\left (-4 \, d x - 4 \, c\right )} - a - b\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^3/(a+b*sech(d*x+c)^2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 2.07, size = 523, normalized size = 7.16 \[ \frac {\ln \left (23\,a\,b^7+8\,a^7\,b-2\,b^8-72\,a^2\,b^6-10\,a^3\,b^5+184\,a^4\,b^4+180\,a^5\,b^3+64\,a^6\,b^2+2\,b^8\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-23\,a\,b^7\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-8\,a^7\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+72\,a^2\,b^6\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+10\,a^3\,b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-184\,a^4\,b^4\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-180\,a^5\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-64\,a^6\,b^2\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\right )\,\left (a+2\,b\right )}{d\,a^2+2\,d\,a\,b+d\,b^2}-\frac {x}{a}-\frac {2}{\left (a\,d+b\,d\right )\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {b^2\,\ln \left (a\,b^4+16\,a^4\,b+4\,a^5-8\,a^2\,b^3+12\,a^3\,b^2+8\,a^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+4\,a^5\,{\mathrm {e}}^{4\,c}\,{\mathrm {e}}^{4\,d\,x}+4\,b^5\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-30\,a\,b^4\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+48\,a^4\,b\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+a\,b^4\,{\mathrm {e}}^{4\,c}\,{\mathrm {e}}^{4\,d\,x}+16\,a^4\,b\,{\mathrm {e}}^{4\,c}\,{\mathrm {e}}^{4\,d\,x}+32\,a^2\,b^3\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}+88\,a^3\,b^2\,{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}-8\,a^2\,b^3\,{\mathrm {e}}^{4\,c}\,{\mathrm {e}}^{4\,d\,x}+12\,a^3\,b^2\,{\mathrm {e}}^{4\,c}\,{\mathrm {e}}^{4\,d\,x}\right )}{2\,d\,a^3+4\,d\,a^2\,b+2\,d\,a\,b^2}-\frac {2\,\left (a^2+b\,a\right )}{a\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )\,\left (a+b\right )\,\left (a\,d+b\,d\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(c + d*x)^3/(a + b/cosh(c + d*x)^2),x)

[Out]

(log(23*a*b^7 + 8*a^7*b - 2*b^8 - 72*a^2*b^6 - 10*a^3*b^5 + 184*a^4*b^4 + 180*a^5*b^3 + 64*a^6*b^2 + 2*b^8*exp
(2*c)*exp(2*d*x) - 23*a*b^7*exp(2*c)*exp(2*d*x) - 8*a^7*b*exp(2*c)*exp(2*d*x) + 72*a^2*b^6*exp(2*c)*exp(2*d*x)
+ 10*a^3*b^5*exp(2*c)*exp(2*d*x) - 184*a^4*b^4*exp(2*c)*exp(2*d*x) - 180*a^5*b^3*exp(2*c)*exp(2*d*x) - 64*a^6
*b^2*exp(2*c)*exp(2*d*x))*(a + 2*b))/(a^2*d + b^2*d + 2*a*b*d) - x/a - 2/((a*d + b*d)*(exp(4*c + 4*d*x) - 2*ex
p(2*c + 2*d*x) + 1)) + (b^2*log(a*b^4 + 16*a^4*b + 4*a^5 - 8*a^2*b^3 + 12*a^3*b^2 + 8*a^5*exp(2*c)*exp(2*d*x)
+ 4*a^5*exp(4*c)*exp(4*d*x) + 4*b^5*exp(2*c)*exp(2*d*x) - 30*a*b^4*exp(2*c)*exp(2*d*x) + 48*a^4*b*exp(2*c)*exp
(2*d*x) + a*b^4*exp(4*c)*exp(4*d*x) + 16*a^4*b*exp(4*c)*exp(4*d*x) + 32*a^2*b^3*exp(2*c)*exp(2*d*x) + 88*a^3*b
^2*exp(2*c)*exp(2*d*x) - 8*a^2*b^3*exp(4*c)*exp(4*d*x) + 12*a^3*b^2*exp(4*c)*exp(4*d*x)))/(2*a^3*d + 2*a*b^2*d
+ 4*a^2*b*d) - (2*(a*b + a^2))/(a*(exp(2*c + 2*d*x) - 1)*(a + b)*(a*d + b*d))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\coth ^{3}{\left (c + d x \right )}}{a + b \operatorname {sech}^{2}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)**3/(a+b*sech(d*x+c)**2),x)

[Out]

Integral(coth(c + d*x)**3/(a + b*sech(c + d*x)**2), x)

________________________________________________________________________________________