3.151 \(\int \coth ^3(c+d x) (a+b \tanh ^2(c+d x))^2 \, dx\)

Optimal. Leaf size=52 \[ -\frac{a^2 \coth ^2(c+d x)}{2 d}+\frac{a (a+2 b) \log (\tanh (c+d x))}{d}+\frac{(a+b)^2 \log (\cosh (c+d x))}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0947188, antiderivative size = 52, 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 = {3670, 446, 88} \[ -\frac{a^2 \coth ^2(c+d x)}{2 d}+\frac{a (a+2 b) \log (\tanh (c+d x))}{d}+\frac{(a+b)^2 \log (\cosh (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3670

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

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

Mathematica [A]  time = 0.152807, size = 50, normalized size = 0.96 \[ \frac{-a^2 \coth ^2(c+d x)+2 a (a+2 b) \log (\tanh (c+d x))+2 (a+b)^2 \log (\cosh (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.058, size = 60, normalized size = 1.2 \begin{align*}{\frac{{a}^{2}\ln \left ( \sinh \left ( dx+c \right ) \right ) }{d}}-{\frac{{a}^{2} \left ({\rm coth} \left (dx+c\right ) \right ) ^{2}}{2\,d}}+2\,{\frac{ab\ln \left ( \sinh \left ( dx+c \right ) \right ) }{d}}+{\frac{{b}^{2}\ln \left ( \cosh \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [B]  time = 1.01771, size = 181, normalized size = 3.48 \begin{align*} a^{2}{\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 )} + \frac{b^{2} \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}{d} + \frac{2 \, a b \log \left (e^{\left (d x + c\right )} - e^{\left (-d x - c\right )}\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [B]  time = 2.03624, size = 1715, normalized size = 32.98 \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*tanh(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

-((a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b + b^2)*d*x*cosh(d*x + c)*sinh(d*x + c)^3 + (a^2 + 2
*a*b + b^2)*d*x*sinh(d*x + c)^4 + (a^2 + 2*a*b + b^2)*d*x - 2*((a^2 + 2*a*b + b^2)*d*x - a^2)*cosh(d*x + c)^2
+ 2*(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)*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))*sinh(d*x + c))*log(2*cosh
(d*x + c)/(cosh(d*x + c) - sinh(d*x + c))) - ((a^2 + 2*a*b)*cosh(d*x + c)^4 + 4*(a^2 + 2*a*b)*cosh(d*x + c)*si
nh(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))) + 4*((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)*cosh(d*x + c))*sinh(d*x + c))/(d*cosh(d*x + c)^4 + 4*d*cosh(d*x +
c)*sinh(d*x + c)^3 + d*sinh(d*x + c)^4 - 2*d*cosh(d*x + c)^2 + 2*(3*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^2 +
4*(d*cosh(d*x + c)^3 - d*cosh(d*x + c))*sinh(d*x + c) + d)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

Giac [B]  time = 1.2946, size = 198, normalized size = 3.81 \begin{align*} \frac{b^{2} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} + 2\right )}{2 \, d} + \frac{{\left (a^{2} + 2 \, a b\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2\right )}{2 \, d} - \frac{a^{2}{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a b{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )}\right )} + 2 \, a^{2} - 4 \, a b}{2 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} + e^{\left (-2 \, d x - 2 \, c\right )} - 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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