3.13 \(\int \text{csch}(c+d x) (a+b \text{sech}^2(c+d x))^2 \, dx\)

Optimal. Leaf size=52 \[ \frac{b (2 a+b) \text{sech}(c+d x)}{d}-\frac{(a+b)^2 \tanh ^{-1}(\cosh (c+d x))}{d}+\frac{b^2 \text{sech}^3(c+d x)}{3 d} \]

[Out]

-(((a + b)^2*ArcTanh[Cosh[c + d*x]])/d) + (b*(2*a + b)*Sech[c + d*x])/d + (b^2*Sech[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.0771389, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4133, 461, 207} \[ \frac{b (2 a+b) \text{sech}(c+d x)}{d}-\frac{(a+b)^2 \tanh ^{-1}(\cosh (c+d x))}{d}+\frac{b^2 \text{sech}^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a + b)^2*ArcTanh[Cosh[c + d*x]])/d) + (b*(2*a + b)*Sech[c + d*x])/d + (b^2*Sech[c + d*x]^3)/(3*d)

Rule 4133

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = F
reeFactors[Cos[e + f*x], x]}, -Dist[ff/f, Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*(ff*x)^n)^p)/(ff*x)^(n*
p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n] && IntegerQ[p
]

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \text{csch}(c+d x) \left (a+b \text{sech}^2(c+d x)\right )^2 \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{\left (b+a x^2\right )^2}{x^4 \left (1-x^2\right )} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{b^2}{x^4}+\frac{b (2 a+b)}{x^2}-\frac{(a+b)^2}{-1+x^2}\right ) \, dx,x,\cosh (c+d x)\right )}{d}\\ &=\frac{b (2 a+b) \text{sech}(c+d x)}{d}+\frac{b^2 \text{sech}^3(c+d x)}{3 d}+\frac{(a+b)^2 \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac{(a+b)^2 \tanh ^{-1}(\cosh (c+d x))}{d}+\frac{b (2 a+b) \text{sech}(c+d x)}{d}+\frac{b^2 \text{sech}^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [B]  time = 0.553278, size = 108, normalized size = 2.08 \[ -\frac{4 \text{sech}^3(c+d x) \left (a \cosh ^2(c+d x)+b\right )^2 \left (-3 b (2 a+b) \cosh ^2(c+d x)+3 (a+b)^2 \cosh ^3(c+d x) \left (\log \left (\cosh \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sinh \left (\frac{1}{2} (c+d x)\right )\right )\right )-b^2\right )}{3 d (a \cosh (2 (c+d x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(b + a*Cosh[c + d*x]^2)^2*(-b^2 - 3*b*(2*a + b)*Cosh[c + d*x]^2 + 3*(a + b)^2*Cosh[c + d*x]^3*(Log[Cosh[(c
+ d*x)/2]] - Log[Sinh[(c + d*x)/2]]))*Sech[c + d*x]^3)/(3*d*(a + 2*b + a*Cosh[2*(c + d*x)])^2)

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Maple [A]  time = 0.036, size = 72, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ( -2\,{a}^{2}{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) +2\,ab \left ( \left ( \cosh \left ( dx+c \right ) \right ) ^{-1}-2\,{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) \right ) +{b}^{2} \left ({\frac{1}{3\, \left ( \cosh \left ( dx+c \right ) \right ) ^{3}}}+ \left ( \cosh \left ( dx+c \right ) \right ) ^{-1}-2\,{\it Artanh} \left ({{\rm e}^{dx+c}} \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)*(a+b*sech(d*x+c)^2)^2,x)

[Out]

1/d*(-2*a^2*arctanh(exp(d*x+c))+2*a*b*(1/cosh(d*x+c)-2*arctanh(exp(d*x+c)))+b^2*(1/3/cosh(d*x+c)^3+1/cosh(d*x+
c)-2*arctanh(exp(d*x+c))))

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Maxima [B]  time = 1.05542, size = 266, normalized size = 5.12 \begin{align*} -\frac{1}{3} \, b^{2}{\left (\frac{3 \, \log \left (e^{\left (-d x - c\right )} + 1\right )}{d} - \frac{3 \, \log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac{2 \,{\left (3 \, e^{\left (-d x - c\right )} + 10 \, e^{\left (-3 \, d x - 3 \, c\right )} + 3 \, e^{\left (-5 \, d x - 5 \, c\right )}\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 )} - 2 \, a b{\left (\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 (-d x - c\right )}}{d{\left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}}\right )} + \frac{a^{2} \log \left (\tanh \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.67833, size = 3016, normalized size = 58. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(6*(2*a*b + b^2)*cosh(d*x + c)^5 + 30*(2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^4 + 6*(2*a*b + b^2)*sinh(d
*x + c)^5 + 4*(6*a*b + 5*b^2)*cosh(d*x + c)^3 + 4*(15*(2*a*b + b^2)*cosh(d*x + c)^2 + 6*a*b + 5*b^2)*sinh(d*x
+ c)^3 + 12*(5*(2*a*b + b^2)*cosh(d*x + c)^3 + (6*a*b + 5*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 + 6*(2*a*b + b^2
)*cosh(d*x + c) - 3*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 6*(a^2 + 2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^5
+ (a^2 + 2*a*b + b^2)*sinh(d*x + c)^6 + 3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 3*(5*(a^2 + 2*a*b + b^2)*cosh
(d*x + c)^2 + a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 + 4*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + 3*(a^2 + 2*a*b +
b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + 3*(5*(a^2 + 2*a*b + b^2)*cosh(d
*x + c)^4 + 6*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b + b^2)*sinh(d*x + c)^2 + a^2 + 2*a*b + b^2 + 6
*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^5 + 2*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^3 + (a^2 + 2*a*b + b^2)*cosh(d*x +
c))*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) + 1) + 3*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^6 + 6*(a^2 +
2*a*b + b^2)*cosh(d*x + c)*sinh(d*x + c)^5 + (a^2 + 2*a*b + b^2)*sinh(d*x + c)^6 + 3*(a^2 + 2*a*b + b^2)*cosh
(d*x + c)^4 + 3*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b + b^2)*sinh(d*x + c)^4 + 4*(5*(a^2 + 2*a*
b + b^2)*cosh(d*x + c)^3 + 3*(a^2 + 2*a*b + b^2)*cosh(d*x + c))*sinh(d*x + c)^3 + 3*(a^2 + 2*a*b + b^2)*cosh(d
*x + c)^2 + 3*(5*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^4 + 6*(a^2 + 2*a*b + b^2)*cosh(d*x + c)^2 + a^2 + 2*a*b + b
^2)*sinh(d*x + c)^2 + a^2 + 2*a*b + b^2 + 6*((a^2 + 2*a*b + b^2)*cosh(d*x + c)^5 + 2*(a^2 + 2*a*b + b^2)*cosh(
d*x + c)^3 + (a^2 + 2*a*b + b^2)*cosh(d*x + c))*sinh(d*x + c))*log(cosh(d*x + c) + sinh(d*x + c) - 1) + 6*(5*(
2*a*b + b^2)*cosh(d*x + c)^4 + 2*(6*a*b + 5*b^2)*cosh(d*x + c)^2 + 2*a*b + b^2)*sinh(d*x + c))/(d*cosh(d*x + c
)^6 + 6*d*cosh(d*x + c)*sinh(d*x + c)^5 + d*sinh(d*x + c)^6 + 3*d*cosh(d*x + c)^4 + 3*(5*d*cosh(d*x + c)^2 + d
)*sinh(d*x + c)^4 + 4*(5*d*cosh(d*x + c)^3 + 3*d*cosh(d*x + c))*sinh(d*x + c)^3 + 3*d*cosh(d*x + c)^2 + 3*(5*d
*cosh(d*x + c)^4 + 6*d*cosh(d*x + c)^2 + d)*sinh(d*x + c)^2 + 6*(d*cosh(d*x + c)^5 + 2*d*cosh(d*x + c)^3 + d*c
osh(d*x + c))*sinh(d*x + c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \operatorname{sech}^{2}{\left (c + d x \right )}\right )^{2} \operatorname{csch}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)*(a+b*sech(d*x+c)**2)**2,x)

[Out]

Integral((a + b*sech(c + d*x)**2)**2*csch(c + d*x), x)

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Giac [B]  time = 1.19978, size = 193, normalized size = 3.71 \begin{align*} -\frac{{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )} + 2\right )}{2 \, d} + \frac{{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )} - 2\right )}{2 \, d} + \frac{2 \,{\left (6 \, a b{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} + 3 \, b^{2}{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{2} + 4 \, b^{2}\right )}}{3 \, d{\left (e^{\left (d x + c\right )} + e^{\left (-d x - c\right )}\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(a^2 + 2*a*b + b^2)*log(e^(d*x + c) + e^(-d*x - c) + 2)/d + 1/2*(a^2 + 2*a*b + b^2)*log(e^(d*x + c) + e^(
-d*x - c) - 2)/d + 2/3*(6*a*b*(e^(d*x + c) + e^(-d*x - c))^2 + 3*b^2*(e^(d*x + c) + e^(-d*x - c))^2 + 4*b^2)/(
d*(e^(d*x + c) + e^(-d*x - c))^3)