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

Optimal. Leaf size=53 \[ \frac {(a+b)^2 \log (\sinh (c+d x))}{d}-\frac {b (2 a+b) \log (\cosh (c+d x))}{d}+\frac {b^2 \text {sech}^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 53, normalized size of antiderivative = 1.00, 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 = {4138, 446, 88} \[ \frac {(a+b)^2 \log (\sinh (c+d x))}{d}-\frac {b (2 a+b) \log (\cosh (c+d x))}{d}+\frac {b^2 \text {sech}^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(2*a + b)*Log[Cosh[c + d*x]])/d) + ((a + b)^2*Log[Sinh[c + d*x]])/d + (b^2*Sech[c + d*x]^2)/(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 \coth (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^3 \left (1-x^2\right )} \, dx,x,\cosh (c+d x)\right )}{d}\\ &=-\frac {\operatorname {Subst}\left (\int \frac {(b+a x)^2}{(1-x) x^2} \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=-\frac {\operatorname {Subst}\left (\int \left (-\frac {(a+b)^2}{-1+x}+\frac {b^2}{x^2}+\frac {b (2 a+b)}{x}\right ) \, dx,x,\cosh ^2(c+d x)\right )}{2 d}\\ &=-\frac {b (2 a+b) \log (\cosh (c+d x))}{d}+\frac {(a+b)^2 \log (\sinh (c+d x))}{d}+\frac {b^2 \text {sech}^2(c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 84, normalized size = 1.58 \[ \frac {2 (a \cosh (c+d x)+b \text {sech}(c+d x))^2 \left (2 \cosh ^2(c+d x) \left ((a+b)^2 \log (\sinh (c+d x))-b (2 a+b) \log (\cosh (c+d x))\right )+b^2\right )}{d (a \cosh (2 (c+d x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 665, normalized size = 12.55 \[ -\frac {a^{2} d x \cosh \left (d x + c\right )^{4} + 4 \, a^{2} d x \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + a^{2} d x \sinh \left (d x + c\right )^{4} + a^{2} d x + 2 \, {\left (a^{2} d x - b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, a^{2} d x \cosh \left (d x + c\right )^{2} + a^{2} d x - b^{2}\right )} \sinh \left (d x + c\right )^{2} + {\left ({\left (2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{4} + 4 \, {\left (2 \, a b + b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{4} + 2 \, {\left (2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{2} + 2 \, a b + b^{2} + 4 \, {\left ({\left (2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{3} + {\left (2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \log \left (\frac {2 \, \cosh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) - {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{4} + 4 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + {\left (a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{4} + 2 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2}\right )} \sinh \left (d x + c\right )^{2} + a^{2} + 2 \, a b + b^{2} + 4 \, {\left ({\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )^{3} + {\left (a^{2} + 2 \, a b + b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )\right )} \log \left (\frac {2 \, \sinh \left (d x + c\right )}{\cosh \left (d x + c\right ) - \sinh \left (d x + c\right )}\right ) + 4 \, {\left (a^{2} d x \cosh \left (d x + c\right )^{3} + {\left (a^{2} d x - b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{d \cosh \left (d x + c\right )^{4} + 4 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{3} + d \sinh \left (d x + c\right )^{4} + 2 \, d \cosh \left (d x + c\right )^{2} + 2 \, {\left (3 \, d \cosh \left (d x + c\right )^{2} + d\right )} \sinh \left (d x + c\right )^{2} + 4 \, {\left (d \cosh \left (d x + c\right )^{3} + d \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right ) + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.17, size = 171, normalized size = 3.23 \[ -\frac {2 \, a^{2} d x + 2 \, {\left (2 \, a b e^{\left (2 \, c\right )} + b^{2} e^{\left (2 \, c\right )}\right )} e^{\left (-2 \, c\right )} \log \left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right ) - 2 \, {\left (a^{2} e^{\left (2 \, c\right )} + 2 \, a b e^{\left (2 \, c\right )} + b^{2} e^{\left (2 \, c\right )}\right )} e^{\left (-2 \, c\right )} \log \left ({\left | e^{\left (2 \, d x + 2 \, c\right )} - 1 \right |}\right ) - \frac {6 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 12 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 10 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 6 \, a b + 3 \, b^{2}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*a^2*d*x + 2*(2*a*b*e^(2*c) + b^2*e^(2*c))*e^(-2*c)*log(e^(2*d*x + 2*c) + 1) - 2*(a^2*e^(2*c) + 2*a*b*e
^(2*c) + b^2*e^(2*c))*e^(-2*c)*log(abs(e^(2*d*x + 2*c) - 1)) - (6*a*b*e^(4*d*x + 4*c) + 3*b^2*e^(4*d*x + 4*c)
+ 12*a*b*e^(2*d*x + 2*c) + 10*b^2*e^(2*d*x + 2*c) + 6*a*b + 3*b^2)/(e^(2*d*x + 2*c) + 1)^2)/d

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maple [A]  time = 0.26, size = 60, normalized size = 1.13 \[ \frac {a^{2} \ln \left (\sinh \left (d x +c \right )\right )}{d}+\frac {2 a b \ln \left (\tanh \left (d x +c \right )\right )}{d}+\frac {b^{2}}{2 d \cosh \left (d x +c \right )^{2}}+\frac {b^{2} \ln \left (\tanh \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.70, size = 161, normalized size = 3.04 \[ b^{2} {\left (\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d} + \frac {2 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} + 2 \, a b {\left (\frac {\log \left (e^{\left (-d x - c\right )} + 1\right )}{d} + \frac {\log \left (e^{\left (-d x - c\right )} - 1\right )}{d} - \frac {\log \left (e^{\left (-2 \, d x - 2 \, c\right )} + 1\right )}{d}\right )} + \frac {a^{2} \log \left (\sinh \left (d x + c\right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.30, size = 308, normalized size = 5.81 \[ \frac {2\,b^2}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}-a^2\,x+\frac {a^2\,\ln \left ({\mathrm {e}}^{4\,c+4\,d\,x}-1\right )}{2\,d}-\frac {2\,b^2}{d\,\left (2\,{\mathrm {e}}^{2\,c+2\,d\,x}+{\mathrm {e}}^{4\,c+4\,d\,x}+1\right )}-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{2\,c}\,{\mathrm {e}}^{2\,d\,x}\,\left (a^4\,\sqrt {-d^2}+4\,b^4\,\sqrt {-d^2}+16\,a\,b^3\,\sqrt {-d^2}+8\,a^3\,b\,\sqrt {-d^2}+20\,a^2\,b^2\,\sqrt {-d^2}\right )}{a^2\,d\,\sqrt {a^4+8\,a^3\,b+20\,a^2\,b^2+16\,a\,b^3+4\,b^4}+2\,b^2\,d\,\sqrt {a^4+8\,a^3\,b+20\,a^2\,b^2+16\,a\,b^3+4\,b^4}+4\,a\,b\,d\,\sqrt {a^4+8\,a^3\,b+20\,a^2\,b^2+16\,a\,b^3+4\,b^4}}\right )\,\sqrt {a^4+8\,a^3\,b+20\,a^2\,b^2+16\,a\,b^3+4\,b^4}}{\sqrt {-d^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b^2)/(d*(exp(2*c + 2*d*x) + 1)) - a^2*x + (a^2*log(exp(4*c + 4*d*x) - 1))/(2*d) - (2*b^2)/(d*(2*exp(2*c + 2
*d*x) + exp(4*c + 4*d*x) + 1)) - (atan((exp(2*c)*exp(2*d*x)*(a^4*(-d^2)^(1/2) + 4*b^4*(-d^2)^(1/2) + 16*a*b^3*
(-d^2)^(1/2) + 8*a^3*b*(-d^2)^(1/2) + 20*a^2*b^2*(-d^2)^(1/2)))/(a^2*d*(16*a*b^3 + 8*a^3*b + a^4 + 4*b^4 + 20*
a^2*b^2)^(1/2) + 2*b^2*d*(16*a*b^3 + 8*a^3*b + a^4 + 4*b^4 + 20*a^2*b^2)^(1/2) + 4*a*b*d*(16*a*b^3 + 8*a^3*b +
a^4 + 4*b^4 + 20*a^2*b^2)^(1/2)))*(16*a*b^3 + 8*a^3*b + a^4 + 4*b^4 + 20*a^2*b^2)^(1/2))/(-d^2)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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