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

Optimal. Leaf size=36 \[ a^2 x-\frac {(a+b)^2 \coth (c+d x)}{d}-\frac {b^2 \tanh (c+d x)}{d} \]

[Out]

a^2*x-(a+b)^2*coth(d*x+c)/d-b^2*tanh(d*x+c)/d

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Rubi [A]  time = 0.09, antiderivative size = 36, 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 = {4141, 1802, 207} \[ a^2 x-\frac {(a+b)^2 \coth (c+d x)}{d}-\frac {b^2 \tanh (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

Rule 1802

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 4141

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((d*ff*x)^m*(a + b*(1 + ff^2*x^2)^(n/2))^p)/(1 + ff^
2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] && IntegerQ[n/2] && (IntegerQ[m/2] ||
EqQ[n, 2])

Rubi steps

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

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Mathematica [B]  time = 0.74, size = 82, normalized size = 2.28 \[ \frac {4 \text {sech}(c+d x) \left (a \cosh ^2(c+d x)+b\right )^2 \left (a^2 d x \cosh (c+d x)+\sinh (d x) \left ((a+b)^2 \text {csch}(c) \coth (c+d x)-b^2 \text {sech}(c)\right )\right )}{d (a \cosh (2 (c+d x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.44, size = 106, normalized size = 2.94 \[ -\frac {{\left (a^{2} + 2 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} - 2 \, {\left (a^{2} d x + a^{2} + 2 \, a b + 2 \, b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + {\left (a^{2} + 2 \, a b + 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{2} + a^{2} + 2 \, a b}{2 \, d \cosh \left (d x + c\right ) \sinh \left (d x + c\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 65, normalized size = 1.81 \[ \frac {a^{2} d x - \frac {2 \, {\left (a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a b e^{\left (2 \, d x + 2 \, c\right )} + a^{2} + 2 \, a b + 2 \, b^{2}\right )}}{e^{\left (4 \, d x + 4 \, c\right )} - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.39, size = 64, normalized size = 1.78 \[ \frac {a^{2} \left (d x +c -\coth \left (d x +c \right )\right )-2 a b \coth \left (d x +c \right )+b^{2} \left (-\frac {1}{\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}-2 \tanh \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.55, size = 71, normalized size = 1.97 \[ a^{2} {\left (x + \frac {c}{d} + \frac {2}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}}\right )} + \frac {4 \, a b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} + \frac {4 \, b^{2}}{d {\left (e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.42, size = 60, normalized size = 1.67 \[ a^2\,x-\frac {\frac {2\,\left (a^2+2\,a\,b+2\,b^2\right )}{d}+\frac {2\,a\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (a+2\,b\right )}{d}}{{\mathrm {e}}^{4\,c+4\,d\,x}-1} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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