3.120 \(\int \coth ^4(c+d x) (a+b \text{sech}^2(c+d x))^2 \, dx\)

Optimal. Leaf size=46 \[ -\frac{\left (a^2-b^2\right ) \coth (c+d x)}{d}+a^2 x-\frac{(a+b)^2 \coth ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0917147, antiderivative size = 46, 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 = {4141, 1802, 207} \[ -\frac{\left (a^2-b^2\right ) \coth (c+d x)}{d}+a^2 x-\frac{(a+b)^2 \coth ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Coth[c + d*x]^4*(a + b*Sech[c + d*x]^2)^2,x]

[Out]

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

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])

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

Mathematica [B]  time = 0.818269, size = 160, normalized size = 3.48 \[ \frac{\text{csch}(c) \text{csch}^3(c+d x) \left (-12 a^2 \sinh (2 c+d x)+8 a^2 \sinh (2 c+3 d x)-9 a^2 d x \cosh (2 c+d x)-3 a^2 d x \cosh (2 c+3 d x)+3 a^2 d x \cosh (4 c+3 d x)-12 a^2 \sinh (d x)+9 a^2 d x \cosh (d x)-12 a b \sinh (2 c+d x)+4 a b \sinh (2 c+3 d x)-4 b^2 \sinh (2 c+3 d x)+12 b^2 \sinh (d x)\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Coth[c + d*x]^4*(a + b*Sech[c + d*x]^2)^2,x]

[Out]

(Csch[c]*Csch[c + d*x]^3*(9*a^2*d*x*Cosh[d*x] - 9*a^2*d*x*Cosh[2*c + d*x] - 3*a^2*d*x*Cosh[2*c + 3*d*x] + 3*a^
2*d*x*Cosh[4*c + 3*d*x] - 12*a^2*Sinh[d*x] + 12*b^2*Sinh[d*x] - 12*a^2*Sinh[2*c + d*x] - 12*a*b*Sinh[2*c + d*x
] + 8*a^2*Sinh[2*c + 3*d*x] + 4*a*b*Sinh[2*c + 3*d*x] - 4*b^2*Sinh[2*c + 3*d*x]))/(24*d)

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Maple [B]  time = 0.04, size = 96, normalized size = 2.1 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( dx+c-{\rm coth} \left (dx+c\right )-{\frac{ \left ({\rm coth} \left (dx+c\right ) \right ) ^{3}}{3}} \right ) +2\,ab \left ( -1/2\,{\frac{\cosh \left ( dx+c \right ) }{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}}-1/2\, \left ( 2/3-1/3\, \left ({\rm csch} \left (dx+c\right ) \right ) ^{2} \right ){\rm coth} \left (dx+c\right ) \right ) +{b}^{2} \left ({\frac{2}{3}}-{\frac{ \left ({\rm csch} \left (dx+c\right ) \right ) ^{2}}{3}} \right ){\rm coth} \left (dx+c\right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(coth(d*x+c)^4*(a+b*sech(d*x+c)^2)^2,x)

[Out]

1/d*(a^2*(d*x+c-coth(d*x+c)-1/3*coth(d*x+c)^3)+2*a*b*(-1/2/sinh(d*x+c)^3*cosh(d*x+c)-1/2*(2/3-1/3*csch(d*x+c)^
2)*coth(d*x+c))+b^2*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c))

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Maxima [B]  time = 1.16095, size = 362, normalized size = 7.87 \begin{align*} \frac{1}{3} \, a^{2}{\left (3 \, x + \frac{3 \, c}{d} - \frac{4 \,{\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} - 2\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 )} + \frac{4}{3} \, b^{2}{\left (\frac{3 \, e^{\left (-2 \, d x - 2 \, c\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 )}} - \frac{1}{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 )} + \frac{4}{3} \, a b{\left (\frac{3 \, e^{\left (-4 \, d x - 4 \, c\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 )}} + \frac{1}{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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^4*(a+b*sech(d*x+c)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.06739, size = 478, normalized size = 10.39 \begin{align*} -\frac{2 \,{\left (2 \, a^{2} + a b - b^{2}\right )} \cosh \left (d x + c\right )^{3} + 6 \,{\left (2 \, a^{2} + a b - b^{2}\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} -{\left (3 \, a^{2} d x + 4 \, a^{2} + 2 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )^{3} + 6 \,{\left (a b + b^{2}\right )} \cosh \left (d x + c\right ) + 3 \,{\left (3 \, a^{2} d x -{\left (3 \, a^{2} d x + 4 \, a^{2} + 2 \, a b - 2 \, b^{2}\right )} \cosh \left (d x + c\right )^{2} + 4 \, a^{2} + 2 \, a b - 2 \, b^{2}\right )} \sinh \left (d x + c\right )}{3 \,{\left (d \sinh \left (d x + c\right )^{3} + 3 \,{\left (d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^4*(a+b*sech(d*x+c)^2)^2,x, algorithm="fricas")

[Out]

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

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

[Out]

Timed out

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Giac [B]  time = 1.32449, size = 131, normalized size = 2.85 \begin{align*} \frac{3 \, a^{2} d x - \frac{4 \,{\left (3 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 3 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 3 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a^{2} + a b - b^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(coth(d*x+c)^4*(a+b*sech(d*x+c)^2)^2,x, algorithm="giac")

[Out]

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