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

Optimal. Leaf size=34 \[ -\frac{(a+b) \coth ^3(c+d x)}{3 d}-\frac{a \coth (c+d x)}{d}+a x \]

[Out]

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

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Rubi [A]  time = 0.0628673, antiderivative size = 34, 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 = {4141, 1802, 207} \[ -\frac{(a+b) \coth ^3(c+d x)}{3 d}-\frac{a \coth (c+d x)}{d}+a x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [C]  time = 0.0226054, size = 49, normalized size = 1.44 \[ -\frac{a \coth ^3(c+d x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\tanh ^2(c+d x)\right )}{3 d}-\frac{b \coth ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*Coth[c + d*x]^3)/(3*d) - (a*Coth[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, Tanh[c + d*x]^2])/(3*d)

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Maple [B]  time = 0.04, size = 70, normalized size = 2.1 \begin{align*}{\frac{1}{d} \left ( a \left ( dx+c-{\rm coth} \left (dx+c\right )-{\frac{ \left ({\rm coth} \left (dx+c\right ) \right ) ^{3}}{3}} \right ) +b \left ( -{\frac{\cosh \left ( dx+c \right ) }{2\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}}-{\frac{{\rm coth} \left (dx+c\right )}{2} \left ({\frac{2}{3}}-{\frac{ \left ({\rm csch} \left (dx+c\right ) \right ) ^{2}}{3}} \right ) } \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),x)

[Out]

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

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Maxima [B]  time = 1.15464, size = 230, normalized size = 6.76 \begin{align*} \frac{1}{3} \, a{\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{2}{3} \, 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),x, algorithm="maxima")

[Out]

1/3*a*(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))) + 2/3*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 = 2.04397, size = 370, normalized size = 10.88 \begin{align*} -\frac{{\left (4 \, a + b\right )} \cosh \left (d x + c\right )^{3} + 3 \,{\left (4 \, a + b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} -{\left (3 \, a d x + 4 \, a + b\right )} \sinh \left (d x + c\right )^{3} + 3 \, b \cosh \left (d x + c\right ) + 3 \,{\left (3 \, a d x -{\left (3 \, a d x + 4 \, a + b\right )} \cosh \left (d x + c\right )^{2} + 4 \, a + b\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),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.24198, size = 90, normalized size = 2.65 \begin{align*} \frac{3 \, a d x - \frac{2 \,{\left (6 \, a e^{\left (4 \, d x + 4 \, c\right )} + 3 \, b e^{\left (4 \, d x + 4 \, c\right )} - 6 \, a e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a + b\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),x, algorithm="giac")

[Out]

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