3.142 \(\int \coth ^4(c+d x) (a+b \tanh ^2(c+d x)) \, dx\)

Optimal. Leaf size=36 \[ -\frac{(a+b) \coth (c+d x)}{d}+x (a+b)-\frac{a \coth ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.038836, antiderivative size = 36, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3629, 12, 3473, 8} \[ -\frac{(a+b) \coth (c+d x)}{d}+x (a+b)-\frac{a \coth ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3629

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[
((A*b^2 + a^2*C)*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*
Tan[e + f*x])^(m + 1)*Simp[a*(A - C) - (A*b - b*C)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C}, x] &&
NeQ[A*b^2 + a^2*C, 0] && LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [C]  time = 0.0360232, size = 61, normalized size = 1.69 \[ -\frac{a \coth ^3(c+d x) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\tanh ^2(c+d x)\right )}{3 d}-\frac{b \coth (c+d x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\tanh ^2(c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Coth[c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, Tanh[c + d*x]^2])/(3*d) - (b*Coth[c + d*x]*Hypergeometric
2F1[-1/2, 1, 1/2, Tanh[c + d*x]^2])/d

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Maple [A]  time = 0.042, size = 46, normalized size = 1.3 \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 ( dx+c-{\rm coth} \left (dx+c\right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(d*x+c-coth(d*x+c)-1/3*coth(d*x+c)^3)+b*(d*x+c-coth(d*x+c)))

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Maxima [B]  time = 1.17206, size = 142, normalized size = 3.94 \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 )} + b{\left (x + \frac{c}{d} + \frac{2}{d{\left (e^{\left (-2 \, d x - 2 \, 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*tanh(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))) + b*(x + c/d + 2/(d*(e^(-2*d*x - 2*c) - 1)))

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Fricas [B]  time = 1.91492, size = 408, normalized size = 11.33 \begin{align*} -\frac{{\left (4 \, a + 3 \, b\right )} \cosh \left (d x + c\right )^{3} + 3 \,{\left (4 \, a + 3 \, b\right )} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{2} -{\left (3 \,{\left (a + b\right )} d x + 4 \, a + 3 \, b\right )} \sinh \left (d x + c\right )^{3} - 3 \, b \cosh \left (d x + c\right ) + 3 \,{\left (3 \,{\left (a + b\right )} d x -{\left (3 \,{\left (a + b\right )} d x + 4 \, a + 3 \, b\right )} \cosh \left (d x + c\right )^{2} + 4 \, a + 3 \, 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*tanh(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.19785, size = 116, normalized size = 3.22 \begin{align*} \frac{{\left (d x + c\right )}{\left (a + b\right )}}{d} - \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 )} - 6 \, b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a + 3 \, b\right )}}{3 \, d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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