3.24 \(\int \text{csch}^4(c+d x) (a+b \tanh ^2(c+d x))^3 \, dx\)

Optimal. Leaf size=98 \[ \frac{a^2 (a-3 b) \coth (c+d x)}{d}-\frac{a^3 \coth ^3(c+d x)}{3 d}-\frac{b^2 (3 a-b) \tanh ^3(c+d x)}{3 d}-\frac{3 a b (a-b) \tanh (c+d x)}{d}-\frac{b^3 \tanh ^5(c+d x)}{5 d} \]

[Out]

(a^2*(a - 3*b)*Coth[c + d*x])/d - (a^3*Coth[c + d*x]^3)/(3*d) - (3*a*(a - b)*b*Tanh[c + d*x])/d - ((3*a - b)*b
^2*Tanh[c + d*x]^3)/(3*d) - (b^3*Tanh[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.0965718, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.087, Rules used = {3663, 448} \[ \frac{a^2 (a-3 b) \coth (c+d x)}{d}-\frac{a^3 \coth ^3(c+d x)}{3 d}-\frac{b^2 (3 a-b) \tanh ^3(c+d x)}{3 d}-\frac{3 a b (a-b) \tanh (c+d x)}{d}-\frac{b^3 \tanh ^5(c+d x)}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Csch[c + d*x]^4*(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

(a^2*(a - 3*b)*Coth[c + d*x])/d - (a^3*Coth[c + d*x]^3)/(3*d) - (3*a*(a - b)*b*Tanh[c + d*x])/d - ((3*a - b)*b
^2*Tanh[c + d*x]^3)/(3*d) - (b^3*Tanh[c + d*x]^5)/(5*d)

Rule 3663

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

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

\begin{align*} \int \text{csch}^4(c+d x) \left (a+b \tanh ^2(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right ) \left (a+b x^2\right )^3}{x^4} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (-3 a (a-b) b+\frac{a^3}{x^4}-\frac{a^2 (a-3 b)}{x^2}-(3 a-b) b^2 x^2-b^3 x^4\right ) \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{a^2 (a-3 b) \coth (c+d x)}{d}-\frac{a^3 \coth ^3(c+d x)}{3 d}-\frac{3 a (a-b) b \tanh (c+d x)}{d}-\frac{(3 a-b) b^2 \tanh ^3(c+d x)}{3 d}-\frac{b^3 \tanh ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.21802, size = 87, normalized size = 0.89 \[ \frac{b \tanh (c+d x) \left (-45 a^2+b (15 a+b) \text{sech}^2(c+d x)+30 a b-3 b^2 \text{sech}^4(c+d x)+2 b^2\right )-5 a^2 \coth (c+d x) \left (a \text{csch}^2(c+d x)-2 a+9 b\right )}{15 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Csch[c + d*x]^4*(a + b*Tanh[c + d*x]^2)^3,x]

[Out]

(-5*a^2*Coth[c + d*x]*(-2*a + 9*b + a*Csch[c + d*x]^2) + b*(-45*a^2 + 30*a*b + 2*b^2 + b*(15*a + b)*Sech[c + d
*x]^2 - 3*b^2*Sech[c + d*x]^4)*Tanh[c + d*x])/(15*d)

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Maple [A]  time = 0.058, size = 136, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{2}{3}}-{\frac{ \left ({\rm csch} \left (dx+c\right ) \right ) ^{2}}{3}} \right ){\rm coth} \left (dx+c\right )+3\,{a}^{2}b \left ( -{\frac{1}{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }}-2\,\tanh \left ( dx+c \right ) \right ) +3\,a{b}^{2} \left ( 2/3+1/3\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2} \right ) \tanh \left ( dx+c \right ) +{b}^{3} \left ( -{\frac{\sinh \left ( dx+c \right ) }{4\, \left ( \cosh \left ( dx+c \right ) \right ) ^{5}}}+{\frac{\tanh \left ( dx+c \right ) }{4} \left ({\frac{8}{15}}+{\frac{ \left ({\rm sech} \left (dx+c\right ) \right ) ^{4}}{5}}+{\frac{4\, \left ({\rm sech} \left (dx+c\right ) \right ) ^{2}}{15}} \right ) } \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csch(d*x+c)^4*(a+b*tanh(d*x+c)^2)^3,x)

[Out]

1/d*(a^3*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c)+3*a^2*b*(-1/sinh(d*x+c)/cosh(d*x+c)-2*tanh(d*x+c))+3*a*b^2*(2/3+1
/3*sech(d*x+c)^2)*tanh(d*x+c)+b^3*(-1/4*sinh(d*x+c)/cosh(d*x+c)^5+1/4*(8/15+1/5*sech(d*x+c)^4+4/15*sech(d*x+c)
^2)*tanh(d*x+c)))

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Maxima [B]  time = 1.15839, size = 666, normalized size = 6.8 \begin{align*} \frac{4}{15} \, b^{3}{\left (\frac{5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} - \frac{5 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{15 \, e^{\left (-6 \, d x - 6 \, c\right )}}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac{1}{d{\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + 4 \, a 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^{3}{\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{12 \, a^{2} b}{d{\left (e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4*(a+b*tanh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

4/15*b^3*(5*e^(-2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x
- 8*c) + e^(-10*d*x - 10*c) + 1)) - 5*e^(-4*d*x - 4*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6
*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 15*e^(-6*d*x - 6*c)/(d*(5*e^(-2*d*x - 2*c) + 10*
e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1)) + 1/(d*(5*e^(-2*d*x - 2
*c) + 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1))) + 4*a*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^3*(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)
)) + 12*a^2*b/(d*(e^(-4*d*x - 4*c) - 1))

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Fricas [B]  time = 2.1397, size = 2385, normalized size = 24.34 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4*(a+b*tanh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

-8/15*((5*a^3 + 45*a^2*b + 15*a*b^2 + 7*b^3)*cosh(d*x + c)^6 + 12*(5*a^3 + 15*a*b^2 + 4*b^3)*cosh(d*x + c)*sin
h(d*x + c)^5 + (5*a^3 + 45*a^2*b + 15*a*b^2 + 7*b^3)*sinh(d*x + c)^6 + 2*(15*a^3 + 45*a^2*b - 15*a*b^2 - 13*b^
3)*cosh(d*x + c)^4 + (30*a^3 + 90*a^2*b - 30*a*b^2 - 26*b^3 + 15*(5*a^3 + 45*a^2*b + 15*a*b^2 + 7*b^3)*cosh(d*
x + c)^2)*sinh(d*x + c)^4 + 8*(5*(5*a^3 + 15*a*b^2 + 4*b^3)*cosh(d*x + c)^3 + 4*(5*a^3 - 3*b^3)*cosh(d*x + c))
*sinh(d*x + c)^3 + 50*a^3 - 90*a^2*b + 30*a*b^2 - 22*b^3 + (75*a^3 - 45*a^2*b - 15*a*b^2 + 41*b^3)*cosh(d*x +
c)^2 + (15*(5*a^3 + 45*a^2*b + 15*a*b^2 + 7*b^3)*cosh(d*x + c)^4 + 75*a^3 - 45*a^2*b - 15*a*b^2 + 41*b^3 + 12*
(15*a^3 + 45*a^2*b - 15*a*b^2 - 13*b^3)*cosh(d*x + c)^2)*sinh(d*x + c)^2 + 4*(3*(5*a^3 + 15*a*b^2 + 4*b^3)*cos
h(d*x + c)^5 + 8*(5*a^3 - 3*b^3)*cosh(d*x + c)^3 + (25*a^3 - 45*a*b^2 + 12*b^3)*cosh(d*x + c))*sinh(d*x + c))/
(d*cosh(d*x + c)^10 + 10*d*cosh(d*x + c)*sinh(d*x + c)^9 + d*sinh(d*x + c)^10 + 2*d*cosh(d*x + c)^8 + (45*d*co
sh(d*x + c)^2 + 2*d)*sinh(d*x + c)^8 + 8*(15*d*cosh(d*x + c)^3 + 2*d*cosh(d*x + c))*sinh(d*x + c)^7 - 3*d*cosh
(d*x + c)^6 + (210*d*cosh(d*x + c)^4 + 56*d*cosh(d*x + c)^2 - 3*d)*sinh(d*x + c)^6 + 2*(126*d*cosh(d*x + c)^5
+ 56*d*cosh(d*x + c)^3 - 3*d*cosh(d*x + c))*sinh(d*x + c)^5 - 8*d*cosh(d*x + c)^4 + (210*d*cosh(d*x + c)^6 + 1
40*d*cosh(d*x + c)^4 - 45*d*cosh(d*x + c)^2 - 8*d)*sinh(d*x + c)^4 + 4*(30*d*cosh(d*x + c)^7 + 28*d*cosh(d*x +
c)^5 - 5*d*cosh(d*x + c)^3 - 4*d*cosh(d*x + c))*sinh(d*x + c)^3 + 2*d*cosh(d*x + c)^2 + (45*d*cosh(d*x + c)^8
+ 56*d*cosh(d*x + c)^6 - 45*d*cosh(d*x + c)^4 - 48*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c)^2 + 2*(5*d*cosh(d*x
+ c)^9 + 8*d*cosh(d*x + c)^7 - 3*d*cosh(d*x + c)^5 - 8*d*cosh(d*x + c)^3 - 2*d*cosh(d*x + c))*sinh(d*x + c) +
6*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tanh ^{2}{\left (c + d x \right )}\right )^{3} \operatorname{csch}^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)**4*(a+b*tanh(d*x+c)**2)**3,x)

[Out]

Integral((a + b*tanh(c + d*x)**2)**3*csch(c + d*x)**4, x)

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Giac [B]  time = 1.78319, size = 347, normalized size = 3.54 \begin{align*} -\frac{2 \,{\left (\frac{5 \,{\left (9 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} + 6 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - 18 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 2 \, a^{3} + 9 \, a^{2} b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}} - \frac{45 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 180 \, a^{2} b e^{\left (6 \, d x + 6 \, c\right )} - 90 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 30 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 270 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 210 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 10 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 180 \, a^{2} b e^{\left (2 \, d x + 2 \, c\right )} - 150 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 10 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, a^{2} b - 30 \, a b^{2} - 2 \, b^{3}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}\right )}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csch(d*x+c)^4*(a+b*tanh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

-2/15*(5*(9*a^2*b*e^(4*d*x + 4*c) + 6*a^3*e^(2*d*x + 2*c) - 18*a^2*b*e^(2*d*x + 2*c) - 2*a^3 + 9*a^2*b)/(e^(2*
d*x + 2*c) - 1)^3 - (45*a^2*b*e^(8*d*x + 8*c) + 180*a^2*b*e^(6*d*x + 6*c) - 90*a*b^2*e^(6*d*x + 6*c) - 30*b^3*
e^(6*d*x + 6*c) + 270*a^2*b*e^(4*d*x + 4*c) - 210*a*b^2*e^(4*d*x + 4*c) + 10*b^3*e^(4*d*x + 4*c) + 180*a^2*b*e
^(2*d*x + 2*c) - 150*a*b^2*e^(2*d*x + 2*c) - 10*b^3*e^(2*d*x + 2*c) + 45*a^2*b - 30*a*b^2 - 2*b^3)/(e^(2*d*x +
2*c) + 1)^5)/d