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

Optimal. Leaf size=113 \[ \frac{a \left (2 a^2-5 a b-2 b^2\right ) \coth (c+d x)}{2 d}-\frac{a^2 (2 a+3 b) \coth ^3(c+d x)}{6 d}+\frac{1}{2} b^2 x (6 a-b)+\frac{b \cosh ^2(c+d x) \coth ^3(c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{2 d} \]

[Out]

((6*a - b)*b^2*x)/2 + (a*(2*a^2 - 5*a*b - 2*b^2)*Coth[c + d*x])/(2*d) - (a^2*(2*a + 3*b)*Coth[c + d*x]^3)/(6*d
) + (b*Cosh[c + d*x]^2*Coth[c + d*x]^3*(a - (a - b)*Tanh[c + d*x]^2)^2)/(2*d)

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Rubi [A]  time = 0.141387, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3187, 468, 570, 207} \[ \frac{a \left (2 a^2-5 a b-2 b^2\right ) \coth (c+d x)}{2 d}-\frac{a^2 (2 a+3 b) \coth ^3(c+d x)}{6 d}+\frac{1}{2} b^2 x (6 a-b)+\frac{b \cosh ^2(c+d x) \coth ^3(c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*a - b)*b^2*x)/2 + (a*(2*a^2 - 5*a*b - 2*b^2)*Coth[c + d*x])/(2*d) - (a^2*(2*a + 3*b)*Coth[c + d*x]^3)/(6*d
) + (b*Cosh[c + d*x]^2*Coth[c + d*x]^3*(a - (a - b)*Tanh[c + d*x]^2)^2)/(2*d)

Rule 3187

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

Rule 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[((c*b -
a*d)*(e*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1))/(a*b*e*n*(p + 1)), x] + Dist[1/(a*b*n*(p + 1)), I
nt[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c*b - a*d)*(m + 1)) + d*(c*b*n*(p
+ 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 570

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^
(r_.), x_Symbol] :> Int[ExpandIntegrand[(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a,
b, c, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]

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 \text{csch}^4(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-(a-b) x^2\right )^3}{x^4 \left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b \cosh ^2(c+d x) \coth ^3(c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{2 d}+\frac{\operatorname{Subst}\left (\int \frac{\left (a (2 a+3 b)-(a-b) (2 a-b) x^2\right ) \left (a+(-a+b) x^2\right )}{x^4 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac{b \cosh ^2(c+d x) \coth ^3(c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{2 d}+\frac{\operatorname{Subst}\left (\int \left (\frac{a^2 (2 a+3 b)}{x^4}-\frac{a \left (2 a^2-5 a b-2 b^2\right )}{x^2}+\frac{b^2 (-6 a+b)}{-1+x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac{a \left (2 a^2-5 a b-2 b^2\right ) \coth (c+d x)}{2 d}-\frac{a^2 (2 a+3 b) \coth ^3(c+d x)}{6 d}+\frac{b \cosh ^2(c+d x) \coth ^3(c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{2 d}-\frac{\left ((6 a-b) b^2\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\tanh (c+d x)\right )}{2 d}\\ &=\frac{1}{2} (6 a-b) b^2 x+\frac{a \left (2 a^2-5 a b-2 b^2\right ) \coth (c+d x)}{2 d}-\frac{a^2 (2 a+3 b) \coth ^3(c+d x)}{6 d}+\frac{b \cosh ^2(c+d x) \coth ^3(c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{2 d}\\ \end{align*}

Mathematica [A]  time = 2.5041, size = 107, normalized size = 0.95 \[ \frac{2 \sinh ^6(c+d x) \left (a \text{csch}^2(c+d x)+b\right )^3 \left (3 b^2 (2 (6 a-b) (c+d x)+b \sinh (2 (c+d x)))-4 a^2 \coth (c+d x) \left (a \text{csch}^2(c+d x)-2 a+9 b\right )\right )}{3 d (2 a+b \cosh (2 (c+d x))-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b + a*Csch[c + d*x]^2)^3*Sinh[c + d*x]^6*(-4*a^2*Coth[c + d*x]*(-2*a + 9*b + a*Csch[c + d*x]^2) + 3*b^2*(2
*(6*a - b)*(c + d*x) + b*Sinh[2*(c + d*x)])))/(3*d*(2*a - b + b*Cosh[2*(c + d*x)])^3)

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Maple [A]  time = 0.039, size = 77, normalized size = 0.7 \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{\rm coth} \left (dx+c\right )+3\,a{b}^{2} \left ( dx+c \right ) +{b}^{3} \left ({\frac{\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) }{2}}-{\frac{dx}{2}}-{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]  time = 1.03046, size = 217, normalized size = 1.92 \begin{align*} -\frac{1}{8} \, b^{3}{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} + 3 \, a b^{2} x + \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{6 \, a^{2} b}{d{\left (e^{\left (-2 \, d x - 2 \, 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*sinh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.7636, size = 668, normalized size = 5.91 \begin{align*} \frac{3 \, b^{3} \cosh \left (d x + c\right )^{5} + 15 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} +{\left (16 \, a^{3} - 72 \, a^{2} b - 9 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} - 4 \,{\left (4 \, a^{3} - 18 \, a^{2} b - 3 \,{\left (6 \, a b^{2} - b^{3}\right )} d x\right )} \sinh \left (d x + c\right )^{3} + 3 \,{\left (10 \, b^{3} \cosh \left (d x + c\right )^{3} +{\left (16 \, a^{3} - 72 \, a^{2} b - 9 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 6 \,{\left (8 \, a^{3} - 12 \, a^{2} b - b^{3}\right )} \cosh \left (d x + c\right ) + 12 \,{\left (4 \, a^{3} - 18 \, a^{2} b - 3 \,{\left (6 \, a b^{2} - b^{3}\right )} d x -{\left (4 \, a^{3} - 18 \, a^{2} b - 3 \,{\left (6 \, a b^{2} - b^{3}\right )} d x\right )} \cosh \left (d x + c\right )^{2}\right )} \sinh \left (d x + c\right )}{24 \,{\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(csch(d*x+c)^4*(a+b*sinh(d*x+c)^2)^3,x, algorithm="fricas")

[Out]

1/24*(3*b^3*cosh(d*x + c)^5 + 15*b^3*cosh(d*x + c)*sinh(d*x + c)^4 + (16*a^3 - 72*a^2*b - 9*b^3)*cosh(d*x + c)
^3 - 4*(4*a^3 - 18*a^2*b - 3*(6*a*b^2 - b^3)*d*x)*sinh(d*x + c)^3 + 3*(10*b^3*cosh(d*x + c)^3 + (16*a^3 - 72*a
^2*b - 9*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 - 6*(8*a^3 - 12*a^2*b - b^3)*cosh(d*x + c) + 12*(4*a^3 - 18*a^2*b
- 3*(6*a*b^2 - b^3)*d*x - (4*a^3 - 18*a^2*b - 3*(6*a*b^2 - b^3)*d*x)*cosh(d*x + c)^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(csch(d*x+c)**4*(a+b*sinh(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.36781, size = 217, normalized size = 1.92 \begin{align*} \frac{b^{3} e^{\left (2 \, d x + 2 \, c\right )}}{8 \, d} + \frac{{\left (6 \, a b^{2} - b^{3}\right )}{\left (d x + c\right )}}{2 \, d} - \frac{{\left (12 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 2 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + b^{3}\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, d} - \frac{2 \,{\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 )}}{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(csch(d*x+c)^4*(a+b*sinh(d*x+c)^2)^3,x, algorithm="giac")

[Out]

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