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

Optimal. Leaf size=137 \[ \frac{3}{8} b x \left (8 a^2-4 a b+b^2\right )-\frac{a (2 a+b) (4 a+b) \coth (c+d x)}{8 d}+\frac{b \cosh ^4(c+d x) \coth (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{4 d}+\frac{b \cosh ^2(c+d x) \coth (c+d x) \left (a (4 a+b)-(4 a-3 b) (a-b) \tanh ^2(c+d x)\right )}{8 d} \]

[Out]

(3*b*(8*a^2 - 4*a*b + b^2)*x)/8 - (a*(2*a + b)*(4*a + b)*Coth[c + d*x])/(8*d) + (b*Cosh[c + d*x]^4*Coth[c + d*
x]*(a - (a - b)*Tanh[c + d*x]^2)^2)/(4*d) + (b*Cosh[c + d*x]^2*Coth[c + d*x]*(a*(4*a + b) - (4*a - 3*b)*(a - b
)*Tanh[c + d*x]^2))/(8*d)

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Rubi [A]  time = 0.191937, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3187, 468, 577, 453, 206} \[ \frac{3}{8} b x \left (8 a^2-4 a b+b^2\right )-\frac{a (2 a+b) (4 a+b) \coth (c+d x)}{8 d}+\frac{b \cosh ^4(c+d x) \coth (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{4 d}+\frac{b \cosh ^2(c+d x) \coth (c+d x) \left (a (4 a+b)-(4 a-3 b) (a-b) \tanh ^2(c+d x)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*(8*a^2 - 4*a*b + b^2)*x)/8 - (a*(2*a + b)*(4*a + b)*Coth[c + d*x])/(8*d) + (b*Cosh[c + d*x]^4*Coth[c + d*
x]*(a - (a - b)*Tanh[c + d*x]^2)^2)/(4*d) + (b*Cosh[c + d*x]^2*Coth[c + d*x]*(a*(4*a + b) - (4*a - 3*b)*(a - b
)*Tanh[c + d*x]^2))/(8*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 577

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*b*g*n*(p + 1)), x] + Dist[
1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(m
+ 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] &&
IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \text{csch}^2(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^2 \left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b \cosh ^4(c+d x) \coth (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{\left (a (4 a+b)-(4 a-3 b) (a-b) x^2\right ) \left (a+(-a+b) x^2\right )}{x^2 \left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=\frac{b \cosh ^4(c+d x) \coth (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{4 d}+\frac{b \cosh ^2(c+d x) \coth (c+d x) \left (a (4 a+b)-(4 a-3 b) (a-b) \tanh ^2(c+d x)\right )}{8 d}+\frac{\operatorname{Subst}\left (\int \frac{a (2 a+b) (4 a+b)-(4 a-3 b) (a-b) (2 a-b) x^2}{x^2 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{a (2 a+b) (4 a+b) \coth (c+d x)}{8 d}+\frac{b \cosh ^4(c+d x) \coth (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{4 d}+\frac{b \cosh ^2(c+d x) \coth (c+d x) \left (a (4 a+b)-(4 a-3 b) (a-b) \tanh ^2(c+d x)\right )}{8 d}+\frac{\left (3 b \left (8 a^2-4 a b+b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac{3}{8} b \left (8 a^2-4 a b+b^2\right ) x-\frac{a (2 a+b) (4 a+b) \coth (c+d x)}{8 d}+\frac{b \cosh ^4(c+d x) \coth (c+d x) \left (a-(a-b) \tanh ^2(c+d x)\right )^2}{4 d}+\frac{b \cosh ^2(c+d x) \coth (c+d x) \left (a (4 a+b)-(4 a-3 b) (a-b) \tanh ^2(c+d x)\right )}{8 d}\\ \end{align*}

Mathematica [A]  time = 1.91726, size = 113, normalized size = 0.82 \[ \frac{\sinh ^6(c+d x) \left (a \text{csch}^2(c+d x)+b\right )^3 \left (12 b \left (8 a^2-4 a b+b^2\right ) (c+d x)-32 a^3 \coth (c+d x)+8 b^2 (3 a-b) \sinh (2 (c+d x))+b^3 \sinh (4 (c+d x))\right )}{4 d (2 a+b \cosh (2 (c+d x))-b)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b + a*Csch[c + d*x]^2)^3*Sinh[c + d*x]^6*(12*b*(8*a^2 - 4*a*b + b^2)*(c + d*x) - 32*a^3*Coth[c + d*x] + 8*(3
*a - b)*b^2*Sinh[2*(c + d*x)] + b^3*Sinh[4*(c + d*x)]))/(4*d*(2*a - b + b*Cosh[2*(c + d*x)])^3)

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Maple [A]  time = 0.03, size = 94, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ( -{a}^{3}{\rm coth} \left (dx+c\right )+3\,{a}^{2}b \left ( dx+c \right ) +3\,a{b}^{2} \left ( 1/2\,\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) -1/2\,dx-c/2 \right ) +{b}^{3} \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{4}}-{\frac{3\,\sinh \left ( dx+c \right ) }{8}} \right ) \cosh \left ( dx+c \right ) +{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.06488, size = 176, normalized size = 1.28 \begin{align*} \frac{1}{64} \, b^{3}{\left (24 \, x + \frac{e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac{8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac{e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} - \frac{3}{8} \, a b^{2}{\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^{2} b x + \frac{2 \, a^{3}}{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)^2*(a+b*sinh(d*x+c)^2)^3,x, algorithm="maxima")

[Out]

1/64*b^3*(24*x + e^(4*d*x + 4*c)/d - 8*e^(2*d*x + 2*c)/d + 8*e^(-2*d*x - 2*c)/d - e^(-4*d*x - 4*c)/d) - 3/8*a*
b^2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) + 3*a^2*b*x + 2*a^3/(d*(e^(-2*d*x - 2*c) - 1))

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Fricas [A]  time = 1.84056, size = 409, normalized size = 2.99 \begin{align*} \frac{b^{3} \cosh \left (d x + c\right )^{5} + 5 \, b^{3} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{4} + 3 \,{\left (8 \, a b^{2} - 3 \, b^{3}\right )} \cosh \left (d x + c\right )^{3} +{\left (10 \, b^{3} \cosh \left (d x + c\right )^{3} + 9 \,{\left (8 \, a b^{2} - 3 \, b^{3}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 8 \,{\left (8 \, a^{3} + 3 \, a b^{2} - b^{3}\right )} \cosh \left (d x + c\right ) + 8 \,{\left (8 \, a^{3} + 3 \,{\left (8 \, a^{2} b - 4 \, a b^{2} + b^{3}\right )} d x\right )} \sinh \left (d x + c\right )}{64 \, d \sinh \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/64*(b^3*cosh(d*x + c)^5 + 5*b^3*cosh(d*x + c)*sinh(d*x + c)^4 + 3*(8*a*b^2 - 3*b^3)*cosh(d*x + c)^3 + (10*b^
3*cosh(d*x + c)^3 + 9*(8*a*b^2 - 3*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 - 8*(8*a^3 + 3*a*b^2 - b^3)*cosh(d*x +
c) + 8*(8*a^3 + 3*(8*a^2*b - 4*a*b^2 + b^3)*d*x)*sinh(d*x + c))/(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)**2*(a+b*sinh(d*x+c)**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.36316, size = 257, normalized size = 1.88 \begin{align*} \frac{3 \,{\left (8 \, a^{2} b - 4 \, a b^{2} + b^{3}\right )}{\left (d x + c\right )}}{8 \, d} - \frac{2 \, a^{3}}{d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}} - \frac{{\left (144 \, a^{2} b e^{\left (4 \, d x + 4 \, c\right )} - 72 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 18 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} + 24 \, a b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 8 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + b^{3}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} + \frac{b^{3} d e^{\left (4 \, d x + 4 \, c\right )} + 24 \, a b^{2} d e^{\left (2 \, d x + 2 \, c\right )} - 8 \, b^{3} d e^{\left (2 \, d x + 2 \, c\right )}}{64 \, d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/8*(8*a^2*b - 4*a*b^2 + b^3)*(d*x + c)/d - 2*a^3/(d*(e^(2*d*x + 2*c) - 1)) - 1/64*(144*a^2*b*e^(4*d*x + 4*c)
- 72*a*b^2*e^(4*d*x + 4*c) + 18*b^3*e^(4*d*x + 4*c) + 24*a*b^2*e^(2*d*x + 2*c) - 8*b^3*e^(2*d*x + 2*c) + b^3)*
e^(-4*d*x - 4*c)/d + 1/64*(b^3*d*e^(4*d*x + 4*c) + 24*a*b^2*d*e^(2*d*x + 2*c) - 8*b^3*d*e^(2*d*x + 2*c))/d^2