3.1.25 \(\int \text {csch}^2(c+d x) (a+b \sinh ^2(c+d x))^3 \, dx\) [25]

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

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 137, normalized size of antiderivative = 1.00, 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 = {3266, 479, 591, 464, 212} \begin {gather*} \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} \end {gather*}

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 212

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

Rule 464

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 479

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)),
Int[(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 591

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] + Dis
t[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 3266

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]

Rubi steps

\begin {align*} \int \text {csch}^2(c+d x) \left (a+b \sinh ^2(c+d x)\right )^3 \, dx &=\frac {\text {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 {\text {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 {\text {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 ) \text {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*}

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Mathematica [A]
time = 1.24, size = 113, normalized size = 0.82 \begin {gather*} \frac {\left (b+a \text {csch}^2(c+d x)\right )^3 \sinh ^6(c+d x) \left (12 b \left (8 a^2-4 a b+b^2\right ) (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))\right )}{4 d (2 a-b+b \cosh (2 (c+d x)))^3} \end {gather*}

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 = 1.14, size = 147, normalized size = 1.07

method	result	size
risch	\(3 a^{2} b x -\frac {3 a \,b^{2} x}{2}+\frac {3 b^{3} x}{8}+\frac {{\mathrm e}^{4 d x +4 c} b^{3}}{64 d}+\frac {3 \,{\mathrm e}^{2 d x +2 c} a \,b^{2}}{8 d}-\frac {{\mathrm e}^{2 d x +2 c} b^{3}}{8 d}-\frac {3 \,{\mathrm e}^{-2 d x -2 c} a \,b^{2}}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} b^{3}}{8 d}-\frac {{\mathrm e}^{-4 d x -4 c} b^{3}}{64 d}-\frac {2 a^{3}}{d \left ({\mathrm e}^{2 d x +2 c}-1\right )}\)	\(147\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.28, size = 130, normalized size = 0.95 \begin {gather*} \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 {gather*}

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 = 0.40, size = 169, normalized size = 1.23 \begin {gather*} \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 {gather*}

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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 4369 deep

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Giac [A]
time = 0.46, size = 177, normalized size = 1.29 \begin {gather*} \frac {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 )} + 24 \, {\left (8 \, a^{2} b - 4 \, a b^{2} + b^{3}\right )} {\left (d x + c\right )} - \frac {128 \, a^{3}}{e^{\left (2 \, d x + 2 \, c\right )} - 1} - {\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} \end {gather*}

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]

1/64*(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) + 24*(8*a^2*b - 4*a*b^2 + b^3)*(d
*x + c) - 128*a^3/(e^(2*d*x + 2*c) - 1) - (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

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Mupad [B]
time = 0.76, size = 121, normalized size = 0.88 \begin {gather*} \frac {3\,b\,x\,\left (8\,a^2-4\,a\,b+b^2\right )}{8}-\frac {2\,a^3}{d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )}-\frac {b^3\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d}+\frac {b^3\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d}-\frac {b^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}\,\left (3\,a-b\right )}{8\,d}+\frac {b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (3\,a-b\right )}{8\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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