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

Optimal. Leaf size=91 \[ -\frac{a^2 \coth ^3(c+d x)}{3 d}+\frac{a^2 \coth (c+d x)}{d}+\frac{1}{8} b x (16 a+3 b)+\frac{b^2 \sinh (c+d x) \cosh ^3(c+d x)}{4 d}-\frac{5 b^2 \sinh (c+d x) \cosh (c+d x)}{8 d} \]

[Out]

(b*(16*a + 3*b)*x)/8 + (a^2*Coth[c + d*x])/d - (a^2*Coth[c + d*x]^3)/(3*d) - (5*b^2*Cosh[c + d*x]*Sinh[c + d*x
])/(8*d) + (b^2*Cosh[c + d*x]^3*Sinh[c + d*x])/(4*d)

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Rubi [A]  time = 0.164625, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3217, 1259, 1805, 1261, 207} \[ -\frac{a^2 \coth ^3(c+d x)}{3 d}+\frac{a^2 \coth (c+d x)}{d}+\frac{1}{8} b x (16 a+3 b)+\frac{b^2 \sinh (c+d x) \cosh ^3(c+d x)}{4 d}-\frac{5 b^2 \sinh (c+d x) \cosh (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(16*a + 3*b)*x)/8 + (a^2*Coth[c + d*x])/d - (a^2*Coth[c + d*x]^3)/(3*d) - (5*b^2*Cosh[c + d*x]*Sinh[c + d*x
])/(8*d) + (b^2*Cosh[c + d*x]^3*Sinh[c + d*x])/(4*d)

Rule 3217

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p)/(1 + ff^2
*x^2)^(m/2 + 2*p + 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 1259

Int[(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[((-d)^(
m/2 - 1)*(c*d^2 - b*d*e + a*e^2)^p*x*(d + e*x^2)^(q + 1))/(2*e^(2*p + m/2)*(q + 1)), x] + Dist[(-d)^(m/2 - 1)/
(2*e^(2*p)*(q + 1)), Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1*(2*(-d)^(-(m/2) + 1)*e^(2*p)*(q + 1)*
(a + b*x^2 + c*x^4)^p - ((c*d^2 - b*d*e + a*e^2)^p/(e^(m/2)*x^m))*(d + e*(2*q + 3)*x^2)))/(d + e*x^2)], x], x]
, x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]

Rule 1805

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[(c*x)^m*Pq,
a + b*x^2, x], f = Coeff[PolynomialRemainder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[
(c*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[((a*g - b*f*x)*(a + b*x^2)^(p + 1))/(2*a*b*(p + 1)), x] + Dist[1/(2*a*
(p + 1)), Int[(c*x)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x],
x]] /; FreeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]

Rule 1261

Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> In
t[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] &&
NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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 ^4(c+d x)\right )^2 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-2 a x^2+(a+b) x^4\right )^2}{x^4 \left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b^2 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}+\frac{\operatorname{Subst}\left (\int \frac{4 a^2-12 a^2 x^2+\left (12 a^2+8 a b-b^2\right ) x^4-4 (a+b)^2 x^6}{x^4 \left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{4 d}\\ &=-\frac{5 b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{b^2 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac{\operatorname{Subst}\left (\int \frac{-8 a^2+16 a^2 x^2+\left (-8 a^2-16 a b-3 b^2\right ) x^4}{x^4 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{5 b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{b^2 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac{\operatorname{Subst}\left (\int \left (-\frac{8 a^2}{x^4}+\frac{8 a^2}{x^2}+\frac{b (16 a+3 b)}{-1+x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac{a^2 \coth (c+d x)}{d}-\frac{a^2 \coth ^3(c+d x)}{3 d}-\frac{5 b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{b^2 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}-\frac{(b (16 a+3 b)) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=\frac{1}{8} b (16 a+3 b) x+\frac{a^2 \coth (c+d x)}{d}-\frac{a^2 \coth ^3(c+d x)}{3 d}-\frac{5 b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac{b^2 \cosh ^3(c+d x) \sinh (c+d x)}{4 d}\\ \end{align*}

Mathematica [A]  time = 0.324689, size = 68, normalized size = 0.75 \[ \frac{3 b (64 a d x-8 b \sinh (2 (c+d x))+b \sinh (4 (c+d x))+12 b c+12 b d x)-32 a^2 \coth (c+d x) \left (\text{csch}^2(c+d x)-2\right )}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-32*a^2*Coth[c + d*x]*(-2 + Csch[c + d*x]^2) + 3*b*(12*b*c + 64*a*d*x + 12*b*d*x - 8*b*Sinh[2*(c + d*x)] + b*
Sinh[4*(c + d*x)]))/(96*d)

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Maple [A]  time = 0.044, size = 75, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{2}{3}}-{\frac{ \left ({\rm csch} \left (dx+c\right ) \right ) ^{2}}{3}} \right ){\rm coth} \left (dx+c\right )+2\,ab \left ( dx+c \right ) +{b}^{2} \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)^4*(a+b*sinh(d*x+c)^4)^2,x)

[Out]

1/d*(a^2*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c)+2*a*b*(d*x+c)+b^2*((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 = 1.04836, size = 223, normalized size = 2.45 \begin{align*} \frac{1}{64} \, b^{2}{\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 )} + 2 \, a b x + \frac{4}{3} \, a^{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 )} \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)^4)^2,x, algorithm="maxima")

[Out]

1/64*b^2*(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) + 2*a*b*
x + 4/3*a^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)))

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Fricas [B]  time = 1.74776, size = 761, normalized size = 8.36 \begin{align*} \frac{3 \, b^{2} \cosh \left (d x + c\right )^{7} + 21 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} - 33 \, b^{2} \cosh \left (d x + c\right )^{5} + 15 \,{\left (7 \, b^{2} \cosh \left (d x + c\right )^{3} - 11 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} +{\left (128 \, a^{2} + 81 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 8 \,{\left (3 \,{\left (16 \, a b + 3 \, b^{2}\right )} d x - 16 \, a^{2}\right )} \sinh \left (d x + c\right )^{3} + 3 \,{\left (21 \, b^{2} \cosh \left (d x + c\right )^{5} - 110 \, b^{2} \cosh \left (d x + c\right )^{3} +{\left (128 \, a^{2} + 81 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 3 \,{\left (128 \, a^{2} + 17 \, b^{2}\right )} \cosh \left (d x + c\right ) - 24 \,{\left (3 \,{\left (16 \, a b + 3 \, b^{2}\right )} d x -{\left (3 \,{\left (16 \, a b + 3 \, b^{2}\right )} d x - 16 \, a^{2}\right )} \cosh \left (d x + c\right )^{2} - 16 \, a^{2}\right )} \sinh \left (d x + c\right )}{192 \,{\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)^4)^2,x, algorithm="fricas")

[Out]

1/192*(3*b^2*cosh(d*x + c)^7 + 21*b^2*cosh(d*x + c)*sinh(d*x + c)^6 - 33*b^2*cosh(d*x + c)^5 + 15*(7*b^2*cosh(
d*x + c)^3 - 11*b^2*cosh(d*x + c))*sinh(d*x + c)^4 + (128*a^2 + 81*b^2)*cosh(d*x + c)^3 + 8*(3*(16*a*b + 3*b^2
)*d*x - 16*a^2)*sinh(d*x + c)^3 + 3*(21*b^2*cosh(d*x + c)^5 - 110*b^2*cosh(d*x + c)^3 + (128*a^2 + 81*b^2)*cos
h(d*x + c))*sinh(d*x + c)^2 - 3*(128*a^2 + 17*b^2)*cosh(d*x + c) - 24*(3*(16*a*b + 3*b^2)*d*x - (3*(16*a*b + 3
*b^2)*d*x - 16*a^2)*cosh(d*x + c)^2 - 16*a^2)*sinh(d*x + c))/(d*sinh(d*x + c)^3 + 3*(d*cosh(d*x + c)^2 - d)*si
nh(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)**4)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.31744, size = 207, normalized size = 2.27 \begin{align*} \frac{{\left (16 \, a b + 3 \, b^{2}\right )}{\left (d x + c\right )}}{8 \, d} - \frac{{\left (96 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 18 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )}}{64 \, d} + \frac{b^{2} d e^{\left (4 \, d x + 4 \, c\right )} - 8 \, b^{2} d e^{\left (2 \, d x + 2 \, c\right )}}{64 \, d^{2}} - \frac{4 \,{\left (3 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - a^{2}\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)^4)^2,x, algorithm="giac")

[Out]

1/8*(16*a*b + 3*b^2)*(d*x + c)/d - 1/64*(96*a*b*e^(4*d*x + 4*c) + 18*b^2*e^(4*d*x + 4*c) - 8*b^2*e^(2*d*x + 2*
c) + b^2)*e^(-4*d*x - 4*c)/d + 1/64*(b^2*d*e^(4*d*x + 4*c) - 8*b^2*d*e^(2*d*x + 2*c))/d^2 - 4/3*(3*a^2*e^(2*d*
x + 2*c) - a^2)/(d*(e^(2*d*x + 2*c) - 1)^3)