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

Optimal. Leaf size=161 \[ \frac{1}{128} b x \left (384 a^2+144 a b+35 b^2\right )-\frac{a^3 \coth ^3(c+d x)}{3 d}+\frac{a^3 \coth (c+d x)}{d}+\frac{b^2 (144 a+163 b) \sinh (c+d x) \cosh ^3(c+d x)}{192 d}-\frac{3 b^2 (80 a+31 b) \sinh (c+d x) \cosh (c+d x)}{128 d}+\frac{b^3 \sinh (c+d x) \cosh ^7(c+d x)}{8 d}-\frac{25 b^3 \sinh (c+d x) \cosh ^5(c+d x)}{48 d} \]

[Out]

(b*(384*a^2 + 144*a*b + 35*b^2)*x)/128 + (a^3*Coth[c + d*x])/d - (a^3*Coth[c + d*x]^3)/(3*d) - (3*b^2*(80*a +
31*b)*Cosh[c + d*x]*Sinh[c + d*x])/(128*d) + (b^2*(144*a + 163*b)*Cosh[c + d*x]^3*Sinh[c + d*x])/(192*d) - (25
*b^3*Cosh[c + d*x]^5*Sinh[c + d*x])/(48*d) + (b^3*Cosh[c + d*x]^7*Sinh[c + d*x])/(8*d)

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Rubi [A]  time = 0.384284, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{1}{128} b x \left (384 a^2+144 a b+35 b^2\right )-\frac{a^3 \coth ^3(c+d x)}{3 d}+\frac{a^3 \coth (c+d x)}{d}+\frac{b^2 (144 a+163 b) \sinh (c+d x) \cosh ^3(c+d x)}{192 d}-\frac{3 b^2 (80 a+31 b) \sinh (c+d x) \cosh (c+d x)}{128 d}+\frac{b^3 \sinh (c+d x) \cosh ^7(c+d x)}{8 d}-\frac{25 b^3 \sinh (c+d x) \cosh ^5(c+d x)}{48 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*(384*a^2 + 144*a*b + 35*b^2)*x)/128 + (a^3*Coth[c + d*x])/d - (a^3*Coth[c + d*x]^3)/(3*d) - (3*b^2*(80*a +
31*b)*Cosh[c + d*x]*Sinh[c + d*x])/(128*d) + (b^2*(144*a + 163*b)*Cosh[c + d*x]^3*Sinh[c + d*x])/(192*d) - (25
*b^3*Cosh[c + d*x]^5*Sinh[c + d*x])/(48*d) + (b^3*Cosh[c + d*x]^7*Sinh[c + d*x])/(8*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 )^3 \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a-2 a x^2+(a+b) x^4\right )^3}{x^4 \left (1-x^2\right )^5} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}+\frac{\operatorname{Subst}\left (\int \frac{8 a^3-40 a^3 x^2+\left (80 a^3+24 a^2 b-b^3\right ) x^4-8 \left (10 a^3+9 a^2 b+b^3\right ) x^6+8 (5 a-b) (a+b)^2 x^8-8 (a+b)^3 x^{10}}{x^4 \left (1-x^2\right )^4} \, dx,x,\tanh (c+d x)\right )}{8 d}\\ &=-\frac{25 b^3 \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac{b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}-\frac{\operatorname{Subst}\left (\int \frac{-48 a^3+192 a^3 x^2-\left (288 a^3+144 a^2 b+19 b^3\right ) x^4+96 (2 a-b) (a+b)^2 x^6-48 (a+b)^3 x^8}{x^4 \left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{48 d}\\ &=\frac{b^2 (144 a+163 b) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}-\frac{25 b^3 \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac{b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}+\frac{\operatorname{Subst}\left (\int \frac{192 a^3-576 a^3 x^2+3 \left (192 a^3+192 a^2 b-48 a b^2-29 b^3\right ) x^4-192 (a+b)^3 x^6}{x^4 \left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{192 d}\\ &=-\frac{3 b^2 (80 a+31 b) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac{b^2 (144 a+163 b) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}-\frac{25 b^3 \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac{b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}-\frac{\operatorname{Subst}\left (\int \frac{-384 a^3+768 a^3 x^2-3 \left (128 a^3+384 a^2 b+144 a b^2+35 b^3\right ) x^4}{x^4 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{384 d}\\ &=-\frac{3 b^2 (80 a+31 b) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac{b^2 (144 a+163 b) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}-\frac{25 b^3 \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac{b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}-\frac{\operatorname{Subst}\left (\int \left (-\frac{384 a^3}{x^4}+\frac{384 a^3}{x^2}+\frac{3 b \left (384 a^2+144 a b+35 b^2\right )}{-1+x^2}\right ) \, dx,x,\tanh (c+d x)\right )}{384 d}\\ &=\frac{a^3 \coth (c+d x)}{d}-\frac{a^3 \coth ^3(c+d x)}{3 d}-\frac{3 b^2 (80 a+31 b) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac{b^2 (144 a+163 b) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}-\frac{25 b^3 \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac{b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}-\frac{\left (b \left (384 a^2+144 a b+35 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-1+x^2} \, dx,x,\tanh (c+d x)\right )}{128 d}\\ &=\frac{1}{128} b \left (384 a^2+144 a b+35 b^2\right ) x+\frac{a^3 \coth (c+d x)}{d}-\frac{a^3 \coth ^3(c+d x)}{3 d}-\frac{3 b^2 (80 a+31 b) \cosh (c+d x) \sinh (c+d x)}{128 d}+\frac{b^2 (144 a+163 b) \cosh ^3(c+d x) \sinh (c+d x)}{192 d}-\frac{25 b^3 \cosh ^5(c+d x) \sinh (c+d x)}{48 d}+\frac{b^3 \cosh ^7(c+d x) \sinh (c+d x)}{8 d}\\ \end{align*}

Mathematica [A]  time = 0.782114, size = 131, normalized size = 0.81 \[ \frac{b \left (9216 a^2 c+9216 a^2 d x-96 b (24 a+7 b) \sinh (2 (c+d x))+24 b (12 a+7 b) \sinh (4 (c+d x))+3456 a b c+3456 a b d x-32 b^2 \sinh (6 (c+d x))+3 b^2 \sinh (8 (c+d x))+840 b^2 c+840 b^2 d x\right )-1024 a^3 \coth (c+d x) \left (\text{csch}^2(c+d x)-2\right )}{3072 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1024*a^3*Coth[c + d*x]*(-2 + Csch[c + d*x]^2) + b*(9216*a^2*c + 3456*a*b*c + 840*b^2*c + 9216*a^2*d*x + 3456
*a*b*d*x + 840*b^2*d*x - 96*b*(24*a + 7*b)*Sinh[2*(c + d*x)] + 24*b*(12*a + 7*b)*Sinh[4*(c + d*x)] - 32*b^2*Si
nh[6*(c + d*x)] + 3*b^2*Sinh[8*(c + d*x)]))/(3072*d)

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Maple [A]  time = 0.048, size = 137, normalized size = 0.9 \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 ( dx+c \right ) +3\,a{b}^{2} \left ( \left ( 1/4\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}-3/8\,\sinh \left ( dx+c \right ) \right ) \cosh \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{b}^{3} \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{7}}{8}}-{\frac{7\, \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{48}}+{\frac{35\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{192}}-{\frac{35\,\sinh \left ( dx+c \right ) }{128}} \right ) \cosh \left ( dx+c \right ) +{\frac{35\,dx}{128}}+{\frac{35\,c}{128}} \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)^3,x)

[Out]

1/d*(a^3*(2/3-1/3*csch(d*x+c)^2)*coth(d*x+c)+3*a^2*b*(d*x+c)+3*a*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)+b^3*((1/8*sinh(d*x+c)^7-7/48*sinh(d*x+c)^5+35/192*sinh(d*x+c)^3-35/128*sinh(d*x+c))*cos
h(d*x+c)+35/128*d*x+35/128*c))

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Maxima [A]  time = 1.10793, size = 381, normalized size = 2.37 \begin{align*} \frac{3}{64} \, a 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 )} + 3 \, a^{2} b x - \frac{1}{6144} \, b^{3}{\left (\frac{{\left (32 \, e^{\left (-2 \, d x - 2 \, c\right )} - 168 \, e^{\left (-4 \, d x - 4 \, c\right )} + 672 \, e^{\left (-6 \, d x - 6 \, c\right )} - 3\right )} e^{\left (8 \, d x + 8 \, c\right )}}{d} - \frac{1680 \,{\left (d x + c\right )}}{d} - \frac{672 \, e^{\left (-2 \, d x - 2 \, c\right )} - 168 \, e^{\left (-4 \, d x - 4 \, c\right )} + 32 \, e^{\left (-6 \, d x - 6 \, c\right )} - 3 \, e^{\left (-8 \, d x - 8 \, c\right )}}{d}\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 )} \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)^3,x, algorithm="maxima")

[Out]

3/64*a*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) + 3*a^
2*b*x - 1/6144*b^3*((32*e^(-2*d*x - 2*c) - 168*e^(-4*d*x - 4*c) + 672*e^(-6*d*x - 6*c) - 3)*e^(8*d*x + 8*c)/d
- 1680*(d*x + c)/d - (672*e^(-2*d*x - 2*c) - 168*e^(-4*d*x - 4*c) + 32*e^(-6*d*x - 6*c) - 3*e^(-8*d*x - 8*c))/
d) + 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)))

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Fricas [B]  time = 1.36759, size = 1476, normalized size = 9.17 \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*sinh(d*x+c)^4)^3,x, algorithm="fricas")

[Out]

1/6144*(3*b^3*cosh(d*x + c)^11 + 33*b^3*cosh(d*x + c)*sinh(d*x + c)^10 - 41*b^3*cosh(d*x + c)^9 + 9*(55*b^3*co
sh(d*x + c)^3 - 41*b^3*cosh(d*x + c))*sinh(d*x + c)^8 + 3*(96*a*b^2 + 91*b^3)*cosh(d*x + c)^7 + 21*(66*b^3*cos
h(d*x + c)^5 - 164*b^3*cosh(d*x + c)^3 + (96*a*b^2 + 91*b^3)*cosh(d*x + c))*sinh(d*x + c)^6 - 3*(1056*a*b^2 +
425*b^3)*cosh(d*x + c)^5 + 3*(330*b^3*cosh(d*x + c)^7 - 1722*b^3*cosh(d*x + c)^5 + 35*(96*a*b^2 + 91*b^3)*cosh
(d*x + c)^3 - 5*(1056*a*b^2 + 425*b^3)*cosh(d*x + c))*sinh(d*x + c)^4 + 8*(512*a^3 + 972*a*b^2 + 319*b^3)*cosh
(d*x + c)^3 - 16*(256*a^3 - 3*(384*a^2*b + 144*a*b^2 + 35*b^3)*d*x)*sinh(d*x + c)^3 + 3*(55*b^3*cosh(d*x + c)^
9 - 492*b^3*cosh(d*x + c)^7 + 21*(96*a*b^2 + 91*b^3)*cosh(d*x + c)^5 - 10*(1056*a*b^2 + 425*b^3)*cosh(d*x + c)
^3 + 8*(512*a^3 + 972*a*b^2 + 319*b^3)*cosh(d*x + c))*sinh(d*x + c)^2 - 24*(512*a^3 + 204*a*b^2 + 63*b^3)*cosh
(d*x + c) + 48*(256*a^3 - 3*(384*a^2*b + 144*a*b^2 + 35*b^3)*d*x - (256*a^3 - 3*(384*a^2*b + 144*a*b^2 + 35*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)**4)**3,x)

[Out]

Timed out

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Giac [B]  time = 1.70406, size = 423, normalized size = 2.63 \begin{align*} \frac{{\left (384 \, a^{2} b + 144 \, a b^{2} + 35 \, b^{3}\right )}{\left (d x + c\right )}}{128 \, d} - \frac{{\left (19200 \, a^{2} b e^{\left (8 \, d x + 8 \, c\right )} + 7200 \, a b^{2} e^{\left (8 \, d x + 8 \, c\right )} + 1750 \, b^{3} e^{\left (8 \, d x + 8 \, c\right )} - 2304 \, a b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 672 \, b^{3} e^{\left (6 \, d x + 6 \, c\right )} + 288 \, a b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 168 \, b^{3} e^{\left (4 \, d x + 4 \, c\right )} - 32 \, b^{3} e^{\left (2 \, d x + 2 \, c\right )} + 3 \, b^{3}\right )} e^{\left (-8 \, d x - 8 \, c\right )}}{6144 \, d} + \frac{3 \, b^{3} d^{3} e^{\left (8 \, d x + 8 \, c\right )} - 32 \, b^{3} d^{3} e^{\left (6 \, d x + 6 \, c\right )} + 288 \, a b^{2} d^{3} e^{\left (4 \, d x + 4 \, c\right )} + 168 \, b^{3} d^{3} e^{\left (4 \, d x + 4 \, c\right )} - 2304 \, a b^{2} d^{3} e^{\left (2 \, d x + 2 \, c\right )} - 672 \, b^{3} d^{3} e^{\left (2 \, d x + 2 \, c\right )}}{6144 \, d^{4}} - \frac{4 \,{\left (3 \, a^{3} e^{\left (2 \, d x + 2 \, c\right )} - a^{3}\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)^3,x, algorithm="giac")

[Out]

1/128*(384*a^2*b + 144*a*b^2 + 35*b^3)*(d*x + c)/d - 1/6144*(19200*a^2*b*e^(8*d*x + 8*c) + 7200*a*b^2*e^(8*d*x
+ 8*c) + 1750*b^3*e^(8*d*x + 8*c) - 2304*a*b^2*e^(6*d*x + 6*c) - 672*b^3*e^(6*d*x + 6*c) + 288*a*b^2*e^(4*d*x
+ 4*c) + 168*b^3*e^(4*d*x + 4*c) - 32*b^3*e^(2*d*x + 2*c) + 3*b^3)*e^(-8*d*x - 8*c)/d + 1/6144*(3*b^3*d^3*e^(
8*d*x + 8*c) - 32*b^3*d^3*e^(6*d*x + 6*c) + 288*a*b^2*d^3*e^(4*d*x + 4*c) + 168*b^3*d^3*e^(4*d*x + 4*c) - 2304
*a*b^2*d^3*e^(2*d*x + 2*c) - 672*b^3*d^3*e^(2*d*x + 2*c))/d^4 - 4/3*(3*a^3*e^(2*d*x + 2*c) - a^3)/(d*(e^(2*d*x
+ 2*c) - 1)^3)