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

Optimal. Leaf size=103 \[ -\frac{a^2 \coth (c+d x)}{d}+\frac{b (16 a+11 b) \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac{1}{16} b x (16 a+5 b)+\frac{b^2 \sinh (c+d x) \cosh ^5(c+d x)}{6 d}-\frac{13 b^2 \sinh (c+d x) \cosh ^3(c+d x)}{24 d} \]

[Out]

-(b*(16*a + 5*b)*x)/16 - (a^2*Coth[c + d*x])/d + (b*(16*a + 11*b)*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) - (13*b^
2*Cosh[c + d*x]^3*Sinh[c + d*x])/(24*d) + (b^2*Cosh[c + d*x]^5*Sinh[c + d*x])/(6*d)

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Rubi [A]  time = 0.194466, antiderivative size = 103, 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, 453, 206} \[ -\frac{a^2 \coth (c+d x)}{d}+\frac{b (16 a+11 b) \sinh (c+d x) \cosh (c+d x)}{16 d}-\frac{1}{16} b x (16 a+5 b)+\frac{b^2 \sinh (c+d x) \cosh ^5(c+d x)}{6 d}-\frac{13 b^2 \sinh (c+d x) \cosh ^3(c+d x)}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(16*a + 5*b)*x)/16 - (a^2*Coth[c + d*x])/d + (b*(16*a + 11*b)*Cosh[c + d*x]*Sinh[c + d*x])/(16*d) - (13*b^
2*Cosh[c + d*x]^3*Sinh[c + d*x])/(24*d) + (b^2*Cosh[c + d*x]^5*Sinh[c + d*x])/(6*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 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 ^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^2 \left (1-x^2\right )^4} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{b^2 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}-\frac{\operatorname{Subst}\left (\int \frac{-6 a^2+\left (18 a^2+b^2\right ) x^2-6 (3 a-b) (a+b) x^4+6 (a+b)^2 x^6}{x^2 \left (1-x^2\right )^3} \, dx,x,\tanh (c+d x)\right )}{6 d}\\ &=-\frac{13 b^2 \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b^2 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}+\frac{\operatorname{Subst}\left (\int \frac{24 a^2-3 \left (16 a^2-3 b^2\right ) x^2+24 (a+b)^2 x^4}{x^2 \left (1-x^2\right )^2} \, dx,x,\tanh (c+d x)\right )}{24 d}\\ &=\frac{b (16 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{13 b^2 \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b^2 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}-\frac{\operatorname{Subst}\left (\int \frac{-48 a^2+3 \left (16 a^2+16 a b+5 b^2\right ) x^2}{x^2 \left (1-x^2\right )} \, dx,x,\tanh (c+d x)\right )}{48 d}\\ &=-\frac{a^2 \coth (c+d x)}{d}+\frac{b (16 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{13 b^2 \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b^2 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}-\frac{(b (16 a+5 b)) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{16 d}\\ &=-\frac{1}{16} b (16 a+5 b) x-\frac{a^2 \coth (c+d x)}{d}+\frac{b (16 a+11 b) \cosh (c+d x) \sinh (c+d x)}{16 d}-\frac{13 b^2 \cosh ^3(c+d x) \sinh (c+d x)}{24 d}+\frac{b^2 \cosh ^5(c+d x) \sinh (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 0.306401, size = 77, normalized size = 0.75 \[ \frac{b ((96 a+45 b) \sinh (2 (c+d x))-192 a c-192 a d x-9 b \sinh (4 (c+d x))+b \sinh (6 (c+d x))-60 b c-60 b d x)-192 a^2 \coth (c+d x)}{192 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-192*a^2*Coth[c + d*x] + b*(-192*a*c - 60*b*c - 192*a*d*x - 60*b*d*x + (96*a + 45*b)*Sinh[2*(c + d*x)] - 9*b*
Sinh[4*(c + d*x)] + b*Sinh[6*(c + d*x)]))/(192*d)

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Maple [A]  time = 0.038, size = 91, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ( -{a}^{2}{\rm coth} \left (dx+c\right )+2\,ab \left ( 1/2\,\cosh \left ( dx+c \right ) \sinh \left ( dx+c \right ) -1/2\,dx-c/2 \right ) +{b}^{2} \left ( \left ({\frac{ \left ( \sinh \left ( dx+c \right ) \right ) ^{5}}{6}}-{\frac{5\, \left ( \sinh \left ( dx+c \right ) \right ) ^{3}}{24}}+{\frac{5\,\sinh \left ( dx+c \right ) }{16}} \right ) \cosh \left ( dx+c \right ) -{\frac{5\,dx}{16}}-{\frac{5\,c}{16}} \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)^4)^2,x)

[Out]

1/d*(-a^2*coth(d*x+c)+2*a*b*(1/2*cosh(d*x+c)*sinh(d*x+c)-1/2*d*x-1/2*c)+b^2*((1/6*sinh(d*x+c)^5-5/24*sinh(d*x+
c)^3+5/16*sinh(d*x+c))*cosh(d*x+c)-5/16*d*x-5/16*c))

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Maxima [A]  time = 1.08813, size = 197, normalized size = 1.91 \begin{align*} -\frac{1}{4} \, a b{\left (4 \, x - \frac{e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac{e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac{1}{384} \, b^{2}{\left (\frac{{\left (9 \, e^{\left (-2 \, d x - 2 \, c\right )} - 45 \, e^{\left (-4 \, d x - 4 \, c\right )} - 1\right )} e^{\left (6 \, d x + 6 \, c\right )}}{d} + \frac{120 \,{\left (d x + c\right )}}{d} + \frac{45 \, e^{\left (-2 \, d x - 2 \, c\right )} - 9 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )}}{d}\right )} + \frac{2 \, a^{2}}{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)^4)^2,x, algorithm="maxima")

[Out]

-1/4*a*b*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 1/384*b^2*((9*e^(-2*d*x - 2*c) - 45*e^(-4*d*x - 4*c)
- 1)*e^(6*d*x + 6*c)/d + 120*(d*x + c)/d + (45*e^(-2*d*x - 2*c) - 9*e^(-4*d*x - 4*c) + e^(-6*d*x - 6*c))/d) +
2*a^2/(d*(e^(-2*d*x - 2*c) - 1))

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Fricas [B]  time = 1.70794, size = 562, normalized size = 5.46 \begin{align*} \frac{b^{2} \cosh \left (d x + c\right )^{7} + 7 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} - 10 \, b^{2} \cosh \left (d x + c\right )^{5} + 5 \,{\left (7 \, b^{2} \cosh \left (d x + c\right )^{3} - 10 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 6 \,{\left (16 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} +{\left (21 \, b^{2} \cosh \left (d x + c\right )^{5} - 100 \, b^{2} \cosh \left (d x + c\right )^{3} + 18 \,{\left (16 \, a b + 9 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 3 \,{\left (128 \, a^{2} + 32 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right ) - 24 \,{\left ({\left (16 \, a b + 5 \, b^{2}\right )} d x - 16 \, a^{2}\right )} \sinh \left (d x + c\right )}{384 \, 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)^4)^2,x, algorithm="fricas")

[Out]

1/384*(b^2*cosh(d*x + c)^7 + 7*b^2*cosh(d*x + c)*sinh(d*x + c)^6 - 10*b^2*cosh(d*x + c)^5 + 5*(7*b^2*cosh(d*x
+ c)^3 - 10*b^2*cosh(d*x + c))*sinh(d*x + c)^4 + 6*(16*a*b + 9*b^2)*cosh(d*x + c)^3 + (21*b^2*cosh(d*x + c)^5
- 100*b^2*cosh(d*x + c)^3 + 18*(16*a*b + 9*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 - 3*(128*a^2 + 32*a*b + 15*b^2)
*cosh(d*x + c) - 24*((16*a*b + 5*b^2)*d*x - 16*a^2)*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)**4)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.29338, size = 273, normalized size = 2.65 \begin{align*} -\frac{{\left (16 \, a b + 5 \, b^{2}\right )}{\left (d x + c\right )}}{16 \, d} + \frac{{\left (352 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 110 \, b^{2} e^{\left (6 \, d x + 6 \, c\right )} - 96 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 45 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} + 9 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{2}\right )} e^{\left (-6 \, d x - 6 \, c\right )}}{384 \, d} - \frac{2 \, a^{2}}{d{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}} + \frac{b^{2} d^{2} e^{\left (6 \, d x + 6 \, c\right )} - 9 \, b^{2} d^{2} e^{\left (4 \, d x + 4 \, c\right )} + 96 \, a b d^{2} e^{\left (2 \, d x + 2 \, c\right )} + 45 \, b^{2} d^{2} e^{\left (2 \, d x + 2 \, c\right )}}{384 \, d^{3}} \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)^4)^2,x, algorithm="giac")

[Out]

-1/16*(16*a*b + 5*b^2)*(d*x + c)/d + 1/384*(352*a*b*e^(6*d*x + 6*c) + 110*b^2*e^(6*d*x + 6*c) - 96*a*b*e^(4*d*
x + 4*c) - 45*b^2*e^(4*d*x + 4*c) + 9*b^2*e^(2*d*x + 2*c) - b^2)*e^(-6*d*x - 6*c)/d - 2*a^2/(d*(e^(2*d*x + 2*c
) - 1)) + 1/384*(b^2*d^2*e^(6*d*x + 6*c) - 9*b^2*d^2*e^(4*d*x + 4*c) + 96*a*b*d^2*e^(2*d*x + 2*c) + 45*b^2*d^2
*e^(2*d*x + 2*c))/d^3