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]

1/8*b*(16*a+3*b)*x+a^2*coth(d*x+c)/d-1/3*a^2*coth(d*x+c)^3/d-5/8*b^2*cosh(d*x+c)*sinh(d*x+c)/d+1/4*b^2*cosh(d*
x+c)^3*sinh(d*x+c)/d

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Rubi [A]  time = 0.16, antiderivative size = 91, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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]

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*}

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Mathematica [A]  time = 0.33, 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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fricas [B]  time = 2.06, size = 300, normalized size = 3.30 \[ \frac {3 \, b^{2} \cosh \left (d x + c\right )^{7} + 21 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} - 33 \, b^{2} \cosh \left (d x + c\right )^{5} + 15 \, {\left (7 \, b^{2} \cosh \left (d x + c\right )^{3} - 11 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + {\left (128 \, a^{2} + 81 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 8 \, {\left (3 \, {\left (16 \, a b + 3 \, b^{2}\right )} d x - 16 \, a^{2}\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (21 \, b^{2} \cosh \left (d x + c\right )^{5} - 110 \, b^{2} \cosh \left (d x + c\right )^{3} + {\left (128 \, a^{2} + 81 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 3 \, {\left (128 \, a^{2} + 17 \, b^{2}\right )} \cosh \left (d x + c\right ) - 24 \, {\left (3 \, {\left (16 \, a b + 3 \, b^{2}\right )} d x - {\left (3 \, {\left (16 \, a b + 3 \, b^{2}\right )} d x - 16 \, a^{2}\right )} \cosh \left (d x + c\right )^{2} - 16 \, a^{2}\right )} \sinh \left (d x + c\right )}{192 \, {\left (d \sinh \left (d x + c\right )^{3} + 3 \, {\left (d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )\right )}} \]

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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giac [A]  time = 0.24, size = 142, normalized size = 1.56 \[ \frac {3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 24 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, {\left (16 \, a b + 3 \, b^{2}\right )} {\left (d x + c\right )} - 3 \, {\left (96 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 18 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - \frac {256 \, {\left (3 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - a^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{192 \, d} \]

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]

1/192*(3*b^2*e^(4*d*x + 4*c) - 24*b^2*e^(2*d*x + 2*c) + 24*(16*a*b + 3*b^2)*(d*x + c) - 3*(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) - 256*(3*a^2*e^(2*d*x + 2*c) - a^
2)/(e^(2*d*x + 2*c) - 1)^3)/d

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maple [A]  time = 0.09, size = 75, normalized size = 0.82 \[ \frac {a^{2} \left (\frac {2}{3}-\frac {\mathrm {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )+2 a b \left (d x +c \right )+b^{2} \left (\left (\frac {\left (\sinh ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d} \]

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 = 0.34, size = 165, normalized size = 1.81 \[ \frac {1}{64} \, b^{2} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + 2 \, a b x + \frac {4}{3} \, a^{2} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} - \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} \]

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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mupad [B]  time = 0.17, size = 164, normalized size = 1.80 \[ \frac {b\,x\,\left (16\,a+3\,b\right )}{8}-\frac {4\,a^2}{3\,d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {b^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}-\frac {b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {b^2\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d}+\frac {b^2\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d}-\frac {8\,a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sinh(c + d*x)^4)^2/sinh(c + d*x)^4,x)

[Out]

(b*x*(16*a + 3*b))/8 - (4*a^2)/(3*d*(exp(4*c + 4*d*x) - 2*exp(2*c + 2*d*x) + 1)) + (b^2*exp(- 2*c - 2*d*x))/(8
*d) - (b^2*exp(2*c + 2*d*x))/(8*d) - (b^2*exp(- 4*c - 4*d*x))/(64*d) + (b^2*exp(4*c + 4*d*x))/(64*d) - (8*a^2*
exp(2*c + 2*d*x))/(3*d*(3*exp(2*c + 2*d*x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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