Integrand size = 23, antiderivative size = 91 \[ \int \text {csch}^4(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=\frac {1}{8} b (16 a+3 b) x+\frac {a^2 \coth (c+d x)}{d}-\frac {a^2 \coth ^3(c+d x)}{3 d}-\frac {5 b^2 \cosh (c+d x) \sinh (c+d x)}{8 d}+\frac {b^2 \cosh ^3(c+d x) \sinh (c+d x)}{4 d} \]
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
Time = 0.23 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.75 \[ \int \text {csch}^4(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=\frac {-32 a^2 \coth (c+d x) \left (-2+\text {csch}^2(c+d x)\right )+3 b (12 b c+64 a d x+12 b d x-8 b \sinh (2 (c+d x))+b \sinh (4 (c+d x)))}{96 d} \]
(-32*a^2*Coth[c + d*x]*(-2 + Csch[c + d*x]^2) + 3*b*(12*b*c + 64*a*d*x + 1 2*b*d*x - 8*b*Sinh[2*(c + d*x)] + b*Sinh[4*(c + d*x)]))/(96*d)
Time = 0.47 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.22, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3696, 1582, 2336, 25, 1584, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \text {csch}^4(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \sin (i c+i d x)^4\right )^2}{\sin (i c+i d x)^4}dx\) |
| \(\Big \downarrow \) 3696 |
| \(\displaystyle \frac {\int \frac {\coth ^4(c+d x) \left ((a+b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}{\left (1-\tanh ^2(c+d x)\right )^3}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 1582 |
| \(\displaystyle \frac {\frac {1}{4} \int \frac {\coth ^4(c+d x) \left (-4 (a+b)^2 \tanh ^6(c+d x)+\left (12 a^2+8 b a-b^2\right ) \tanh ^4(c+d x)-12 a^2 \tanh ^2(c+d x)+4 a^2\right )}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)+\frac {b^2 \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 2336 |
| \(\displaystyle \frac {\frac {1}{4} \left (-\frac {1}{2} \int -\frac {\coth ^4(c+d x) \left (\left (8 a^2+16 b a+3 b^2\right ) \tanh ^4(c+d x)-16 a^2 \tanh ^2(c+d x)+8 a^2\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\frac {5 b^2 \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {b^2 \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \frac {\coth ^4(c+d x) \left (\left (8 a^2+16 b a+3 b^2\right ) \tanh ^4(c+d x)-16 a^2 \tanh ^2(c+d x)+8 a^2\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)-\frac {5 b^2 \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {b^2 \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \int \left (8 a^2 \coth ^4(c+d x)-8 a^2 \coth ^2(c+d x)-\frac {b (16 a+3 b)}{\tanh ^2(c+d x)-1}\right )d\tanh (c+d x)-\frac {5 b^2 \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {b^2 \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{4} \left (\frac {1}{2} \left (-\frac {8}{3} a^2 \coth ^3(c+d x)+8 a^2 \coth (c+d x)+b (16 a+3 b) \text {arctanh}(\tanh (c+d x))\right )-\frac {5 b^2 \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}\right )+\frac {b^2 \tanh (c+d x)}{4 \left (1-\tanh ^2(c+d x)\right )^2}}{d}\) |
((b^2*Tanh[c + d*x])/(4*(1 - Tanh[c + d*x]^2)^2) + ((b*(16*a + 3*b)*ArcTan h[Tanh[c + d*x]] + 8*a^2*Coth[c + d*x] - (8*a^2*Coth[c + d*x]^3)/3)/2 - (5 *b^2*Tanh[c + d*x])/(2*(1 - Tanh[c + d*x]^2)))/4)/d
3.3.3.3.1 Defintions of rubi rules used
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] + Simp[(-d)^(m/2 - 1)/(2*e^ (2*p)*(q + 1)) Int[x^m*(d + e*x^2)^(q + 1)*ExpandToSum[Together[(1/(d + e *x^2))*(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))], x], x], x] /; Fre eQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && ILtQ[q, -1] && ILtQ[m/2, 0]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[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]
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[PolynomialRema inder[(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] + Simp[1/(2*a*(p + 1)) Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^( p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[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]
Time = 0.62 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.82
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (\frac {2}{3}-\frac {\operatorname {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )+2 a b \left (d x +c \right )+b^{2} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(75\) |
| default | \(\frac {a^{2} \left (\frac {2}{3}-\frac {\operatorname {csch}\left (d x +c \right )^{2}}{3}\right ) \coth \left (d x +c \right )+2 a b \left (d x +c \right )+b^{2} \left (\left (\frac {\sinh \left (d x +c \right )^{3}}{4}-\frac {3 \sinh \left (d x +c \right )}{8}\right ) \cosh \left (d x +c \right )+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}\) | \(75\) |
| parallelrisch | \(\frac {-\operatorname {sech}\left (\frac {d x}{2}+\frac {c}{2}\right )^{3} \left (\cosh \left (d x +c \right )-\frac {\cosh \left (3 d x +3 c \right )}{3}\right ) a^{2} \operatorname {csch}\left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+32 \left (-\frac {b \sinh \left (2 d x +2 c \right )}{8}+\frac {b \sinh \left (4 d x +4 c \right )}{64}+d x \left (a +\frac {3 b}{16}\right )\right ) b}{16 d}\) | \(88\) |
| risch | \(2 a b x +\frac {3 b^{2} x}{8}+\frac {{\mathrm e}^{4 d x +4 c} b^{2}}{64 d}-\frac {{\mathrm e}^{2 d x +2 c} b^{2}}{8 d}+\frac {{\mathrm e}^{-2 d x -2 c} b^{2}}{8 d}-\frac {{\mathrm e}^{-4 d x -4 c} b^{2}}{64 d}-\frac {4 a^{2} \left (3 \,{\mathrm e}^{2 d x +2 c}-1\right )}{3 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{3}}\) | \(115\) |
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))
Leaf count of result is larger than twice the leaf count of optimal. 300 vs. \(2 (83) = 166\).
Time = 0.27 (sec) , antiderivative size = 300, normalized size of antiderivative = 3.30 \[ \int \text {csch}^4(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=\frac {3 \, b^{2} \cosh \left (d x + c\right )^{7} + 21 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} - 33 \, b^{2} \cosh \left (d x + c\right )^{5} + 15 \, {\left (7 \, b^{2} \cosh \left (d x + c\right )^{3} - 11 \, b^{2} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + {\left (128 \, a^{2} + 81 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 8 \, {\left (3 \, {\left (16 \, a b + 3 \, b^{2}\right )} d x - 16 \, a^{2}\right )} \sinh \left (d x + c\right )^{3} + 3 \, {\left (21 \, b^{2} \cosh \left (d x + c\right )^{5} - 110 \, b^{2} \cosh \left (d x + c\right )^{3} + {\left (128 \, a^{2} + 81 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 3 \, {\left (128 \, a^{2} + 17 \, b^{2}\right )} \cosh \left (d x + c\right ) - 24 \, {\left (3 \, {\left (16 \, a b + 3 \, b^{2}\right )} d x - {\left (3 \, {\left (16 \, a b + 3 \, b^{2}\right )} d x - 16 \, a^{2}\right )} \cosh \left (d x + c\right )^{2} - 16 \, a^{2}\right )} \sinh \left (d x + c\right )}{192 \, {\left (d \sinh \left (d x + c\right )^{3} + 3 \, {\left (d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )\right )}} \]
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))*sin h(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)*cosh(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)*sinh(d*x + c))
Timed out. \[ \int \text {csch}^4(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=\text {Timed out} \]
Time = 0.20 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.81 \[ \int \text {csch}^4(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=\frac {1}{64} \, b^{2} {\left (24 \, x + \frac {e^{\left (4 \, d x + 4 \, c\right )}}{d} - \frac {8 \, e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {8 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d} - \frac {e^{\left (-4 \, d x - 4 \, c\right )}}{d}\right )} + 2 \, a b x + \frac {4}{3} \, a^{2} {\left (\frac {3 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}} - \frac {1}{d {\left (3 \, e^{\left (-2 \, d x - 2 \, c\right )} - 3 \, e^{\left (-4 \, d x - 4 \, c\right )} + e^{\left (-6 \, d x - 6 \, c\right )} - 1\right )}}\right )} \]
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)))
Time = 0.35 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.56 \[ \int \text {csch}^4(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=\frac {3 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 24 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + 24 \, {\left (16 \, a b + 3 \, b^{2}\right )} {\left (d x + c\right )} - 3 \, {\left (96 \, a b e^{\left (4 \, d x + 4 \, c\right )} + 18 \, b^{2} e^{\left (4 \, d x + 4 \, c\right )} - 8 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} + b^{2}\right )} e^{\left (-4 \, d x - 4 \, c\right )} - \frac {256 \, {\left (3 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - a^{2}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{3}}}{192 \, d} \]
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
Time = 0.17 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.80 \[ \int \text {csch}^4(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=\frac {b\,x\,\left (16\,a+3\,b\right )}{8}-\frac {4\,a^2}{3\,d\,\left ({\mathrm {e}}^{4\,c+4\,d\,x}-2\,{\mathrm {e}}^{2\,c+2\,d\,x}+1\right )}+\frac {b^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}-\frac {b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {b^2\,{\mathrm {e}}^{-4\,c-4\,d\,x}}{64\,d}+\frac {b^2\,{\mathrm {e}}^{4\,c+4\,d\,x}}{64\,d}-\frac {8\,a^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{3\,d\,\left (3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1\right )} \]
(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*ex p(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))