Integrand size = 23, antiderivative size = 84 \[ \int \text {csch}^6(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=-\frac {b^2 x}{2}-\frac {a (a+2 b) \coth (c+d x)}{d}+\frac {2 a^2 \coth ^3(c+d x)}{3 d}-\frac {a^2 \coth ^5(c+d x)}{5 d}+\frac {b^2 \cosh (c+d x) \sinh (c+d x)}{2 d} \]
-1/2*b^2*x-a*(a+2*b)*coth(d*x+c)/d+2/3*a^2*coth(d*x+c)^3/d-1/5*a^2*coth(d* x+c)^5/d+1/2*b^2*cosh(d*x+c)*sinh(d*x+c)/d
Time = 0.52 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.80 \[ \int \text {csch}^6(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=\frac {-4 a \coth (c+d x) \left (8 a+30 b-4 a \text {csch}^2(c+d x)+3 a \text {csch}^4(c+d x)\right )+15 b^2 (-2 (c+d x)+\sinh (2 (c+d x)))}{60 d} \]
(-4*a*Coth[c + d*x]*(8*a + 30*b - 4*a*Csch[c + d*x]^2 + 3*a*Csch[c + d*x]^ 4) + 15*b^2*(-2*(c + d*x) + Sinh[2*(c + d*x)]))/(60*d)
Time = 0.44 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 25, 3696, 1582, 25, 2333, 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.
| \(\displaystyle \int \text {csch}^6(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\left (a+b \sin (i c+i d x)^4\right )^2}{\sin (i c+i d x)^6}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\left (b \sin (i c+i d x)^4+a\right )^2}{\sin (i c+i d x)^6}dx\) |
| \(\Big \downarrow \) 3696 |
| \(\displaystyle \frac {\int \frac {\coth ^6(c+d x) \left ((a+b) \tanh ^4(c+d x)-2 a \tanh ^2(c+d x)+a\right )^2}{\left (1-\tanh ^2(c+d x)\right )^2}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 1582 |
| \(\displaystyle \frac {\frac {b^2 \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}-\frac {1}{2} \int -\frac {\coth ^6(c+d x) \left (-\left (\left (2 a^2+4 b a+b^2\right ) \tanh ^6(c+d x)\right )+2 a (3 a+2 b) \tanh ^4(c+d x)-6 a^2 \tanh ^2(c+d x)+2 a^2\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {1}{2} \int \frac {\coth ^6(c+d x) \left (-\left (\left (2 a^2+4 b a+b^2\right ) \tanh ^6(c+d x)\right )+2 a (3 a+2 b) \tanh ^4(c+d x)-6 a^2 \tanh ^2(c+d x)+2 a^2\right )}{1-\tanh ^2(c+d x)}d\tanh (c+d x)+\frac {b^2 \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 2333 |
| \(\displaystyle \frac {\frac {1}{2} \int \left (2 a^2 \coth ^6(c+d x)-4 a^2 \coth ^4(c+d x)+2 a (a+2 b) \coth ^2(c+d x)+\frac {b^2}{\tanh ^2(c+d x)-1}\right )d\tanh (c+d x)+\frac {b^2 \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {1}{2} \left (-\frac {2}{5} a^2 \coth ^5(c+d x)+\frac {4}{3} a^2 \coth ^3(c+d x)-2 a (a+2 b) \coth (c+d x)-b^2 \text {arctanh}(\tanh (c+d x))\right )+\frac {b^2 \tanh (c+d x)}{2 \left (1-\tanh ^2(c+d x)\right )}}{d}\) |
((-(b^2*ArcTanh[Tanh[c + d*x]]) - 2*a*(a + 2*b)*Coth[c + d*x] + (4*a^2*Cot h[c + d*x]^3)/3 - (2*a^2*Coth[c + d*x]^5)/5)/2 + (b^2*Tanh[c + d*x])/(2*(1 - Tanh[c + d*x]^2)))/d
3.3.5.3.1 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[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ ExpandIntegrand[(c*x)^m*Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]
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 = 0.56 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {a^{2} \left (-\frac {8}{15}-\frac {\operatorname {csch}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {csch}\left (d x +c \right )^{2}}{15}\right ) \coth \left (d x +c \right )-2 a b \coth \left (d x +c \right )+b^{2} \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )}{d}\) | \(74\) |
| default | \(\frac {a^{2} \left (-\frac {8}{15}-\frac {\operatorname {csch}\left (d x +c \right )^{4}}{5}+\frac {4 \operatorname {csch}\left (d x +c \right )^{2}}{15}\right ) \coth \left (d x +c \right )-2 a b \coth \left (d x +c \right )+b^{2} \left (\frac {\sinh \left (d x +c \right ) \cosh \left (d x +c \right )}{2}-\frac {d x}{2}-\frac {c}{2}\right )}{d}\) | \(74\) |
| parallelrisch | \(\frac {-\left (\cosh \left (d x +c \right )-\frac {\cosh \left (3 d x +3 c \right )}{2}+\frac {\cosh \left (5 d x +5 c \right )}{10}\right ) \operatorname {sech}\left (\frac {d x}{2}+\frac {c}{2}\right )^{5} a^{2} \operatorname {csch}\left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+96 \,\operatorname {sech}\left (\frac {d x}{2}+\frac {c}{2}\right ) \operatorname {csch}\left (\frac {d x}{2}+\frac {c}{2}\right ) a b -192 a b \coth \left (\frac {d x}{2}+\frac {c}{2}\right )-48 \left (d x -\frac {\sinh \left (2 d x +2 c \right )}{2}\right ) b^{2}}{96 d}\) | \(118\) |
| risch | \(-\frac {b^{2} x}{2}+\frac {{\mathrm e}^{2 d x +2 c} b^{2}}{8 d}-\frac {{\mathrm e}^{-2 d x -2 c} b^{2}}{8 d}-\frac {4 a \left (15 b \,{\mathrm e}^{8 d x +8 c}-60 b \,{\mathrm e}^{6 d x +6 c}+40 \,{\mathrm e}^{4 d x +4 c} a +90 b \,{\mathrm e}^{4 d x +4 c}-20 a \,{\mathrm e}^{2 d x +2 c}-60 b \,{\mathrm e}^{2 d x +2 c}+4 a +15 b \right )}{15 d \left ({\mathrm e}^{2 d x +2 c}-1\right )^{5}}\) | \(140\) |
1/d*(a^2*(-8/15-1/5*csch(d*x+c)^4+4/15*csch(d*x+c)^2)*coth(d*x+c)-2*a*b*co th(d*x+c)+b^2*(1/2*sinh(d*x+c)*cosh(d*x+c)-1/2*d*x-1/2*c))
Leaf count of result is larger than twice the leaf count of optimal. 457 vs. \(2 (76) = 152\).
Time = 0.27 (sec) , antiderivative size = 457, normalized size of antiderivative = 5.44 \[ \int \text {csch}^6(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=\frac {15 \, b^{2} \cosh \left (d x + c\right )^{7} + 105 \, b^{2} \cosh \left (d x + c\right ) \sinh \left (d x + c\right )^{6} - {\left (64 \, a^{2} + 240 \, a b + 75 \, b^{2}\right )} \cosh \left (d x + c\right )^{5} - 4 \, {\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \sinh \left (d x + c\right )^{5} + 5 \, {\left (105 \, b^{2} \cosh \left (d x + c\right )^{3} - {\left (64 \, a^{2} + 240 \, a b + 75 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{4} + 5 \, {\left (64 \, a^{2} + 144 \, a b + 27 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 20 \, {\left (15 \, b^{2} d x - 2 \, {\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \cosh \left (d x + c\right )^{2} - 16 \, a^{2} - 60 \, a b\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (63 \, b^{2} \cosh \left (d x + c\right )^{5} - 2 \, {\left (64 \, a^{2} + 240 \, a b + 75 \, b^{2}\right )} \cosh \left (d x + c\right )^{3} + 3 \, {\left (64 \, a^{2} + 144 \, a b + 27 \, b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )^{2} - 5 \, {\left (128 \, a^{2} + 96 \, a b + 15 \, b^{2}\right )} \cosh \left (d x + c\right ) - 20 \, {\left ({\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \cosh \left (d x + c\right )^{4} + 30 \, b^{2} d x - 3 \, {\left (15 \, b^{2} d x - 16 \, a^{2} - 60 \, a b\right )} \cosh \left (d x + c\right )^{2} - 32 \, a^{2} - 120 \, a b\right )} \sinh \left (d x + c\right )}{120 \, {\left (d \sinh \left (d x + c\right )^{5} + 5 \, {\left (2 \, d \cosh \left (d x + c\right )^{2} - d\right )} \sinh \left (d x + c\right )^{3} + 5 \, {\left (d \cosh \left (d x + c\right )^{4} - 3 \, d \cosh \left (d x + c\right )^{2} + 2 \, d\right )} \sinh \left (d x + c\right )\right )}} \]
1/120*(15*b^2*cosh(d*x + c)^7 + 105*b^2*cosh(d*x + c)*sinh(d*x + c)^6 - (6 4*a^2 + 240*a*b + 75*b^2)*cosh(d*x + c)^5 - 4*(15*b^2*d*x - 16*a^2 - 60*a* b)*sinh(d*x + c)^5 + 5*(105*b^2*cosh(d*x + c)^3 - (64*a^2 + 240*a*b + 75*b ^2)*cosh(d*x + c))*sinh(d*x + c)^4 + 5*(64*a^2 + 144*a*b + 27*b^2)*cosh(d* x + c)^3 + 20*(15*b^2*d*x - 2*(15*b^2*d*x - 16*a^2 - 60*a*b)*cosh(d*x + c) ^2 - 16*a^2 - 60*a*b)*sinh(d*x + c)^3 + 5*(63*b^2*cosh(d*x + c)^5 - 2*(64* a^2 + 240*a*b + 75*b^2)*cosh(d*x + c)^3 + 3*(64*a^2 + 144*a*b + 27*b^2)*co sh(d*x + c))*sinh(d*x + c)^2 - 5*(128*a^2 + 96*a*b + 15*b^2)*cosh(d*x + c) - 20*((15*b^2*d*x - 16*a^2 - 60*a*b)*cosh(d*x + c)^4 + 30*b^2*d*x - 3*(15 *b^2*d*x - 16*a^2 - 60*a*b)*cosh(d*x + c)^2 - 32*a^2 - 120*a*b)*sinh(d*x + c))/(d*sinh(d*x + c)^5 + 5*(2*d*cosh(d*x + c)^2 - d)*sinh(d*x + c)^3 + 5* (d*cosh(d*x + c)^4 - 3*d*cosh(d*x + c)^2 + 2*d)*sinh(d*x + c))
Timed out. \[ \int \text {csch}^6(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=\text {Timed out} \]
Leaf count of result is larger than twice the leaf count of optimal. 267 vs. \(2 (76) = 152\).
Time = 0.19 (sec) , antiderivative size = 267, normalized size of antiderivative = 3.18 \[ \int \text {csch}^6(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=-\frac {1}{8} \, b^{2} {\left (4 \, x - \frac {e^{\left (2 \, d x + 2 \, c\right )}}{d} + \frac {e^{\left (-2 \, d x - 2 \, c\right )}}{d}\right )} - \frac {16}{15} \, a^{2} {\left (\frac {5 \, e^{\left (-2 \, d x - 2 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac {10 \, e^{\left (-4 \, d x - 4 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}} - \frac {1}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} - 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} - 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} - 1\right )}}\right )} + \frac {4 \, a b}{d {\left (e^{\left (-2 \, d x - 2 \, c\right )} - 1\right )}} \]
-1/8*b^2*(4*x - e^(2*d*x + 2*c)/d + e^(-2*d*x - 2*c)/d) - 16/15*a^2*(5*e^( -2*d*x - 2*c)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1)) - 10*e^(-4*d*x - 4* c)/(d*(5*e^(-2*d*x - 2*c) - 10*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5* e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1)) - 1/(d*(5*e^(-2*d*x - 2*c) - 1 0*e^(-4*d*x - 4*c) + 10*e^(-6*d*x - 6*c) - 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) - 1))) + 4*a*b/(d*(e^(-2*d*x - 2*c) - 1))
Leaf count of result is larger than twice the leaf count of optimal. 166 vs. \(2 (76) = 152\).
Time = 0.36 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.98 \[ \int \text {csch}^6(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=-\frac {60 \, {\left (d x + c\right )} b^{2} - 15 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - 15 \, {\left (2 \, b^{2} e^{\left (2 \, d x + 2 \, c\right )} - b^{2}\right )} e^{\left (-2 \, d x - 2 \, c\right )} + \frac {32 \, {\left (15 \, a b e^{\left (8 \, d x + 8 \, c\right )} - 60 \, a b e^{\left (6 \, d x + 6 \, c\right )} + 40 \, a^{2} e^{\left (4 \, d x + 4 \, c\right )} + 90 \, a b e^{\left (4 \, d x + 4 \, c\right )} - 20 \, a^{2} e^{\left (2 \, d x + 2 \, c\right )} - 60 \, a b e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a^{2} + 15 \, a b\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} - 1\right )}^{5}}}{120 \, d} \]
-1/120*(60*(d*x + c)*b^2 - 15*b^2*e^(2*d*x + 2*c) - 15*(2*b^2*e^(2*d*x + 2 *c) - b^2)*e^(-2*d*x - 2*c) + 32*(15*a*b*e^(8*d*x + 8*c) - 60*a*b*e^(6*d*x + 6*c) + 40*a^2*e^(4*d*x + 4*c) + 90*a*b*e^(4*d*x + 4*c) - 20*a^2*e^(2*d* x + 2*c) - 60*a*b*e^(2*d*x + 2*c) + 4*a^2 + 15*a*b)/(e^(2*d*x + 2*c) - 1)^ 5)/d
Time = 1.46 (sec) , antiderivative size = 397, normalized size of antiderivative = 4.73 \[ \int \text {csch}^6(c+d x) \left (a+b \sinh ^4(c+d x)\right )^2 \, dx=\frac {b^2\,{\mathrm {e}}^{2\,c+2\,d\,x}}{8\,d}-\frac {\frac {4\,{\mathrm {e}}^{2\,c+2\,d\,x}\,\left (4\,a^2+3\,b\,a\right )}{5\,d}-\frac {4\,a\,b}{5\,d}-\frac {12\,a\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}+\frac {4\,a\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}}{6\,{\mathrm {e}}^{4\,c+4\,d\,x}-4\,{\mathrm {e}}^{2\,c+2\,d\,x}-4\,{\mathrm {e}}^{6\,c+6\,d\,x}+{\mathrm {e}}^{8\,c+8\,d\,x}+1}-\frac {b^2\,x}{2}-\frac {\frac {4\,\left (4\,a^2+3\,b\,a\right )}{15\,d}-\frac {8\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}+\frac {4\,a\,b\,{\mathrm {e}}^{4\,c+4\,d\,x}}{5\,d}}{3\,{\mathrm {e}}^{2\,c+2\,d\,x}-3\,{\mathrm {e}}^{4\,c+4\,d\,x}+{\mathrm {e}}^{6\,c+6\,d\,x}-1}-\frac {b^2\,{\mathrm {e}}^{-2\,c-2\,d\,x}}{8\,d}-\frac {\frac {8\,{\mathrm {e}}^{4\,c+4\,d\,x}\,\left (4\,a^2+3\,b\,a\right )}{5\,d}+\frac {4\,a\,b}{5\,d}-\frac {16\,a\,b\,{\mathrm {e}}^{2\,c+2\,d\,x}}{5\,d}-\frac {16\,a\,b\,{\mathrm {e}}^{6\,c+6\,d\,x}}{5\,d}+\frac {4\,a\,b\,{\mathrm {e}}^{8\,c+8\,d\,x}}{5\,d}}{5\,{\mathrm {e}}^{2\,c+2\,d\,x}-10\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,{\mathrm {e}}^{6\,c+6\,d\,x}-5\,{\mathrm {e}}^{8\,c+8\,d\,x}+{\mathrm {e}}^{10\,c+10\,d\,x}-1}-\frac {8\,a\,b}{5\,d\,\left ({\mathrm {e}}^{2\,c+2\,d\,x}-1\right )} \]
(b^2*exp(2*c + 2*d*x))/(8*d) - ((4*exp(2*c + 2*d*x)*(3*a*b + 4*a^2))/(5*d) - (4*a*b)/(5*d) - (12*a*b*exp(4*c + 4*d*x))/(5*d) + (4*a*b*exp(6*c + 6*d* x))/(5*d))/(6*exp(4*c + 4*d*x) - 4*exp(2*c + 2*d*x) - 4*exp(6*c + 6*d*x) + exp(8*c + 8*d*x) + 1) - (b^2*x)/2 - ((4*(3*a*b + 4*a^2))/(15*d) - (8*a*b* exp(2*c + 2*d*x))/(5*d) + (4*a*b*exp(4*c + 4*d*x))/(5*d))/(3*exp(2*c + 2*d *x) - 3*exp(4*c + 4*d*x) + exp(6*c + 6*d*x) - 1) - (b^2*exp(- 2*c - 2*d*x) )/(8*d) - ((8*exp(4*c + 4*d*x)*(3*a*b + 4*a^2))/(5*d) + (4*a*b)/(5*d) - (1 6*a*b*exp(2*c + 2*d*x))/(5*d) - (16*a*b*exp(6*c + 6*d*x))/(5*d) + (4*a*b*e xp(8*c + 8*d*x))/(5*d))/(5*exp(2*c + 2*d*x) - 10*exp(4*c + 4*d*x) + 10*exp (6*c + 6*d*x) - 5*exp(8*c + 8*d*x) + exp(10*c + 10*d*x) - 1) - (8*a*b)/(5* d*(exp(2*c + 2*d*x) - 1))