Integrand size = 21, antiderivative size = 98 \[ \int \text {csch}(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {a b \arctan (\sinh (c+d x))}{d}-\frac {a^2 \text {arctanh}(\cosh (c+d x))}{d}-\frac {b^2 \text {sech}(c+d x)}{d}+\frac {2 b^2 \text {sech}^3(c+d x)}{3 d}-\frac {b^2 \text {sech}^5(c+d x)}{5 d}-\frac {a b \text {sech}(c+d x) \tanh (c+d x)}{d} \]
a*b*arctan(sinh(d*x+c))/d-a^2*arctanh(cosh(d*x+c))/d-b^2*sech(d*x+c)/d+2/3 *b^2*sech(d*x+c)^3/d-1/5*b^2*sech(d*x+c)^5/d-a*b*sech(d*x+c)*tanh(d*x+c)/d
Time = 0.96 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.28 \[ \int \text {csch}(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {2 a b \arctan \left (\tanh \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {a^2 \log \left (\cosh \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {a^2 \log \left (\sinh \left (\frac {1}{2} (c+d x)\right )\right )}{d}-\frac {b^2 \text {sech}(c+d x)}{d}+\frac {2 b^2 \text {sech}^3(c+d x)}{3 d}-\frac {b^2 \text {sech}^5(c+d x)}{5 d}-\frac {a b \text {sech}(c+d x) \tanh (c+d x)}{d} \]
(2*a*b*ArcTan[Tanh[(c + d*x)/2]])/d - (a^2*Log[Cosh[(c + d*x)/2]])/d + (a^ 2*Log[Sinh[(c + d*x)/2]])/d - (b^2*Sech[c + d*x])/d + (2*b^2*Sech[c + d*x] ^3)/(3*d) - (b^2*Sech[c + d*x]^5)/(5*d) - (a*b*Sech[c + d*x]*Tanh[c + d*x] )/d
Result contains complex when optimal does not.
Time = 0.36 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.17, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 26, 4149, 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.
| \(\displaystyle \int \text {csch}(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {i \left (a+i b \tan (i c+i d x)^3\right )^2}{\sin (i c+i d x)}dx\) |
| \(\Big \downarrow \) 26 |
| \(\displaystyle i \int \frac {\left (i b \tan (i c+i d x)^3+a\right )^2}{\sin (i c+i d x)}dx\) |
| \(\Big \downarrow \) 4149 |
| \(\displaystyle i \int \left (-i b^2 \text {sech}(c+d x) \tanh ^5(c+d x)-2 i a b \text {sech}(c+d x) \tanh ^2(c+d x)-i a^2 \text {csch}(c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle i \left (\frac {i a^2 \text {arctanh}(\cosh (c+d x))}{d}-\frac {i a b \arctan (\sinh (c+d x))}{d}+\frac {i a b \tanh (c+d x) \text {sech}(c+d x)}{d}+\frac {i b^2 \text {sech}^5(c+d x)}{5 d}-\frac {2 i b^2 \text {sech}^3(c+d x)}{3 d}+\frac {i b^2 \text {sech}(c+d x)}{d}\right )\) |
I*(((-I)*a*b*ArcTan[Sinh[c + d*x]])/d + (I*a^2*ArcTanh[Cosh[c + d*x]])/d + (I*b^2*Sech[c + d*x])/d - (((2*I)/3)*b^2*Sech[c + d*x]^3)/d + ((I/5)*b^2* Sech[c + d*x]^5)/d + (I*a*b*Sech[c + d*x]*Tanh[c + d*x])/d)
3.1.61.3.1 Defintions of rubi rules used
Int[(Complex[0, a_])*(Fx_), x_Symbol] :> Simp[(Complex[Identity[0], a]) I nt[Fx, x], x] /; FreeQ[a, x] && EqQ[a^2, 1]
Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^m*(a + b*(c*tan[e + f*x])^n)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0]
Time = 1.04 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.13
| method | result | size |
| derivativedivides | \(\frac {-2 a^{2} \operatorname {arctanh}\left ({\mathrm e}^{d x +c}\right )+2 a b \left (-\frac {\sinh \left (d x +c \right )}{\cosh \left (d x +c \right )^{2}}+\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )+b^{2} \left (-\frac {\sinh \left (d x +c \right )^{4}}{\cosh \left (d x +c \right )^{5}}-\frac {4 \sinh \left (d x +c \right )^{2}}{3 \cosh \left (d x +c \right )^{5}}-\frac {8}{15 \cosh \left (d x +c \right )^{5}}\right )}{d}\) | \(111\) |
| default | \(\frac {-2 a^{2} \operatorname {arctanh}\left ({\mathrm e}^{d x +c}\right )+2 a b \left (-\frac {\sinh \left (d x +c \right )}{\cosh \left (d x +c \right )^{2}}+\frac {\operatorname {sech}\left (d x +c \right ) \tanh \left (d x +c \right )}{2}+\arctan \left ({\mathrm e}^{d x +c}\right )\right )+b^{2} \left (-\frac {\sinh \left (d x +c \right )^{4}}{\cosh \left (d x +c \right )^{5}}-\frac {4 \sinh \left (d x +c \right )^{2}}{3 \cosh \left (d x +c \right )^{5}}-\frac {8}{15 \cosh \left (d x +c \right )^{5}}\right )}{d}\) | \(111\) |
| risch | \(-\frac {2 b \,{\mathrm e}^{d x +c} \left (15 a \,{\mathrm e}^{8 d x +8 c}+15 b \,{\mathrm e}^{8 d x +8 c}+30 a \,{\mathrm e}^{6 d x +6 c}+20 b \,{\mathrm e}^{6 d x +6 c}+58 b \,{\mathrm e}^{4 d x +4 c}-30 \,{\mathrm e}^{2 d x +2 c} a +20 b \,{\mathrm e}^{2 d x +2 c}-15 a +15 b \right )}{15 d \left ({\mathrm e}^{2 d x +2 c}+1\right )^{5}}+\frac {i b a \ln \left ({\mathrm e}^{d x +c}+i\right )}{d}-\frac {i b a \ln \left ({\mathrm e}^{d x +c}-i\right )}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{d x +c}-1\right )}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{d x +c}+1\right )}{d}\) | \(187\) |
1/d*(-2*a^2*arctanh(exp(d*x+c))+2*a*b*(-sinh(d*x+c)/cosh(d*x+c)^2+1/2*sech (d*x+c)*tanh(d*x+c)+arctan(exp(d*x+c)))+b^2*(-sinh(d*x+c)^4/cosh(d*x+c)^5- 4/3*sinh(d*x+c)^2/cosh(d*x+c)^5-8/15/cosh(d*x+c)^5))
Leaf count of result is larger than twice the leaf count of optimal. 2498 vs. \(2 (94) = 188\).
Time = 0.29 (sec) , antiderivative size = 2498, normalized size of antiderivative = 25.49 \[ \int \text {csch}(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\text {Too large to display} \]
-1/15*(30*(a*b + b^2)*cosh(d*x + c)^9 + 270*(a*b + b^2)*cosh(d*x + c)*sinh (d*x + c)^8 + 30*(a*b + b^2)*sinh(d*x + c)^9 + 20*(3*a*b + 2*b^2)*cosh(d*x + c)^7 + 20*(54*(a*b + b^2)*cosh(d*x + c)^2 + 3*a*b + 2*b^2)*sinh(d*x + c )^7 + 116*b^2*cosh(d*x + c)^5 + 140*(18*(a*b + b^2)*cosh(d*x + c)^3 + (3*a *b + 2*b^2)*cosh(d*x + c))*sinh(d*x + c)^6 + 4*(945*(a*b + b^2)*cosh(d*x + c)^4 + 105*(3*a*b + 2*b^2)*cosh(d*x + c)^2 + 29*b^2)*sinh(d*x + c)^5 + 20 *(189*(a*b + b^2)*cosh(d*x + c)^5 + 35*(3*a*b + 2*b^2)*cosh(d*x + c)^3 + 2 9*b^2*cosh(d*x + c))*sinh(d*x + c)^4 - 20*(3*a*b - 2*b^2)*cosh(d*x + c)^3 + 20*(126*(a*b + b^2)*cosh(d*x + c)^6 + 35*(3*a*b + 2*b^2)*cosh(d*x + c)^4 + 58*b^2*cosh(d*x + c)^2 - 3*a*b + 2*b^2)*sinh(d*x + c)^3 + 20*(54*(a*b + b^2)*cosh(d*x + c)^7 + 21*(3*a*b + 2*b^2)*cosh(d*x + c)^5 + 58*b^2*cosh(d *x + c)^3 - 3*(3*a*b - 2*b^2)*cosh(d*x + c))*sinh(d*x + c)^2 - 30*(a*b*cos h(d*x + c)^10 + 10*a*b*cosh(d*x + c)*sinh(d*x + c)^9 + a*b*sinh(d*x + c)^1 0 + 5*a*b*cosh(d*x + c)^8 + 5*(9*a*b*cosh(d*x + c)^2 + a*b)*sinh(d*x + c)^ 8 + 10*a*b*cosh(d*x + c)^6 + 40*(3*a*b*cosh(d*x + c)^3 + a*b*cosh(d*x + c) )*sinh(d*x + c)^7 + 10*(21*a*b*cosh(d*x + c)^4 + 14*a*b*cosh(d*x + c)^2 + a*b)*sinh(d*x + c)^6 + 10*a*b*cosh(d*x + c)^4 + 4*(63*a*b*cosh(d*x + c)^5 + 70*a*b*cosh(d*x + c)^3 + 15*a*b*cosh(d*x + c))*sinh(d*x + c)^5 + 10*(21* a*b*cosh(d*x + c)^6 + 35*a*b*cosh(d*x + c)^4 + 15*a*b*cosh(d*x + c)^2 + a* b)*sinh(d*x + c)^4 + 5*a*b*cosh(d*x + c)^2 + 40*(3*a*b*cosh(d*x + c)^7 ...
\[ \int \text {csch}(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\int \left (a + b \tanh ^{3}{\left (c + d x \right )}\right )^{2} \operatorname {csch}{\left (c + d x \right )}\, dx \]
Leaf count of result is larger than twice the leaf count of optimal. 447 vs. \(2 (94) = 188\).
Time = 0.29 (sec) , antiderivative size = 447, normalized size of antiderivative = 4.56 \[ \int \text {csch}(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=-2 \, a b {\left (\frac {\arctan \left (e^{\left (-d x - c\right )}\right )}{d} + \frac {e^{\left (-d x - c\right )} - e^{\left (-3 \, d x - 3 \, c\right )}}{d {\left (2 \, e^{\left (-2 \, d x - 2 \, c\right )} + e^{\left (-4 \, d x - 4 \, c\right )} + 1\right )}}\right )} - \frac {2}{15} \, b^{2} {\left (\frac {15 \, e^{\left (-d x - c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {20 \, e^{\left (-3 \, d x - 3 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {58 \, e^{\left (-5 \, d x - 5 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {20 \, e^{\left (-7 \, d x - 7 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}} + \frac {15 \, e^{\left (-9 \, d x - 9 \, c\right )}}{d {\left (5 \, e^{\left (-2 \, d x - 2 \, c\right )} + 10 \, e^{\left (-4 \, d x - 4 \, c\right )} + 10 \, e^{\left (-6 \, d x - 6 \, c\right )} + 5 \, e^{\left (-8 \, d x - 8 \, c\right )} + e^{\left (-10 \, d x - 10 \, c\right )} + 1\right )}}\right )} + \frac {a^{2} \log \left (\tanh \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}{d} \]
-2*a*b*(arctan(e^(-d*x - c))/d + (e^(-d*x - c) - e^(-3*d*x - 3*c))/(d*(2*e ^(-2*d*x - 2*c) + e^(-4*d*x - 4*c) + 1))) - 2/15*b^2*(15*e^(-d*x - 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)) + 20*e^(-3*d*x - 3*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)) + 58*e^(-5*d*x - 5*c)/(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)) + 20*e^(-7*d*x - 7*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)) + 15*e^(-9*d*x - 9*c)/(d*(5*e^(-2*d*x - 2*c) + 10*e^(-4*d*x - 4*c) + 1 0*e^(-6*d*x - 6*c) + 5*e^(-8*d*x - 8*c) + e^(-10*d*x - 10*c) + 1))) + a^2* log(tanh(1/2*d*x + 1/2*c))/d
Time = 0.38 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.82 \[ \int \text {csch}(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {30 \, a b \arctan \left (e^{\left (d x + c\right )}\right ) - 15 \, a^{2} \log \left (e^{\left (d x + c\right )} + 1\right ) + 15 \, a^{2} \log \left ({\left | e^{\left (d x + c\right )} - 1 \right |}\right ) - \frac {2 \, {\left (15 \, a b e^{\left (9 \, d x + 9 \, c\right )} + 15 \, b^{2} e^{\left (9 \, d x + 9 \, c\right )} + 30 \, a b e^{\left (7 \, d x + 7 \, c\right )} + 20 \, b^{2} e^{\left (7 \, d x + 7 \, c\right )} + 58 \, b^{2} e^{\left (5 \, d x + 5 \, c\right )} - 30 \, a b e^{\left (3 \, d x + 3 \, c\right )} + 20 \, b^{2} e^{\left (3 \, d x + 3 \, c\right )} - 15 \, a b e^{\left (d x + c\right )} + 15 \, b^{2} e^{\left (d x + c\right )}\right )}}{{\left (e^{\left (2 \, d x + 2 \, c\right )} + 1\right )}^{5}}}{15 \, d} \]
1/15*(30*a*b*arctan(e^(d*x + c)) - 15*a^2*log(e^(d*x + c) + 1) + 15*a^2*lo g(abs(e^(d*x + c) - 1)) - 2*(15*a*b*e^(9*d*x + 9*c) + 15*b^2*e^(9*d*x + 9* c) + 30*a*b*e^(7*d*x + 7*c) + 20*b^2*e^(7*d*x + 7*c) + 58*b^2*e^(5*d*x + 5 *c) - 30*a*b*e^(3*d*x + 3*c) + 20*b^2*e^(3*d*x + 3*c) - 15*a*b*e^(d*x + c) + 15*b^2*e^(d*x + c))/(e^(2*d*x + 2*c) + 1)^5)/d
Time = 4.05 (sec) , antiderivative size = 522, normalized size of antiderivative = 5.33 \[ \int \text {csch}(c+d x) \left (a+b \tanh ^3(c+d x)\right )^2 \, dx=\frac {a^2\,\ln \left (32\,a^6+32\,a^4\,b^2-32\,a^6\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-32\,a^4\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{d}-\frac {176\,b^2\,{\mathrm {e}}^{c+d\,x}}{15\,\left (d+3\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+3\,d\,{\mathrm {e}}^{4\,c+4\,d\,x}+d\,{\mathrm {e}}^{6\,c+6\,d\,x}\right )}-\frac {32\,b^2\,{\mathrm {e}}^{c+d\,x}}{5\,\left (d+5\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+10\,d\,{\mathrm {e}}^{4\,c+4\,d\,x}+10\,d\,{\mathrm {e}}^{6\,c+6\,d\,x}+5\,d\,{\mathrm {e}}^{8\,c+8\,d\,x}+d\,{\mathrm {e}}^{10\,c+10\,d\,x}\right )}-\frac {a^2\,\ln \left (-32\,a^6-32\,a^4\,b^2-32\,a^6\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-32\,a^4\,b^2\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c\right )}{d}-\frac {2\,b^2\,{\mathrm {e}}^{c+d\,x}}{d+d\,{\mathrm {e}}^{2\,c+2\,d\,x}}+\frac {16\,b^2\,{\mathrm {e}}^{c+d\,x}}{3\,\left (d+2\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{4\,c+4\,d\,x}\right )}+\frac {64\,b^2\,{\mathrm {e}}^{c+d\,x}}{5\,\left (d+4\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+6\,d\,{\mathrm {e}}^{4\,c+4\,d\,x}+4\,d\,{\mathrm {e}}^{6\,c+6\,d\,x}+d\,{\mathrm {e}}^{8\,c+8\,d\,x}\right )}-\frac {2\,a\,b\,{\mathrm {e}}^{c+d\,x}}{d+d\,{\mathrm {e}}^{2\,c+2\,d\,x}}+\frac {4\,a\,b\,{\mathrm {e}}^{c+d\,x}}{d+2\,d\,{\mathrm {e}}^{2\,c+2\,d\,x}+d\,{\mathrm {e}}^{4\,c+4\,d\,x}}-\frac {a\,b\,\left (\ln \left (32\,a^3\,b^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+32\,a^5\,b\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c-a^5\,b\,32{}\mathrm {i}-a^3\,b^3\,32{}\mathrm {i}\right )\,1{}\mathrm {i}-\ln \left (32\,a^3\,b^3\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+32\,a^5\,b\,{\mathrm {e}}^{d\,x}\,{\mathrm {e}}^c+a^5\,b\,32{}\mathrm {i}+a^3\,b^3\,32{}\mathrm {i}\right )\,1{}\mathrm {i}\right )}{d} \]
(a^2*log(32*a^6 + 32*a^4*b^2 - 32*a^6*exp(d*x)*exp(c) - 32*a^4*b^2*exp(d*x )*exp(c)))/d - (176*b^2*exp(c + d*x))/(15*(d + 3*d*exp(2*c + 2*d*x) + 3*d* exp(4*c + 4*d*x) + d*exp(6*c + 6*d*x))) - (32*b^2*exp(c + d*x))/(5*(d + 5* d*exp(2*c + 2*d*x) + 10*d*exp(4*c + 4*d*x) + 10*d*exp(6*c + 6*d*x) + 5*d*e xp(8*c + 8*d*x) + d*exp(10*c + 10*d*x))) - (a^2*log(- 32*a^6 - 32*a^4*b^2 - 32*a^6*exp(d*x)*exp(c) - 32*a^4*b^2*exp(d*x)*exp(c)))/d - (2*b^2*exp(c + d*x))/(d + d*exp(2*c + 2*d*x)) + (16*b^2*exp(c + d*x))/(3*(d + 2*d*exp(2* c + 2*d*x) + d*exp(4*c + 4*d*x))) + (64*b^2*exp(c + d*x))/(5*(d + 4*d*exp( 2*c + 2*d*x) + 6*d*exp(4*c + 4*d*x) + 4*d*exp(6*c + 6*d*x) + d*exp(8*c + 8 *d*x))) - (2*a*b*exp(c + d*x))/(d + d*exp(2*c + 2*d*x)) + (4*a*b*exp(c + d *x))/(d + 2*d*exp(2*c + 2*d*x) + d*exp(4*c + 4*d*x)) - (a*b*(log(32*a^3*b^ 3*exp(d*x)*exp(c) - a^3*b^3*32i - a^5*b*32i + 32*a^5*b*exp(d*x)*exp(c))*1i - log(a^5*b*32i + a^3*b^3*32i + 32*a^3*b^3*exp(d*x)*exp(c) + 32*a^5*b*exp (d*x)*exp(c))*1i))/d