Integrand size = 23, antiderivative size = 384 \[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\frac {a^{2/3} \sqrt [3]{b} \left (a^2-3 a^{2/3} b^{4/3}+2 b^2\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \left (a^2-b^2\right )^2 d}+\frac {(a-2 b) \log (1-\tanh (c+d x))}{4 (a+b)^2 d}-\frac {(a+2 b) \log (1+\tanh (c+d x))}{4 (a-b)^2 d}+\frac {a^{2/3} \sqrt [3]{b} \left (a^2+3 a^{2/3} b^{4/3}+2 b^2\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}-\frac {a^{2/3} \sqrt [3]{b} \left (a^2+3 a^{2/3} b^{4/3}+2 b^2\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 \left (a^2-b^2\right )^2 d}+\frac {b \left (2 a^2+b^2\right ) \log \left (a+b \tanh ^3(c+d x)\right )}{3 \left (a^2-b^2\right )^2 d}+\frac {1}{4 (a+b) d (1-\tanh (c+d x))}-\frac {1}{4 (a-b) d (1+\tanh (c+d x))} \] Output:
1/3*a^(2/3)*b^(1/3)*(a^2-3*a^(2/3)*b^(4/3)+2*b^2)*arctan(1/3*(a^(1/3)-2*b^ (1/3)*tanh(d*x+c))*3^(1/2)/a^(1/3))*3^(1/2)/(a^2-b^2)^2/d+1/4*(a-2*b)*ln(1 -tanh(d*x+c))/(a+b)^2/d-1/4*(a+2*b)*ln(1+tanh(d*x+c))/(a-b)^2/d+1/3*a^(2/3 )*b^(1/3)*(a^2+3*a^(2/3)*b^(4/3)+2*b^2)*ln(a^(1/3)+b^(1/3)*tanh(d*x+c))/(a ^2-b^2)^2/d-1/6*a^(2/3)*b^(1/3)*(a^2+3*a^(2/3)*b^(4/3)+2*b^2)*ln(a^(2/3)-a ^(1/3)*b^(1/3)*tanh(d*x+c)+b^(2/3)*tanh(d*x+c)^2)/(a^2-b^2)^2/d+1/3*b*(2*a ^2+b^2)*ln(a+b*tanh(d*x+c)^3)/(a^2-b^2)^2/d+1/4/(a+b)/d/(1-tanh(d*x+c))-1/ 4/(a-b)/d/(1+tanh(d*x+c))
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.80 (sec) , antiderivative size = 423, normalized size of antiderivative = 1.10 \[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=-\frac {6 \left (a^2-3 a b+2 b^2\right ) (c+d x)+3 b (a+b) \cosh (2 (c+d x))+4 b \text {RootSum}\left [a-b+3 a \text {$\#$1}+3 b \text {$\#$1}+3 a \text {$\#$1}^2-3 b \text {$\#$1}^2+a \text {$\#$1}^3+b \text {$\#$1}^3\&,\frac {4 a^2 c+2 b^2 c+4 a^2 d x+2 b^2 d x-2 a^2 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right )-b^2 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right )+4 a^2 c \text {$\#$1}-4 b^2 c \text {$\#$1}+4 a^2 d x \text {$\#$1}-4 b^2 d x \text {$\#$1}-2 a^2 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right ) \text {$\#$1}+2 b^2 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right ) \text {$\#$1}+8 a^2 c \text {$\#$1}^2-8 a b c \text {$\#$1}^2+2 b^2 c \text {$\#$1}^2+8 a^2 d x \text {$\#$1}^2-8 a b d x \text {$\#$1}^2+2 b^2 d x \text {$\#$1}^2-4 a^2 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right ) \text {$\#$1}^2+4 a b \log \left (e^{2 (c+d x)}-\text {$\#$1}\right ) \text {$\#$1}^2-b^2 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right ) \text {$\#$1}^2}{a-b+2 a \text {$\#$1}+2 b \text {$\#$1}+a \text {$\#$1}^2-b \text {$\#$1}^2}\&\right ]-3 a (a+b) \sinh (2 (c+d x))}{12 (a-b) (a+b)^2 d} \] Input:
Integrate[Sinh[c + d*x]^2/(a + b*Tanh[c + d*x]^3),x]
Output:
-1/12*(6*(a^2 - 3*a*b + 2*b^2)*(c + d*x) + 3*b*(a + b)*Cosh[2*(c + d*x)] + 4*b*RootSum[a - b + 3*a*#1 + 3*b*#1 + 3*a*#1^2 - 3*b*#1^2 + a*#1^3 + b*#1 ^3 & , (4*a^2*c + 2*b^2*c + 4*a^2*d*x + 2*b^2*d*x - 2*a^2*Log[E^(2*(c + d* x)) - #1] - b^2*Log[E^(2*(c + d*x)) - #1] + 4*a^2*c*#1 - 4*b^2*c*#1 + 4*a^ 2*d*x*#1 - 4*b^2*d*x*#1 - 2*a^2*Log[E^(2*(c + d*x)) - #1]*#1 + 2*b^2*Log[E ^(2*(c + d*x)) - #1]*#1 + 8*a^2*c*#1^2 - 8*a*b*c*#1^2 + 2*b^2*c*#1^2 + 8*a ^2*d*x*#1^2 - 8*a*b*d*x*#1^2 + 2*b^2*d*x*#1^2 - 4*a^2*Log[E^(2*(c + d*x)) - #1]*#1^2 + 4*a*b*Log[E^(2*(c + d*x)) - #1]*#1^2 - b^2*Log[E^(2*(c + d*x) ) - #1]*#1^2)/(a - b + 2*a*#1 + 2*b*#1 + a*#1^2 - b*#1^2) & ] - 3*a*(a + b )*Sinh[2*(c + d*x)])/((a - b)*(a + b)^2*d)
Time = 0.91 (sec) , antiderivative size = 364, normalized size of antiderivative = 0.95, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 25, 4146, 7276, 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.
| \(\displaystyle \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {\sin (i c+i d x)^2}{a+i b \tan (i c+i d x)^3}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {\sin (i c+i d x)^2}{i b \tan (i c+i d x)^3+a}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tanh ^2(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^2 \left (b \tanh ^3(c+d x)+a\right )}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \frac {\int \left (\frac {-a-2 b}{4 (a-b)^2 (\tanh (c+d x)+1)}+\frac {a-2 b}{4 (a+b)^2 (\tanh (c+d x)-1)}+\frac {b \left (3 b a^2-\left (a^2+2 b^2\right ) \tanh (c+d x) a+b \left (2 a^2+b^2\right ) \tanh ^2(c+d x)\right )}{\left (a^2-b^2\right )^2 \left (b \tanh ^3(c+d x)+a\right )}+\frac {1}{4 (a+b) (\tanh (c+d x)-1)^2}+\frac {1}{4 (a-b) (\tanh (c+d x)+1)^2}\right )d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\frac {b \left (2 a^2+b^2\right ) \log \left (a+b \tanh ^3(c+d x)\right )}{3 \left (a^2-b^2\right )^2}+\frac {a^{2/3} \sqrt [3]{b} \left (-3 a^{2/3} b^{4/3}+a^2+2 b^2\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \left (a^2-b^2\right )^2}-\frac {a^{2/3} \sqrt [3]{b} \left (3 a^{2/3} b^{4/3}+a^2+2 b^2\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 \left (a^2-b^2\right )^2}+\frac {a^{2/3} \sqrt [3]{b} \left (3 a^{2/3} b^{4/3}+a^2+2 b^2\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 \left (a^2-b^2\right )^2}+\frac {1}{4 (a+b) (1-\tanh (c+d x))}-\frac {1}{4 (a-b) (\tanh (c+d x)+1)}+\frac {(a-2 b) \log (1-\tanh (c+d x))}{4 (a+b)^2}-\frac {(a+2 b) \log (\tanh (c+d x)+1)}{4 (a-b)^2}}{d}\) |
Input:
Int[Sinh[c + d*x]^2/(a + b*Tanh[c + d*x]^3),x]
Output:
((a^(2/3)*b^(1/3)*(a^2 - 3*a^(2/3)*b^(4/3) + 2*b^2)*ArcTan[(a^(1/3) - 2*b^ (1/3)*Tanh[c + d*x])/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*(a^2 - b^2)^2) + ((a - 2 *b)*Log[1 - Tanh[c + d*x]])/(4*(a + b)^2) - ((a + 2*b)*Log[1 + Tanh[c + d* x]])/(4*(a - b)^2) + (a^(2/3)*b^(1/3)*(a^2 + 3*a^(2/3)*b^(4/3) + 2*b^2)*Lo g[a^(1/3) + b^(1/3)*Tanh[c + d*x]])/(3*(a^2 - b^2)^2) - (a^(2/3)*b^(1/3)*( a^2 + 3*a^(2/3)*b^(4/3) + 2*b^2)*Log[a^(2/3) - a^(1/3)*b^(1/3)*Tanh[c + d* x] + b^(2/3)*Tanh[c + d*x]^2])/(6*(a^2 - b^2)^2) + (b*(2*a^2 + b^2)*Log[a + b*Tanh[c + d*x]^3])/(3*(a^2 - b^2)^2) + 1/(4*(a + b)*(1 - Tanh[c + d*x]) ) - 1/(4*(a - b)*(1 + Tanh[c + d*x])))/d
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_ )])^(n_))^(p_.), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Sim p[c*(ff^(m + 1)/f) Subst[Int[x^m*((a + b*(ff*x)^n)^p/(c^2 + ff^2*x^2)^(m/ 2 + 1)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, e, f, n, p}, x ] && IntegerQ[m/2]
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 4.36 (sec) , antiderivative size = 301, normalized size of antiderivative = 0.78
| method | result | size |
| derivativedivides | \(\frac {\frac {4}{\left (8 a +8 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (a -2 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {4}{\left (8 a -8 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{\left (16 a -16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-a -2 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 \left (a -b \right )^{2}}+\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (a \left (2 a^{2}+b^{2}\right ) \textit {\_R}^{5}-3 a^{2} b \,\textit {\_R}^{4}+6 a \left (a^{2}+b^{2}\right ) \textit {\_R}^{3}+4 b \left (2 a^{2}+b^{2}\right ) \textit {\_R}^{2}-3 a \textit {\_R} \,b^{2}+3 a^{2} b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 \left (a -b \right )^{2} \left (a +b \right )^{2}}}{d}\) | \(301\) |
| default | \(\frac {\frac {4}{\left (8 a +8 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\frac {8}{\left (16 a +16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\left (a -2 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 \left (a +b \right )^{2}}-\frac {4}{\left (8 a -8 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {8}{\left (16 a -16 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {\left (-a -2 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 \left (a -b \right )^{2}}+\frac {b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (a \left (2 a^{2}+b^{2}\right ) \textit {\_R}^{5}-3 a^{2} b \,\textit {\_R}^{4}+6 a \left (a^{2}+b^{2}\right ) \textit {\_R}^{3}+4 b \left (2 a^{2}+b^{2}\right ) \textit {\_R}^{2}-3 a \textit {\_R} \,b^{2}+3 a^{2} b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 \left (a -b \right )^{2} \left (a +b \right )^{2}}}{d}\) | \(301\) |
| risch | \(-\frac {a x}{2 \left (a +b \right )^{2}}+\frac {x b}{\left (a +b \right )^{2}}+\frac {{\mathrm e}^{2 d x +2 c}}{8 \left (a +b \right ) d}-\frac {{\mathrm e}^{-2 d x -2 c}}{8 \left (a -b \right ) d}-\frac {4 a^{2} b \,d^{3} x}{a^{4} d^{3}-2 a^{2} b^{2} d^{3}+b^{4} d^{3}}-\frac {2 b^{3} d^{3} x}{a^{4} d^{3}-2 a^{2} b^{2} d^{3}+b^{4} d^{3}}-\frac {4 a^{2} b c \,d^{2}}{a^{4} d^{3}-2 a^{2} b^{2} d^{3}+b^{4} d^{3}}-\frac {2 b^{3} c \,d^{2}}{a^{4} d^{3}-2 a^{2} b^{2} d^{3}+b^{4} d^{3}}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (27 a^{4} d^{3}-54 a^{2} b^{2} d^{3}+27 b^{4} d^{3}\right ) \textit {\_Z}^{3}+\left (-54 a^{2} b \,d^{2}-27 b^{3} d^{2}\right ) \textit {\_Z}^{2}+9 b^{2} d \textit {\_Z} -b \right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 d x +2 c}+\left (\frac {18 d^{2} a^{5}}{a^{3}+8 b^{2} a}+\frac {36 d^{2} b \,a^{4}}{a^{3}+8 b^{2} a}-\frac {36 d^{2} b^{2} a^{3}}{a^{3}+8 b^{2} a}-\frac {72 d^{2} b^{3} a^{2}}{a^{3}+8 b^{2} a}+\frac {18 d^{2} b^{4} a}{a^{3}+8 b^{2} a}+\frac {36 d^{2} b^{5}}{a^{3}+8 b^{2} a}\right ) \textit {\_R}^{2}+\left (-\frac {6 a^{4} d}{a^{3}+8 b^{2} a}-\frac {18 a^{3} b d}{a^{3}+8 b^{2} a}-\frac {78 a^{2} b^{2} d}{a^{3}+8 b^{2} a}-\frac {36 a \,b^{3} d}{a^{3}+8 b^{2} a}-\frac {24 b^{4} d}{a^{3}+8 b^{2} a}\right ) \textit {\_R} +\frac {a^{3}}{a^{3}+8 b^{2} a}+\frac {2 a^{2} b}{a^{3}+8 b^{2} a}+\frac {2 b^{2} a}{a^{3}+8 b^{2} a}+\frac {4 b^{3}}{a^{3}+8 b^{2} a}\right )\right )\) | \(591\) |
Input:
int(sinh(d*x+c)^2/(a+b*tanh(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
1/d*(4/(8*a+8*b)/(tanh(1/2*d*x+1/2*c)-1)^2+8/(16*a+16*b)/(tanh(1/2*d*x+1/2 *c)-1)+1/2*(a-2*b)/(a+b)^2*ln(tanh(1/2*d*x+1/2*c)-1)-4/(8*a-8*b)/(tanh(1/2 *d*x+1/2*c)+1)^2+8/(16*a-16*b)/(tanh(1/2*d*x+1/2*c)+1)+1/2/(a-b)^2*(-a-2*b )*ln(tanh(1/2*d*x+1/2*c)+1)+1/3*b/(a-b)^2/(a+b)^2*sum((a*(2*a^2+b^2)*_R^5- 3*a^2*b*_R^4+6*a*(a^2+b^2)*_R^3+4*b*(2*a^2+b^2)*_R^2-3*a*_R*b^2+3*a^2*b)/( _R^5*a+2*_R^3*a+4*_R^2*b+_R*a)*ln(tanh(1/2*d*x+1/2*c)-_R),_R=RootOf(_Z^6*a +3*_Z^4*a+8*_Z^3*b+3*_Z^2*a+a)))
Result contains complex when optimal does not.
Time = 1.00 (sec) , antiderivative size = 10695, normalized size of antiderivative = 27.85 \[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^2/(a+b*tanh(d*x+c)^3),x, algorithm="fricas")
Output:
Too large to include
\[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\int \frac {\sinh ^{2}{\left (c + d x \right )}}{a + b \tanh ^{3}{\left (c + d x \right )}}\, dx \] Input:
integrate(sinh(d*x+c)**2/(a+b*tanh(d*x+c)**3),x)
Output:
Integral(sinh(c + d*x)**2/(a + b*tanh(c + d*x)**3), x)
\[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\int { \frac {\sinh \left (d x + c\right )^{2}}{b \tanh \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(sinh(d*x+c)^2/(a+b*tanh(d*x+c)^3),x, algorithm="maxima")
Output:
4*a^2*b*(integrate(((a + b)*e^(4*d*x + 4*c) + 3*(a - b)*e^(2*d*x + 2*c) + 3*a + 3*b)*e^(2*d*x + 2*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^4 - 2*a^2*b^2 + b^4) - ( d*x + c)/((a^4 - 2*a^2*b^2 + b^4)*d)) + 2*b^3*(integrate(((a + b)*e^(4*d*x + 4*c) + 3*(a - b)*e^(2*d*x + 2*c) + 3*a + 3*b)*e^(2*d*x + 2*c)/((a + b)* e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^4 - 2*a^2*b^2 + b^4) - (d*x + c)/((a^4 - 2*a^2*b^2 + b^4)*d) ) - 8*a^2*b*integrate(e^(4*d*x + 4*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b) *e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^3 + a^2*b - a *b^2 - b^3) + 8*a*b^2*integrate(e^(4*d*x + 4*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^3 + a^2*b - a*b^2 - b^3) - 2*b^3*integrate(e^(4*d*x + 4*c)/((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x )/(a^3 + a^2*b - a*b^2 - b^3) - 4*a^2*b*integrate(e^(2*d*x + 2*c)/((a + b) *e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^3 + a^2*b - a*b^2 - b^3) + 4*b^3*integrate(e^(2*d*x + 2*c)/ ((a + b)*e^(6*d*x + 6*c) + 3*(a - b)*e^(4*d*x + 4*c) + 3*(a + b)*e^(2*d*x + 2*c) + a - b), x)/(a^3 + a^2*b - a*b^2 - b^3) - 1/8*(4*(a^2*d*e^(2*c) - 3*a*b*d*e^(2*c) + 2*b^2*d*e^(2*c))*x*e^(2*d*x) + a^2 + 2*a*b + b^2 - (a^2* e^(4*c) - b^2*e^(4*c))*e^(4*d*x))*e^(-2*d*x)/(a^3*d*e^(2*c) + a^2*b*d*e...
\[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\int { \frac {\sinh \left (d x + c\right )^{2}}{b \tanh \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(sinh(d*x+c)^2/(a+b*tanh(d*x+c)^3),x, algorithm="giac")
Output:
sage0*x
Time = 3.88 (sec) , antiderivative size = 2100, normalized size of antiderivative = 5.47 \[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
int(sinh(c + d*x)^2/(a + b*tanh(c + d*x)^3),x)
Output:
symsum(log(root(54*a^2*b^2*d^3*z^3 - 27*b^4*d^3*z^3 - 27*a^4*d^3*z^3 + 54* a^2*b*d^2*z^2 + 27*b^3*d^2*z^2 - 9*b^2*d*z + b, z, k)*((2304*root(54*a^2*b ^2*d^3*z^3 - 27*b^4*d^3*z^3 - 27*a^4*d^3*z^3 + 54*a^2*b*d^2*z^2 + 27*b^3*d ^2*z^2 - 9*b^2*d*z + b, z, k)*(146*a^5*b^5*d^2 - 133*a^4*b^6*d^2 - 24*a^3* b^7*d^2 - 12*a^6*b^4*d^2 + 22*a^7*b^3*d^2 + a^8*b^2*d^2 + 32*a^3*b^7*d^2*e xp(2*root(54*a^2*b^2*d^3*z^3 - 27*b^4*d^3*z^3 - 27*a^4*d^3*z^3 + 54*a^2*b* d^2*z^2 + 27*b^3*d^2*z^2 - 9*b^2*d*z + b, z, k))*exp(2*d*x) + 577*a^4*b^6* d^2*exp(2*root(54*a^2*b^2*d^3*z^3 - 27*b^4*d^3*z^3 - 27*a^4*d^3*z^3 + 54*a ^2*b*d^2*z^2 + 27*b^3*d^2*z^2 - 9*b^2*d*z + b, z, k))*exp(2*d*x) + 548*a^5 *b^5*d^2*exp(2*root(54*a^2*b^2*d^3*z^3 - 27*b^4*d^3*z^3 - 27*a^4*d^3*z^3 + 54*a^2*b*d^2*z^2 + 27*b^3*d^2*z^2 - 9*b^2*d*z + b, z, k))*exp(2*d*x) + 70 *a^6*b^4*d^2*exp(2*root(54*a^2*b^2*d^3*z^3 - 27*b^4*d^3*z^3 - 27*a^4*d^3*z ^3 + 54*a^2*b*d^2*z^2 + 27*b^3*d^2*z^2 - 9*b^2*d*z + b, z, k))*exp(2*d*x) + 68*a^7*b^3*d^2*exp(2*root(54*a^2*b^2*d^3*z^3 - 27*b^4*d^3*z^3 - 27*a^4*d ^3*z^3 + 54*a^2*b*d^2*z^2 + 27*b^3*d^2*z^2 - 9*b^2*d*z + b, z, k))*exp(2*d *x) + a^8*b^2*d^2*exp(2*root(54*a^2*b^2*d^3*z^3 - 27*b^4*d^3*z^3 - 27*a^4* d^3*z^3 + 54*a^2*b*d^2*z^2 + 27*b^3*d^2*z^2 - 9*b^2*d*z + b, z, k))*exp(2* d*x)))/((a + b)^8*(a - b)^2*(a^2 - 2*a*b + b^2)) + (1536*(24*a^3*b^8*d + 1 05*a^4*b^7*d - 156*a^5*b^6*d + 51*a^6*b^5*d - 30*a^7*b^4*d + 6*a^8*b^3*d - 32*a^3*b^8*d*exp(2*root(54*a^2*b^2*d^3*z^3 - 27*b^4*d^3*z^3 - 27*a^4*d...
\[ \int \frac {\sinh ^2(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\text {too large to display} \] Input:
int(sinh(d*x+c)^2/(a+b*tanh(d*x+c)^3),x)
Output:
(3*e**(4*c + 4*d*x)*a**3 - 27*e**(4*c + 4*d*x)*a**2*b - 27*e**(4*c + 4*d*x )*a*b**2 + 3*e**(4*c + 4*d*x)*b**3 - 288*e**(2*c + 2*d*x)*int(1/(e**(8*c + 8*d*x)*a**5 - 7*e**(8*c + 8*d*x)*a**4*b - 26*e**(8*c + 8*d*x)*a**3*b**2 - 26*e**(8*c + 8*d*x)*a**2*b**3 - 7*e**(8*c + 8*d*x)*a*b**4 + e**(8*c + 8*d *x)*b**5 + 3*e**(6*c + 6*d*x)*a**5 - 27*e**(6*c + 6*d*x)*a**4*b - 30*e**(6 *c + 6*d*x)*a**3*b**2 + 30*e**(6*c + 6*d*x)*a**2*b**3 + 27*e**(6*c + 6*d*x )*a*b**4 - 3*e**(6*c + 6*d*x)*b**5 + 3*e**(4*c + 4*d*x)*a**5 - 21*e**(4*c + 4*d*x)*a**4*b - 78*e**(4*c + 4*d*x)*a**3*b**2 - 78*e**(4*c + 4*d*x)*a**2 *b**3 - 21*e**(4*c + 4*d*x)*a*b**4 + 3*e**(4*c + 4*d*x)*b**5 + e**(2*c + 2 *d*x)*a**5 - 9*e**(2*c + 2*d*x)*a**4*b - 10*e**(2*c + 2*d*x)*a**3*b**2 + 1 0*e**(2*c + 2*d*x)*a**2*b**3 + 9*e**(2*c + 2*d*x)*a*b**4 - e**(2*c + 2*d*x )*b**5),x)*a**7*b*d + 2208*e**(2*c + 2*d*x)*int(1/(e**(8*c + 8*d*x)*a**5 - 7*e**(8*c + 8*d*x)*a**4*b - 26*e**(8*c + 8*d*x)*a**3*b**2 - 26*e**(8*c + 8*d*x)*a**2*b**3 - 7*e**(8*c + 8*d*x)*a*b**4 + e**(8*c + 8*d*x)*b**5 + 3*e **(6*c + 6*d*x)*a**5 - 27*e**(6*c + 6*d*x)*a**4*b - 30*e**(6*c + 6*d*x)*a* *3*b**2 + 30*e**(6*c + 6*d*x)*a**2*b**3 + 27*e**(6*c + 6*d*x)*a*b**4 - 3*e **(6*c + 6*d*x)*b**5 + 3*e**(4*c + 4*d*x)*a**5 - 21*e**(4*c + 4*d*x)*a**4* b - 78*e**(4*c + 4*d*x)*a**3*b**2 - 78*e**(4*c + 4*d*x)*a**2*b**3 - 21*e** (4*c + 4*d*x)*a*b**4 + 3*e**(4*c + 4*d*x)*b**5 + e**(2*c + 2*d*x)*a**5 - 9 *e**(2*c + 2*d*x)*a**4*b - 10*e**(2*c + 2*d*x)*a**3*b**2 + 10*e**(2*c +...