Integrand size = 23, antiderivative size = 491 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=-\frac {a^{2/3} \sqrt [3]{b} \left (a^2+3 a^{4/3} b^{2/3}-b^2\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \left (a^{4/3}+a^{2/3} b^{2/3}+b^{4/3}\right )^3 d}-\frac {3 a (a-5 b) \log (1-\tanh (c+d x))}{16 (a+b)^3 d}+\frac {3 a (a+5 b) \log (1+\tanh (c+d x))}{16 (a-b)^3 d}-\frac {a^{2/3} \sqrt [3]{b} \left (a^4+7 a^2 b^2+b^4+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 \left (a^2-b^2\right )^3 d}+\frac {a^{2/3} \sqrt [3]{b} \left (a^4+7 a^2 b^2+b^4+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 \left (a^2-b^2\right )^3 d}-\frac {a^2 b \left (a^2+2 b^2\right ) \log \left (a+b \tanh ^3(c+d x)\right )}{\left (a^2-b^2\right )^3 d}+\frac {1}{16 (a+b) d (1-\tanh (c+d x))^2}-\frac {5 a-b}{16 (a+b)^2 d (1-\tanh (c+d x))}-\frac {1}{16 (a-b) d (1+\tanh (c+d x))^2}+\frac {5 a+b}{16 (a-b)^2 d (1+\tanh (c+d x))} \] Output:
-1/3*a^(2/3)*b^(1/3)*(a^2+3*a^(4/3)*b^(2/3)-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^(4/3)+a^(2/3)*b^(2/3)+b^(4/3 ))^3/d-3/16*a*(a-5*b)*ln(1-tanh(d*x+c))/(a+b)^3/d+3/16*a*(a+5*b)*ln(1+tanh (d*x+c))/(a-b)^3/d-1/3*a^(2/3)*b^(1/3)*(a^4+7*a^2*b^2+b^4+3*a^(2/3)*b^(4/3 )*(2*a^2+b^2))*ln(a^(1/3)+b^(1/3)*tanh(d*x+c))/(a^2-b^2)^3/d+1/6*a^(2/3)*b ^(1/3)*(a^4+7*a^2*b^2+b^4+3*a^(2/3)*b^(4/3)*(2*a^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)^3/d-a^2*b*(a^2+2*b^ 2)*ln(a+b*tanh(d*x+c)^3)/(a^2-b^2)^3/d+1/16/(a+b)/d/(1-tanh(d*x+c))^2-1/16 *(5*a-b)/(a+b)^2/d/(1-tanh(d*x+c))-1/16/(a-b)/d/(1+tanh(d*x+c))^2+1/16*(5* a+b)/(a-b)^2/d/(1+tanh(d*x+c))
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 1.30 (sec) , antiderivative size = 645, normalized size of antiderivative = 1.31 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\frac {-32 a b \text {RootSum}\left [a-b+3 a \text {$\#$1}+3 b \text {$\#$1}+3 a \text {$\#$1}^2-3 b \text {$\#$1}^2+a \text {$\#$1}^3+b \text {$\#$1}^3\&,\frac {-6 a^3 c-12 a b^2 c-6 a^3 d x-12 a b^2 d x+3 a^3 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right )+6 a b^2 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right )-8 a^3 c \text {$\#$1}+4 a^2 b c \text {$\#$1}+8 a b^2 c \text {$\#$1}-4 b^3 c \text {$\#$1}-8 a^3 d x \text {$\#$1}+4 a^2 b d x \text {$\#$1}+8 a b^2 d x \text {$\#$1}-4 b^3 d x \text {$\#$1}+4 a^3 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right ) \text {$\#$1}-2 a^2 b \log \left (e^{2 (c+d x)}-\text {$\#$1}\right ) \text {$\#$1}-4 a b^2 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right ) \text {$\#$1}+2 b^3 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right ) \text {$\#$1}-10 a^3 c \text {$\#$1}^2+20 a^2 b c \text {$\#$1}^2-20 a b^2 c \text {$\#$1}^2+4 b^3 c \text {$\#$1}^2-10 a^3 d x \text {$\#$1}^2+20 a^2 b d x \text {$\#$1}^2-20 a b^2 d x \text {$\#$1}^2+4 b^3 d x \text {$\#$1}^2+5 a^3 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right ) \text {$\#$1}^2-10 a^2 b \log \left (e^{2 (c+d x)}-\text {$\#$1}\right ) \text {$\#$1}^2+10 a b^2 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right ) \text {$\#$1}^2-2 b^3 \log \left (e^{2 (c+d x)}-\text {$\#$1}\right ) \text {$\#$1}^2}{a-b+2 a \text {$\#$1}+2 b \text {$\#$1}+a \text {$\#$1}^2-b \text {$\#$1}^2}\&\right ]+3 \left (4 b \left (5 a^3+5 a^2 b+a b^2+b^3\right ) \cosh (2 (c+d x))-(a-b) b (a+b)^2 \cosh (4 (c+d x))-8 a \left (a^3+a^2 b+2 a b^2+2 b^3\right ) \sinh (2 (c+d x))+a (a-b) \left (12 \left (a^2-6 a b+5 b^2\right ) (c+d x)+(a+b)^2 \sinh (4 (c+d x))\right )\right )}{96 (a-b)^2 (a+b)^3 d} \] Input:
Integrate[Sinh[c + d*x]^4/(a + b*Tanh[c + d*x]^3),x]
Output:
(-32*a*b*RootSum[a - b + 3*a*#1 + 3*b*#1 + 3*a*#1^2 - 3*b*#1^2 + a*#1^3 + b*#1^3 & , (-6*a^3*c - 12*a*b^2*c - 6*a^3*d*x - 12*a*b^2*d*x + 3*a^3*Log[E ^(2*(c + d*x)) - #1] + 6*a*b^2*Log[E^(2*(c + d*x)) - #1] - 8*a^3*c*#1 + 4* a^2*b*c*#1 + 8*a*b^2*c*#1 - 4*b^3*c*#1 - 8*a^3*d*x*#1 + 4*a^2*b*d*x*#1 + 8 *a*b^2*d*x*#1 - 4*b^3*d*x*#1 + 4*a^3*Log[E^(2*(c + d*x)) - #1]*#1 - 2*a^2* b*Log[E^(2*(c + d*x)) - #1]*#1 - 4*a*b^2*Log[E^(2*(c + d*x)) - #1]*#1 + 2* b^3*Log[E^(2*(c + d*x)) - #1]*#1 - 10*a^3*c*#1^2 + 20*a^2*b*c*#1^2 - 20*a* b^2*c*#1^2 + 4*b^3*c*#1^2 - 10*a^3*d*x*#1^2 + 20*a^2*b*d*x*#1^2 - 20*a*b^2 *d*x*#1^2 + 4*b^3*d*x*#1^2 + 5*a^3*Log[E^(2*(c + d*x)) - #1]*#1^2 - 10*a^2 *b*Log[E^(2*(c + d*x)) - #1]*#1^2 + 10*a*b^2*Log[E^(2*(c + d*x)) - #1]*#1^ 2 - 2*b^3*Log[E^(2*(c + d*x)) - #1]*#1^2)/(a - b + 2*a*#1 + 2*b*#1 + a*#1^ 2 - b*#1^2) & ] + 3*(4*b*(5*a^3 + 5*a^2*b + a*b^2 + b^3)*Cosh[2*(c + d*x)] - (a - b)*b*(a + b)^2*Cosh[4*(c + d*x)] - 8*a*(a^3 + a^2*b + 2*a*b^2 + 2* b^3)*Sinh[2*(c + d*x)] + a*(a - b)*(12*(a^2 - 6*a*b + 5*b^2)*(c + d*x) + ( a + b)^2*Sinh[4*(c + d*x)])))/(96*(a - b)^2*(a + b)^3*d)
Time = 1.17 (sec) , antiderivative size = 465, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 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.
| \(\displaystyle \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^3(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (i c+i d x)^4}{a+i b \tan (i c+i d x)^3}dx\) |
| \(\Big \downarrow \) 4146 |
| \(\displaystyle \frac {\int \frac {\tanh ^4(c+d x)}{\left (1-\tanh ^2(c+d x)\right )^3 \left (b \tanh ^3(c+d x)+a\right )}d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \frac {\int \left (-\frac {3 a (a-5 b)}{16 (a+b)^3 (\tanh (c+d x)-1)}+\frac {3 a (a+5 b)}{16 (a-b)^3 (\tanh (c+d x)+1)}+\frac {a b \left (-3 a b \left (a^2+2 b^2\right ) \tanh ^2(c+d x)+\left (a^4+7 b^2 a^2+b^4\right ) \tanh (c+d x)-3 a b \left (2 a^2+b^2\right )\right )}{\left (a^2-b^2\right )^3 \left (b \tanh ^3(c+d x)+a\right )}+\frac {b-5 a}{16 (a+b)^2 (\tanh (c+d x)-1)^2}+\frac {-5 a-b}{16 (a-b)^2 (\tanh (c+d x)+1)^2}-\frac {1}{8 (a+b) (\tanh (c+d x)-1)^3}+\frac {1}{8 (a-b) (\tanh (c+d x)+1)^3}\right )d\tanh (c+d x)}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {a^2 b \left (a^2+2 b^2\right ) \log \left (a+b \tanh ^3(c+d x)\right )}{\left (a^2-b^2\right )^3}-\frac {a^{2/3} \sqrt [3]{b} \left (3 a^{4/3} b^{2/3}+a^2-b^2\right ) \arctan \left (\frac {\sqrt [3]{a}-2 \sqrt [3]{b} \tanh (c+d x)}{\sqrt {3} \sqrt [3]{a}}\right )}{\sqrt {3} \left (a^{2/3} b^{2/3}+a^{4/3}+b^{4/3}\right )^3}+\frac {a^{2/3} \sqrt [3]{b} \left (a^4+7 a^2 b^2+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )+b^4\right ) \log \left (a^{2/3}-\sqrt [3]{a} \sqrt [3]{b} \tanh (c+d x)+b^{2/3} \tanh ^2(c+d x)\right )}{6 \left (a^2-b^2\right )^3}-\frac {a^{2/3} \sqrt [3]{b} \left (a^4+7 a^2 b^2+3 a^{2/3} b^{4/3} \left (2 a^2+b^2\right )+b^4\right ) \log \left (\sqrt [3]{a}+\sqrt [3]{b} \tanh (c+d x)\right )}{3 \left (a^2-b^2\right )^3}-\frac {5 a-b}{16 (a+b)^2 (1-\tanh (c+d x))}+\frac {5 a+b}{16 (a-b)^2 (\tanh (c+d x)+1)}+\frac {1}{16 (a+b) (1-\tanh (c+d x))^2}-\frac {1}{16 (a-b) (\tanh (c+d x)+1)^2}-\frac {3 a (a-5 b) \log (1-\tanh (c+d x))}{16 (a+b)^3}+\frac {3 a (a+5 b) \log (\tanh (c+d x)+1)}{16 (a-b)^3}}{d}\) |
Input:
Int[Sinh[c + d*x]^4/(a + b*Tanh[c + d*x]^3),x]
Output:
(-((a^(2/3)*b^(1/3)*(a^2 + 3*a^(4/3)*b^(2/3) - b^2)*ArcTan[(a^(1/3) - 2*b^ (1/3)*Tanh[c + d*x])/(Sqrt[3]*a^(1/3))])/(Sqrt[3]*(a^(4/3) + a^(2/3)*b^(2/ 3) + b^(4/3))^3)) - (3*a*(a - 5*b)*Log[1 - Tanh[c + d*x]])/(16*(a + b)^3) + (3*a*(a + 5*b)*Log[1 + Tanh[c + d*x]])/(16*(a - b)^3) - (a^(2/3)*b^(1/3) *(a^4 + 7*a^2*b^2 + b^4 + 3*a^(2/3)*b^(4/3)*(2*a^2 + b^2))*Log[a^(1/3) + b ^(1/3)*Tanh[c + d*x]])/(3*(a^2 - b^2)^3) + (a^(2/3)*b^(1/3)*(a^4 + 7*a^2*b ^2 + b^4 + 3*a^(2/3)*b^(4/3)*(2*a^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)^3) - (a^2*b*(a^2 + 2*b^2)*Log[a + b*Tanh[c + d*x]^3])/(a^2 - b^2)^3 + 1/(16*(a + b)*(1 - Ta nh[c + d*x])^2) - (5*a - b)/(16*(a + b)^2*(1 - Tanh[c + d*x])) - 1/(16*(a - b)*(1 + Tanh[c + d*x])^2) + (5*a + b)/(16*(a - b)^2*(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 = 35.32 (sec) , antiderivative size = 452, normalized size of antiderivative = 0.92
| method | result | size |
| derivativedivides | \(\frac {-\frac {8}{\left (32 a -32 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {32}{\left (64 a -64 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a -5 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {3 a +3 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 a \left (a +5 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 \left (a -b \right )^{3}}+\frac {8}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {32}{\left (64 a +64 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a -5 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3 a -3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 a \left (a -5 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a +b \right )^{3}}-\frac {a b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (3 a^{2} \left (a^{2}+2 b^{2}\right ) \textit {\_R}^{5}+3 a b \left (-2 a^{2}-b^{2}\right ) \textit {\_R}^{4}+2 \left (4 a^{4}+13 a^{2} b^{2}+b^{4}\right ) \textit {\_R}^{3}+12 a b \left (a^{2}+2 b^{2}\right ) \textit {\_R}^{2}+\left (a^{4}-8 a^{2} b^{2}-2 b^{4}\right ) \textit {\_R} +6 a^{3} b +3 a \,b^{3}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 \left (a -b \right )^{3} \left (a +b \right )^{3}}}{d}\) | \(452\) |
| default | \(\frac {-\frac {8}{\left (32 a -32 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {32}{\left (64 a -64 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}-\frac {-a -5 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {3 a +3 b}{8 \left (a -b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}+\frac {3 a \left (a +5 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 \left (a -b \right )^{3}}+\frac {8}{\left (32 a +32 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {32}{\left (64 a +64 b \right ) \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a -5 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {3 a -3 b}{8 \left (a +b \right )^{2} \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {3 a \left (a -5 b \right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a +b \right )^{3}}-\frac {a b \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (a \,\textit {\_Z}^{6}+3 a \,\textit {\_Z}^{4}+8 b \,\textit {\_Z}^{3}+3 a \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {\left (3 a^{2} \left (a^{2}+2 b^{2}\right ) \textit {\_R}^{5}+3 a b \left (-2 a^{2}-b^{2}\right ) \textit {\_R}^{4}+2 \left (4 a^{4}+13 a^{2} b^{2}+b^{4}\right ) \textit {\_R}^{3}+12 a b \left (a^{2}+2 b^{2}\right ) \textit {\_R}^{2}+\left (a^{4}-8 a^{2} b^{2}-2 b^{4}\right ) \textit {\_R} +6 a^{3} b +3 a \,b^{3}\right ) \ln \left (\tanh \left (\frac {d x}{2}+\frac {c}{2}\right )-\textit {\_R} \right )}{\textit {\_R}^{5} a +2 \textit {\_R}^{3} a +4 \textit {\_R}^{2} b +\textit {\_R} a}\right )}{3 \left (a -b \right )^{3} \left (a +b \right )^{3}}}{d}\) | \(452\) |
| risch | \(\text {Expression too large to display}\) | \(1384\) |
Input:
int(sinh(d*x+c)^4/(a+b*tanh(d*x+c)^3),x,method=_RETURNVERBOSE)
Output:
1/d*(-8/(32*a-32*b)/(tanh(1/2*d*x+1/2*c)+1)^4+32/(64*a-64*b)/(tanh(1/2*d*x +1/2*c)+1)^3-1/8*(-a-5*b)/(a-b)^2/(tanh(1/2*d*x+1/2*c)+1)^2-1/8*(3*a+3*b)/ (a-b)^2/(tanh(1/2*d*x+1/2*c)+1)+3/8*a*(a+5*b)/(a-b)^3*ln(tanh(1/2*d*x+1/2* c)+1)+8/(32*a+32*b)/(tanh(1/2*d*x+1/2*c)-1)^4+32/(64*a+64*b)/(tanh(1/2*d*x +1/2*c)-1)^3-1/8*(a-5*b)/(a+b)^2/(tanh(1/2*d*x+1/2*c)-1)^2-1/8*(3*a-3*b)/( a+b)^2/(tanh(1/2*d*x+1/2*c)-1)-3/8*a*(a-5*b)/(a+b)^3*ln(tanh(1/2*d*x+1/2*c )-1)-1/3/(a-b)^3*a/(a+b)^3*b*sum((3*a^2*(a^2+2*b^2)*_R^5+3*a*b*(-2*a^2-b^2 )*_R^4+2*(4*a^4+13*a^2*b^2+b^4)*_R^3+12*a*b*(a^2+2*b^2)*_R^2+(a^4-8*a^2*b^ 2-2*b^4)*_R+6*a^3*b+3*a*b^3)/(_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 = 1.33 (sec) , antiderivative size = 17123, normalized size of antiderivative = 34.87 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate(sinh(d*x+c)^4/(a+b*tanh(d*x+c)^3),x, algorithm="fricas")
Output:
Too large to include
Timed out. \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\text {Timed out} \] Input:
integrate(sinh(d*x+c)**4/(a+b*tanh(d*x+c)**3),x)
Output:
Timed out
\[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\int { \frac {\sinh \left (d x + c\right )^{4}}{b \tanh \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(sinh(d*x+c)^4/(a+b*tanh(d*x+c)^3),x, algorithm="maxima")
Output:
-6*a^4*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^6 - 3*a^4*b^2 + 3*a^2*b ^4 - b^6) - (d*x + c)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d)) - 12*a^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^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^ 6) - (d*x + c)/((a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*d)) + 10*a^4*b*integra te(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^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^ 3 + a*b^4 + b^5) - 20*a^3*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^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5) + 20*a^2*b^3*integra te(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^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^ 3 + a*b^4 + b^5) - 4*a*b^4*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^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 + a*b^4 + b^5) + 8*a^4*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^5 + a^4*b - 2*a^3*b^2 - 2*a^2*b^3 +...
\[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\int { \frac {\sinh \left (d x + c\right )^{4}}{b \tanh \left (d x + c\right )^{3} + a} \,d x } \] Input:
integrate(sinh(d*x+c)^4/(a+b*tanh(d*x+c)^3),x, algorithm="giac")
Output:
sage0*x
Time = 5.77 (sec) , antiderivative size = 3313, normalized size of antiderivative = 6.75 \[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\text {Too large to display} \] Input:
int(sinh(c + d*x)^4/(a + b*tanh(c + d*x)^3),x)
Output:
symsum(log(- root(81*a^4*b^2*d^3*z^3 - 81*a^2*b^4*d^3*z^3 + 27*b^6*d^3*z^3 - 27*a^6*d^3*z^3 - 162*a^2*b^3*d^2*z^2 - 81*a^4*b*d^2*z^2 - 27*a^2*b^2*d* z - a^2*b, z, k)*((96*(a^2*b^10*d + 20*a^3*b^9*d - 89*a^4*b^8*d + 270*a^5* b^7*d - 417*a^6*b^6*d + 408*a^7*b^5*d - 190*a^8*b^4*d + 58*a^9*b^3*d - 7*a ^10*b^2*d - a^2*b^10*d*exp(2*root(81*a^4*b^2*d^3*z^3 - 81*a^2*b^4*d^3*z^3 + 27*b^6*d^3*z^3 - 27*a^6*d^3*z^3 - 162*a^2*b^3*d^2*z^2 - 81*a^4*b*d^2*z^2 - 27*a^2*b^2*d*z - a^2*b, z, k))*exp(2*d*x) - 52*a^3*b^9*d*exp(2*root(81* a^4*b^2*d^3*z^3 - 81*a^2*b^4*d^3*z^3 + 27*b^6*d^3*z^3 - 27*a^6*d^3*z^3 - 1 62*a^2*b^3*d^2*z^2 - 81*a^4*b*d^2*z^2 - 27*a^2*b^2*d*z - a^2*b, z, k))*exp (2*d*x) + 59*a^4*b^8*d*exp(2*root(81*a^4*b^2*d^3*z^3 - 81*a^2*b^4*d^3*z^3 + 27*b^6*d^3*z^3 - 27*a^6*d^3*z^3 - 162*a^2*b^3*d^2*z^2 - 81*a^4*b*d^2*z^2 - 27*a^2*b^2*d*z - a^2*b, z, k))*exp(2*d*x) - 218*a^5*b^7*d*exp(2*root(81 *a^4*b^2*d^3*z^3 - 81*a^2*b^4*d^3*z^3 + 27*b^6*d^3*z^3 - 27*a^6*d^3*z^3 - 162*a^2*b^3*d^2*z^2 - 81*a^4*b*d^2*z^2 - 27*a^2*b^2*d*z - a^2*b, z, k))*ex p(2*d*x) + 241*a^6*b^6*d*exp(2*root(81*a^4*b^2*d^3*z^3 - 81*a^2*b^4*d^3*z^ 3 + 27*b^6*d^3*z^3 - 27*a^6*d^3*z^3 - 162*a^2*b^3*d^2*z^2 - 81*a^4*b*d^2*z ^2 - 27*a^2*b^2*d*z - a^2*b, z, k))*exp(2*d*x) + 220*a^7*b^5*d*exp(2*root( 81*a^4*b^2*d^3*z^3 - 81*a^2*b^4*d^3*z^3 + 27*b^6*d^3*z^3 - 27*a^6*d^3*z^3 - 162*a^2*b^3*d^2*z^2 - 81*a^4*b*d^2*z^2 - 27*a^2*b^2*d*z - a^2*b, z, k))* exp(2*d*x) - 298*a^8*b^4*d*exp(2*root(81*a^4*b^2*d^3*z^3 - 81*a^2*b^4*d...
\[ \int \frac {\sinh ^4(c+d x)}{a+b \tanh ^3(c+d x)} \, dx=\text {too large to display} \] Input:
int(sinh(d*x+c)^4/(a+b*tanh(d*x+c)^3),x)
Output:
(e**(8*c + 8*d*x)*a**5 - 33*e**(8*c + 8*d*x)*a**4*b - 34*e**(8*c + 8*d*x)* a**3*b**2 + 34*e**(8*c + 8*d*x)*a**2*b**3 + 33*e**(8*c + 8*d*x)*a*b**4 - e **(8*c + 8*d*x)*b**5 - 8*e**(6*c + 6*d*x)*a**5 + 276*e**(6*c + 6*d*x)*a**4 *b - 136*e**(6*c + 6*d*x)*a**3*b**2 - 272*e**(6*c + 6*d*x)*a**2*b**3 + 144 *e**(6*c + 6*d*x)*a*b**4 - 4*e**(6*c + 6*d*x)*b**5 - 1792*e**(6*c + 4*d*x) *int(1/(e**(10*c + 8*d*x)*a**7 - 31*e**(10*c + 8*d*x)*a**6*b - 99*e**(10*c + 8*d*x)*a**5*b**2 - 67*e**(10*c + 8*d*x)*a**4*b**3 + 67*e**(10*c + 8*d*x )*a**3*b**4 + 99*e**(10*c + 8*d*x)*a**2*b**5 + 31*e**(10*c + 8*d*x)*a*b**6 - e**(10*c + 8*d*x)*b**7 + 3*e**(8*c + 6*d*x)*a**7 - 99*e**(8*c + 6*d*x)* a**6*b - 105*e**(8*c + 6*d*x)*a**5*b**2 + 201*e**(8*c + 6*d*x)*a**4*b**3 + 201*e**(8*c + 6*d*x)*a**3*b**4 - 105*e**(8*c + 6*d*x)*a**2*b**5 - 99*e**( 8*c + 6*d*x)*a*b**6 + 3*e**(8*c + 6*d*x)*b**7 + 3*e**(6*c + 4*d*x)*a**7 - 93*e**(6*c + 4*d*x)*a**6*b - 297*e**(6*c + 4*d*x)*a**5*b**2 - 201*e**(6*c + 4*d*x)*a**4*b**3 + 201*e**(6*c + 4*d*x)*a**3*b**4 + 297*e**(6*c + 4*d*x) *a**2*b**5 + 93*e**(6*c + 4*d*x)*a*b**6 - 3*e**(6*c + 4*d*x)*b**7 + e**(4* c + 2*d*x)*a**7 - 33*e**(4*c + 2*d*x)*a**6*b - 35*e**(4*c + 2*d*x)*a**5*b* *2 + 67*e**(4*c + 2*d*x)*a**4*b**3 + 67*e**(4*c + 2*d*x)*a**3*b**4 - 35*e* *(4*c + 2*d*x)*a**2*b**5 - 33*e**(4*c + 2*d*x)*a*b**6 + e**(4*c + 2*d*x)*b **7),x)*a**11*b*d + 32256*e**(6*c + 4*d*x)*int(1/(e**(10*c + 8*d*x)*a**7 - 31*e**(10*c + 8*d*x)*a**6*b - 99*e**(10*c + 8*d*x)*a**5*b**2 - 67*e**(...