Integrand size = 7, antiderivative size = 84 \[ \int \sinh (x) \tanh (5 x) \, dx=-\frac {1}{5} \arctan (\sinh (x))-\frac {1}{5} \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \arctan \left (\sqrt {2 \left (3-\sqrt {5}\right )} \sinh (x)\right )-\frac {1}{5} \sqrt {\frac {2}{3+\sqrt {5}}} \arctan \left (\sqrt {2 \left (3+\sqrt {5}\right )} \sinh (x)\right )+\sinh (x) \] Output:
-1/5*arctan(sinh(x))-1/5*(1/2+1/2*5^(1/2))*arctan((5^(1/2)-1)*sinh(x))-1/5 *2^(1/2)/(3+5^(1/2))^(1/2)*arctan((5^(1/2)+1)*sinh(x))+sinh(x)
Time = 0.14 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.96 \[ \int \sinh (x) \tanh (5 x) \, dx=\frac {1}{10} \left (-2 \arctan (\sinh (x))-\sqrt {2 \left (3+\sqrt {5}\right )} \arctan \left (2 \sqrt {\frac {2}{3+\sqrt {5}}} \sinh (x)\right )-\sqrt {6-2 \sqrt {5}} \arctan \left (\sqrt {2 \left (3+\sqrt {5}\right )} \sinh (x)\right )+10 \sinh (x)\right ) \] Input:
Integrate[Sinh[x]*Tanh[5*x],x]
Output:
(-2*ArcTan[Sinh[x]] - Sqrt[2*(3 + Sqrt[5])]*ArcTan[2*Sqrt[2/(3 + Sqrt[5])] *Sinh[x]] - Sqrt[6 - 2*Sqrt[5]]*ArcTan[Sqrt[2*(3 + Sqrt[5])]*Sinh[x]] + 10 *Sinh[x])/10
Time = 0.42 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.857, Rules used = {3042, 25, 4878, 25, 2460, 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.
\(\displaystyle \int \sinh (x) \tanh (5 x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\sin (i x) \tan (5 i x)dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \sin (i x) \tan (5 i x)dx\) |
\(\Big \downarrow \) 4878 |
\(\displaystyle -\int -\frac {\sinh ^2(x) \left (16 \sinh ^4(x)+20 \sinh ^2(x)+5\right )}{16 \sinh ^6(x)+28 \sinh ^4(x)+13 \sinh ^2(x)+1}d\sinh (x)\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {\sinh ^2(x) \left (16 \sinh ^4(x)+20 \sinh ^2(x)+5\right )}{16 \sinh ^6(x)+28 \sinh ^4(x)+13 \sinh ^2(x)+1}d\sinh (x)\) |
\(\Big \downarrow \) 2460 |
\(\displaystyle \int \left (-\frac {1}{5 \left (\sinh ^2(x)+1\right )}-\frac {4 \left (6 \sinh ^2(x)+1\right )}{5 \left (16 \sinh ^4(x)+12 \sinh ^2(x)+1\right )}+1\right )d\sinh (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {1}{5} \arctan (\sinh (x))-\frac {1}{5} \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \arctan \left (2 \sqrt {\frac {2}{3+\sqrt {5}}} \sinh (x)\right )-\frac {1}{5} \sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} \arctan \left (\sqrt {2 \left (3+\sqrt {5}\right )} \sinh (x)\right )+\sinh (x)\) |
Input:
Int[Sinh[x]*Tanh[5*x],x]
Output:
-1/5*ArcTan[Sinh[x]] - (Sqrt[(3 + Sqrt[5])/2]*ArcTan[2*Sqrt[2/(3 + Sqrt[5] )]*Sinh[x]])/5 - (Sqrt[(3 - Sqrt[5])/2]*ArcTan[Sqrt[2*(3 + Sqrt[5])]*Sinh[ x]])/5 + Sinh[x]
Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px /. x -> Sqrt[x]]}, Int[ExpandIntegrand[u*(Qx /. x -> x^2)^p, x], x] /; !SumQ[NonfreeFactors[Q x, x]]] /; PolyQ[Px, x^2] && GtQ[Expon[Px, x], 2] && !BinomialQ[Px, x] && !TrinomialQ[Px, x] && ILtQ[p, 0] && RationalFunctionQ[u, x]
Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa ctors[Sin[v], x]}, d/Coefficient[v, x, 1] Subst[Int[SubstFor[1, Sin[v]/d, u/Cos[v], x], x], x, Sin[v]/d]], x] /; !FalseQ[v] && FunctionOfQ[NonfreeF actors[Sin[v], x], u/Cos[v], x]]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.10 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.71
method | result | size |
risch | \(\frac {{\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{-x}}{2}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{5}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{5}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (10000 \textit {\_Z}^{4}+300 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-10 \textit {\_R} \,{\mathrm e}^{x}+{\mathrm e}^{2 x}-1\right )\right )\) | \(60\) |
Input:
int(sinh(x)*tanh(5*x),x,method=_RETURNVERBOSE)
Output:
1/2*exp(x)-1/2*exp(-x)+1/5*I*ln(exp(x)-I)-1/5*I*ln(exp(x)+I)+sum(_R*ln(-10 *_R*exp(x)+exp(2*x)-1),_R=RootOf(10000*_Z^4+300*_Z^2+1))
Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (45) = 90\).
Time = 0.10 (sec) , antiderivative size = 246, normalized size of antiderivative = 2.93 \[ \int \sinh (x) \tanh (5 x) \, dx=-\frac {{\left ({\left (\sqrt {5} - 1\right )} \cosh \left (x\right ) + {\left (\sqrt {5} - 1\right )} \sinh \left (x\right )\right )} \arctan \left (\frac {1}{2} \, {\left (\sqrt {5} + 1\right )} \cosh \left (x\right ) + \frac {1}{2} \, {\left (\sqrt {5} + 1\right )} \sinh \left (x\right )\right ) + {\left ({\left (\sqrt {5} + 1\right )} \cosh \left (x\right ) + {\left (\sqrt {5} + 1\right )} \sinh \left (x\right )\right )} \arctan \left (\frac {1}{2} \, {\left (\sqrt {5} - 1\right )} \cosh \left (x\right ) + \frac {1}{2} \, {\left (\sqrt {5} - 1\right )} \sinh \left (x\right )\right ) - {\left ({\left (\sqrt {5} - 1\right )} \cosh \left (x\right ) + {\left (\sqrt {5} - 1\right )} \sinh \left (x\right )\right )} \arctan \left (-\frac {{\left (\sqrt {5} + 1\right )} \cosh \left (x\right )^{2} + 2 \, {\left (\sqrt {5} + 1\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (\sqrt {5} + 1\right )} \sinh \left (x\right )^{2} - 2}{2 \, {\left (\cosh \left (x\right ) - \sinh \left (x\right )\right )}}\right ) - {\left ({\left (\sqrt {5} + 1\right )} \cosh \left (x\right ) + {\left (\sqrt {5} + 1\right )} \sinh \left (x\right )\right )} \arctan \left (-\frac {{\left (\sqrt {5} - 1\right )} \cosh \left (x\right )^{2} + 2 \, {\left (\sqrt {5} - 1\right )} \cosh \left (x\right ) \sinh \left (x\right ) + {\left (\sqrt {5} - 1\right )} \sinh \left (x\right )^{2} + 2}{2 \, {\left (\cosh \left (x\right ) - \sinh \left (x\right )\right )}}\right ) + 4 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \arctan \left (\cosh \left (x\right ) + \sinh \left (x\right )\right ) - 5 \, \cosh \left (x\right )^{2} - 10 \, \cosh \left (x\right ) \sinh \left (x\right ) - 5 \, \sinh \left (x\right )^{2} + 5}{10 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \] Input:
integrate(sinh(x)*tanh(5*x),x, algorithm="fricas")
Output:
-1/10*(((sqrt(5) - 1)*cosh(x) + (sqrt(5) - 1)*sinh(x))*arctan(1/2*(sqrt(5) + 1)*cosh(x) + 1/2*(sqrt(5) + 1)*sinh(x)) + ((sqrt(5) + 1)*cosh(x) + (sqr t(5) + 1)*sinh(x))*arctan(1/2*(sqrt(5) - 1)*cosh(x) + 1/2*(sqrt(5) - 1)*si nh(x)) - ((sqrt(5) - 1)*cosh(x) + (sqrt(5) - 1)*sinh(x))*arctan(-1/2*((sqr t(5) + 1)*cosh(x)^2 + 2*(sqrt(5) + 1)*cosh(x)*sinh(x) + (sqrt(5) + 1)*sinh (x)^2 - 2)/(cosh(x) - sinh(x))) - ((sqrt(5) + 1)*cosh(x) + (sqrt(5) + 1)*s inh(x))*arctan(-1/2*((sqrt(5) - 1)*cosh(x)^2 + 2*(sqrt(5) - 1)*cosh(x)*sin h(x) + (sqrt(5) - 1)*sinh(x)^2 + 2)/(cosh(x) - sinh(x))) + 4*(cosh(x) + si nh(x))*arctan(cosh(x) + sinh(x)) - 5*cosh(x)^2 - 10*cosh(x)*sinh(x) - 5*si nh(x)^2 + 5)/(cosh(x) + sinh(x))
\[ \int \sinh (x) \tanh (5 x) \, dx=\int \sinh {\left (x \right )} \tanh {\left (5 x \right )}\, dx \] Input:
integrate(sinh(x)*tanh(5*x),x)
Output:
Integral(sinh(x)*tanh(5*x), x)
\[ \int \sinh (x) \tanh (5 x) \, dx=\int { \sinh \left (x\right ) \tanh \left (5 \, x\right ) \,d x } \] Input:
integrate(sinh(x)*tanh(5*x),x, algorithm="maxima")
Output:
1/2*(e^(2*x) - 1)*e^(-x) - 2/5*arctan(e^x) - 1/2*integrate(2/5*(3*e^(7*x) - e^(5*x) - e^(3*x) + 3*e^x)/(e^(8*x) - e^(6*x) + e^(4*x) - e^(2*x) + 1), x)
Time = 0.13 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.96 \[ \int \sinh (x) \tanh (5 x) \, dx=-\frac {1}{10} \, \pi - \frac {1}{10} \, {\left (\sqrt {5} + 1\right )} \arctan \left (-\frac {2 \, {\left (e^{\left (-x\right )} - e^{x}\right )}}{\sqrt {5} + 1}\right ) - \frac {1}{10} \, {\left (\sqrt {5} - 1\right )} \arctan \left (-\frac {2 \, {\left (e^{\left (-x\right )} - e^{x}\right )}}{\sqrt {5} - 1}\right ) - \frac {1}{5} \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right ) - \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \] Input:
integrate(sinh(x)*tanh(5*x),x, algorithm="giac")
Output:
-1/10*pi - 1/10*(sqrt(5) + 1)*arctan(-2*(e^(-x) - e^x)/(sqrt(5) + 1)) - 1/ 10*(sqrt(5) - 1)*arctan(-2*(e^(-x) - e^x)/(sqrt(5) - 1)) - 1/5*arctan(1/2* (e^(2*x) - 1)*e^(-x)) - 1/2*e^(-x) + 1/2*e^x
Time = 2.22 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.98 \[ \int \sinh (x) \tanh (5 x) \, dx=\frac {{\mathrm {e}}^x}{2}-\frac {2\,\mathrm {atan}\left ({\mathrm {e}}^x\right )}{5}-\frac {{\mathrm {e}}^{-x}}{2}-2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{2\,x}-1\right )}{10\,\sqrt {\frac {3}{200}-\frac {\sqrt {5}}{200}}}\right )\,\sqrt {\frac {3}{200}-\frac {\sqrt {5}}{200}}-2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{2\,x}-1\right )}{10\,\sqrt {\frac {\sqrt {5}}{200}+\frac {3}{200}}}\right )\,\sqrt {\frac {\sqrt {5}}{200}+\frac {3}{200}} \] Input:
int(tanh(5*x)*sinh(x),x)
Output:
exp(x)/2 - (2*atan(exp(x)))/5 - exp(-x)/2 - 2*atan((exp(-x)*(exp(2*x) - 1) )/(10*(3/200 - 5^(1/2)/200)^(1/2)))*(3/200 - 5^(1/2)/200)^(1/2) - 2*atan(( exp(-x)*(exp(2*x) - 1))/(10*(5^(1/2)/200 + 3/200)^(1/2)))*(5^(1/2)/200 + 3 /200)^(1/2)
\[ \int \sinh (x) \tanh (5 x) \, dx=\frac {e^{2 x}-2 e^{x} \left (\int \frac {e^{x}}{e^{10 x}+1}d x \right )+2 e^{x} \left (\int \frac {1}{e^{11 x}+e^{x}}d x \right )+1}{2 e^{x}} \] Input:
int(sinh(x)*tanh(5*x),x)
Output:
(e**(2*x) - 2*e**x*int(e**x/(e**(10*x) + 1),x) + 2*e**x*int(1/(e**(11*x) + e**x),x) + 1)/(2*e**x)