Integrand size = 7, antiderivative size = 75 \[ \int \sinh (x) \tanh (4 x) \, dx=-\frac {\arctan \left (\sqrt {2 \left (2-\sqrt {2}\right )} \sinh (x)\right )}{2 \sqrt {2 \left (2-\sqrt {2}\right )}}-\frac {\arctan \left (\sqrt {2 \left (2+\sqrt {2}\right )} \sinh (x)\right )}{2 \sqrt {2 \left (2+\sqrt {2}\right )}}+\sinh (x) \] Output:
-1/2*arctan((4-2*2^(1/2))^(1/2)*sinh(x))/(4-2*2^(1/2))^(1/2)-1/2*arctan((4 +2*2^(1/2))^(1/2)*sinh(x))/(4+2*2^(1/2))^(1/2)+sinh(x)
Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.92 \[ \int \sinh (x) \tanh (4 x) \, dx=-\frac {1}{4} \sqrt {2-\sqrt {2}} \arctan \left (\frac {2 \sinh (x)}{\sqrt {2-\sqrt {2}}}\right )-\frac {1}{4} \sqrt {2+\sqrt {2}} \arctan \left (\frac {2 \sinh (x)}{\sqrt {2+\sqrt {2}}}\right )+\sinh (x) \] Input:
Integrate[Sinh[x]*Tanh[4*x],x]
Output:
-1/4*(Sqrt[2 - Sqrt[2]]*ArcTan[(2*Sinh[x])/Sqrt[2 - Sqrt[2]]]) - (Sqrt[2 + Sqrt[2]]*ArcTan[(2*Sinh[x])/Sqrt[2 + Sqrt[2]]])/4 + Sinh[x]
Time = 0.31 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 1.143, Rules used = {3042, 25, 4878, 27, 1602, 27, 1480, 216}
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.
\(\displaystyle \int \sinh (x) \tanh (4 x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\sin (i x) \tan (4 i x)dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \sin (i x) \tan (4 i x)dx\) |
\(\Big \downarrow \) 4878 |
\(\displaystyle -\int -\frac {4 \sinh ^2(x) \left (2 \sinh ^2(x)+1\right )}{8 \sinh ^4(x)+8 \sinh ^2(x)+1}d\sinh (x)\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \int \frac {\sinh ^2(x) \left (2 \sinh ^2(x)+1\right )}{8 \sinh ^4(x)+8 \sinh ^2(x)+1}d\sinh (x)\) |
\(\Big \downarrow \) 1602 |
\(\displaystyle 4 \left (\frac {\sinh (x)}{4}-\frac {1}{8} \int \frac {2 \left (4 \sinh ^2(x)+1\right )}{8 \sinh ^4(x)+8 \sinh ^2(x)+1}d\sinh (x)\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle 4 \left (\frac {\sinh (x)}{4}-\frac {1}{4} \int \frac {4 \sinh ^2(x)+1}{8 \sinh ^4(x)+8 \sinh ^2(x)+1}d\sinh (x)\right )\) |
\(\Big \downarrow \) 1480 |
\(\displaystyle 4 \left (\frac {1}{4} \left (-\left (\left (2-\sqrt {2}\right ) \int \frac {1}{8 \sinh ^2(x)+2 \left (2-\sqrt {2}\right )}d\sinh (x)\right )-\left (2+\sqrt {2}\right ) \int \frac {1}{8 \sinh ^2(x)+2 \left (2+\sqrt {2}\right )}d\sinh (x)\right )+\frac {\sinh (x)}{4}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle 4 \left (\frac {1}{4} \left (-\frac {1}{4} \sqrt {2-\sqrt {2}} \arctan \left (\frac {2 \sinh (x)}{\sqrt {2-\sqrt {2}}}\right )-\frac {1}{4} \sqrt {2+\sqrt {2}} \arctan \left (\frac {2 \sinh (x)}{\sqrt {2+\sqrt {2}}}\right )\right )+\frac {\sinh (x)}{4}\right )\) |
Input:
Int[Sinh[x]*Tanh[4*x],x]
Output:
4*((-1/4*(Sqrt[2 - Sqrt[2]]*ArcTan[(2*Sinh[x])/Sqrt[2 - Sqrt[2]]]) - (Sqrt [2 + Sqrt[2]]*ArcTan[(2*Sinh[x])/Sqrt[2 + Sqrt[2]]])/4)/4 + Sinh[x]/4)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] : > With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(e/2 + (2*c*d - b*e)/(2*q)) Int[1/( b/2 - q/2 + c*x^2), x], x] + Simp[(e/2 - (2*c*d - b*e)/(2*q)) Int[1/(b/2 + q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^2 - 4*a*c]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)*((a_) + (b_.)*(x_)^2 + (c_.)*( x_)^4)^(p_), x_Symbol] :> Simp[e*f*(f*x)^(m - 1)*((a + b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 3))), x] - Simp[f^2/(c*(m + 4*p + 3)) Int[(f*x)^(m - 2)* (a + b*x^2 + c*x^4)^p*Simp[a*e*(m - 1) + (b*e*(m + 2*p + 1) - c*d*(m + 4*p + 3))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && NeQ[b^2 - 4*a*c , 0] && GtQ[m, 1] && NeQ[m + 4*p + 3, 0] && IntegerQ[2*p] && (IntegerQ[p] | | IntegerQ[m])
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 = 0.08 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.56
method | result | size |
risch | \(\frac {{\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{-x}}{2}+\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (2048 \textit {\_Z}^{4}+128 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-8 \textit {\_R} \,{\mathrm e}^{x}+{\mathrm e}^{2 x}-1\right )\right )\) | \(42\) |
Input:
int(sinh(x)*tanh(4*x),x,method=_RETURNVERBOSE)
Output:
1/2*exp(x)-1/2*exp(-x)+sum(_R*ln(-8*_R*exp(x)+exp(2*x)-1),_R=RootOf(2048*_ Z^4+128*_Z^2+1))
Leaf count of result is larger than twice the leaf count of optimal. 287 vs. \(2 (51) = 102\).
Time = 0.10 (sec) , antiderivative size = 287, normalized size of antiderivative = 3.83 \[ \int \sinh (x) \tanh (4 x) \, dx=\frac {\sqrt {\sqrt {2} + 2} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \arctan \left (\frac {1}{2} \, {\left ({\left (\sqrt {2} - 2\right )} \cosh \left (x\right )^{3} + 3 \, {\left (\sqrt {2} - 2\right )} \cosh \left (x\right ) \sinh \left (x\right )^{2} + {\left (\sqrt {2} - 2\right )} \sinh \left (x\right )^{3} + {\left (3 \, {\left (\sqrt {2} - 2\right )} \cosh \left (x\right )^{2} - \sqrt {2}\right )} \sinh \left (x\right ) - \sqrt {2} \cosh \left (x\right )\right )} \sqrt {\sqrt {2} + 2}\right ) + \sqrt {\sqrt {2} + 2} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \arctan \left (\frac {1}{2} \, {\left ({\left (\sqrt {2} - 2\right )} \cosh \left (x\right ) + {\left (\sqrt {2} - 2\right )} \sinh \left (x\right )\right )} \sqrt {\sqrt {2} + 2}\right ) - \sqrt {-\sqrt {2} + 2} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \arctan \left (\frac {1}{2} \, {\left ({\left (\sqrt {2} + 2\right )} \cosh \left (x\right )^{3} + 3 \, {\left (\sqrt {2} + 2\right )} \cosh \left (x\right ) \sinh \left (x\right )^{2} + {\left (\sqrt {2} + 2\right )} \sinh \left (x\right )^{3} + {\left (3 \, {\left (\sqrt {2} + 2\right )} \cosh \left (x\right )^{2} - \sqrt {2}\right )} \sinh \left (x\right ) - \sqrt {2} \cosh \left (x\right )\right )} \sqrt {-\sqrt {2} + 2}\right ) - \sqrt {-\sqrt {2} + 2} {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )} \arctan \left (\frac {1}{2} \, {\left ({\left (\sqrt {2} + 2\right )} \cosh \left (x\right ) + {\left (\sqrt {2} + 2\right )} \sinh \left (x\right )\right )} \sqrt {-\sqrt {2} + 2}\right ) + 2 \, \cosh \left (x\right )^{2} + 4 \, \cosh \left (x\right ) \sinh \left (x\right ) + 2 \, \sinh \left (x\right )^{2} - 2}{4 \, {\left (\cosh \left (x\right ) + \sinh \left (x\right )\right )}} \] Input:
integrate(sinh(x)*tanh(4*x),x, algorithm="fricas")
Output:
1/4*(sqrt(sqrt(2) + 2)*(cosh(x) + sinh(x))*arctan(1/2*((sqrt(2) - 2)*cosh( x)^3 + 3*(sqrt(2) - 2)*cosh(x)*sinh(x)^2 + (sqrt(2) - 2)*sinh(x)^3 + (3*(s qrt(2) - 2)*cosh(x)^2 - sqrt(2))*sinh(x) - sqrt(2)*cosh(x))*sqrt(sqrt(2) + 2)) + sqrt(sqrt(2) + 2)*(cosh(x) + sinh(x))*arctan(1/2*((sqrt(2) - 2)*cos h(x) + (sqrt(2) - 2)*sinh(x))*sqrt(sqrt(2) + 2)) - sqrt(-sqrt(2) + 2)*(cos h(x) + sinh(x))*arctan(1/2*((sqrt(2) + 2)*cosh(x)^3 + 3*(sqrt(2) + 2)*cosh (x)*sinh(x)^2 + (sqrt(2) + 2)*sinh(x)^3 + (3*(sqrt(2) + 2)*cosh(x)^2 - sqr t(2))*sinh(x) - sqrt(2)*cosh(x))*sqrt(-sqrt(2) + 2)) - sqrt(-sqrt(2) + 2)* (cosh(x) + sinh(x))*arctan(1/2*((sqrt(2) + 2)*cosh(x) + (sqrt(2) + 2)*sinh (x))*sqrt(-sqrt(2) + 2)) + 2*cosh(x)^2 + 4*cosh(x)*sinh(x) + 2*sinh(x)^2 - 2)/(cosh(x) + sinh(x))
\[ \int \sinh (x) \tanh (4 x) \, dx=\int \sinh {\left (x \right )} \tanh {\left (4 x \right )}\, dx \] Input:
integrate(sinh(x)*tanh(4*x),x)
Output:
Integral(sinh(x)*tanh(4*x), x)
\[ \int \sinh (x) \tanh (4 x) \, dx=\int { \sinh \left (x\right ) \tanh \left (4 \, x\right ) \,d x } \] Input:
integrate(sinh(x)*tanh(4*x),x, algorithm="maxima")
Output:
1/2*(e^(2*x) - 1)*e^(-x) - 1/2*integrate(2*(e^(7*x) + e^x)/(e^(8*x) + 1), x)
Time = 0.15 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \sinh (x) \tanh (4 x) \, dx=-\frac {1}{4} \, \sqrt {\sqrt {2} + 2} \arctan \left (-\frac {e^{\left (-x\right )} - e^{x}}{\sqrt {\sqrt {2} + 2}}\right ) - \frac {1}{4} \, \sqrt {-\sqrt {2} + 2} \arctan \left (-\frac {e^{\left (-x\right )} - e^{x}}{\sqrt {-\sqrt {2} + 2}}\right ) - \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \] Input:
integrate(sinh(x)*tanh(4*x),x, algorithm="giac")
Output:
-1/4*sqrt(sqrt(2) + 2)*arctan(-(e^(-x) - e^x)/sqrt(sqrt(2) + 2)) - 1/4*sqr t(-sqrt(2) + 2)*arctan(-(e^(-x) - e^x)/sqrt(-sqrt(2) + 2)) - 1/2*e^(-x) + 1/2*e^x
Time = 1.76 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.95 \[ \int \sinh (x) \tanh (4 x) \, dx=\frac {{\mathrm {e}}^x}{2}-\frac {{\mathrm {e}}^{-x}}{2}-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{2\,x}-1\right )}{\sqrt {\sqrt {2}+2}}\right )\,\sqrt {\sqrt {2}+2}}{4}-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{2\,x}-1\right )}{\sqrt {2-\sqrt {2}}}\right )\,\sqrt {2-\sqrt {2}}}{4} \] Input:
int(tanh(4*x)*sinh(x),x)
Output:
exp(x)/2 - exp(-x)/2 - (atan((exp(-x)*(exp(2*x) - 1))/(2^(1/2) + 2)^(1/2)) *(2^(1/2) + 2)^(1/2))/4 - (atan((exp(-x)*(exp(2*x) - 1))/(2 - 2^(1/2))^(1/ 2))*(2 - 2^(1/2))^(1/2))/4
Time = 0.27 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.95 \[ \int \sinh (x) \tanh (4 x) \, dx=\frac {e^{x} \sqrt {\sqrt {2}+2}\, \mathit {atan} \left (\frac {\sqrt {-\sqrt {2}+2}-2 e^{x}}{\sqrt {\sqrt {2}+2}}\right )-e^{x} \sqrt {\sqrt {2}+2}\, \mathit {atan} \left (\frac {\sqrt {-\sqrt {2}+2}+2 e^{x}}{\sqrt {\sqrt {2}+2}}\right )+e^{x} \sqrt {-\sqrt {2}+2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}+2}-2 e^{x}}{\sqrt {-\sqrt {2}+2}}\right )-e^{x} \sqrt {-\sqrt {2}+2}\, \mathit {atan} \left (\frac {\sqrt {\sqrt {2}+2}+2 e^{x}}{\sqrt {-\sqrt {2}+2}}\right )+2 e^{2 x}-2}{4 e^{x}} \] Input:
int(sinh(x)*tanh(4*x),x)
Output:
(e**x*sqrt(sqrt(2) + 2)*atan((sqrt( - sqrt(2) + 2) - 2*e**x)/sqrt(sqrt(2) + 2)) - e**x*sqrt(sqrt(2) + 2)*atan((sqrt( - sqrt(2) + 2) + 2*e**x)/sqrt(s qrt(2) + 2)) + e**x*sqrt( - sqrt(2) + 2)*atan((sqrt(sqrt(2) + 2) - 2*e**x) /sqrt( - sqrt(2) + 2)) - e**x*sqrt( - sqrt(2) + 2)*atan((sqrt(sqrt(2) + 2) + 2*e**x)/sqrt( - sqrt(2) + 2)) + 2*e**(2*x) - 2)/(4*e**x)