3.203 \(\int \sinh (x) \tanh (5 x) \, dx\)

Optimal. Leaf size=87 \[ \sinh (x)-\frac {1}{5} \tan ^{-1}(\sinh (x))-\frac {1}{5} \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \tan ^{-1}\left (2 \sqrt {\frac {2}{3+\sqrt {5}}} \sinh (x)\right )-\frac {1}{5} \sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {2 \left (3+\sqrt {5}\right )} \sinh (x)\right ) \]

[Out]

-1/5*arctan(sinh(x))+sinh(x)-1/5*arctan(sinh(x)*(5^(1/2)+1))*(1/2*5^(1/2)-1/2)-1/5*arctan(2*sinh(x)*2^(1/2)/(3
+5^(1/2))^(1/2))*(1/2+1/2*5^(1/2))

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Rubi [A]  time = 0.29, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {6742, 2073, 203, 1166} \[ \sinh (x)-\frac {1}{5} \tan ^{-1}(\sinh (x))-\frac {1}{5} \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \tan ^{-1}\left (2 \sqrt {\frac {2}{3+\sqrt {5}}} \sinh (x)\right )-\frac {1}{5} \sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {2 \left (3+\sqrt {5}\right )} \sinh (x)\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]*Tanh[5*x],x]

[Out]

-ArcTan[Sinh[x]]/5 - (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]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[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]

Rule 2073

Int[(P_)^(p_)*(Q_)^(q_.), x_Symbol] :> With[{PP = Factor[P /. x -> Sqrt[x]]}, Int[ExpandIntegrand[(PP /. x ->
x^2)^p*Q^q, x], x] /;  !SumQ[NonfreeFactors[PP, x]]] /; FreeQ[q, x] && PolyQ[P, x^2] && PolyQ[Q, x] && ILtQ[p,
0]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {align*} \int \sinh (x) \tanh (5 x) \, dx &=-\operatorname {Subst}\left (\int \frac {x^2 \left (-5-20 x^2-16 x^4\right )}{1+13 x^2+28 x^4+16 x^6} \, dx,x,\sinh (x)\right )\\ &=-\operatorname {Subst}\left (\int \left (-1+\frac {1+8 x^2+8 x^4}{1+13 x^2+28 x^4+16 x^6}\right ) \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-\operatorname {Subst}\left (\int \frac {1+8 x^2+8 x^4}{1+13 x^2+28 x^4+16 x^6} \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-\operatorname {Subst}\left (\int \left (\frac {1}{5 \left (1+x^2\right )}+\frac {4 \left (1+6 x^2\right )}{5 \left (1+12 x^2+16 x^4\right )}\right ) \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sinh (x)\right )-\frac {4}{5} \operatorname {Subst}\left (\int \frac {1+6 x^2}{1+12 x^2+16 x^4} \, dx,x,\sinh (x)\right )\\ &=-\frac {1}{5} \tan ^{-1}(\sinh (x))+\sinh (x)-\frac {1}{5} \left (4 \left (3-\sqrt {5}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{6-2 \sqrt {5}+16 x^2} \, dx,x,\sinh (x)\right )-\frac {1}{5} \left (4 \left (3+\sqrt {5}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{6+2 \sqrt {5}+16 x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac {1}{5} \tan ^{-1}(\sinh (x))-\frac {1}{5} \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \tan ^{-1}\left (2 \sqrt {\frac {2}{3+\sqrt {5}}} \sinh (x)\right )-\frac {1}{5} \sqrt {\frac {1}{2} \left (3-\sqrt {5}\right )} \tan ^{-1}\left (\sqrt {2 \left (3+\sqrt {5}\right )} \sinh (x)\right )+\sinh (x)\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 81, normalized size = 0.93 \[ \frac {1}{10} \left (10 \sinh (x)-2 \tan ^{-1}(\sinh (x))-\sqrt {2 \left (3+\sqrt {5}\right )} \tan ^{-1}\left (2 \sqrt {\frac {2}{3+\sqrt {5}}} \sinh (x)\right )-\sqrt {6-2 \sqrt {5}} \tan ^{-1}\left (\sqrt {2 \left (3+\sqrt {5}\right )} \sinh (x)\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]*Tanh[5*x],x]

[Out]

(-2*ArcTan[Sinh[x]] - Sqrt[2*(3 + Sqrt[5])]*ArcTan[2*Sqrt[2/(3 + Sqrt[5])]*Sinh[x]] - Sqrt[6 - 2*Sqrt[5]]*ArcT
an[Sqrt[2*(3 + Sqrt[5])]*Sinh[x]] + 10*Sinh[x])/10

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fricas [B]  time = 0.46, size = 237, normalized size = 2.72 \[ -\frac {1}{10} \, {\left (2 \, \sqrt {2} \sqrt {\sqrt {5} + 3} \arctan \left (\frac {1}{8} \, {\left (\sqrt {2 \, {\left (\sqrt {5} - 1\right )} e^{\left (2 \, x\right )} + 4 \, e^{\left (4 \, x\right )} + 4} {\left (\sqrt {5} \sqrt {2} - 3 \, \sqrt {2}\right )} \sqrt {\sqrt {5} + 3} - 2 \, {\left ({\left (\sqrt {5} \sqrt {2} - 3 \, \sqrt {2}\right )} e^{\left (2 \, x\right )} - \sqrt {5} \sqrt {2} + 3 \, \sqrt {2}\right )} \sqrt {\sqrt {5} + 3}\right )} e^{\left (-x\right )}\right ) e^{x} - 2 \, \sqrt {2} \sqrt {-\sqrt {5} + 3} \arctan \left (\frac {1}{8} \, {\left (\sqrt {-2 \, {\left (\sqrt {5} + 1\right )} e^{\left (2 \, x\right )} + 4 \, e^{\left (4 \, x\right )} + 4} {\left (\sqrt {5} \sqrt {2} + 3 \, \sqrt {2}\right )} \sqrt {-\sqrt {5} + 3} - 2 \, {\left ({\left (\sqrt {5} \sqrt {2} + 3 \, \sqrt {2}\right )} e^{\left (2 \, x\right )} - \sqrt {5} \sqrt {2} - 3 \, \sqrt {2}\right )} \sqrt {-\sqrt {5} + 3}\right )} e^{\left (-x\right )}\right ) e^{x} + 4 \, \arctan \left (e^{x}\right ) e^{x} - 5 \, e^{\left (2 \, x\right )} + 5\right )} e^{\left (-x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(5*x),x, algorithm="fricas")

[Out]

-1/10*(2*sqrt(2)*sqrt(sqrt(5) + 3)*arctan(1/8*(sqrt(2*(sqrt(5) - 1)*e^(2*x) + 4*e^(4*x) + 4)*(sqrt(5)*sqrt(2)
- 3*sqrt(2))*sqrt(sqrt(5) + 3) - 2*((sqrt(5)*sqrt(2) - 3*sqrt(2))*e^(2*x) - sqrt(5)*sqrt(2) + 3*sqrt(2))*sqrt(
sqrt(5) + 3))*e^(-x))*e^x - 2*sqrt(2)*sqrt(-sqrt(5) + 3)*arctan(1/8*(sqrt(-2*(sqrt(5) + 1)*e^(2*x) + 4*e^(4*x)
+ 4)*(sqrt(5)*sqrt(2) + 3*sqrt(2))*sqrt(-sqrt(5) + 3) - 2*((sqrt(5)*sqrt(2) + 3*sqrt(2))*e^(2*x) - sqrt(5)*sq
rt(2) - 3*sqrt(2))*sqrt(-sqrt(5) + 3))*e^(-x))*e^x + 4*arctan(e^x)*e^x - 5*e^(2*x) + 5)*e^(-x)

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giac [A]  time = 0.15, size = 81, normalized size = 0.93 \[ -\frac {1}{10} \, \pi - \frac {1}{10} \, {\left (\sqrt {5} + 1\right )} \arctan \left (-\frac {2 \, {\left (e^{\left (-x\right )} - e^{x}\right )}}{\sqrt {5} + 1}\right ) - \frac {1}{10} \, {\left (\sqrt {5} - 1\right )} \arctan \left (-\frac {2 \, {\left (e^{\left (-x\right )} - e^{x}\right )}}{\sqrt {5} - 1}\right ) - \frac {1}{5} \, \arctan \left (\frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right ) - \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(5*x),x, algorithm="giac")

[Out]

-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

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maple [C]  time = 0.34, size = 60, normalized size = 0.69 \[ \frac {{\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{-x}}{2}+\frac {i \ln \left ({\mathrm e}^{x}-i\right )}{5}-\frac {i \ln \left ({\mathrm e}^{x}+i\right )}{5}+\left (\munderset {\textit {\_R} =\RootOf \left (10000 \textit {\_Z}^{4}+300 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-10 \textit {\_R} \,{\mathrm e}^{x}+{\mathrm e}^{2 x}-1\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)*tanh(5*x),x)

[Out]

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(100
00*_Z^4+300*_Z^2+1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )} - \frac {2}{5} \, \arctan \left (e^{x}\right ) - \frac {1}{2} \, \int \frac {2 \, {\left (3 \, e^{\left (7 \, x\right )} - e^{\left (5 \, x\right )} - e^{\left (3 \, x\right )} + 3 \, e^{x}\right )}}{5 \, {\left (e^{\left (8 \, x\right )} - e^{\left (6 \, x\right )} + e^{\left (4 \, x\right )} - e^{\left (2 \, x\right )} + 1\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(5*x),x, algorithm="maxima")

[Out]

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)

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mupad [B]  time = 2.66, size = 82, normalized size = 0.94 \[ \frac {{\mathrm {e}}^x}{2}-\frac {2\,\mathrm {atan}\left ({\mathrm {e}}^x\right )}{5}-\frac {{\mathrm {e}}^{-x}}{2}-2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{2\,x}-1\right )}{10\,\sqrt {\frac {3}{200}-\frac {\sqrt {5}}{200}}}\right )\,\sqrt {\frac {3}{200}-\frac {\sqrt {5}}{200}}-2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{2\,x}-1\right )}{10\,\sqrt {\frac {\sqrt {5}}{200}+\frac {3}{200}}}\right )\,\sqrt {\frac {\sqrt {5}}{200}+\frac {3}{200}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(5*x)*sinh(x),x)

[Out]

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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh {\relax (x )} \tanh {\left (5 x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(5*x),x)

[Out]

Integral(sinh(x)*tanh(5*x), x)

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