3.204 \(\int \sinh (x) \tanh (6 x) \, dx\)

Optimal. Leaf size=87 \[ \sinh (x)-\frac {\tan ^{-1}\left (\sqrt {2} \sinh (x)\right )}{3 \sqrt {2}}-\frac {1}{6} \sqrt {2-\sqrt {3}} \tan ^{-1}\left (\frac {2 \sinh (x)}{\sqrt {2-\sqrt {3}}}\right )-\frac {1}{6} \sqrt {2+\sqrt {3}} \tan ^{-1}\left (\frac {2 \sinh (x)}{\sqrt {2+\sqrt {3}}}\right ) \]

[Out]

sinh(x)-1/6*arctan(sinh(x)*2^(1/2))*2^(1/2)-1/6*arctan(2*sinh(x)/(1/2*6^(1/2)-1/2*2^(1/2)))*(1/2*6^(1/2)-1/2*2
^(1/2))-1/6*arctan(2*sinh(x)/(1/2*6^(1/2)+1/2*2^(1/2)))*(1/2*6^(1/2)+1/2*2^(1/2))

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Rubi [A]  time = 0.26, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {12, 6742, 2073, 203, 1166} \[ \sinh (x)-\frac {\tan ^{-1}\left (\sqrt {2} \sinh (x)\right )}{3 \sqrt {2}}-\frac {1}{6} \sqrt {2-\sqrt {3}} \tan ^{-1}\left (\frac {2 \sinh (x)}{\sqrt {2-\sqrt {3}}}\right )-\frac {1}{6} \sqrt {2+\sqrt {3}} \tan ^{-1}\left (\frac {2 \sinh (x)}{\sqrt {2+\sqrt {3}}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sinh[x]*Tanh[6*x],x]

[Out]

-ArcTan[Sqrt[2]*Sinh[x]]/(3*Sqrt[2]) - (Sqrt[2 - Sqrt[3]]*ArcTan[(2*Sinh[x])/Sqrt[2 - Sqrt[3]]])/6 - (Sqrt[2 +
Sqrt[3]]*ArcTan[(2*Sinh[x])/Sqrt[2 + Sqrt[3]]])/6 + Sinh[x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 (6 x) \, dx &=-\operatorname {Subst}\left (\int \frac {2 x^2 \left (-3-16 x^2-16 x^4\right )}{1+18 x^2+48 x^4+32 x^6} \, dx,x,\sinh (x)\right )\\ &=-\left (2 \operatorname {Subst}\left (\int \frac {x^2 \left (-3-16 x^2-16 x^4\right )}{1+18 x^2+48 x^4+32 x^6} \, dx,x,\sinh (x)\right )\right )\\ &=-\left (2 \operatorname {Subst}\left (\int \left (-\frac {1}{2}+\frac {1+12 x^2+16 x^4}{2 \left (1+18 x^2+48 x^4+32 x^6\right )}\right ) \, dx,x,\sinh (x)\right )\right )\\ &=\sinh (x)-\operatorname {Subst}\left (\int \frac {1+12 x^2+16 x^4}{1+18 x^2+48 x^4+32 x^6} \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-\operatorname {Subst}\left (\int \left (\frac {1}{3 \left (1+2 x^2\right )}+\frac {2 \left (1+8 x^2\right )}{3 \left (1+16 x^2+16 x^4\right )}\right ) \, dx,x,\sinh (x)\right )\\ &=\sinh (x)-\frac {1}{3} \operatorname {Subst}\left (\int \frac {1}{1+2 x^2} \, dx,x,\sinh (x)\right )-\frac {2}{3} \operatorname {Subst}\left (\int \frac {1+8 x^2}{1+16 x^2+16 x^4} \, dx,x,\sinh (x)\right )\\ &=-\frac {\tan ^{-1}\left (\sqrt {2} \sinh (x)\right )}{3 \sqrt {2}}+\sinh (x)-\frac {1}{3} \left (4 \left (2-\sqrt {3}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{8-4 \sqrt {3}+16 x^2} \, dx,x,\sinh (x)\right )-\frac {1}{3} \left (4 \left (2+\sqrt {3}\right )\right ) \operatorname {Subst}\left (\int \frac {1}{8+4 \sqrt {3}+16 x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac {\tan ^{-1}\left (\sqrt {2} \sinh (x)\right )}{3 \sqrt {2}}-\frac {1}{6} \sqrt {2-\sqrt {3}} \tan ^{-1}\left (\frac {2 \sinh (x)}{\sqrt {2-\sqrt {3}}}\right )-\frac {1}{6} \sqrt {2+\sqrt {3}} \tan ^{-1}\left (\frac {2 \sinh (x)}{\sqrt {2+\sqrt {3}}}\right )+\sinh (x)\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 87, normalized size = 1.00 \[ \sinh (x)-\frac {\tan ^{-1}\left (\sqrt {2} \sinh (x)\right )}{3 \sqrt {2}}-\frac {1}{6} \sqrt {2-\sqrt {3}} \tan ^{-1}\left (\frac {2 \sinh (x)}{\sqrt {2-\sqrt {3}}}\right )-\frac {1}{6} \sqrt {2+\sqrt {3}} \tan ^{-1}\left (\frac {2 \sinh (x)}{\sqrt {2+\sqrt {3}}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[x]*Tanh[6*x],x]

[Out]

-1/3*ArcTan[Sqrt[2]*Sinh[x]]/Sqrt[2] - (Sqrt[2 - Sqrt[3]]*ArcTan[(2*Sinh[x])/Sqrt[2 - Sqrt[3]]])/6 - (Sqrt[2 +
Sqrt[3]]*ArcTan[(2*Sinh[x])/Sqrt[2 + Sqrt[3]]])/6 + Sinh[x]

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fricas [B]  time = 0.58, size = 205, normalized size = 2.36 \[ -\frac {1}{6} \, {\left (2 \, \sqrt {\sqrt {3} + 2} \arctan \left ({\left (\sqrt {\sqrt {3} e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )} + 1} \sqrt {\sqrt {3} + 2} {\left (\sqrt {3} - 2\right )} - {\left ({\left (\sqrt {3} - 2\right )} e^{\left (2 \, x\right )} - \sqrt {3} + 2\right )} \sqrt {\sqrt {3} + 2}\right )} e^{\left (-x\right )}\right ) e^{x} - 2 \, \sqrt {-\sqrt {3} + 2} \arctan \left ({\left (\sqrt {-\sqrt {3} e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )} + 1} {\left (\sqrt {3} + 2\right )} \sqrt {-\sqrt {3} + 2} - {\left ({\left (\sqrt {3} + 2\right )} e^{\left (2 \, x\right )} - \sqrt {3} - 2\right )} \sqrt {-\sqrt {3} + 2}\right )} e^{\left (-x\right )}\right ) e^{x} + \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} e^{\left (3 \, x\right )} + \frac {1}{2} \, \sqrt {2} e^{x}\right ) e^{x} + \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} e^{x}\right ) e^{x} - 3 \, e^{\left (2 \, x\right )} + 3\right )} e^{\left (-x\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(6*x),x, algorithm="fricas")

[Out]

-1/6*(2*sqrt(sqrt(3) + 2)*arctan((sqrt(sqrt(3)*e^(2*x) + e^(4*x) + 1)*sqrt(sqrt(3) + 2)*(sqrt(3) - 2) - ((sqrt
(3) - 2)*e^(2*x) - sqrt(3) + 2)*sqrt(sqrt(3) + 2))*e^(-x))*e^x - 2*sqrt(-sqrt(3) + 2)*arctan((sqrt(-sqrt(3)*e^
(2*x) + e^(4*x) + 1)*(sqrt(3) + 2)*sqrt(-sqrt(3) + 2) - ((sqrt(3) + 2)*e^(2*x) - sqrt(3) - 2)*sqrt(-sqrt(3) +
2))*e^(-x))*e^x + sqrt(2)*arctan(1/2*sqrt(2)*e^(3*x) + 1/2*sqrt(2)*e^x)*e^x + sqrt(2)*arctan(1/2*sqrt(2)*e^x)*
e^x - 3*e^(2*x) + 3)*e^(-x)

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giac [A]  time = 0.15, size = 100, normalized size = 1.15 \[ -\frac {1}{12} \, {\left (\sqrt {6} + \sqrt {2}\right )} \arctan \left (-\frac {2 \, {\left (e^{\left (-x\right )} - e^{x}\right )}}{\sqrt {6} + \sqrt {2}}\right ) - \frac {1}{12} \, {\left (\sqrt {6} - \sqrt {2}\right )} \arctan \left (-\frac {2 \, {\left (e^{\left (-x\right )} - e^{x}\right )}}{\sqrt {6} - \sqrt {2}}\right ) - \frac {1}{12} \, \sqrt {2} {\left (\pi + 2 \, \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (e^{\left (2 \, x\right )} - 1\right )} e^{\left (-x\right )}\right )\right )} - \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(6*x),x, algorithm="giac")

[Out]

-1/12*(sqrt(6) + sqrt(2))*arctan(-2*(e^(-x) - e^x)/(sqrt(6) + sqrt(2))) - 1/12*(sqrt(6) - sqrt(2))*arctan(-2*(
e^(-x) - e^x)/(sqrt(6) - sqrt(2))) - 1/12*sqrt(2)*(pi + 2*arctan(1/2*sqrt(2)*(e^(2*x) - 1)*e^(-x))) - 1/2*e^(-
x) + 1/2*e^x

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maple [C]  time = 0.36, size = 84, normalized size = 0.97 \[ \frac {{\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{-x}}{2}+\left (\munderset {\textit {\_R} =\RootOf \left (20736 \textit {\_Z}^{4}+576 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-12 \textit {\_R} \,{\mathrm e}^{x}+{\mathrm e}^{2 x}-1\right )\right )+\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 x}-i \sqrt {2}\, {\mathrm e}^{x}-1\right )}{12}-\frac {i \sqrt {2}\, \ln \left ({\mathrm e}^{2 x}+i \sqrt {2}\, {\mathrm e}^{x}-1\right )}{12} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(x)*tanh(6*x),x)

[Out]

1/2*exp(x)-1/2*exp(-x)+sum(_R*ln(-12*_R*exp(x)+exp(2*x)-1),_R=RootOf(20736*_Z^4+576*_Z^2+1))+1/12*I*2^(1/2)*ln
(exp(2*x)-I*2^(1/2)*exp(x)-1)-1/12*I*2^(1/2)*ln(exp(2*x)+I*2^(1/2)*exp(x)-1)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(6*x),x, algorithm="maxima")

[Out]

1/2*(e^(2*x) - 1)*e^(-x) - 1/6*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*e^x)) - 1/6*sqrt(2)*arctan(-1/2*sqrt(2)
*(sqrt(2) - 2*e^x)) - 1/2*integrate(2/3*(2*e^(7*x) - e^(5*x) - e^(3*x) + 2*e^x)/(e^(8*x) - e^(4*x) + 1), x)

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mupad [B]  time = 2.67, size = 98, normalized size = 1.13 \[ \frac {{\mathrm {e}}^x}{2}-\frac {{\mathrm {e}}^{-x}}{2}-\frac {\sqrt {2}\,\mathrm {atan}\left (\frac {\sqrt {2}\,{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{2\,x}-1\right )}{2}\right )}{6}-2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{2\,x}-1\right )}{12\,\sqrt {\frac {1}{72}-\frac {\sqrt {3}}{144}}}\right )\,\sqrt {\frac {1}{72}-\frac {\sqrt {3}}{144}}-2\,\mathrm {atan}\left (\frac {{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{2\,x}-1\right )}{12\,\sqrt {\frac {\sqrt {3}}{144}+\frac {1}{72}}}\right )\,\sqrt {\frac {\sqrt {3}}{144}+\frac {1}{72}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(6*x)*sinh(x),x)

[Out]

exp(x)/2 - exp(-x)/2 - (2^(1/2)*atan((2^(1/2)*exp(-x)*(exp(2*x) - 1))/2))/6 - 2*atan((exp(-x)*(exp(2*x) - 1))/
(12*(1/72 - 3^(1/2)/144)^(1/2)))*(1/72 - 3^(1/2)/144)^(1/2) - 2*atan((exp(-x)*(exp(2*x) - 1))/(12*(3^(1/2)/144
+ 1/72)^(1/2)))*(3^(1/2)/144 + 1/72)^(1/2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh {\relax (x )} \tanh {\left (6 x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(6*x),x)

[Out]

Integral(sinh(x)*tanh(6*x), x)

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