### 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]

-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]

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Rubi [A]  time = 0.258683, antiderivative size = 87, normalized size of antiderivative = 1., 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 6742

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

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 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]

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*}

Mathematica [A]  time = 0.123729, size = 87, normalized size = 1. $\sinh (x)-\frac{\tan ^{-1}\left (\sqrt{2} \sinh (x)\right )}{3 \sqrt{2}}-\frac{1}{6} \sqrt{2-\sqrt{3}} \tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2-\sqrt{3}}}\right )-\frac{1}{6} \sqrt{2+\sqrt{3}} \tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2+\sqrt{3}}}\right )$

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

-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]

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Maple [C]  time = 0.105, size = 84, normalized size = 1. \begin{align*}{\frac{{{\rm e}^{x}}}{2}}-{\frac{{{\rm e}^{-x}}}{2}}+{\frac{i}{12}}\sqrt{2}\ln \left ({{\rm e}^{2\,x}}-i\sqrt{2}{{\rm e}^{x}}-1 \right ) -{\frac{i}{12}}\sqrt{2}\ln \left ({{\rm e}^{2\,x}}+i\sqrt{2}{{\rm e}^{x}}-1 \right ) +\sum _{{\it \_R}={\it RootOf} \left ( 20736\,{{\it \_Z}}^{4}+576\,{{\it \_Z}}^{2}+1 \right ) }{\it \_R}\,\ln \left ( -12\,{\it \_R}\,{{\rm e}^{x}}+{{\rm e}^{2\,x}}-1 \right ) \end{align*}

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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Fricas [B]  time = 2.37504, size = 633, normalized size = 7.28 \begin{align*} -\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 )} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh{\left (x \right )} \tanh{\left (6 x \right )}\, dx \end{align*}

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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Giac [A]  time = 1.19229, size = 135, normalized size = 1.55 \begin{align*} -\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} \end{align*}

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