### 3.202 $$\int \sinh (x) \tanh (4 x) \, dx$$

Optimal. Leaf size=69 $\sinh (x)-\frac{1}{4} \sqrt{2-\sqrt{2}} \tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2-\sqrt{2}}}\right )-\frac{1}{4} \sqrt{2+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2+\sqrt{2}}}\right )$

[Out]

-(Sqrt[2 - Sqrt[2]]*ArcTan[(2*Sinh[x])/Sqrt[2 - Sqrt[2]]])/4 - (Sqrt[2 + Sqrt[2]]*ArcTan[(2*Sinh[x])/Sqrt[2 +
Sqrt[2]]])/4 + Sinh[x]

________________________________________________________________________________________

Rubi [A]  time = 0.0962809, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 7, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.571, Rules used = {12, 1279, 1166, 203} $\sinh (x)-\frac{1}{4} \sqrt{2-\sqrt{2}} \tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2-\sqrt{2}}}\right )-\frac{1}{4} \sqrt{2+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2+\sqrt{2}}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Int[Sinh[x]*Tanh[4*x],x]

[Out]

-(Sqrt[2 - Sqrt[2]]*ArcTan[(2*Sinh[x])/Sqrt[2 - Sqrt[2]]])/4 - (Sqrt[2 + Sqrt[2]]*ArcTan[(2*Sinh[x])/Sqrt[2 +
Sqrt[2]]])/4 + 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 1279

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] - Dist[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] && (Inte
gerQ[p] || IntegerQ[m])

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 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])

Rubi steps

\begin{align*} \int \sinh (x) \tanh (4 x) \, dx &=-\operatorname{Subst}\left (\int \frac{4 x^2 \left (-1-2 x^2\right )}{1+8 x^2+8 x^4} \, dx,x,\sinh (x)\right )\\ &=-\left (4 \operatorname{Subst}\left (\int \frac{x^2 \left (-1-2 x^2\right )}{1+8 x^2+8 x^4} \, dx,x,\sinh (x)\right )\right )\\ &=\sinh (x)+\frac{1}{2} \operatorname{Subst}\left (\int \frac{-2-8 x^2}{1+8 x^2+8 x^4} \, dx,x,\sinh (x)\right )\\ &=\sinh (x)+\left (-2+\sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{4-2 \sqrt{2}+8 x^2} \, dx,x,\sinh (x)\right )-\left (2+\sqrt{2}\right ) \operatorname{Subst}\left (\int \frac{1}{4+2 \sqrt{2}+8 x^2} \, dx,x,\sinh (x)\right )\\ &=-\frac{1}{4} \sqrt{2-\sqrt{2}} \tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2-\sqrt{2}}}\right )-\frac{1}{4} \sqrt{2+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2+\sqrt{2}}}\right )+\sinh (x)\\ \end{align*}

Mathematica [A]  time = 0.134683, size = 69, normalized size = 1. $\sinh (x)-\frac{1}{4} \sqrt{2-\sqrt{2}} \tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2-\sqrt{2}}}\right )-\frac{1}{4} \sqrt{2+\sqrt{2}} \tan ^{-1}\left (\frac{2 \sinh (x)}{\sqrt{2+\sqrt{2}}}\right )$

Antiderivative was successfully veriﬁed.

[In]

Integrate[Sinh[x]*Tanh[4*x],x]

[Out]

-(Sqrt[2 - Sqrt[2]]*ArcTan[(2*Sinh[x])/Sqrt[2 - Sqrt[2]]])/4 - (Sqrt[2 + Sqrt[2]]*ArcTan[(2*Sinh[x])/Sqrt[2 +
Sqrt[2]]])/4 + Sinh[x]

________________________________________________________________________________________

Maple [C]  time = 0.083, size = 42, normalized size = 0.6 \begin{align*}{\frac{{{\rm e}^{x}}}{2}}-{\frac{{{\rm e}^{-x}}}{2}}+\sum _{{\it \_R}={\it RootOf} \left ( 2048\,{{\it \_Z}}^{4}+128\,{{\it \_Z}}^{2}+1 \right ) }{\it \_R}\,\ln \left ( -8\,{\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(4*x),x)

[Out]

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))

________________________________________________________________________________________

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}{2} \, \int \frac{2 \,{\left (e^{\left (7 \, x\right )} + e^{x}\right )}}{e^{\left (8 \, x\right )} + 1}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(4*x),x, algorithm="maxima")

[Out]

1/2*(e^(2*x) - 1)*e^(-x) - 1/2*integrate(2*(e^(7*x) + e^x)/(e^(8*x) + 1), x)

________________________________________________________________________________________

Fricas [B]  time = 2.35069, size = 504, normalized size = 7.3 \begin{align*} -\frac{1}{2} \,{\left (\sqrt{\sqrt{2} + 2} \arctan \left (\frac{1}{2} \,{\left (\sqrt{\sqrt{2} e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )} + 1} \sqrt{\sqrt{2} + 2}{\left (\sqrt{2} - 2\right )} -{\left ({\left (\sqrt{2} - 2\right )} e^{\left (2 \, x\right )} - \sqrt{2} + 2\right )} \sqrt{\sqrt{2} + 2}\right )} e^{\left (-x\right )}\right ) e^{x} - \sqrt{-\sqrt{2} + 2} \arctan \left (\frac{1}{2} \,{\left (\sqrt{-\sqrt{2} e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )} + 1}{\left (\sqrt{2} + 2\right )} \sqrt{-\sqrt{2} + 2} -{\left ({\left (\sqrt{2} + 2\right )} e^{\left (2 \, x\right )} - \sqrt{2} - 2\right )} \sqrt{-\sqrt{2} + 2}\right )} e^{\left (-x\right )}\right ) e^{x} - e^{\left (2 \, x\right )} + 1\right )} e^{\left (-x\right )} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(4*x),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sinh{\left (x \right )} \tanh{\left (4 x \right )}\, dx \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(4*x),x)

[Out]

Integral(sinh(x)*tanh(4*x), x)

________________________________________________________________________________________

Giac [A]  time = 1.25248, size = 96, normalized size = 1.39 \begin{align*} -\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} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(x)*tanh(4*x),x, algorithm="giac")

[Out]

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