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]

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

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Rubi [A]  time = 0.10, antiderivative size = 69, normalized size of antiderivative = 1.00, 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 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 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])

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*}

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Mathematica [A]  time = 0.15, size = 69, normalized size = 1.00 \[ \sinh (x)-\frac {1}{4} \sqrt {2-\sqrt {2}} \tan ^{-1}\left (\frac {2 \sinh (x)}{\sqrt {2-\sqrt {2}}}\right )-\frac {1}{4} \sqrt {2+\sqrt {2}} \tan ^{-1}\left (\frac {2 \sinh (x)}{\sqrt {2+\sqrt {2}}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-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]

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fricas [B]  time = 0.49, size = 168, normalized size = 2.43 \[ -\frac {1}{2} \, {\left (\sqrt {\sqrt {2} + 2} \arctan \left (\frac {1}{2} \, {\left (\sqrt {\sqrt {2} e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )} + 1} \sqrt {\sqrt {2} + 2} {\left (\sqrt {2} - 2\right )} - {\left ({\left (\sqrt {2} - 2\right )} e^{\left (2 \, x\right )} - \sqrt {2} + 2\right )} \sqrt {\sqrt {2} + 2}\right )} e^{\left (-x\right )}\right ) e^{x} - \sqrt {-\sqrt {2} + 2} \arctan \left (\frac {1}{2} \, {\left (\sqrt {-\sqrt {2} e^{\left (2 \, x\right )} + e^{\left (4 \, x\right )} + 1} {\left (\sqrt {2} + 2\right )} \sqrt {-\sqrt {2} + 2} - {\left ({\left (\sqrt {2} + 2\right )} e^{\left (2 \, x\right )} - \sqrt {2} - 2\right )} \sqrt {-\sqrt {2} + 2}\right )} e^{\left (-x\right )}\right ) e^{x} - e^{\left (2 \, x\right )} + 1\right )} e^{\left (-x\right )} \]

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)

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giac [A]  time = 0.22, size = 71, normalized size = 1.03 \[ -\frac {1}{4} \, \sqrt {\sqrt {2} + 2} \arctan \left (-\frac {e^{\left (-x\right )} - e^{x}}{\sqrt {\sqrt {2} + 2}}\right ) - \frac {1}{4} \, \sqrt {-\sqrt {2} + 2} \arctan \left (-\frac {e^{\left (-x\right )} - e^{x}}{\sqrt {-\sqrt {2} + 2}}\right ) - \frac {1}{2} \, e^{\left (-x\right )} + \frac {1}{2} \, e^{x} \]

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

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maple [C]  time = 0.32, size = 42, normalized size = 0.61 \[ \frac {{\mathrm e}^{x}}{2}-\frac {{\mathrm e}^{-x}}{2}+\left (\munderset {\textit {\_R} =\RootOf \left (2048 \textit {\_Z}^{4}+128 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left (-8 \textit {\_R} \,{\mathrm e}^{x}+{\mathrm e}^{2 x}-1\right )\right ) \]

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))

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

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)

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mupad [B]  time = 0.77, size = 71, normalized size = 1.03 \[ \frac {{\mathrm {e}}^x}{2}-\frac {{\mathrm {e}}^{-x}}{2}-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{2\,x}-1\right )}{\sqrt {\sqrt {2}+2}}\right )\,\sqrt {\sqrt {2}+2}}{4}-\frac {\mathrm {atan}\left (\frac {{\mathrm {e}}^{-x}\,\left ({\mathrm {e}}^{2\,x}-1\right )}{\sqrt {2-\sqrt {2}}}\right )\,\sqrt {2-\sqrt {2}}}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sinh {\relax (x )} \tanh {\left (4 x \right )}\, dx \]

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)

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