3.77 \(\int \sin (x) \tan (5 x) \, dx\)

Optimal. Leaf size=112 \[ -\sin (x)-\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (-4 \sin (x)-\sqrt {5}+1\right )-\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (-4 \sin (x)+\sqrt {5}+1\right )+\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (4 \sin (x)-\sqrt {5}+1\right )+\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (4 \sin (x)+\sqrt {5}+1\right )+\frac {1}{5} \tanh ^{-1}(\sin (x)) \]

[Out]

1/5*arctanh(sin(x))-sin(x)-1/20*ln(1-4*sin(x)-5^(1/2))*(-5^(1/2)+1)+1/20*ln(1+4*sin(x)-5^(1/2))*(-5^(1/2)+1)-1
/20*ln(1-4*sin(x)+5^(1/2))*(5^(1/2)+1)+1/20*ln(1+4*sin(x)+5^(1/2))*(5^(1/2)+1)

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Rubi [A]  time = 0.17, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 4, integrand size = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {2075, 207, 632, 31} \[ -\sin (x)-\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (-4 \sin (x)-\sqrt {5}+1\right )-\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (-4 \sin (x)+\sqrt {5}+1\right )+\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (4 \sin (x)-\sqrt {5}+1\right )+\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (4 \sin (x)+\sqrt {5}+1\right )+\frac {1}{5} \tanh ^{-1}(\sin (x)) \]

Antiderivative was successfully verified.

[In]

Int[Sin[x]*Tan[5*x],x]

[Out]

ArcTanh[Sin[x]]/5 - ((1 - Sqrt[5])*Log[1 - Sqrt[5] - 4*Sin[x]])/20 - ((1 + Sqrt[5])*Log[1 + Sqrt[5] - 4*Sin[x]
])/20 + ((1 - Sqrt[5])*Log[1 - Sqrt[5] + 4*Sin[x]])/20 + ((1 + Sqrt[5])*Log[1 + Sqrt[5] + 4*Sin[x]])/20 - Sin[
x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 207

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

Rule 632

Int[((d_.) + (e_.)*(x_))/((a_) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[
(c*d - e*(b/2 - q/2))/q, Int[1/(b/2 - q/2 + c*x), x], x] - Dist[(c*d - e*(b/2 + q/2))/q, Int[1/(b/2 + q/2 + c*
x), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[2*c*d - b*e, 0] && NeQ[b^2 - 4*a*c, 0] && NiceSqrtQ[b^2 - 4*a*
c]

Rule 2075

Int[(P_)^(p_)*(Qm_), x_Symbol] :> With[{PP = Factor[P]}, Int[ExpandIntegrand[PP^p*Qm, x], x] /; QuadraticProdu
ctQ[PP, x]] /; PolyQ[Qm, x] && PolyQ[P, x] && ILtQ[p, 0]

Rubi steps

\begin {align*} \int \sin (x) \tan (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,\sin (x)\right )\\ &=\operatorname {Subst}\left (\int \left (-1-\frac {1}{5 \left (-1+x^2\right )}-\frac {2 (1+x)}{5 \left (-1-2 x+4 x^2\right )}+\frac {2 (-1+x)}{5 \left (-1+2 x+4 x^2\right )}\right ) \, dx,x,\sin (x)\right )\\ &=-\sin (x)-\frac {1}{5} \operatorname {Subst}\left (\int \frac {1}{-1+x^2} \, dx,x,\sin (x)\right )-\frac {2}{5} \operatorname {Subst}\left (\int \frac {1+x}{-1-2 x+4 x^2} \, dx,x,\sin (x)\right )+\frac {2}{5} \operatorname {Subst}\left (\int \frac {-1+x}{-1+2 x+4 x^2} \, dx,x,\sin (x)\right )\\ &=\frac {1}{5} \tanh ^{-1}(\sin (x))-\sin (x)+\frac {1}{5} \left (1-\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {5}+4 x} \, dx,x,\sin (x)\right )-\frac {1}{5} \left (1-\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{-1+\sqrt {5}+4 x} \, dx,x,\sin (x)\right )-\frac {1}{5} \left (1+\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{-1-\sqrt {5}+4 x} \, dx,x,\sin (x)\right )+\frac {1}{5} \left (1+\sqrt {5}\right ) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {5}+4 x} \, dx,x,\sin (x)\right )\\ &=\frac {1}{5} \tanh ^{-1}(\sin (x))-\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (1-\sqrt {5}-4 \sin (x)\right )-\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (1+\sqrt {5}-4 \sin (x)\right )+\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (1-\sqrt {5}+4 \sin (x)\right )+\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (1+\sqrt {5}+4 \sin (x)\right )-\sin (x)\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 100, normalized size = 0.89 \[ \frac {1}{20} \left (-20 \sin (x)+\left (\sqrt {5}-1\right ) \log \left (-4 \sin (x)-\sqrt {5}+1\right )-\left (1+\sqrt {5}\right ) \log \left (-4 \sin (x)+\sqrt {5}+1\right )-\left (\sqrt {5}-1\right ) \log \left (4 \sin (x)-\sqrt {5}+1\right )+\left (1+\sqrt {5}\right ) \log \left (4 \sin (x)+\sqrt {5}+1\right )+4 \tanh ^{-1}(\sin (x))\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[x]*Tan[5*x],x]

[Out]

(4*ArcTanh[Sin[x]] + (-1 + Sqrt[5])*Log[1 - Sqrt[5] - 4*Sin[x]] - (1 + Sqrt[5])*Log[1 + Sqrt[5] - 4*Sin[x]] -
(-1 + Sqrt[5])*Log[1 - Sqrt[5] + 4*Sin[x]] + (1 + Sqrt[5])*Log[1 + Sqrt[5] + 4*Sin[x]] - 20*Sin[x])/20

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fricas [A]  time = 0.84, size = 136, normalized size = 1.21 \[ \frac {1}{20} \, \sqrt {5} \log \left (\frac {8 \, \cos \relax (x)^{2} - 4 \, {\left (\sqrt {5} - 1\right )} \sin \relax (x) + \sqrt {5} - 11}{4 \, \cos \relax (x)^{2} + 2 \, \sin \relax (x) - 3}\right ) + \frac {1}{20} \, \sqrt {5} \log \left (-\frac {8 \, \cos \relax (x)^{2} - 4 \, {\left (\sqrt {5} + 1\right )} \sin \relax (x) - \sqrt {5} - 11}{4 \, \cos \relax (x)^{2} - 2 \, \sin \relax (x) - 3}\right ) - \frac {1}{20} \, \log \left (4 \, \cos \relax (x)^{2} + 2 \, \sin \relax (x) - 3\right ) + \frac {1}{20} \, \log \left (4 \, \cos \relax (x)^{2} - 2 \, \sin \relax (x) - 3\right ) + \frac {1}{10} \, \log \left (\sin \relax (x) + 1\right ) - \frac {1}{10} \, \log \left (-\sin \relax (x) + 1\right ) - \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(5*x),x, algorithm="fricas")

[Out]

1/20*sqrt(5)*log((8*cos(x)^2 - 4*(sqrt(5) - 1)*sin(x) + sqrt(5) - 11)/(4*cos(x)^2 + 2*sin(x) - 3)) + 1/20*sqrt
(5)*log(-(8*cos(x)^2 - 4*(sqrt(5) + 1)*sin(x) - sqrt(5) - 11)/(4*cos(x)^2 - 2*sin(x) - 3)) - 1/20*log(4*cos(x)
^2 + 2*sin(x) - 3) + 1/20*log(4*cos(x)^2 - 2*sin(x) - 3) + 1/10*log(sin(x) + 1) - 1/10*log(-sin(x) + 1) - sin(
x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sin \relax (x) \tan \left (5 \, x\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(5*x),x, algorithm="giac")

[Out]

integrate(sin(x)*tan(5*x), x)

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maple [A]  time = 0.41, size = 84, normalized size = 0.75 \[ \frac {\ln \left (4 \left (\sin ^{2}\relax (x )\right )+2 \sin \relax (x )-1\right )}{20}+\frac {\sqrt {5}\, \arctanh \left (\frac {\left (8 \sin \relax (x )+2\right ) \sqrt {5}}{10}\right )}{10}-\frac {\ln \left (\sin \relax (x )-1\right )}{10}-\frac {\ln \left (4 \left (\sin ^{2}\relax (x )\right )-2 \sin \relax (x )-1\right )}{20}+\frac {\sqrt {5}\, \arctanh \left (\frac {\left (8 \sin \relax (x )-2\right ) \sqrt {5}}{10}\right )}{10}+\frac {\ln \left (1+\sin \relax (x )\right )}{10}-\sin \relax (x ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(x)*tan(5*x),x)

[Out]

1/20*ln(4*sin(x)^2+2*sin(x)-1)+1/10*5^(1/2)*arctanh(1/10*(8*sin(x)+2)*5^(1/2))-1/10*ln(sin(x)-1)-1/20*ln(4*sin
(x)^2-2*sin(x)-1)+1/10*5^(1/2)*arctanh(1/10*(8*sin(x)-2)*5^(1/2))+1/10*ln(1+sin(x))-sin(x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (3 \, \cos \left (7 \, x\right ) - \cos \left (5 \, x\right ) - \cos \left (3 \, x\right ) + 3 \, \cos \relax (x)\right )} \cos \left (8 \, x\right ) - 3 \, {\left (\cos \left (6 \, x\right ) - \cos \left (4 \, x\right ) + \cos \left (2 \, x\right ) - 1\right )} \cos \left (7 \, x\right ) + {\left (\cos \left (5 \, x\right ) + \cos \left (3 \, x\right ) - 3 \, \cos \relax (x)\right )} \cos \left (6 \, x\right ) - {\left (\cos \left (4 \, x\right ) - \cos \left (2 \, x\right ) + 1\right )} \cos \left (5 \, x\right ) - {\left (\cos \left (3 \, x\right ) - 3 \, \cos \relax (x)\right )} \cos \left (4 \, x\right ) + {\left (\cos \left (2 \, x\right ) - 1\right )} \cos \left (3 \, x\right ) - 3 \, \cos \left (2 \, x\right ) \cos \relax (x) + {\left (3 \, \sin \left (7 \, x\right ) - \sin \left (5 \, x\right ) - \sin \left (3 \, x\right ) + 3 \, \sin \relax (x)\right )} \sin \left (8 \, x\right ) - 3 \, {\left (\sin \left (6 \, x\right ) - \sin \left (4 \, x\right ) + \sin \left (2 \, x\right )\right )} \sin \left (7 \, x\right ) + {\left (\sin \left (5 \, x\right ) + \sin \left (3 \, x\right ) - 3 \, \sin \relax (x)\right )} \sin \left (6 \, x\right ) - {\left (\sin \left (4 \, x\right ) - \sin \left (2 \, x\right )\right )} \sin \left (5 \, x\right ) - {\left (\sin \left (3 \, x\right ) - 3 \, \sin \relax (x)\right )} \sin \left (4 \, x\right ) + \sin \left (3 \, x\right ) \sin \left (2 \, x\right ) - 3 \, \sin \left (2 \, x\right ) \sin \relax (x) + 3 \, \cos \relax (x)}{5 \, {\left (2 \, {\left (\cos \left (6 \, x\right ) - \cos \left (4 \, x\right ) + \cos \left (2 \, x\right ) - 1\right )} \cos \left (8 \, x\right ) - \cos \left (8 \, x\right )^{2} + 2 \, {\left (\cos \left (4 \, x\right ) - \cos \left (2 \, x\right ) + 1\right )} \cos \left (6 \, x\right ) - \cos \left (6 \, x\right )^{2} + 2 \, {\left (\cos \left (2 \, x\right ) - 1\right )} \cos \left (4 \, x\right ) - \cos \left (4 \, x\right )^{2} - \cos \left (2 \, x\right )^{2} + 2 \, {\left (\sin \left (6 \, x\right ) - \sin \left (4 \, x\right ) + \sin \left (2 \, x\right )\right )} \sin \left (8 \, x\right ) - \sin \left (8 \, x\right )^{2} + 2 \, {\left (\sin \left (4 \, x\right ) - \sin \left (2 \, x\right )\right )} \sin \left (6 \, x\right ) - \sin \left (6 \, x\right )^{2} - \sin \left (4 \, x\right )^{2} + 2 \, \sin \left (4 \, x\right ) \sin \left (2 \, x\right ) - \sin \left (2 \, x\right )^{2} + 2 \, \cos \left (2 \, x\right ) - 1\right )}}\,{d x} + \frac {1}{10} \, \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} + 2 \, \sin \relax (x) + 1\right ) - \frac {1}{10} \, \log \left (\cos \relax (x)^{2} + \sin \relax (x)^{2} - 2 \, \sin \relax (x) + 1\right ) - \sin \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(5*x),x, algorithm="maxima")

[Out]

integrate(-1/5*((3*cos(7*x) - cos(5*x) - cos(3*x) + 3*cos(x))*cos(8*x) - 3*(cos(6*x) - cos(4*x) + cos(2*x) - 1
)*cos(7*x) + (cos(5*x) + cos(3*x) - 3*cos(x))*cos(6*x) - (cos(4*x) - cos(2*x) + 1)*cos(5*x) - (cos(3*x) - 3*co
s(x))*cos(4*x) + (cos(2*x) - 1)*cos(3*x) - 3*cos(2*x)*cos(x) + (3*sin(7*x) - sin(5*x) - sin(3*x) + 3*sin(x))*s
in(8*x) - 3*(sin(6*x) - sin(4*x) + sin(2*x))*sin(7*x) + (sin(5*x) + sin(3*x) - 3*sin(x))*sin(6*x) - (sin(4*x)
- sin(2*x))*sin(5*x) - (sin(3*x) - 3*sin(x))*sin(4*x) + sin(3*x)*sin(2*x) - 3*sin(2*x)*sin(x) + 3*cos(x))/(2*(
cos(6*x) - cos(4*x) + cos(2*x) - 1)*cos(8*x) - cos(8*x)^2 + 2*(cos(4*x) - cos(2*x) + 1)*cos(6*x) - cos(6*x)^2
+ 2*(cos(2*x) - 1)*cos(4*x) - cos(4*x)^2 - cos(2*x)^2 + 2*(sin(6*x) - sin(4*x) + sin(2*x))*sin(8*x) - sin(8*x)
^2 + 2*(sin(4*x) - sin(2*x))*sin(6*x) - sin(6*x)^2 - sin(4*x)^2 + 2*sin(4*x)*sin(2*x) - sin(2*x)^2 + 2*cos(2*x
) - 1), x) + 1/10*log(cos(x)^2 + sin(x)^2 + 2*sin(x) + 1) - 1/10*log(cos(x)^2 + sin(x)^2 - 2*sin(x) + 1) - sin
(x)

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mupad [B]  time = 2.89, size = 107, normalized size = 0.96 \[ \frac {2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{5}+\frac {\mathrm {atan}\left (\frac {\sin \relax (x)\,1042{}\mathrm {i}-\sqrt {5}\,\sin \relax (x)\,466{}\mathrm {i}}{377\,\sqrt {5}-843}\right )\,1{}\mathrm {i}}{10}-\frac {\mathrm {atanh}\left (\sin \relax (x)-\sqrt {5}\,\sin \relax (x)\right )}{10}-\sin \relax (x)-\frac {\sqrt {5}\,\mathrm {atanh}\left (\sin \relax (x)-\sqrt {5}\,\sin \relax (x)\right )}{10}-\frac {\sqrt {5}\,\mathrm {atan}\left (\frac {\sin \relax (x)\,1042{}\mathrm {i}-\sqrt {5}\,\sin \relax (x)\,466{}\mathrm {i}}{377\,\sqrt {5}-843}\right )\,1{}\mathrm {i}}{10} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(5*x)*sin(x),x)

[Out]

(atan((sin(x)*1042i - 5^(1/2)*sin(x)*466i)/(377*5^(1/2) - 843))*1i)/10 - atanh(sin(x) - 5^(1/2)*sin(x))/10 + (
2*atanh(sin(x/2)/cos(x/2)))/5 - sin(x) - (5^(1/2)*atanh(sin(x) - 5^(1/2)*sin(x)))/10 - (5^(1/2)*atan((sin(x)*1
042i - 5^(1/2)*sin(x)*466i)/(377*5^(1/2) - 843))*1i)/10

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(x)*tan(5*x),x)

[Out]

Integral(sin(x)*tan(5*x), x)

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