3.2.0 ∫ cos(x) tan(6x) dx [109]

Integrand size = 7, antiderivative size = 89 \[ \int \cos (x) \tan (6 x) \, dx=\frac {\text {arctanh}\left (\sqrt {2} \cos (x)\right )}{3 \sqrt {2}}+\frac {1}{6} \sqrt {2-\sqrt {3}} \text {arctanh}\left (\frac {2 \cos (x)}{\sqrt {2-\sqrt {3}}}\right )+\frac {1}{6} \sqrt {2+\sqrt {3}} \text {arctanh}\left (\frac {2 \cos (x)}{\sqrt {2+\sqrt {3}}}\right )-\cos (x) \]

-cos(x)+1/6*arctanh(cos(x)*2^(1/2))*2^(1/2)+1/6*arctanh(2*cos(x)/(1/2*6^(1/2)-1/2*2^(1/2)))*(1/2*6^(1/2)-1/2*2

Rubi [A] (veriﬁed)

Time = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.714, Rules used = {12, 6874, 2098, 213, 1180} \[ \int \cos (x) \tan (6 x) \, dx=\frac {\text {arctanh}\left (\sqrt {2} \cos (x)\right )}{3 \sqrt {2}}+\frac {1}{6} \sqrt {2-\sqrt {3}} \text {arctanh}\left (\frac {2 \cos (x)}{\sqrt {2-\sqrt {3}}}\right )+\frac {1}{6} \sqrt {2+\sqrt {3}} \text {arctanh}\left (\frac {2 \cos (x)}{\sqrt {2+\sqrt {3}}}\right )-\cos (x) \]

ArcTanh[Sqrt[2]*Cos[x]]/(3*Sqrt[2]) + (Sqrt[2 - Sqrt[3]]*ArcTanh[(2*Cos[x])/Sqrt[2 - Sqrt[3]]])/6 + (Sqrt[2 +

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[b, 2])^(-1))*ArcTanh[Rt[b, 2]*(x/Rt[-a, 2])]

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^

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,

Rubi steps \begin{align*} \text {integral}& = -\text {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,\cos (x)\right ) \\ & = -\left (2 \text {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,\cos (x)\right )\right ) \\ & = -\left (2 \text {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,\cos (x)\right )\right ) \\ & = -\cos (x)+\text {Subst}\left (\int \frac {1-12 x^2+16 x^4}{1-18 x^2+48 x^4-32 x^6} \, dx,x,\cos (x)\right ) \\ & = -\cos (x)+\text {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,\cos (x)\right ) \\ & = -\cos (x)-\frac {1}{3} \text {Subst}\left (\int \frac {1}{-1+2 x^2} \, dx,x,\cos (x)\right )-\frac {2}{3} \text {Subst}\left (\int \frac {-1+8 x^2}{1-16 x^2+16 x^4} \, dx,x,\cos (x)\right ) \\ & = \frac {\text {arctanh}\left (\sqrt {2} \cos (x)\right )}{3 \sqrt {2}}-\cos (x)-\frac {1}{3} \left (4 \left (2-\sqrt {3}\right )\right ) \text {Subst}\left (\int \frac {1}{-8+4 \sqrt {3}+16 x^2} \, dx,x,\cos (x)\right )-\frac {1}{3} \left (4 \left (2+\sqrt {3}\right )\right ) \text {Subst}\left (\int \frac {1}{-8-4 \sqrt {3}+16 x^2} \, dx,x,\cos (x)\right ) \\ & = \frac {\text {arctanh}\left (\sqrt {2} \cos (x)\right )}{3 \sqrt {2}}+\frac {1}{6} \sqrt {2-\sqrt {3}} \text {arctanh}\left (\frac {2 \cos (x)}{\sqrt {2-\sqrt {3}}}\right )+\frac {1}{6} \sqrt {2+\sqrt {3}} \text {arctanh}\left (\frac {2 \cos (x)}{\sqrt {2+\sqrt {3}}}\right )-\cos (x) \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Time = 8.21 (sec) , antiderivative size = 628, normalized size of antiderivative = 7.06 \[ \int \cos (x) \tan (6 x) \, dx=\frac {1}{24} \left ((4+4 i) (-1)^{3/4} \text {arctanh}\left (\frac {-1+\tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right )+(4-4 i) \sqrt [4]{-1} \text {arctanh}\left (\frac {1+\tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right )-24 \cos (x)-\frac {2 \left (1+\sqrt {2}\right ) \left (x-2 \sqrt {3} \text {arctanh}\left (\frac {2+\left (2+\sqrt {2}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {6}}\right )-\log \left (\sec ^2\left (\frac {x}{2}\right )\right )+\log \left (-\sec ^2\left (\frac {x}{2}\right ) \left (\sqrt {2}-2 \cos (x)+2 \sin (x)\right )\right )\right )}{2+\sqrt {2}}+\sqrt {2} \left (x+2 \sqrt {3} \text {arctanh}\left (\frac {\sqrt {2}+\left (-1+\sqrt {2}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {3}}\right )-\log \left (\sec ^2\left (\frac {x}{2}\right )\right )+\log \left (\sec ^2\left (\frac {x}{2}\right ) \left (1+\sqrt {2} \cos (x)-\sqrt {2} \sin (x)\right )\right )\right )-\frac {2 \left (2 \left (-2+\sqrt {6}\right ) \text {arctanh}\left (\sqrt {2}+\left (\sqrt {2}-\sqrt {3}\right ) \tan \left (\frac {x}{2}\right )\right )+\left (3 \sqrt {2}-2 \sqrt {3}\right ) \left (x-\log \left (\sec ^2\left (\frac {x}{2}\right )\right )+\log \left (-\sec ^2\left (\frac {x}{2}\right ) \left (\sqrt {3}+\sqrt {2} \cos (x)-\sqrt {2} \sin (x)\right )\right )\right )\right ) \left (\sqrt {2}-\sqrt {3} \sin (x)\right ) \left (-3+\sqrt {6}-\left (-2+\sqrt {6}\right ) \cos (x)+\left (-2+\sqrt {6}\right ) \sin (x)\right )}{-36+15 \sqrt {6}+\left (20-8 \sqrt {6}\right ) \cos (x)+\left (12-5 \sqrt {6}\right ) \cos (2 x)-50 \sin (x)+20 \sqrt {6} \sin (x)+12 \sin (2 x)-5 \sqrt {6} \sin (2 x)}+\frac {2 \left (-2 \left (\sqrt {2}+\sqrt {3}\right ) \text {arctanh}\left (\frac {2+\left (2+\sqrt {6}\right ) \tan \left (\frac {x}{2}\right )}{\sqrt {2}}\right )+\left (3+\sqrt {6}\right ) \left (x-\log \left (\sec ^2\left (\frac {x}{2}\right )\right )+\log \left (-\sec ^2\left (\frac {x}{2}\right ) \left (\sqrt {6}-2 \cos (x)+2 \sin (x)\right )\right )\right )\right ) \left (2+\sqrt {6} \sin (x)\right ) \left (3+\sqrt {6}-\left (2+\sqrt {6}\right ) \cos (x)+\left (2+\sqrt {6}\right ) \sin (x)\right )}{-36-15 \sqrt {6}+4 \left (5+2 \sqrt {6}\right ) \cos (x)+\left (12+5 \sqrt {6}\right ) \cos (2 x)-50 \sin (x)-20 \sqrt {6} \sin (x)+12 \sin (2 x)+5 \sqrt {6} \sin (2 x)}\right ) \]

((4 + 4*I)*(-1)^(3/4)*ArcTanh[(-1 + Tan[x/2])/Sqrt[2]] + (4 - 4*I)*(-1)^(1/4)*ArcTanh[(1 + Tan[x/2])/Sqrt[2]]

- 24*Cos[x] - (2*(1 + Sqrt[2])*(x - 2*Sqrt[3]*ArcTanh[(2 + (2 + Sqrt[2])*Tan[x/2])/Sqrt[6]] - Log[Sec[x/2]^2]

+ Log[-(Sec[x/2]^2*(Sqrt[2] - 2*Cos[x] + 2*Sin[x]))]))/(2 + Sqrt[2]) + Sqrt[2]*(x + 2*Sqrt[3]*ArcTanh[(Sqrt[2]

+ (-1 + Sqrt[2])*Tan[x/2])/Sqrt[3]] - Log[Sec[x/2]^2] + Log[Sec[x/2]^2*(1 + Sqrt[2]*Cos[x] - Sqrt[2]*Sin[x])]

) - (2*(2*(-2 + Sqrt[6])*ArcTanh[Sqrt[2] + (Sqrt[2] - Sqrt[3])*Tan[x/2]] + (3*Sqrt[2] - 2*Sqrt[3])*(x - Log[Se

c[x/2]^2] + Log[-(Sec[x/2]^2*(Sqrt[3] + Sqrt[2]*Cos[x] - Sqrt[2]*Sin[x]))]))*(Sqrt[2] - Sqrt[3]*Sin[x])*(-3 +

Sqrt[6] - (-2 + Sqrt[6])*Cos[x] + (-2 + Sqrt[6])*Sin[x]))/(-36 + 15*Sqrt[6] + (20 - 8*Sqrt[6])*Cos[x] + (12 -

5*Sqrt[6])*Cos[2*x] - 50*Sin[x] + 20*Sqrt[6]*Sin[x] + 12*Sin[2*x] - 5*Sqrt[6]*Sin[2*x]) + (2*(-2*(Sqrt[2] + Sq

rt[3])*ArcTanh[(2 + (2 + Sqrt[6])*Tan[x/2])/Sqrt[2]] + (3 + Sqrt[6])*(x - Log[Sec[x/2]^2] + Log[-(Sec[x/2]^2*(

Sqrt[6] - 2*Cos[x] + 2*Sin[x]))]))*(2 + Sqrt[6]*Sin[x])*(3 + Sqrt[6] - (2 + Sqrt[6])*Cos[x] + (2 + Sqrt[6])*Si

n[x]))/(-36 - 15*Sqrt[6] + 4*(5 + 2*Sqrt[6])*Cos[x] + (12 + 5*Sqrt[6])*Cos[2*x] - 50*Sin[x] - 20*Sqrt[6]*Sin[x

Maple [C] (veriﬁed)

Time = 1.17 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.11


method	result	size

risch	\(-\frac {{\mathrm e}^{i x}}{2}-\frac {{\mathrm e}^{-i x}}{2}-i \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (20736 \textit {\_Z}^{4}+576 \textit {\_Z}^{2}+1\right )}{\sum }\textit {\_R} \ln \left ({\mathrm e}^{2 i x}-12 i \textit {\_R} \,{\mathrm e}^{i x}+1\right )\right )+\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}+\sqrt {2}\, {\mathrm e}^{i x}+1\right )}{12}-\frac {\sqrt {2}\, \ln \left ({\mathrm e}^{2 i x}-\sqrt {2}\, {\mathrm e}^{i x}+1\right )}{12}\)	\(99\)

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

Fricas [A] (veriﬁcation not implemented)

Time = 0.26 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.51 \[ \int \cos (x) \tan (6 x) \, dx=\frac {1}{12} \, \sqrt {\sqrt {3} + 2} \log \left (\sqrt {\sqrt {3} + 2} + 2 \, \cos \left (x\right )\right ) - \frac {1}{12} \, \sqrt {\sqrt {3} + 2} \log \left (\sqrt {\sqrt {3} + 2} - 2 \, \cos \left (x\right )\right ) + \frac {1}{12} \, \sqrt {-\sqrt {3} + 2} \log \left (\sqrt {-\sqrt {3} + 2} + 2 \, \cos \left (x\right )\right ) - \frac {1}{12} \, \sqrt {-\sqrt {3} + 2} \log \left (\sqrt {-\sqrt {3} + 2} - 2 \, \cos \left (x\right )\right ) + \frac {1}{12} \, \sqrt {2} \log \left (-\frac {2 \, \cos \left (x\right )^{2} + 2 \, \sqrt {2} \cos \left (x\right ) + 1}{2 \, \cos \left (x\right )^{2} - 1}\right ) - \cos \left (x\right ) \]

1/12*sqrt(sqrt(3) + 2)*log(sqrt(sqrt(3) + 2) + 2*cos(x)) - 1/12*sqrt(sqrt(3) + 2)*log(sqrt(sqrt(3) + 2) - 2*co

s(x)) + 1/12*sqrt(-sqrt(3) + 2)*log(sqrt(-sqrt(3) + 2) + 2*cos(x)) - 1/12*sqrt(-sqrt(3) + 2)*log(sqrt(-sqrt(3)

Sympy [F]

\[ \int \cos (x) \tan (6 x) \, dx=\int \cos {\left (x \right )} \tan {\left (6 x \right )}\, dx \]

Maxima [F]

\[ \int \cos (x) \tan (6 x) \, dx=\int { \cos \left (x\right ) \tan \left (6 \, x\right ) \,d x } \]

1/24*sqrt(2)*log(2*sqrt(2)*sin(2*x)*sin(x) + 2*(sqrt(2)*cos(x) + 1)*cos(2*x) + cos(2*x)^2 + 2*cos(x)^2 + sin(2

*x)^2 + 2*sin(x)^2 + 2*sqrt(2)*cos(x) + 1) - 1/24*sqrt(2)*log(-2*sqrt(2)*sin(2*x)*sin(x) - 2*(sqrt(2)*cos(x) -

1)*cos(2*x) + cos(2*x)^2 + 2*cos(x)^2 + sin(2*x)^2 + 2*sin(x)^2 - 2*sqrt(2)*cos(x) + 1) - cos(x) - integrate(

1/3*((2*sin(7*x) + sin(5*x) - sin(3*x) - 2*sin(x))*cos(8*x) + (sin(3*x) + 2*sin(x))*cos(4*x) - (2*cos(7*x) + c

os(5*x) - cos(3*x) - 2*cos(x))*sin(8*x) - 2*(cos(4*x) - 1)*sin(7*x) - (cos(4*x) - 1)*sin(5*x) - (cos(3*x) + 2*

cos(x))*sin(4*x) + 2*cos(7*x)*sin(4*x) + cos(5*x)*sin(4*x) - sin(3*x) - 2*sin(x))/(2*(cos(4*x) - 1)*cos(8*x) -

Giac [F]

\[ \int \cos (x) \tan (6 x) \, dx=\int { \cos \left (x\right ) \tan \left (6 \, x\right ) \,d x } \]

Mupad [B] (veriﬁcation not implemented)

Time = 29.06 (sec) , antiderivative size = 787, normalized size of antiderivative = 8.84 \[ \int \cos (x) \tan (6 x) \, dx=\text {Too large to display} \]

(6^(1/2)*(atan((2^(1/2)*321030945816576i)/(213254896304333030400*tan(x/2)^4 - 129275829262795438080*tan(x/2)^2

+ 2176593611144037376) + (6^(1/2)*888405273481134080i)/(213254896304333030400*tan(x/2)^4 - 129275829262795438

080*tan(x/2)^2 + 2176593611144037376) - (2^(1/2)*tan(x/2)^2*18711054724802560i)/(213254896304333030400*tan(x/2

)^4 - 129275829262795438080*tan(x/2)^2 + 2176593611144037376) + (2^(1/2)*tan(x/2)^4*10905601889064960i)/(21325

4896304333030400*tan(x/2)^4 - 129275829262795438080*tan(x/2)^2 + 2176593611144037376) - (6^(1/2)*tan(x/2)^2*52

765833462352287744i)/(213254896304333030400*tan(x/2)^4 - 129275829262795438080*tan(x/2)^2 + 217659361114403737

6) + (6^(1/2)*tan(x/2)^4*87054650497106012160i)/(213254896304333030400*tan(x/2)^4 - 129275829262795438080*tan(

x/2)^2 + 2176593611144037376)) + atan((2^(1/2)*1443325504589801788190484332544i)/(5892324042622606506545538662

40*2^(1/2)*6^(1/2) + 119129717169909888440949339586560*tan(x/2)^2 - 34367271726987959946466862039040*2^(1/2)*6

^(1/2)*tan(x/2)^2 - 2087090309450798997834557292544) - (6^(1/2)*852047139771204346616741888000i)/(589232404262

260650654553866240*2^(1/2)*6^(1/2) + 119129717169909888440949339586560*tan(x/2)^2 - 34367271726987959946466862

039040*2^(1/2)*6^(1/2)*tan(x/2)^2 - 2087090309450798997834557292544) - (2^(1/2)*tan(x/2)^2*8418228357130530454

3568582410240i)/(589232404262260650654553866240*2^(1/2)*6^(1/2) + 119129717169909888440949339586560*tan(x/2)^2

- 34367271726987959946466862039040*2^(1/2)*6^(1/2)*tan(x/2)^2 - 2087090309450798997834557292544) + (6^(1/2)*t

an(x/2)^2*48634501075236486504873424060416i)/(589232404262260650654553866240*2^(1/2)*6^(1/2) + 119129717169909

888440949339586560*tan(x/2)^2 - 34367271726987959946466862039040*2^(1/2)*6^(1/2)*tan(x/2)^2 - 2087090309450798

997834557292544)))*1i)/12 - 2/(tan(x/2)^2 + 1) - (2^(1/2)*(2*atan((2^(1/2)*2276803846003180334341033033728i)/(

18766876017666378997952094928896*tan(x/2)^2 - 3219886877884552553529320931328) - (2^(1/2)*tan(x/2)^2*132701852

93778646110081740963840i)/(18766876017666378997952094928896*tan(x/2)^2 - 3219886877884552553529320931328)) - a

tan((2^(1/2)*321030945816576i)/(213254896304333030400*tan(x/2)^4 - 129275829262795438080*tan(x/2)^2 + 21765936

11144037376) + (6^(1/2)*888405273481134080i)/(213254896304333030400*tan(x/2)^4 - 129275829262795438080*tan(x/2

)^2 + 2176593611144037376) - (2^(1/2)*tan(x/2)^2*18711054724802560i)/(213254896304333030400*tan(x/2)^4 - 12927

5829262795438080*tan(x/2)^2 + 2176593611144037376) + (2^(1/2)*tan(x/2)^4*10905601889064960i)/(2132548963043330

30400*tan(x/2)^4 - 129275829262795438080*tan(x/2)^2 + 2176593611144037376) - (6^(1/2)*tan(x/2)^2*5276583346235

2287744i)/(213254896304333030400*tan(x/2)^4 - 129275829262795438080*tan(x/2)^2 + 2176593611144037376) + (6^(1/

2)*tan(x/2)^4*87054650497106012160i)/(213254896304333030400*tan(x/2)^4 - 129275829262795438080*tan(x/2)^2 + 21

76593611144037376)) + atan((2^(1/2)*1443325504589801788190484332544i)/(589232404262260650654553866240*2^(1/2)*

6^(1/2) + 119129717169909888440949339586560*tan(x/2)^2 - 34367271726987959946466862039040*2^(1/2)*6^(1/2)*tan(

x/2)^2 - 2087090309450798997834557292544) - (6^(1/2)*852047139771204346616741888000i)/(58923240426226065065455

3866240*2^(1/2)*6^(1/2) + 119129717169909888440949339586560*tan(x/2)^2 - 34367271726987959946466862039040*2^(1

/2)*6^(1/2)*tan(x/2)^2 - 2087090309450798997834557292544) - (2^(1/2)*tan(x/2)^2*841822835713053045435685824102

40i)/(589232404262260650654553866240*2^(1/2)*6^(1/2) + 119129717169909888440949339586560*tan(x/2)^2 - 34367271

726987959946466862039040*2^(1/2)*6^(1/2)*tan(x/2)^2 - 2087090309450798997834557292544) + (6^(1/2)*tan(x/2)^2*4

8634501075236486504873424060416i)/(589232404262260650654553866240*2^(1/2)*6^(1/2) + 11912971716990988844094933

9586560*tan(x/2)^2 - 34367271726987959946466862039040*2^(1/2)*6^(1/2)*tan(x/2)^2 - 208709030945079899783455729

\(\int \cos (x) \tan (6 x) \, dx\) [109]

Optimal result