\(\int \sin (x) \tan (5 x) \, dx\) [25]

Optimal result

Integrand size = 7, antiderivative size = 86 \[ \int \sin (x) \tan (5 x) \, dx=\frac {1}{5} \text {arctanh}(\sin (x))+\frac {1}{5} \sqrt {\frac {1}{2} \left (3+\sqrt {5}\right )} \text {arctanh}\left (\sqrt {2 \left (3-\sqrt {5}\right )} \sin (x)\right )+\frac {1}{5} \sqrt {\frac {2}{3+\sqrt {5}}} \text {arctanh}\left (\sqrt {2 \left (3+\sqrt {5}\right )} \sin (x)\right )-\sin (x) \] Output:

1/5*arctanh(sin(x))+1/5*(1/2+1/2*5^(1/2))*arctanh((5^(1/2)-1)*sin(x))+1/5* 
2^(1/2)/(3+5^(1/2))^(1/2)*arctanh((5^(1/2)+1)*sin(x))-sin(x)
 

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.16 \[ \int \sin (x) \tan (5 x) \, dx=\frac {1}{20} \left (4 \text {arctanh}(\sin (x))+\left (-1+\sqrt {5}\right ) \log \left (1-\sqrt {5}-4 \sin (x)\right )-\left (1+\sqrt {5}\right ) \log \left (1+\sqrt {5}-4 \sin (x)\right )-\left (-1+\sqrt {5}\right ) \log \left (1-\sqrt {5}+4 \sin (x)\right )+\left (1+\sqrt {5}\right ) \log \left (1+\sqrt {5}+4 \sin (x)\right )-20 \sin (x)\right ) \] Input:

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

Output:

(4*ArcTanh[Sin[x]] + (-1 + Sqrt[5])*Log[1 - Sqrt[5] - 4*Sin[x]] - (1 + Sqr 
t[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
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.30, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.571, Rules used = {3042, 4878, 2460, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (x) \tan (5 x)dx\)

\(\Big \downarrow \) 4878

\(\displaystyle \int \frac {\sin ^2(x) \left (16 \sin ^4(x)-20 \sin ^2(x)+5\right )}{-16 \sin ^6(x)+28 \sin ^4(x)-13 \sin ^2(x)+1}d\sin (x)\)

\(\Big \downarrow \) 2460

\(\displaystyle \int \left (\frac {2 (\sin (x)-1)}{5 \left (4 \sin ^2(x)+2 \sin (x)-1\right )}-\frac {1}{5 \left (\sin ^2(x)-1\right )}-\frac {2 (\sin (x)+1)}{5 \left (4 \sin ^2(x)-2 \sin (x)-1\right )}-1\right )d\sin (x)\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{5} \text {arctanh}(\sin (x))-\sin (x)-\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (-4 \sin (x)-\sqrt {5}+1\right )-\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (-4 \sin (x)+\sqrt {5}+1\right )+\frac {1}{20} \left (1-\sqrt {5}\right ) \log \left (4 \sin (x)-\sqrt {5}+1\right )+\frac {1}{20} \left (1+\sqrt {5}\right ) \log \left (4 \sin (x)+\sqrt {5}+1\right )\)

Input:

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

Output:

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]
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2460

Int[(u_.)*(Px_)^(p_), x_Symbol] :> With[{Qx = Factor[Px /. x -> Sqrt[x]]}, 
Int[ExpandIntegrand[u*(Qx /. x -> x^2)^p, x], x] /;  !SumQ[NonfreeFactors[Q 
x, x]]] /; PolyQ[Px, x^2] && GtQ[Expon[Px, x], 2] &&  !BinomialQ[Px, x] && 
 !TrinomialQ[Px, x] && ILtQ[p, 0] && RationalFunctionQ[u, x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4878

Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa 
ctors[Sin[v], x]}, d/Coefficient[v, x, 1]   Subst[Int[SubstFor[1, Sin[v]/d, 
 u/Cos[v], x], x], x, Sin[v]/d]], x] /;  !FalseQ[v] && FunctionOfQ[NonfreeF 
actors[Sin[v], x], u/Cos[v], x]]
 

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 0.19 (sec) , antiderivative size = 236, normalized size of antiderivative = 2.74

method	result	size
risch	\(\frac {i {\mathrm e}^{i x}}{2}-\frac {i {\mathrm e}^{-i x}}{2}+\frac {\ln \left ({\mathrm e}^{i x}+i\right )}{5}-\frac {\ln \left ({\mathrm e}^{i x}-i\right )}{5}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {i \left (\sqrt {5}+1\right ) {\mathrm e}^{i x}}{2}-1\right )}{20}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {i \left (\sqrt {5}+1\right ) {\mathrm e}^{i x}}{2}-1\right ) \sqrt {5}}{20}-\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {i \left (\sqrt {5}-1\right ) {\mathrm e}^{i x}}{2}-1\right )}{20}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {i \left (\sqrt {5}-1\right ) {\mathrm e}^{i x}}{2}-1\right ) \sqrt {5}}{20}+\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {i \left (\sqrt {5}-1\right ) {\mathrm e}^{i x}}{2}-1\right )}{20}-\frac {\ln \left ({\mathrm e}^{2 i x}-\frac {i \left (\sqrt {5}-1\right ) {\mathrm e}^{i x}}{2}-1\right ) \sqrt {5}}{20}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {i \left (\sqrt {5}+1\right ) {\mathrm e}^{i x}}{2}-1\right )}{20}+\frac {\ln \left ({\mathrm e}^{2 i x}+\frac {i \left (\sqrt {5}+1\right ) {\mathrm e}^{i x}}{2}-1\right ) \sqrt {5}}{20}\)	\(236\)

Input:

int(sin(x)*tan(5*x),x,method=_RETURNVERBOSE)
 

Output:

1/2*I*exp(I*x)-1/2*I*exp(-I*x)+1/5*ln(exp(I*x)+I)-1/5*ln(exp(I*x)-I)-1/20* 
ln(exp(2*I*x)-1/2*I*(5^(1/2)+1)*exp(I*x)-1)-1/20*ln(exp(2*I*x)-1/2*I*(5^(1 
/2)+1)*exp(I*x)-1)*5^(1/2)-1/20*ln(exp(2*I*x)+1/2*I*(5^(1/2)-1)*exp(I*x)-1 
)+1/20*ln(exp(2*I*x)+1/2*I*(5^(1/2)-1)*exp(I*x)-1)*5^(1/2)+1/20*ln(exp(2*I 
*x)-1/2*I*(5^(1/2)-1)*exp(I*x)-1)-1/20*ln(exp(2*I*x)-1/2*I*(5^(1/2)-1)*exp 
(I*x)-1)*5^(1/2)+1/20*ln(exp(2*I*x)+1/2*I*(5^(1/2)+1)*exp(I*x)-1)+1/20*ln( 
exp(2*I*x)+1/2*I*(5^(1/2)+1)*exp(I*x)-1)*5^(1/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 136 vs. \(2 (47) = 94\).

Time = 0.10 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.58 \[ \int \sin (x) \tan (5 x) \, dx=\frac {1}{20} \, \sqrt {5} \log \left (\frac {8 \, \cos \left (x\right )^{2} - 4 \, {\left (\sqrt {5} - 1\right )} \sin \left (x\right ) + \sqrt {5} - 11}{4 \, \cos \left (x\right )^{2} + 2 \, \sin \left (x\right ) - 3}\right ) + \frac {1}{20} \, \sqrt {5} \log \left (-\frac {8 \, \cos \left (x\right )^{2} - 4 \, {\left (\sqrt {5} + 1\right )} \sin \left (x\right ) - \sqrt {5} - 11}{4 \, \cos \left (x\right )^{2} - 2 \, \sin \left (x\right ) - 3}\right ) - \frac {1}{20} \, \log \left (4 \, \cos \left (x\right )^{2} + 2 \, \sin \left (x\right ) - 3\right ) + \frac {1}{20} \, \log \left (4 \, \cos \left (x\right )^{2} - 2 \, \sin \left (x\right ) - 3\right ) + \frac {1}{10} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac {1}{10} \, \log \left (-\sin \left (x\right ) + 1\right ) - \sin \left (x\right ) \] Input:

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

Output:

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

Sympy [F]

\[ \int \sin (x) \tan (5 x) \, dx=\int \sin {\left (x \right )} \tan {\left (5 x \right )}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \sin (x) \tan (5 x) \, dx=\int { \sin \left (x\right ) \tan \left (5 \, x\right ) \,d x } \] Input:

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

Output:

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*cos( 
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))*sin(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) - si 
n(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*(c 
os(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 + si 
n(x)^2 - 2*sin(x) + 1) - sin(x)
 

Giac [F]

\[ \int \sin (x) \tan (5 x) \, dx=\int { \sin \left (x\right ) \tan \left (5 \, x\right ) \,d x } \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 17.45 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.24 \[ \int \sin (x) \tan (5 x) \, dx=\frac {2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{5}+\frac {\mathrm {atan}\left (\frac {\sin \left (x\right )\,1042{}\mathrm {i}-\sqrt {5}\,\sin \left (x\right )\,466{}\mathrm {i}}{377\,\sqrt {5}-843}\right )\,1{}\mathrm {i}}{10}-\frac {\mathrm {atanh}\left (\sin \left (x\right )-\sqrt {5}\,\sin \left (x\right )\right )}{10}-\sin \left (x\right )-\frac {\sqrt {5}\,\mathrm {atanh}\left (\sin \left (x\right )-\sqrt {5}\,\sin \left (x\right )\right )}{10}-\frac {\sqrt {5}\,\mathrm {atan}\left (\frac {\sin \left (x\right )\,1042{}\mathrm {i}-\sqrt {5}\,\sin \left (x\right )\,466{}\mathrm {i}}{377\,\sqrt {5}-843}\right )\,1{}\mathrm {i}}{10} \] Input:

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

Output:

(atan((sin(x)*1042i - 5^(1/2)*sin(x)*466i)/(377*5^(1/2) - 843))*1i)/10 - a 
tanh(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)*104 
2i - 5^(1/2)*sin(x)*466i)/(377*5^(1/2) - 843))*1i)/10
 

Reduce [F]

\[ \int \sin (x) \tan (5 x) \, dx=\int \sin \left (x \right ) \tan \left (5 x \right )d x \] Input:

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

Output:

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