\(\int \sin ^{10}(x) \tan (x) \, dx\) [346]

Optimal result

Integrand size = 7, antiderivative size = 46 \[ \int \sin ^{10}(x) \tan (x) \, dx=\frac {5 \cos ^2(x)}{2}-\frac {5 \cos ^4(x)}{2}+\frac {5 \cos ^6(x)}{3}-\frac {5 \cos ^8(x)}{8}+\frac {\cos ^{10}(x)}{10}-\log (\cos (x)) \]

[Out]

5/2*cos(x)^2-5/2*cos(x)^4+5/3*cos(x)^6-5/8*cos(x)^8+1/10*cos(x)^10-ln(cos(x))

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2670, 272, 45} \[ \int \sin ^{10}(x) \tan (x) \, dx=\frac {\cos ^{10}(x)}{10}-\frac {5 \cos ^8(x)}{8}+\frac {5 \cos ^6(x)}{3}-\frac {5 \cos ^4(x)}{2}+\frac {5 \cos ^2(x)}{2}-\log (\cos (x)) \]

[In]

Int[Sin[x]^10*Tan[x],x]

[Out]

(5*Cos[x]^2)/2 - (5*Cos[x]^4)/2 + (5*Cos[x]^6)/3 - (5*Cos[x]^8)/8 + Cos[x]^10/10 - Log[Cos[x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2670

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rubi steps \begin{align*} \text {integral}& = -\text {Subst}\left (\int \frac {\left (1-x^2\right )^5}{x} \, dx,x,\cos (x)\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \frac {(1-x)^5}{x} \, dx,x,\cos ^2(x)\right )\right ) \\ & = -\left (\frac {1}{2} \text {Subst}\left (\int \left (-5+\frac {1}{x}+10 x-10 x^2+5 x^3-x^4\right ) \, dx,x,\cos ^2(x)\right )\right ) \\ & = \frac {5 \cos ^2(x)}{2}-\frac {5 \cos ^4(x)}{2}+\frac {5 \cos ^6(x)}{3}-\frac {5 \cos ^8(x)}{8}+\frac {\cos ^{10}(x)}{10}-\log (\cos (x)) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.00 \[ \int \sin ^{10}(x) \tan (x) \, dx=\frac {5 \cos ^2(x)}{2}-\frac {5 \cos ^4(x)}{2}+\frac {5 \cos ^6(x)}{3}-\frac {5 \cos ^8(x)}{8}+\frac {\cos ^{10}(x)}{10}-\log (\cos (x)) \]

[In]

Integrate[Sin[x]^10*Tan[x],x]

[Out]

(5*Cos[x]^2)/2 - (5*Cos[x]^4)/2 + (5*Cos[x]^6)/3 - (5*Cos[x]^8)/8 + Cos[x]^10/10 - Log[Cos[x]]

Maple [A] (verified)

Time = 0.53 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.80

method	result	size
default	\(-\frac {\left (\sin ^{10}\left (x \right )\right )}{10}-\frac {\left (\sin ^{8}\left (x \right )\right )}{8}-\frac {\left (\sin ^{6}\left (x \right )\right )}{6}-\frac {\left (\sin ^{4}\left (x \right )\right )}{4}-\frac {\left (\sin ^{2}\left (x \right )\right )}{2}-\ln \left (\cos \left (x \right )\right )\)	\(37\)
risch	\(i x +\frac {281 \,{\mathrm e}^{2 i x}}{1024}+\frac {281 \,{\mathrm e}^{-2 i x}}{1024}-\ln \left ({\mathrm e}^{2 i x}+1\right )+\frac {\cos \left (10 x \right )}{5120}-\frac {3 \cos \left (8 x \right )}{1024}+\frac {67 \cos \left (6 x \right )}{3072}-\frac {29 \cos \left (4 x \right )}{256}\)	\(54\)
parallelrisch	\(-\frac {469}{46080}+\ln \left (\frac {2}{\cos \left (x \right )+1}\right )-\ln \left (-\cot \left (x \right )+1+\csc \left (x \right )\right )-\ln \left (-\cot \left (x \right )+\csc \left (x \right )-1\right )+\frac {\cos \left (10 x \right )}{5120}-\frac {3 \cos \left (8 x \right )}{1024}+\frac {67 \cos \left (6 x \right )}{3072}-\frac {29 \cos \left (4 x \right )}{256}+\frac {281 \cos \left (2 x \right )}{512}\)	\(64\)

[In]

int(sin(x)^11/cos(x),x,method=_RETURNVERBOSE)

[Out]

-1/10*sin(x)^10-1/8*sin(x)^8-1/6*sin(x)^6-1/4*sin(x)^4-1/2*sin(x)^2-ln(cos(x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 38, normalized size of antiderivative = 0.83 \[ \int \sin ^{10}(x) \tan (x) \, dx=\frac {1}{10} \, \cos \left (x\right )^{10} - \frac {5}{8} \, \cos \left (x\right )^{8} + \frac {5}{3} \, \cos \left (x\right )^{6} - \frac {5}{2} \, \cos \left (x\right )^{4} + \frac {5}{2} \, \cos \left (x\right )^{2} - \log \left (-\cos \left (x\right )\right ) \]

[In]

integrate(sin(x)^11/cos(x),x, algorithm="fricas")

[Out]

1/10*cos(x)^10 - 5/8*cos(x)^8 + 5/3*cos(x)^6 - 5/2*cos(x)^4 + 5/2*cos(x)^2 - log(-cos(x))

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.96 \[ \int \sin ^{10}(x) \tan (x) \, dx=- \log {\left (\cos {\left (x \right )} \right )} + \frac {\cos ^{10}{\left (x \right )}}{10} - \frac {5 \cos ^{8}{\left (x \right )}}{8} + \frac {5 \cos ^{6}{\left (x \right )}}{3} - \frac {5 \cos ^{4}{\left (x \right )}}{2} + \frac {5 \cos ^{2}{\left (x \right )}}{2} \]

[In]

integrate(sin(x)**11/cos(x),x)

[Out]

-log(cos(x)) + cos(x)**10/10 - 5*cos(x)**8/8 + 5*cos(x)**6/3 - 5*cos(x)**4/2 + 5*cos(x)**2/2

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.87 \[ \int \sin ^{10}(x) \tan (x) \, dx=-\frac {1}{10} \, \sin \left (x\right )^{10} - \frac {1}{8} \, \sin \left (x\right )^{8} - \frac {1}{6} \, \sin \left (x\right )^{6} - \frac {1}{4} \, \sin \left (x\right )^{4} - \frac {1}{2} \, \sin \left (x\right )^{2} - \frac {1}{2} \, \log \left (\sin \left (x\right )^{2} - 1\right ) \]

[In]

integrate(sin(x)^11/cos(x),x, algorithm="maxima")

[Out]

-1/10*sin(x)^10 - 1/8*sin(x)^8 - 1/6*sin(x)^6 - 1/4*sin(x)^4 - 1/2*sin(x)^2 - 1/2*log(sin(x)^2 - 1)

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.91 \[ \int \sin ^{10}(x) \tan (x) \, dx=-\frac {1}{10} \, \sin \left (x\right )^{10} - \frac {1}{8} \, \sin \left (x\right )^{8} - \frac {1}{6} \, \sin \left (x\right )^{6} - \frac {1}{4} \, \sin \left (x\right )^{4} - \frac {1}{2} \, \sin \left (x\right )^{2} - \frac {1}{2} \, \log \left (-\sin \left (x\right )^{2} + 1\right ) \]

[In]

integrate(sin(x)^11/cos(x),x, algorithm="giac")

[Out]

-1/10*sin(x)^10 - 1/8*sin(x)^8 - 1/6*sin(x)^6 - 1/4*sin(x)^4 - 1/2*sin(x)^2 - 1/2*log(-sin(x)^2 + 1)

Mupad [B] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.78 \[ \int \sin ^{10}(x) \tan (x) \, dx=-\frac {{\sin \left (x\right )}^{10}}{10}-\frac {{\sin \left (x\right )}^8}{8}-\frac {{\sin \left (x\right )}^6}{6}-\frac {{\sin \left (x\right )}^4}{4}-\frac {{\sin \left (x\right )}^2}{2}-\ln \left (\cos \left (x\right )\right ) \]

[In]

int(sin(x)^11/cos(x),x)

[Out]

- log(cos(x)) - sin(x)^2/2 - sin(x)^4/4 - sin(x)^6/6 - sin(x)^8/8 - sin(x)^10/10