\(\int \sin ^3(5 x) \tan ^4(5 x) \, dx\) [891]

Optimal result

Integrand size = 13, antiderivative size = 37 \[ \int \sin ^3(5 x) \tan ^4(5 x) \, dx=-\frac {3}{5} \cos (5 x)+\frac {1}{15} \cos ^3(5 x)-\frac {3}{5} \sec (5 x)+\frac {1}{15} \sec ^3(5 x) \]

[Out]

-3/5*cos(5*x)+1/15*cos(5*x)^3-3/5*sec(5*x)+1/15*sec(5*x)^3

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2670, 276} \[ \int \sin ^3(5 x) \tan ^4(5 x) \, dx=\frac {1}{15} \cos ^3(5 x)-\frac {3}{5} \cos (5 x)+\frac {1}{15} \sec ^3(5 x)-\frac {3}{5} \sec (5 x) \]

[In]

Int[Sin[5*x]^3*Tan[5*x]^4,x]

[Out]

(-3*Cos[5*x])/5 + Cos[5*x]^3/15 - (3*Sec[5*x])/5 + Sec[5*x]^3/15

Rule 276

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

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}& = -\left (\frac {1}{5} \text {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^4} \, dx,x,\cos (5 x)\right )\right ) \\ & = -\left (\frac {1}{5} \text {Subst}\left (\int \left (3+\frac {1}{x^4}-\frac {3}{x^2}-x^2\right ) \, dx,x,\cos (5 x)\right )\right ) \\ & = -\frac {3}{5} \cos (5 x)+\frac {1}{15} \cos ^3(5 x)-\frac {3}{5} \sec (5 x)+\frac {1}{15} \sec ^3(5 x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.95 \[ \int \sin ^3(5 x) \tan ^4(5 x) \, dx=-\frac {11}{20} \cos (5 x)+\frac {1}{60} \cos (15 x)-\frac {3}{5} \sec (5 x)+\frac {1}{15} \sec ^3(5 x) \]

[In]

Integrate[Sin[5*x]^3*Tan[5*x]^4,x]

[Out]

(-11*Cos[5*x])/20 + Cos[15*x]/60 - (3*Sec[5*x])/5 + Sec[5*x]^3/15

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 5.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.54

method	result	size
risch	\(\frac {{\mathrm e}^{45 i x}-30 \,{\mathrm e}^{35 i x}-273 \,{\mathrm e}^{25 i x}-420 \,{\mathrm e}^{15 i x}+{\mathrm e}^{-15 i x}-303 \cos \left (5 x \right )-243 i \sin \left (5 x \right )}{120 \left ({\mathrm e}^{10 i x}+1\right )^{3}}\)	\(57\)
derivativedivides	\(\frac {\sin \left (5 x \right )^{8}}{15 \cos \left (5 x \right )^{3}}-\frac {\sin \left (5 x \right )^{8}}{3 \cos \left (5 x \right )}-\frac {\left (\frac {16}{5}+\sin \left (5 x \right )^{6}+\frac {6 \sin \left (5 x \right )^{4}}{5}+\frac {8 \sin \left (5 x \right )^{2}}{5}\right ) \cos \left (5 x \right )}{3}\)	\(60\)
default	\(\frac {\sin \left (5 x \right )^{8}}{15 \cos \left (5 x \right )^{3}}-\frac {\sin \left (5 x \right )^{8}}{3 \cos \left (5 x \right )}-\frac {\left (\frac {16}{5}+\sin \left (5 x \right )^{6}+\frac {6 \sin \left (5 x \right )^{4}}{5}+\frac {8 \sin \left (5 x \right )^{2}}{5}\right ) \cos \left (5 x \right )}{3}\)	\(60\)

[In]

int(sin(5*x)^3*tan(5*x)^4,x,method=_RETURNVERBOSE)

[Out]

1/120*(exp(45*I*x)-30*exp(35*I*x)-273*exp(25*I*x)-420*exp(15*I*x)+exp(-15*I*x)-303*cos(5*x)-243*I*sin(5*x))/(e
xp(10*I*x)+1)^3

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 32, normalized size of antiderivative = 0.86 \[ \int \sin ^3(5 x) \tan ^4(5 x) \, dx=\frac {\cos \left (5 \, x\right )^{6} - 9 \, \cos \left (5 \, x\right )^{4} - 9 \, \cos \left (5 \, x\right )^{2} + 1}{15 \, \cos \left (5 \, x\right )^{3}} \]

[In]

integrate(sin(5*x)^3*tan(5*x)^4,x, algorithm="fricas")

[Out]

1/15*(cos(5*x)^6 - 9*cos(5*x)^4 - 9*cos(5*x)^2 + 1)/cos(5*x)^3

Sympy [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.92 \[ \int \sin ^3(5 x) \tan ^4(5 x) \, dx=\frac {1 - 9 \cos ^{2}{\left (5 x \right )}}{15 \cos ^{3}{\left (5 x \right )}} + \frac {\cos ^{3}{\left (5 x \right )}}{15} - \frac {3 \cos {\left (5 x \right )}}{5} \]

[In]

integrate(sin(5*x)**3*tan(5*x)**4,x)

[Out]

(1 - 9*cos(5*x)**2)/(15*cos(5*x)**3) + cos(5*x)**3/15 - 3*cos(5*x)/5

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \sin ^3(5 x) \tan ^4(5 x) \, dx=\frac {1}{15} \, \cos \left (5 \, x\right )^{3} - \frac {9 \, \cos \left (5 \, x\right )^{2} - 1}{15 \, \cos \left (5 \, x\right )^{3}} - \frac {3}{5} \, \cos \left (5 \, x\right ) \]

[In]

integrate(sin(5*x)^3*tan(5*x)^4,x, algorithm="maxima")

[Out]

1/15*cos(5*x)^3 - 1/15*(9*cos(5*x)^2 - 1)/cos(5*x)^3 - 3/5*cos(5*x)

Giac [A] (verification not implemented)

none

Time = 0.56 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.89 \[ \int \sin ^3(5 x) \tan ^4(5 x) \, dx=\frac {1}{15} \, \cos \left (5 \, x\right )^{3} - \frac {9 \, \cos \left (5 \, x\right )^{2} - 1}{15 \, \cos \left (5 \, x\right )^{3}} - \frac {3}{5} \, \cos \left (5 \, x\right ) \]

[In]

integrate(sin(5*x)^3*tan(5*x)^4,x, algorithm="giac")

[Out]

1/15*cos(5*x)^3 - 1/15*(9*cos(5*x)^2 - 1)/cos(5*x)^3 - 3/5*cos(5*x)

Mupad [B] (verification not implemented)

Time = 26.45 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.81 \[ \int \sin ^3(5 x) \tan ^4(5 x) \, dx=\frac {{\left (\cos \left (5\,x\right )+1\right )}^4\,\left ({\cos \left (5\,x\right )}^2-4\,\cos \left (5\,x\right )+1\right )}{15\,{\cos \left (5\,x\right )}^3} \]

[In]

int(sin(5*x)^3*tan(5*x)^4,x)

[Out]

((cos(5*x) + 1)^4*(cos(5*x)^2 - 4*cos(5*x) + 1))/(15*cos(5*x)^3)