3.891 \(\int \sin ^3(5 x) \tan ^4(5 x) \, dx\)

Optimal. Leaf size=37 \[ \frac{1}{15} \cos ^3(5 x)-\frac{3}{5} \cos (5 x)+\frac{1}{15} \sec ^3(5 x)-\frac{3}{5} \sec (5 x) \]

[Out]

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

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Rubi [A]  time = 0.0328969, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {2590, 270} \[ \frac{1}{15} \cos ^3(5 x)-\frac{3}{5} \cos (5 x)+\frac{1}{15} \sec ^3(5 x)-\frac{3}{5} \sec (5 x) \]

Antiderivative was successfully verified.

[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 2590

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]

Rule 270

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]

Rubi steps

\begin{align*} \int \sin ^3(5 x) \tan ^4(5 x) \, dx &=-\left (\frac{1}{5} \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{x^4} \, dx,x,\cos (5 x)\right )\right )\\ &=-\left (\frac{1}{5} \operatorname{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]  time = 0.030634, size = 35, normalized size = 0.95 \[ -\frac{11}{20} \cos (5 x)+\frac{1}{60} \cos (15 x)+\frac{1}{15} \sec ^3(5 x)-\frac{3}{5} \sec (5 x) \]

Antiderivative was successfully verified.

[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

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Maple [B]  time = 0.029, size = 60, normalized size = 1.6 \begin{align*}{\frac{ \left ( \sin \left ( 5\,x \right ) \right ) ^{8}}{15\, \left ( \cos \left ( 5\,x \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( 5\,x \right ) \right ) ^{8}}{3\,\cos \left ( 5\,x \right ) }}-{\frac{\cos \left ( 5\,x \right ) }{3} \left ({\frac{16}{5}}+ \left ( \sin \left ( 5\,x \right ) \right ) ^{6}+{\frac{6\, \left ( \sin \left ( 5\,x \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \sin \left ( 5\,x \right ) \right ) ^{2}}{5}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*sin(5*x)^8/cos(5*x)^3-1/3*sin(5*x)^8/cos(5*x)-1/3*(16/5+sin(5*x)^6+6/5*sin(5*x)^4+8/5*sin(5*x)^2)*cos(5*x
)

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Maxima [A]  time = 0.957101, size = 45, normalized size = 1.22 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [A]  time = 2.40515, size = 86, normalized size = 2.32 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Sympy [A]  time = 0.098131, size = 34, normalized size = 0.92 \begin{align*} - \frac{9 \cos ^{2}{\left (5 x \right )} - 1}{15 \cos ^{3}{\left (5 x \right )}} + \frac{\cos ^{3}{\left (5 x \right )}}{15} - \frac{3 \cos{\left (5 x \right )}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.27442, size = 45, normalized size = 1.22 \begin{align*} \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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)