3.894 \(\int \sin (x) \tan ^5(x) \, dx\)

Optimal. Leaf size=34 \[ -\frac{15 \sin (x)}{8}+\frac{1}{4} \sin (x) \tan ^4(x)-\frac{5}{8} \sin (x) \tan ^2(x)+\frac{15}{8} \tanh ^{-1}(\sin (x)) \]

[Out]

(15*ArcTanh[Sin[x]])/8 - (15*Sin[x])/8 - (5*Sin[x]*Tan[x]^2)/8 + (Sin[x]*Tan[x]^4)/4

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Rubi [A]  time = 0.0249877, antiderivative size = 34, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 7, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.571, Rules used = {2592, 288, 321, 206} \[ -\frac{15 \sin (x)}{8}+\frac{1}{4} \sin (x) \tan ^4(x)-\frac{5}{8} \sin (x) \tan ^2(x)+\frac{15}{8} \tanh ^{-1}(\sin (x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*ArcTanh[Sin[x]])/8 - (15*Sin[x])/8 - (5*Sin[x]*Tan[x]^2)/8 + (Sin[x]*Tan[x]^4)/4

Rule 2592

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sin (x) \tan ^5(x) \, dx &=\operatorname{Subst}\left (\int \frac{x^6}{\left (1-x^2\right )^3} \, dx,x,\sin (x)\right )\\ &=\frac{1}{4} \sin (x) \tan ^4(x)-\frac{5}{4} \operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^2} \, dx,x,\sin (x)\right )\\ &=-\frac{5}{8} \sin (x) \tan ^2(x)+\frac{1}{4} \sin (x) \tan ^4(x)+\frac{15}{8} \operatorname{Subst}\left (\int \frac{x^2}{1-x^2} \, dx,x,\sin (x)\right )\\ &=-\frac{15 \sin (x)}{8}-\frac{5}{8} \sin (x) \tan ^2(x)+\frac{1}{4} \sin (x) \tan ^4(x)+\frac{15}{8} \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\sin (x)\right )\\ &=\frac{15}{8} \tanh ^{-1}(\sin (x))-\frac{15 \sin (x)}{8}-\frac{5}{8} \sin (x) \tan ^2(x)+\frac{1}{4} \sin (x) \tan ^4(x)\\ \end{align*}

Mathematica [A]  time = 0.0095064, size = 42, normalized size = 1.24 \[ -\sin (x) \tan ^4(x)+\frac{15}{8} \tanh ^{-1}(\sin (x))-\frac{15}{4} \tan (x) \sec ^3(x)+5 \tan ^3(x) \sec (x)+\frac{15}{8} \tan (x) \sec (x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*ArcTanh[Sin[x]])/8 + (15*Sec[x]*Tan[x])/8 - (15*Sec[x]^3*Tan[x])/4 + 5*Sec[x]*Tan[x]^3 - Sin[x]*Tan[x]^4

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Maple [A]  time = 0.013, size = 46, normalized size = 1.4 \begin{align*}{\frac{ \left ( \sin \left ( x \right ) \right ) ^{7}}{4\, \left ( \cos \left ( x \right ) \right ) ^{4}}}-{\frac{3\, \left ( \sin \left ( x \right ) \right ) ^{7}}{8\, \left ( \cos \left ( x \right ) \right ) ^{2}}}-{\frac{3\, \left ( \sin \left ( x \right ) \right ) ^{5}}{8}}-{\frac{5\, \left ( \sin \left ( x \right ) \right ) ^{3}}{8}}-{\frac{15\,\sin \left ( x \right ) }{8}}+{\frac{15\,\ln \left ( \sec \left ( x \right ) +\tan \left ( x \right ) \right ) }{8}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*sin(x)^7/cos(x)^4-3/8*sin(x)^7/cos(x)^2-3/8*sin(x)^5-5/8*sin(x)^3-15/8*sin(x)+15/8*ln(sec(x)+tan(x))

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Maxima [A]  time = 0.945148, size = 62, normalized size = 1.82 \begin{align*} \frac{9 \, \sin \left (x\right )^{3} - 7 \, \sin \left (x\right )}{8 \,{\left (\sin \left (x\right )^{4} - 2 \, \sin \left (x\right )^{2} + 1\right )}} + \frac{15}{16} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac{15}{16} \, \log \left (\sin \left (x\right ) - 1\right ) - \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(9*sin(x)^3 - 7*sin(x))/(sin(x)^4 - 2*sin(x)^2 + 1) + 15/16*log(sin(x) + 1) - 15/16*log(sin(x) - 1) - sin(
x)

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Fricas [A]  time = 2.59876, size = 158, normalized size = 4.65 \begin{align*} \frac{15 \, \cos \left (x\right )^{4} \log \left (\sin \left (x\right ) + 1\right ) - 15 \, \cos \left (x\right )^{4} \log \left (-\sin \left (x\right ) + 1\right ) - 2 \,{\left (8 \, \cos \left (x\right )^{4} + 9 \, \cos \left (x\right )^{2} - 2\right )} \sin \left (x\right )}{16 \, \cos \left (x\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(15*cos(x)^4*log(sin(x) + 1) - 15*cos(x)^4*log(-sin(x) + 1) - 2*(8*cos(x)^4 + 9*cos(x)^2 - 2)*sin(x))/cos
(x)^4

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Sympy [A]  time = 0.149226, size = 49, normalized size = 1.44 \begin{align*} \frac{9 \sin ^{3}{\left (x \right )} - 7 \sin{\left (x \right )}}{8 \sin ^{4}{\left (x \right )} - 16 \sin ^{2}{\left (x \right )} + 8} - \frac{15 \log{\left (\sin{\left (x \right )} - 1 \right )}}{16} + \frac{15 \log{\left (\sin{\left (x \right )} + 1 \right )}}{16} - \sin{\left (x \right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(9*sin(x)**3 - 7*sin(x))/(8*sin(x)**4 - 16*sin(x)**2 + 8) - 15*log(sin(x) - 1)/16 + 15*log(sin(x) + 1)/16 - si
n(x)

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Giac [A]  time = 1.06898, size = 57, normalized size = 1.68 \begin{align*} \frac{9 \, \sin \left (x\right )^{3} - 7 \, \sin \left (x\right )}{8 \,{\left (\sin \left (x\right )^{2} - 1\right )}^{2}} + \frac{15}{16} \, \log \left (\sin \left (x\right ) + 1\right ) - \frac{15}{16} \, \log \left (-\sin \left (x\right ) + 1\right ) - \sin \left (x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*(9*sin(x)^3 - 7*sin(x))/(sin(x)^2 - 1)^2 + 15/16*log(sin(x) + 1) - 15/16*log(-sin(x) + 1) - sin(x)