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]

15/8*arctanh(sin(x))-15/8*sin(x)-5/8*sin(x)*tan(x)^2+1/4*sin(x)*tan(x)^4

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Rubi [A]  time = 0.02, antiderivative size = 34, normalized size of antiderivative = 1.00, 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 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])

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 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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.97, size = 49, normalized size = 1.44 \[ \frac {15 \, \cos \relax (x)^{4} \log \left (\sin \relax (x) + 1\right ) - 15 \, \cos \relax (x)^{4} \log \left (-\sin \relax (x) + 1\right ) - 2 \, {\left (8 \, \cos \relax (x)^{4} + 9 \, \cos \relax (x)^{2} - 2\right )} \sin \relax (x)}{16 \, \cos \relax (x)^{4}} \]

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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giac [A]  time = 0.15, size = 42, normalized size = 1.24 \[ \frac {9 \, \sin \relax (x)^{3} - 7 \, \sin \relax (x)}{8 \, {\left (\sin \relax (x)^{2} - 1\right )}^{2}} + \frac {15}{16} \, \log \left (\sin \relax (x) + 1\right ) - \frac {15}{16} \, \log \left (-\sin \relax (x) + 1\right ) - \sin \relax (x) \]

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)

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maple [A]  time = 0.06, size = 46, normalized size = 1.35 \[ \frac {\sin ^{7}\relax (x )}{4 \cos \relax (x )^{4}}-\frac {3 \left (\sin ^{7}\relax (x )\right )}{8 \cos \relax (x )^{2}}-\frac {3 \left (\sin ^{5}\relax (x )\right )}{8}-\frac {5 \left (\sin ^{3}\relax (x )\right )}{8}-\frac {15 \sin \relax (x )}{8}+\frac {15 \ln \left (\sec \relax (x )+\tan \relax (x )\right )}{8} \]

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 = 0.31, size = 46, normalized size = 1.35 \[ \frac {9 \, \sin \relax (x)^{3} - 7 \, \sin \relax (x)}{8 \, {\left (\sin \relax (x)^{4} - 2 \, \sin \relax (x)^{2} + 1\right )}} + \frac {15}{16} \, \log \left (\sin \relax (x) + 1\right ) - \frac {15}{16} \, \log \left (\sin \relax (x) - 1\right ) - \sin \relax (x) \]

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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mupad [B]  time = 3.04, size = 69, normalized size = 2.03 \[ \frac {15\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right )}{4}-\frac {\frac {15\,{\mathrm {tan}\left (\frac {x}{2}\right )}^9}{4}-10\,{\mathrm {tan}\left (\frac {x}{2}\right )}^7+\frac {9\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{2}-10\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3+\frac {15\,\mathrm {tan}\left (\frac {x}{2}\right )}{4}}{{\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2-1\right )}^4\,\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(15*atanh(tan(x/2)))/4 - ((15*tan(x/2))/4 - 10*tan(x/2)^3 + (9*tan(x/2)^5)/2 - 10*tan(x/2)^7 + (15*tan(x/2)^9)
/4)/((tan(x/2)^2 - 1)^4*(tan(x/2)^2 + 1))

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sympy [A]  time = 0.15, size = 49, normalized size = 1.44 \[ - \frac {- 9 \sin ^{3}{\relax (x )} + 7 \sin {\relax (x )}}{8 \sin ^{4}{\relax (x )} - 16 \sin ^{2}{\relax (x )} + 8} - \frac {15 \log {\left (\sin {\relax (x )} - 1 \right )}}{16} + \frac {15 \log {\left (\sin {\relax (x )} + 1 \right )}}{16} - \sin {\relax (x )} \]

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 -
sin(x)

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