\(\int (-\cos (x)+\sec (x))^3 \, dx\) [323]

Optimal result

Integrand size = 9, antiderivative size = 34 \[ \int (-\cos (x)+\sec (x))^3 \, dx=-\frac {5}{2} \text {arctanh}(\sin (x))+\frac {5 \sin (x)}{2}+\frac {5 \sin ^3(x)}{6}+\frac {1}{2} \sin ^3(x) \tan ^2(x) \]

[Out]

-5/2*arctanh(sin(x))+5/2*sin(x)+5/6*sin(x)^3+1/2*sin(x)^3*tan(x)^2

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {4482, 2672, 294, 308, 212} \[ \int (-\cos (x)+\sec (x))^3 \, dx=-\frac {5}{2} \text {arctanh}(\sin (x))+\frac {5 \sin ^3(x)}{6}+\frac {5 \sin (x)}{2}+\frac {1}{2} \sin ^3(x) \tan ^2(x) \]

[In]

Int[(-Cos[x] + Sec[x])^3,x]

[Out]

(-5*ArcTanh[Sin[x]])/2 + (5*Sin[x])/2 + (5*Sin[x]^3)/6 + (Sin[x]^3*Tan[x]^2)/2

Rule 212

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

Rule 294

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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 2672

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 4482

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

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

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 38, normalized size of antiderivative = 1.12 \[ \int (-\cos (x)+\sec (x))^3 \, dx=-\frac {5}{2} \text {arctanh}(\sin (x))+\frac {5}{2} \sec (x) \tan (x)-\frac {5}{3} \sin (x) \tan ^2(x)-\frac {1}{3} \sin ^3(x) \tan ^2(x) \]

[In]

Integrate[(-Cos[x] + Sec[x])^3,x]

[Out]

(-5*ArcTanh[Sin[x]])/2 + (5*Sec[x]*Tan[x])/2 - (5*Sin[x]*Tan[x]^2)/3 - (Sin[x]^3*Tan[x]^2)/3

Maple [A] (verified)

Time = 1.66 (sec) , antiderivative size = 30, normalized size of antiderivative = 0.88

method	result	size
default	\(-\frac {\left (2+\cos \left (x \right )^{2}\right ) \sin \left (x \right )}{3}+3 \sin \left (x \right )-\frac {5 \ln \left (\sec \left (x \right )+\tan \left (x \right )\right )}{2}+\frac {\sec \left (x \right ) \tan \left (x \right )}{2}\)	\(30\)
parts	\(-\frac {\left (2+\cos \left (x \right )^{2}\right ) \sin \left (x \right )}{3}+3 \sin \left (x \right )-\frac {5 \ln \left (\sec \left (x \right )+\tan \left (x \right )\right )}{2}+\frac {\sec \left (x \right ) \tan \left (x \right )}{2}\)	\(30\)
parallelrisch	\(-\frac {5 \ln \left (\csc \left (x \right )-\cot \left (x \right )+1\right )}{2}+\frac {5 \ln \left (-\cot \left (x \right )+\csc \left (x \right )-1\right )}{2}-\frac {\sin \left (3 x \right )}{12}+\frac {9 \sin \left (x \right )}{4}+\frac {\sec \left (x \right ) \tan \left (x \right )}{2}\)	\(40\)
norman	\(\frac {\frac {20 \tan \left (\frac {x}{2}\right )^{3}}{3}-\frac {22 \tan \left (\frac {x}{2}\right )^{5}}{3}+\frac {20 \tan \left (\frac {x}{2}\right )^{7}}{3}+5 \tan \left (\frac {x}{2}\right )^{9}+5 \tan \left (\frac {x}{2}\right )}{\left (1+\tan \left (\frac {x}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {x}{2}\right )^{2}-1\right )^{2}}+\frac {5 \ln \left (\tan \left (\frac {x}{2}\right )-1\right )}{2}-\frac {5 \ln \left (\tan \left (\frac {x}{2}\right )+1\right )}{2}\)	\(80\)
risch	\(\frac {i {\mathrm e}^{3 i x}}{24}-\frac {9 i {\mathrm e}^{i x}}{8}+\frac {9 i {\mathrm e}^{-i x}}{8}-\frac {i {\mathrm e}^{-3 i x}}{24}-\frac {i \left ({\mathrm e}^{3 i x}-{\mathrm e}^{i x}\right )}{\left ({\mathrm e}^{2 i x}+1\right )^{2}}+\frac {5 \ln \left ({\mathrm e}^{i x}-i\right )}{2}-\frac {5 \ln \left (i+{\mathrm e}^{i x}\right )}{2}\)	\(81\)

[In]

int((-cos(x)+sec(x))^3,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.44 \[ \int (-\cos (x)+\sec (x))^3 \, dx=-\frac {15 \, \cos \left (x\right )^{2} \log \left (\sin \left (x\right ) + 1\right ) - 15 \, \cos \left (x\right )^{2} \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, {\left (2 \, \cos \left (x\right )^{4} - 14 \, \cos \left (x\right )^{2} - 3\right )} \sin \left (x\right )}{12 \, \cos \left (x\right )^{2}} \]

[In]

integrate((-cos(x)+sec(x))^3,x, algorithm="fricas")

[Out]

-1/12*(15*cos(x)^2*log(sin(x) + 1) - 15*cos(x)^2*log(-sin(x) + 1) + 2*(2*cos(x)^4 - 14*cos(x)^2 - 3)*sin(x))/c
os(x)^2

Sympy [A] (verification not implemented)

Time = 1.63 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.24 \[ \int (-\cos (x)+\sec (x))^3 \, dx=\frac {5 \log {\left (\sin {\left (x \right )} - 1 \right )}}{4} - \frac {5 \log {\left (\sin {\left (x \right )} + 1 \right )}}{4} + \frac {\sin ^{3}{\left (x \right )}}{3} + 2 \sin {\left (x \right )} - \frac {\sin {\left (x \right )}}{2 \sin ^{2}{\left (x \right )} - 2} \]

[In]

integrate((-cos(x)+sec(x))**3,x)

[Out]

5*log(sin(x) - 1)/4 - 5*log(sin(x) + 1)/4 + sin(x)**3/3 + 2*sin(x) - sin(x)/(2*sin(x)**2 - 2)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 1.09 \[ \int (-\cos (x)+\sec (x))^3 \, dx=\frac {1}{3} \, \sin \left (x\right )^{3} - \frac {\sin \left (x\right )}{2 \, {\left (\sin \left (x\right )^{2} - 1\right )}} - \frac {5}{4} \, \log \left (\sin \left (x\right ) + 1\right ) + \frac {5}{4} \, \log \left (\sin \left (x\right ) - 1\right ) + 2 \, \sin \left (x\right ) \]

[In]

integrate((-cos(x)+sec(x))^3,x, algorithm="maxima")

[Out]

1/3*sin(x)^3 - 1/2*sin(x)/(sin(x)^2 - 1) - 5/4*log(sin(x) + 1) + 5/4*log(sin(x) - 1) + 2*sin(x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.15 \[ \int (-\cos (x)+\sec (x))^3 \, dx=\frac {1}{3} \, \sin \left (x\right )^{3} - \frac {\sin \left (x\right )}{2 \, {\left (\sin \left (x\right )^{2} - 1\right )}} - \frac {5}{4} \, \log \left (\sin \left (x\right ) + 1\right ) + \frac {5}{4} \, \log \left (-\sin \left (x\right ) + 1\right ) + 2 \, \sin \left (x\right ) \]

[In]

integrate((-cos(x)+sec(x))^3,x, algorithm="giac")

[Out]

1/3*sin(x)^3 - 1/2*sin(x)/(sin(x)^2 - 1) - 5/4*log(sin(x) + 1) + 5/4*log(-sin(x) + 1) + 2*sin(x)

Mupad [B] (verification not implemented)

Time = 26.10 (sec) , antiderivative size = 68, normalized size of antiderivative = 2.00 \[ \int (-\cos (x)+\sec (x))^3 \, dx=\frac {5\,{\mathrm {tan}\left (\frac {x}{2}\right )}^9+\frac {20\,{\mathrm {tan}\left (\frac {x}{2}\right )}^7}{3}-\frac {22\,{\mathrm {tan}\left (\frac {x}{2}\right )}^5}{3}+\frac {20\,{\mathrm {tan}\left (\frac {x}{2}\right )}^3}{3}+5\,\mathrm {tan}\left (\frac {x}{2}\right )}{{\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2-1\right )}^2\,{\left ({\mathrm {tan}\left (\frac {x}{2}\right )}^2+1\right )}^3}-5\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {x}{2}\right )\right ) \]

[In]

int(-(cos(x) - 1/cos(x))^3,x)

[Out]

(5*tan(x/2) + (20*tan(x/2)^3)/3 - (22*tan(x/2)^5)/3 + (20*tan(x/2)^7)/3 + 5*tan(x/2)^9)/((tan(x/2)^2 - 1)^2*(t
an(x/2)^2 + 1)^3) - 5*atanh(tan(x/2))