\(\int \frac {(-3+\cos (2 x)) \sec ^4(x)}{\sqrt {4-\cot ^2(x)}} \, dx\) [437]

Optimal result

Integrand size = 23, antiderivative size = 39 \[ \int \frac {(-3+\cos (2 x)) \sec ^4(x)}{\sqrt {4-\cot ^2(x)}} \, dx=-\frac {2}{3} \sqrt {4-\cot ^2(x)} \tan (x)-\frac {1}{3} \sqrt {4-\cot ^2(x)} \tan ^3(x) \]

[Out]

-2/3*(4-cot(x)^2)^(1/2)*tan(x)-1/3*(4-cot(x)^2)^(1/2)*tan(x)^3

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {12, 445, 464, 197} \[ \int \frac {(-3+\cos (2 x)) \sec ^4(x)}{\sqrt {4-\cot ^2(x)}} \, dx=-\frac {1}{3} \tan ^3(x) \sqrt {4-\cot ^2(x)}-\frac {2}{3} \tan (x) \sqrt {4-\cot ^2(x)} \]

[In]

Int[((-3 + Cos[2*x])*Sec[x]^4)/Sqrt[4 - Cot[x]^2],x]

[Out]

(-2*Sqrt[4 - Cot[x]^2]*Tan[x])/3 - (Sqrt[4 - Cot[x]^2]*Tan[x]^3)/3

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 445

Int[((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[(a + b*x^n)^p*((d + c*x
^n)^q/x^(n*q)), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] ||  !IntegerQ[p])

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rubi steps \begin{align*} \text {integral}& = \text {Subst}\left (\int \frac {2 \left (-1-2 x^2\right )}{\sqrt {4-\frac {1}{x^2}}} \, dx,x,\tan (x)\right ) \\ & = 2 \text {Subst}\left (\int \frac {-1-2 x^2}{\sqrt {4-\frac {1}{x^2}}} \, dx,x,\tan (x)\right ) \\ & = 2 \text {Subst}\left (\int \frac {\left (-2-\frac {1}{x^2}\right ) x^2}{\sqrt {4-\frac {1}{x^2}}} \, dx,x,\tan (x)\right ) \\ & = -\frac {1}{3} \sqrt {4-\cot ^2(x)} \tan ^3(x)-\frac {8}{3} \text {Subst}\left (\int \frac {1}{\sqrt {4-\frac {1}{x^2}}} \, dx,x,\tan (x)\right ) \\ & = -\frac {2}{3} \sqrt {4-\cot ^2(x)} \tan (x)-\frac {1}{3} \sqrt {4-\cot ^2(x)} \tan ^3(x) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.92 \[ \int \frac {(-3+\cos (2 x)) \sec ^4(x)}{\sqrt {4-\cot ^2(x)}} \, dx=\frac {(3+\cos (2 x)) (-3+5 \cos (2 x)) \csc (x) \sec ^3(x)}{12 \sqrt {4-\cot ^2(x)}} \]

[In]

Integrate[((-3 + Cos[2*x])*Sec[x]^4)/Sqrt[4 - Cot[x]^2],x]

[Out]

((3 + Cos[2*x])*(-3 + 5*Cos[2*x])*Csc[x]*Sec[x]^3)/(12*Sqrt[4 - Cot[x]^2])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(66\) vs. \(2(31)=62\).

Time = 2.73 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.72

method	result	size
default	\(\frac {\left (\sec ^{3}\left (x \right )\right ) \csc \left (x \right ) \left (25 \left (\cos ^{4}\left (x \right )\right )-10 \left (\cos ^{2}\left (x \right )\right )-8\right )}{6 \sqrt {-5 \left (\cot ^{2}\left (x \right )\right )+4 \left (\csc ^{2}\left (x \right )\right )}}-\frac {5 \cot \left (x \right )-4 \sec \left (x \right ) \csc \left (x \right )}{2 \sqrt {-5 \left (\cot ^{2}\left (x \right )\right )+4 \left (\csc ^{2}\left (x \right )\right )}}\)	\(67\)

[In]

int((-3+cos(2*x))/cos(x)^4/(4-cot(x)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/6*sec(x)^3*csc(x)*(25*cos(x)^4-10*cos(x)^2-8)/(-5*cot(x)^2+4*csc(x)^2)^(1/2)-1/2/(-5*cot(x)^2+4*csc(x)^2)^(1
/2)*(5*cot(x)-4*sec(x)*csc(x))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 33, normalized size of antiderivative = 0.85 \[ \int \frac {(-3+\cos (2 x)) \sec ^4(x)}{\sqrt {4-\cot ^2(x)}} \, dx=-\frac {{\left (\cos \left (x\right )^{2} + 1\right )} \sqrt {\frac {5 \, \cos \left (x\right )^{2} - 4}{\cos \left (x\right )^{2} - 1}} \sin \left (x\right )}{3 \, \cos \left (x\right )^{3}} \]

[In]

integrate((-3+cos(2*x))/cos(x)^4/(4-cot(x)^2)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {(-3+\cos (2 x)) \sec ^4(x)}{\sqrt {4-\cot ^2(x)}} \, dx=\int \frac {\cos {\left (2 x \right )} - 3}{\sqrt {- \left (\cot {\left (x \right )} - 2\right ) \left (\cot {\left (x \right )} + 2\right )} \cos ^{4}{\left (x \right )}}\, dx \]

[In]

integrate((-3+cos(2*x))/cos(x)**4/(4-cot(x)**2)**(1/2),x)

[Out]

Integral((cos(2*x) - 3)/(sqrt(-(cot(x) - 2)*(cot(x) + 2))*cos(x)**4), x)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 63 vs. \(2 (31) = 62\).

Time = 0.21 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.62 \[ \int \frac {(-3+\cos (2 x)) \sec ^4(x)}{\sqrt {4-\cot ^2(x)}} \, dx=-\frac {1}{48} \, {\left (-\frac {1}{\tan \left (x\right )^{2}} + 4\right )}^{\frac {3}{2}} \tan \left (x\right )^{3} + \frac {3}{16} \, \sqrt {-\frac {1}{\tan \left (x\right )^{2}} + 4} \tan \left (x\right ) - \frac {8 \, \tan \left (x\right )^{4} + 26 \, \tan \left (x\right )^{2} - 7}{8 \, \sqrt {2 \, \tan \left (x\right ) + 1} \sqrt {2 \, \tan \left (x\right ) - 1}} \]

[In]

integrate((-3+cos(2*x))/cos(x)^4/(4-cot(x)^2)^(1/2),x, algorithm="maxima")

[Out]

-1/48*(-1/tan(x)^2 + 4)^(3/2)*tan(x)^3 + 3/16*sqrt(-1/tan(x)^2 + 4)*tan(x) - 1/8*(8*tan(x)^4 + 26*tan(x)^2 - 7
)/(sqrt(2*tan(x) + 1)*sqrt(2*tan(x) - 1))

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.28 (sec) , antiderivative size = 135, normalized size of antiderivative = 3.46 \[ \int \frac {(-3+\cos (2 x)) \sec ^4(x)}{\sqrt {4-\cot ^2(x)}} \, dx=\frac {\frac {125 \, \sqrt {5} {\left (\frac {21 \, {\left (\sqrt {5} \sqrt {-5 \, \cos \left (x\right )^{2} + 4} - 2 \, \sqrt {5}\right )}^{2}}{\cos \left (x\right )^{2}} + 125\right )} \cos \left (x\right )^{3}}{{\left (\sqrt {5} \sqrt {-5 \, \cos \left (x\right )^{2} + 4} - 2 \, \sqrt {5}\right )}^{3}} - \frac {\sqrt {5} {\left (\sqrt {5} \sqrt {-5 \, \cos \left (x\right )^{2} + 4} - 2 \, \sqrt {5}\right )}^{3}}{\cos \left (x\right )^{3}} - \frac {105 \, \sqrt {5} {\left (\sqrt {5} \sqrt {-5 \, \cos \left (x\right )^{2} + 4} - 2 \, \sqrt {5}\right )}}{\cos \left (x\right )}}{2400 \, \mathrm {sgn}\left (\sin \left (x\right )\right )} + \frac {2}{3} i \, \mathrm {sgn}\left (\sin \left (x\right )\right ) \]

[In]

integrate((-3+cos(2*x))/cos(x)^4/(4-cot(x)^2)^(1/2),x, algorithm="giac")

[Out]

1/2400*(125*sqrt(5)*(21*(sqrt(5)*sqrt(-5*cos(x)^2 + 4) - 2*sqrt(5))^2/cos(x)^2 + 125)*cos(x)^3/(sqrt(5)*sqrt(-
5*cos(x)^2 + 4) - 2*sqrt(5))^3 - sqrt(5)*(sqrt(5)*sqrt(-5*cos(x)^2 + 4) - 2*sqrt(5))^3/cos(x)^3 - 105*sqrt(5)*
(sqrt(5)*sqrt(-5*cos(x)^2 + 4) - 2*sqrt(5))/cos(x))/sgn(sin(x)) + 2/3*I*sgn(sin(x))

Mupad [B] (verification not implemented)

Time = 0.85 (sec) , antiderivative size = 20, normalized size of antiderivative = 0.51 \[ \int \frac {(-3+\cos (2 x)) \sec ^4(x)}{\sqrt {4-\cot ^2(x)}} \, dx=-\frac {\mathrm {tan}\left (x\right )\,\left ({\mathrm {tan}\left (x\right )}^2+2\right )\,\sqrt {4-\frac {1}{{\mathrm {tan}\left (x\right )}^2}}}{3} \]

[In]

int((cos(2*x) - 3)/(cos(x)^4*(4 - cot(x)^2)^(1/2)),x)

[Out]

-(tan(x)*(tan(x)^2 + 2)*(4 - 1/tan(x)^2)^(1/2))/3