Integrand size = 61, antiderivative size = 133 \[ \int \frac {\sec ^2(x) \tan (x) \left (\sqrt [3]{1-3 \sec ^2(x)} \sin ^2(x)+3 \tan ^2(x)\right )}{\left (1-3 \sec ^2(x)\right )^{5/6} \left (1-\sqrt {1-3 \sec ^2(x)}\right )} \, dx=\sqrt {3} \arctan \left (\frac {1+2 \sqrt [6]{1-3 \sec ^2(x)}}{\sqrt {3}}\right )+\frac {1}{4} \log \left (\sec ^2(x)\right )-\frac {3}{2} \log \left (1-\sqrt [6]{1-3 \sec ^2(x)}\right )+\frac {1}{3} \log \left (1-\sqrt {1-3 \sec ^2(x)}\right )-\sqrt [6]{1-3 \sec ^2(x)}-\frac {1}{4} \left (1-3 \sec ^2(x)\right )^{2/3}+\frac {1}{2 \left (1-\sqrt {1-3 \sec ^2(x)}\right )} \]
1/4*ln(sec(x)^2)-3/2*ln(1-(1-3*sec(x)^2)^(1/6))+1/3*ln(1-(1-3*sec(x)^2)^(1 /2))-(1-3*sec(x)^2)^(1/6)-1/4*(1-3*sec(x)^2)^(2/3)+arctan(1/3*(1+2*(1-3*se c(x)^2)^(1/6))*3^(1/2))*3^(1/2)+1/2/(1-(1-3*sec(x)^2)^(1/2))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 25.30 (sec) , antiderivative size = 1447, normalized size of antiderivative = 10.88 \[ \int \frac {\sec ^2(x) \tan (x) \left (\sqrt [3]{1-3 \sec ^2(x)} \sin ^2(x)+3 \tan ^2(x)\right )}{\left (1-3 \sec ^2(x)\right )^{5/6} \left (1-\sqrt {1-3 \sec ^2(x)}\right )} \, dx =\text {Too large to display} \]
Integrate[(Sec[x]^2*Tan[x]*((1 - 3*Sec[x]^2)^(1/3)*Sin[x]^2 + 3*Tan[x]^2)) /((1 - 3*Sec[x]^2)^(5/6)*(1 - Sqrt[1 - 3*Sec[x]^2])),x]
-1/3*((6 + ((-5 + Cos[2*x])/(1 + Cos[2*x]))^(1/3) + Cos[2*x]*((-5 + Cos[2* x])/(1 + Cos[2*x]))^(1/3))*(3*Sec[x]^2 + (1 - 3*Sec[x]^2)^(1/3))*Tan[x]*(- 2 - 3*Tan[x]^2)^(5/6)*(2 + 3*Tan[x]^2)*Sqrt[-(2 + 3*Tan[x]^2)^2]*(-1 + (-2 - 3*Tan[x]^2)^(1/3))*(1 + (-2 - 3*Tan[x]^2)^(1/3) + (-2 - 3*Tan[x]^2)^(2/ 3))*(6*ArcTan[Sqrt[2 + 3*Tan[x]^2]]*Sqrt[-2 - 3*Tan[x]^2] - 5*Sqrt[2 + 3*T an[x]^2] - 4*ArcTanh[Sqrt[-2 - 3*Tan[x]^2]]*Sqrt[2 + 3*Tan[x]^2] + Cos[2*x ]*Sqrt[2 + 3*Tan[x]^2] + 5*Log[Sec[x]^2]*Sqrt[2 + 3*Tan[x]^2] - 9*Log[1 - (-2 - 3*Tan[x]^2)^(1/3)]*Sqrt[2 + 3*Tan[x]^2] - 12*(-2 - 3*Tan[x]^2)^(1/6) *Sqrt[2 + 3*Tan[x]^2] + 36*Hypergeometric2F1[1/6, 1, 7/6, -2 - 3*Tan[x]^2] *(-2 - 3*Tan[x]^2)^(1/6)*Sqrt[2 + 3*Tan[x]^2] - 3*(-2 - 3*Tan[x]^2)^(2/3)* Sqrt[2 + 3*Tan[x]^2] + Sqrt[-(2 + 3*Tan[x]^2)^2] + Cos[2*x]*Sqrt[-(2 + 3*T an[x]^2)^2] - 6*ArcTan[(1 + 2*(-2 - 3*Tan[x]^2)^(1/3))/Sqrt[3]]*Sqrt[6 + 9 *Tan[x]^2]))/((-1 + Sqrt[(-5 + Cos[2*x])/(1 + Cos[2*x])])*(1 - 3*Sec[x]^2) ^(5/6)*(6 + (1 - 3*Sec[x]^2)^(1/3) + Cos[2*x]*(1 - 3*Sec[x]^2)^(1/3))*(12* Csc[x]*Sec[x]*(-2 - 3*Tan[x]^2)^(5/6) + 12*Cos[2*x]*Csc[x]*Sec[x]*(-2 - 3* Tan[x]^2)^(5/6) - 88*Sin[2*x]*(-2 - 3*Tan[x]^2)^(5/6) - 16*Cot[x]^2*Sin[2* x]*(-2 - 3*Tan[x]^2)^(5/6) + 48*Sec[x]^2*Tan[x]*(-2 - 3*Tan[x]^2)^(5/6) + 48*Cos[2*x]*Sec[x]^2*Tan[x]*(-2 - 3*Tan[x]^2)^(5/6) - 180*Sin[2*x]*Tan[x]^ 2*(-2 - 3*Tan[x]^2)^(5/6) + 63*Sec[x]^2*Tan[x]^3*(-2 - 3*Tan[x]^2)^(5/6) + 63*Cos[2*x]*Sec[x]^2*Tan[x]^3*(-2 - 3*Tan[x]^2)^(5/6) - 162*Sin[2*x]*T...
Time = 4.49 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.50, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.066, Rules used = {3042, 4861, 7293, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\tan (x) \sec ^2(x) \left (3 \tan ^2(x)+\sin ^2(x) \sqrt [3]{1-3 \sec ^2(x)}\right )}{\left (1-3 \sec ^2(x)\right )^{5/6} \left (1-\sqrt {1-3 \sec ^2(x)}\right )} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (x) \sec (x)^2 \left (3 \tan (x)^2+\sin (x)^2 \sqrt [3]{1-3 \sec (x)^2}\right )}{\left (1-3 \sec (x)^2\right )^{5/6} \left (1-\sqrt {1-3 \sec (x)^2}\right )}dx\) |
\(\Big \downarrow \) 4861 |
\(\displaystyle -\int \frac {\left (1-\cos ^2(x)\right ) \sec ^5(x) \left (\sqrt [3]{1-3 \sec ^2(x)} \cos ^2(x)+3\right )}{\left (1-3 \sec ^2(x)\right )^{5/6} \left (1-\sqrt {1-3 \sec ^2(x)}\right )}d\cos (x)\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle -\int \left (\frac {\left (-\sqrt [3]{\left (\cos ^2(x)-3\right ) \sec ^2(x)} \cos ^2(x)-3\right ) \sec ^5(x)}{\left (1-3 \sec ^2(x)\right )^{5/6} \left (\sqrt {\left (\cos ^2(x)-3\right ) \sec ^2(x)}-1\right )}+\frac {\left (\sqrt [3]{\left (\cos ^2(x)-3\right ) \sec ^2(x)} \cos ^2(x)+3\right ) \sec ^3(x)}{\left (1-3 \sec ^2(x)\right )^{5/6} \left (\sqrt {\left (\cos ^2(x)-3\right ) \sec ^2(x)}-1\right )}\right )d\cos (x)\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {3-\cos ^2(x)} \sec (x) \arcsin \left (\frac {\cos (x)}{\sqrt {3}}\right )}{2 \sqrt {1-3 \sec ^2(x)}}+\sqrt {3} \arctan \left (\frac {2 \sqrt [6]{1-3 \sec ^2(x)}+1}{\sqrt {3}}\right )+\frac {\cos ^2(x)}{6}-\frac {1}{4} \left (1-3 \sec ^2(x)\right )^{2/3}-\sqrt [6]{1-3 \sec ^2(x)}-\frac {3}{2} \log \left (1-\sqrt [6]{1-3 \sec ^2(x)}\right )+\frac {1}{2} \log \left (1-\sqrt {1-3 \sec ^2(x)}\right )-\frac {3-\cos ^2(x)}{6 \sqrt {1-3 \sec ^2(x)}}+\frac {1}{3} \log \left (1-\sqrt {-\left (\left (3-\cos ^2(x)\right ) \sec ^2(x)\right )}\right )\) |
Int[(Sec[x]^2*Tan[x]*((1 - 3*Sec[x]^2)^(1/3)*Sin[x]^2 + 3*Tan[x]^2))/((1 - 3*Sec[x]^2)^(5/6)*(1 - Sqrt[1 - 3*Sec[x]^2])),x]
Sqrt[3]*ArcTan[(1 + 2*(1 - 3*Sec[x]^2)^(1/6))/Sqrt[3]] + Cos[x]^2/6 + Log[ 1 - Sqrt[-((3 - Cos[x]^2)*Sec[x]^2)]]/3 - (3*Log[1 - (1 - 3*Sec[x]^2)^(1/6 )])/2 + Log[1 - Sqrt[1 - 3*Sec[x]^2]]/2 - (3 - Cos[x]^2)/(6*Sqrt[1 - 3*Sec [x]^2]) + (ArcSin[Cos[x]/Sqrt[3]]*Sqrt[3 - Cos[x]^2]*Sec[x])/(2*Sqrt[1 - 3 *Sec[x]^2]) - (1 - 3*Sec[x]^2)^(1/6) - (1 - 3*Sec[x]^2)^(2/3)/4
3.5.46.3.1 Defintions of rubi rules used
Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFacto rs[Cos[c*(a + b*x)], x]}, Simp[-(b*c)^(-1) Subst[Int[SubstFor[1/x, Cos[c* (a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a + b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Tan] || EqQ[F, tan])
\[\int \frac {\tan \left (x \right ) \left ({\left (1-3 \left (\sec ^{2}\left (x \right )\right )\right )}^{\frac {1}{3}} \left (\sin ^{2}\left (x \right )\right )+3 \left (\tan ^{2}\left (x \right )\right )\right )}{\cos \left (x \right )^{2} {\left (1-3 \left (\sec ^{2}\left (x \right )\right )\right )}^{\frac {5}{6}} \left (1-\sqrt {1-3 \left (\sec ^{2}\left (x \right )\right )}\right )}d x\]
int(tan(x)*((1-3*sec(x)^2)^(1/3)*sin(x)^2+3*tan(x)^2)/cos(x)^2/(1-3*sec(x) ^2)^(5/6)/(1-(1-3*sec(x)^2)^(1/2)),x)
int(tan(x)*((1-3*sec(x)^2)^(1/3)*sin(x)^2+3*tan(x)^2)/cos(x)^2/(1-3*sec(x) ^2)^(5/6)/(1-(1-3*sec(x)^2)^(1/2)),x)
Exception generated. \[ \int \frac {\sec ^2(x) \tan (x) \left (\sqrt [3]{1-3 \sec ^2(x)} \sin ^2(x)+3 \tan ^2(x)\right )}{\left (1-3 \sec ^2(x)\right )^{5/6} \left (1-\sqrt {1-3 \sec ^2(x)}\right )} \, dx=\text {Exception raised: TypeError} \]
integrate(tan(x)*((1-3*sec(x)^2)^(1/3)*sin(x)^2+3*tan(x)^2)/cos(x)^2/(1-3* sec(x)^2)^(5/6)/(1-(1-3*sec(x)^2)^(1/2)),x, algorithm="fricas")
Exception raised: TypeError >> Error detected within library code: Curv e not irreducible after change of variable 0 -> infinity
Timed out. \[ \int \frac {\sec ^2(x) \tan (x) \left (\sqrt [3]{1-3 \sec ^2(x)} \sin ^2(x)+3 \tan ^2(x)\right )}{\left (1-3 \sec ^2(x)\right )^{5/6} \left (1-\sqrt {1-3 \sec ^2(x)}\right )} \, dx=\text {Timed out} \]
integrate(tan(x)*((1-3*sec(x)**2)**(1/3)*sin(x)**2+3*tan(x)**2)/cos(x)**2/ (1-3*sec(x)**2)**(5/6)/(1-(1-3*sec(x)**2)**(1/2)),x)
Timed out. \[ \int \frac {\sec ^2(x) \tan (x) \left (\sqrt [3]{1-3 \sec ^2(x)} \sin ^2(x)+3 \tan ^2(x)\right )}{\left (1-3 \sec ^2(x)\right )^{5/6} \left (1-\sqrt {1-3 \sec ^2(x)}\right )} \, dx=\text {Timed out} \]
integrate(tan(x)*((1-3*sec(x)^2)^(1/3)*sin(x)^2+3*tan(x)^2)/cos(x)^2/(1-3* sec(x)^2)^(5/6)/(1-(1-3*sec(x)^2)^(1/2)),x, algorithm="maxima")
\[ \int \frac {\sec ^2(x) \tan (x) \left (\sqrt [3]{1-3 \sec ^2(x)} \sin ^2(x)+3 \tan ^2(x)\right )}{\left (1-3 \sec ^2(x)\right )^{5/6} \left (1-\sqrt {1-3 \sec ^2(x)}\right )} \, dx=\int { -\frac {{\left ({\left (-3 \, \sec \left (x\right )^{2} + 1\right )}^{\frac {1}{3}} \sin \left (x\right )^{2} + 3 \, \tan \left (x\right )^{2}\right )} \tan \left (x\right )}{{\left (-3 \, \sec \left (x\right )^{2} + 1\right )}^{\frac {5}{6}} {\left (\sqrt {-3 \, \sec \left (x\right )^{2} + 1} - 1\right )} \cos \left (x\right )^{2}} \,d x } \]
integrate(tan(x)*((1-3*sec(x)^2)^(1/3)*sin(x)^2+3*tan(x)^2)/cos(x)^2/(1-3* sec(x)^2)^(5/6)/(1-(1-3*sec(x)^2)^(1/2)),x, algorithm="giac")
Timed out. \[ \int \frac {\sec ^2(x) \tan (x) \left (\sqrt [3]{1-3 \sec ^2(x)} \sin ^2(x)+3 \tan ^2(x)\right )}{\left (1-3 \sec ^2(x)\right )^{5/6} \left (1-\sqrt {1-3 \sec ^2(x)}\right )} \, dx=-\int \frac {\mathrm {tan}\left (x\right )\,\left ({\sin \left (x\right )}^2\,{\left (1-\frac {3}{{\cos \left (x\right )}^2}\right )}^{1/3}+3\,{\mathrm {tan}\left (x\right )}^2\right )}{{\cos \left (x\right )}^2\,\left (\sqrt {1-\frac {3}{{\cos \left (x\right )}^2}}-1\right )\,{\left (1-\frac {3}{{\cos \left (x\right )}^2}\right )}^{5/6}} \,d x \]
int(-(tan(x)*(sin(x)^2*(1 - 3/cos(x)^2)^(1/3) + 3*tan(x)^2))/(cos(x)^2*((1 - 3/cos(x)^2)^(1/2) - 1)*(1 - 3/cos(x)^2)^(5/6)),x)