Integrand size = 25, antiderivative size = 687 \[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\frac {5 a b^{2/3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) \sqrt [3]{d \sec (e+f x)}}{2 \sqrt {3} \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) \sqrt [3]{d \sec (e+f x)}}{2 \sqrt {3} \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \text {arctanh}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) \sqrt [3]{d \sec (e+f x)}}{3 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}+\frac {5 a b^{2/3} \log \left (\sqrt [3]{a^2+b^2}-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sqrt [3]{d \sec (e+f x)}}{12 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}-\frac {5 a b^{2/3} \log \left (\sqrt [3]{a^2+b^2}+\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sqrt [3]{d \sec (e+f x)}}{12 \left (a^2+b^2\right )^{11/6} f \sqrt [6]{\sec ^2(e+f x)}}+\frac {\operatorname {AppellF1}\left (\frac {1}{2},2,\frac {5}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan (e+f x)}{a^2 f \sqrt [6]{\sec ^2(e+f x)}}+\frac {b^2 \operatorname {AppellF1}\left (\frac {3}{2},2,\frac {5}{6},\frac {5}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sqrt [3]{d \sec (e+f x)} \tan ^3(e+f x)}{3 a^4 f \sqrt [6]{\sec ^2(e+f x)}}-\frac {a b \sqrt [3]{d \sec (e+f x)}}{\left (a^2+b^2\right ) f \left (a^2-b^2 \tan ^2(e+f x)\right )} \]
-5/3*a*b^(2/3)*arctanh(b^(1/3)*(sec(f*x+e)^2)^(1/6)/(a^2+b^2)^(1/6))*(d*se c(f*x+e))^(1/3)/(a^2+b^2)^(11/6)/f/(sec(f*x+e)^2)^(1/6)+5/12*a*b^(2/3)*ln( (a^2+b^2)^(1/3)-b^(1/3)*(a^2+b^2)^(1/6)*(sec(f*x+e)^2)^(1/6)+b^(2/3)*(sec( f*x+e)^2)^(1/3))*(d*sec(f*x+e))^(1/3)/(a^2+b^2)^(11/6)/f/(sec(f*x+e)^2)^(1 /6)-5/12*a*b^(2/3)*ln((a^2+b^2)^(1/3)+b^(1/3)*(a^2+b^2)^(1/6)*(sec(f*x+e)^ 2)^(1/6)+b^(2/3)*(sec(f*x+e)^2)^(1/3))*(d*sec(f*x+e))^(1/3)/(a^2+b^2)^(11/ 6)/f/(sec(f*x+e)^2)^(1/6)-5/6*a*b^(2/3)*arctan(-1/3*3^(1/2)+2/3*b^(1/3)*(s ec(f*x+e)^2)^(1/6)/(a^2+b^2)^(1/6)*3^(1/2))*(d*sec(f*x+e))^(1/3)/(a^2+b^2) ^(11/6)/f/(sec(f*x+e)^2)^(1/6)*3^(1/2)-5/6*a*b^(2/3)*arctan(1/3*3^(1/2)+2/ 3*b^(1/3)*(sec(f*x+e)^2)^(1/6)/(a^2+b^2)^(1/6)*3^(1/2))*(d*sec(f*x+e))^(1/ 3)/(a^2+b^2)^(11/6)/f/(sec(f*x+e)^2)^(1/6)*3^(1/2)+AppellF1(1/2,2,5/6,3/2, b^2*tan(f*x+e)^2/a^2,-tan(f*x+e)^2)*(d*sec(f*x+e))^(1/3)*tan(f*x+e)/a^2/f/ (sec(f*x+e)^2)^(1/6)+1/3*b^2*AppellF1(3/2,2,5/6,5/2,b^2*tan(f*x+e)^2/a^2,- tan(f*x+e)^2)*(d*sec(f*x+e))^(1/3)*tan(f*x+e)^3/a^4/f/(sec(f*x+e)^2)^(1/6) -a*b*(d*sec(f*x+e))^(1/3)/(a^2+b^2)/f/(a^2-b^2*tan(f*x+e)^2)
Leaf count is larger than twice the leaf count of optimal. \(7801\) vs. \(2(687)=1374\).
Time = 78.55 (sec) , antiderivative size = 7801, normalized size of antiderivative = 11.36 \[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\text {Result too large to show} \]
Time = 1.16 (sec) , antiderivative size = 545, normalized size of antiderivative = 0.79, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3042, 3994, 505, 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.
| \(\displaystyle \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2}dx\) |
| \(\Big \downarrow \) 3994 |
| \(\displaystyle \frac {\sqrt [3]{d \sec (e+f x)} \int \frac {1}{(a+b \tan (e+f x))^2 \left (\tan ^2(e+f x)+1\right )^{5/6}}d(b \tan (e+f x))}{b f \sqrt [6]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 505 |
| \(\displaystyle \frac {\sqrt [3]{d \sec (e+f x)} \int \left (\frac {a^2}{\left (\tan ^2(e+f x)+1\right )^{5/6} \left (a^2-b^2 \tan ^2(e+f x)\right )^2}-\frac {2 b \tan (e+f x) a}{\left (\tan ^2(e+f x)+1\right )^{5/6} \left (a^2-b^2 \tan ^2(e+f x)\right )^2}+\frac {b^2 \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^{5/6} \left (b^2 \tan ^2(e+f x)-a^2\right )^2}\right )d(b \tan (e+f x))}{b f \sqrt [6]{\sec ^2(e+f x)}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt [3]{d \sec (e+f x)} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},2,\frac {5}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2}+\frac {5 a b^{5/3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [6]{\tan ^2(e+f x)+1}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right )}{2 \sqrt {3} \left (a^2+b^2\right )^{11/6}}-\frac {5 a b^{5/3} \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [6]{\tan ^2(e+f x)+1}}{\sqrt {3} \sqrt [6]{a^2+b^2}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt {3} \left (a^2+b^2\right )^{11/6}}-\frac {5 a b^{5/3} \text {arctanh}\left (\frac {\sqrt [3]{b} \sqrt [6]{\tan ^2(e+f x)+1}}{\sqrt [6]{a^2+b^2}}\right )}{3 \left (a^2+b^2\right )^{11/6}}-\frac {a b^2 \sqrt [6]{\tan ^2(e+f x)+1}}{\left (a^2+b^2\right ) \left (a^2-b^2 \tan ^2(e+f x)\right )}+\frac {5 a b^{5/3} \log \left (-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\tan ^2(e+f x)+1}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\tan ^2(e+f x)+1}\right )}{12 \left (a^2+b^2\right )^{11/6}}-\frac {5 a b^{5/3} \log \left (\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\tan ^2(e+f x)+1}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\tan ^2(e+f x)+1}\right )}{12 \left (a^2+b^2\right )^{11/6}}+\frac {b^3 \tan ^3(e+f x) \operatorname {AppellF1}\left (\frac {3}{2},2,\frac {5}{6},\frac {5}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4}\right )}{b f \sqrt [6]{\sec ^2(e+f x)}}\) |
((d*Sec[e + f*x])^(1/3)*((5*a*b^(5/3)*ArcTan[1/Sqrt[3] - (2*b^(1/3)*(1 + T an[e + f*x]^2)^(1/6))/(Sqrt[3]*(a^2 + b^2)^(1/6))])/(2*Sqrt[3]*(a^2 + b^2) ^(11/6)) - (5*a*b^(5/3)*ArcTan[1/Sqrt[3] + (2*b^(1/3)*(1 + Tan[e + f*x]^2) ^(1/6))/(Sqrt[3]*(a^2 + b^2)^(1/6))])/(2*Sqrt[3]*(a^2 + b^2)^(11/6)) - (5* a*b^(5/3)*ArcTanh[(b^(1/3)*(1 + Tan[e + f*x]^2)^(1/6))/(a^2 + b^2)^(1/6)]) /(3*(a^2 + b^2)^(11/6)) + (5*a*b^(5/3)*Log[(a^2 + b^2)^(1/3) - b^(1/3)*(a^ 2 + b^2)^(1/6)*(1 + Tan[e + f*x]^2)^(1/6) + b^(2/3)*(1 + Tan[e + f*x]^2)^( 1/3)])/(12*(a^2 + b^2)^(11/6)) - (5*a*b^(5/3)*Log[(a^2 + b^2)^(1/3) + b^(1 /3)*(a^2 + b^2)^(1/6)*(1 + Tan[e + f*x]^2)^(1/6) + b^(2/3)*(1 + Tan[e + f* x]^2)^(1/3)])/(12*(a^2 + b^2)^(11/6)) + (b*AppellF1[1/2, 2, 5/6, 3/2, (b^2 *Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*Tan[e + f*x])/a^2 + (b^3*AppellF1[3 /2, 2, 5/6, 5/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*Tan[e + f*x]^3 )/(3*a^4) - (a*b^2*(1 + Tan[e + f*x]^2)^(1/6))/((a^2 + b^2)*(a^2 - b^2*Tan [e + f*x]^2))))/(b*f*(Sec[e + f*x]^2)^(1/6))
3.7.37.3.1 Defintions of rubi rules used
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[E xpandIntegrand[(a + b*x^2)^p, (c/(c^2 - d^2*x^2) - d*(x/(c^2 - d^2*x^2)))^( -n), x], x] /; FreeQ[{a, b, c, d, p}, x] && ILtQ[n, -1] && PosQ[a/b]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[d^(2*IntPart[m/2])*((d*Sec[e + f*x])^(2*FracP art[m/2])/(b*f*(Sec[e + f*x]^2)^FracPart[m/2])) Subst[Int[(a + x)^n*(1 + x^2/b^2)^(m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && NeQ[a^2 + b^2, 0] && !IntegerQ[m] && IntegerQ[n]
\[\int \frac {\left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}}}{\left (a +b \tan \left (f x +e \right )\right )^{2}}d x\]
Timed out. \[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\int \frac {\sqrt [3]{d \sec {\left (e + f x \right )}}}{\left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \]
\[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{d \sec (e+f x)}}{(a+b \tan (e+f x))^2} \, dx=\int \frac {{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{1/3}}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \]