Integrand size = 25, antiderivative size = 717 \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2} \, dx=\frac {11 a b}{5 \left (a^2+b^2\right )^2 f (d \sec (e+f x))^{5/3}}+\frac {11 a b^{8/3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) \sec ^2(e+f x)^{5/6}}{2 \sqrt {3} \left (a^2+b^2\right )^{17/6} f (d \sec (e+f x))^{5/3}}-\frac {11 a b^{8/3} \arctan \left (\frac {1}{\sqrt {3}}+\frac {2 \sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right ) \sec ^2(e+f x)^{5/6}}{2 \sqrt {3} \left (a^2+b^2\right )^{17/6} f (d \sec (e+f x))^{5/3}}-\frac {11 a b^{8/3} \text {arctanh}\left (\frac {\sqrt [3]{b} \sqrt [6]{\sec ^2(e+f x)}}{\sqrt [6]{a^2+b^2}}\right ) \sec ^2(e+f x)^{5/6}}{3 \left (a^2+b^2\right )^{17/6} f (d \sec (e+f x))^{5/3}}+\frac {11 a b^{8/3} \log \left (\sqrt [3]{a^2+b^2}-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sec ^2(e+f x)^{5/6}}{12 \left (a^2+b^2\right )^{17/6} f (d \sec (e+f x))^{5/3}}-\frac {11 a b^{8/3} \log \left (\sqrt [3]{a^2+b^2}+\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\sec ^2(e+f x)}+b^{2/3} \sqrt [3]{\sec ^2(e+f x)}\right ) \sec ^2(e+f x)^{5/6}}{12 \left (a^2+b^2\right )^{17/6} f (d \sec (e+f x))^{5/3}}+\frac {\operatorname {AppellF1}\left (\frac {1}{2},2,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/6} \tan (e+f x)}{a^2 f (d \sec (e+f x))^{5/3}}+\frac {b^2 \operatorname {AppellF1}\left (\frac {3}{2},2,\frac {11}{6},\frac {5}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right ) \sec ^2(e+f x)^{5/6} \tan ^3(e+f x)}{3 a^4 f (d \sec (e+f x))^{5/3}}-\frac {a b}{\left (a^2+b^2\right ) f (d \sec (e+f x))^{5/3} \left (a^2-b^2 \tan ^2(e+f x)\right )} \]
11/5*a*b/(a^2+b^2)^2/f/(d*sec(f*x+e))^(5/3)-11/3*a*b^(8/3)*arctanh(b^(1/3) *(sec(f*x+e)^2)^(1/6)/(a^2+b^2)^(1/6))*(sec(f*x+e)^2)^(5/6)/(a^2+b^2)^(17/ 6)/f/(d*sec(f*x+e))^(5/3)+11/12*a*b^(8/3)*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))*(sec(f*x+e)^ 2)^(5/6)/(a^2+b^2)^(17/6)/f/(d*sec(f*x+e))^(5/3)-11/12*a*b^(8/3)*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))*(sec(f*x+e)^2)^(5/6)/(a^2+b^2)^(17/6)/f/(d*sec(f*x+e))^(5/3)-11 /6*a*b^(8/3)*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))*(sec(f*x+e)^2)^(5/6)/(a^2+b^2)^(17/6)/f/(d*sec(f*x+e))^(5 /3)*3^(1/2)-11/6*a*b^(8/3)*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))*(sec(f*x+e)^2)^(5/6)/(a^2+b^2)^(17/6)/f/(d*s ec(f*x+e))^(5/3)*3^(1/2)+AppellF1(1/2,2,11/6,3/2,b^2*tan(f*x+e)^2/a^2,-tan (f*x+e)^2)*(sec(f*x+e)^2)^(5/6)*tan(f*x+e)/a^2/f/(d*sec(f*x+e))^(5/3)+1/3* b^2*AppellF1(3/2,2,11/6,5/2,b^2*tan(f*x+e)^2/a^2,-tan(f*x+e)^2)*(sec(f*x+e )^2)^(5/6)*tan(f*x+e)^3/a^4/f/(d*sec(f*x+e))^(5/3)-a*b/(a^2+b^2)/f/(d*sec( f*x+e))^(5/3)/(a^2-b^2*tan(f*x+e)^2)
Leaf count is larger than twice the leaf count of optimal. \(11783\) vs. \(2(717)=1434\).
Time = 81.40 (sec) , antiderivative size = 11783, normalized size of antiderivative = 16.43 \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2} \, dx=\text {Result too large to show} \]
Time = 1.23 (sec) , antiderivative size = 576, normalized size of antiderivative = 0.80, 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.
| \(\displaystyle \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2}dx\) |
| \(\Big \downarrow \) 3994 |
| \(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \int \frac {1}{(a+b \tan (e+f x))^2 \left (\tan ^2(e+f x)+1\right )^{11/6}}d(b \tan (e+f x))}{b f (d \sec (e+f x))^{5/3}}\) |
| \(\Big \downarrow \) 505 |
| \(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \int \left (\frac {a^2}{\left (\tan ^2(e+f x)+1\right )^{11/6} \left (a^2-b^2 \tan ^2(e+f x)\right )^2}-\frac {2 b \tan (e+f x) a}{\left (\tan ^2(e+f x)+1\right )^{11/6} \left (a^2-b^2 \tan ^2(e+f x)\right )^2}+\frac {b^2 \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right )^{11/6} \left (b^2 \tan ^2(e+f x)-a^2\right )^2}\right )d(b \tan (e+f x))}{b f (d \sec (e+f x))^{5/3}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sec ^2(e+f x)^{5/6} \left (\frac {b \tan (e+f x) \operatorname {AppellF1}\left (\frac {1}{2},2,\frac {11}{6},\frac {3}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{a^2}+\frac {11 a b^{11/3} \arctan \left (\frac {1}{\sqrt {3}}-\frac {2 \sqrt [3]{b} \sqrt [6]{\tan ^2(e+f x)+1}}{\sqrt {3} \sqrt [6]{a^2+b^2}}\right )}{2 \sqrt {3} \left (a^2+b^2\right )^{17/6}}-\frac {11 a b^{11/3} \arctan \left (\frac {2 \sqrt [3]{b} \sqrt [6]{\tan ^2(e+f x)+1}}{\sqrt {3} \sqrt [6]{a^2+b^2}}+\frac {1}{\sqrt {3}}\right )}{2 \sqrt {3} \left (a^2+b^2\right )^{17/6}}-\frac {11 a b^{11/3} \text {arctanh}\left (\frac {\sqrt [3]{b} \sqrt [6]{\tan ^2(e+f x)+1}}{\sqrt [6]{a^2+b^2}}\right )}{3 \left (a^2+b^2\right )^{17/6}}+\frac {11 a b^2}{5 \left (a^2+b^2\right )^2 \left (\tan ^2(e+f x)+1\right )^{5/6}}-\frac {a b^2}{\left (a^2+b^2\right ) \left (\tan ^2(e+f x)+1\right )^{5/6} \left (a^2-b^2 \tan ^2(e+f x)\right )}+\frac {11 a b^{11/3} \log \left (-\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\tan ^2(e+f x)+1}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\tan ^2(e+f x)+1}\right )}{12 \left (a^2+b^2\right )^{17/6}}-\frac {11 a b^{11/3} \log \left (\sqrt [3]{b} \sqrt [6]{a^2+b^2} \sqrt [6]{\tan ^2(e+f x)+1}+\sqrt [3]{a^2+b^2}+b^{2/3} \sqrt [3]{\tan ^2(e+f x)+1}\right )}{12 \left (a^2+b^2\right )^{17/6}}+\frac {b^3 \tan ^3(e+f x) \operatorname {AppellF1}\left (\frac {3}{2},2,\frac {11}{6},\frac {5}{2},\frac {b^2 \tan ^2(e+f x)}{a^2},-\tan ^2(e+f x)\right )}{3 a^4}\right )}{b f (d \sec (e+f x))^{5/3}}\) |
((Sec[e + f*x]^2)^(5/6)*((11*a*b^(11/3)*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)^(17/6)) - (11*a*b^(11/3)*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)^(17/6)) - (11*a*b^(11/3)*ArcTanh[(b^(1/3)*(1 + Tan[e + f*x]^2)^(1/6))/(a^2 + b^2)^( 1/6)])/(3*(a^2 + b^2)^(17/6)) + (11*a*b^(11/3)*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)^(17/6)) - (11*a*b^(11/3)*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)^(17/6)) + (b*AppellF1[1/2, 2, 11/6 , 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, 11/6, 5/2, (b^2*Tan[e + f*x]^2)/a^2, -Tan[e + f*x]^2]*Ta n[e + f*x]^3)/(3*a^4) + (11*a*b^2)/(5*(a^2 + b^2)^2*(1 + Tan[e + f*x]^2)^( 5/6)) - (a*b^2)/((a^2 + b^2)*(1 + Tan[e + f*x]^2)^(5/6)*(a^2 - b^2*Tan[e + f*x]^2))))/(b*f*(d*Sec[e + f*x])^(5/3))
3.7.39.3.1 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]
\[\int \frac {1}{\left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +b \tan \left (f x +e \right )\right )^{2}}d x\]
Timed out. \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2} \, dx=\int \frac {1}{\left (d \sec {\left (e + f x \right )}\right )^{\frac {5}{3}} \left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \]
\[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
\[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2} \, dx=\int { \frac {1}{\left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \]
Timed out. \[ \int \frac {1}{(d \sec (e+f x))^{5/3} (a+b \tan (e+f x))^2} \, dx=\int \frac {1}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}\,{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \]