Integrand size = 28, antiderivative size = 71 \[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\frac {12 i 2^{5/6} a^2 \operatorname {Hypergeometric2F1}\left (-\frac {11}{6},\frac {5}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) (d \sec (e+f x))^{5/3}}{5 f (1+i \tan (e+f x))^{5/6}} \]
12/5*I*2^(5/6)*a^2*hypergeom([-11/6, 5/6],[11/6],1/2-1/2*I*tan(f*x+e))*(d* sec(f*x+e))^(5/3)/f/(1+I*tan(f*x+e))^(5/6)
Time = 0.47 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.56 \[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\frac {3 i a^2 (d \sec (e+f x))^{5/3} \left (i \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {5}{6},\frac {11}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+i \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{6},\frac {11}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+2 \sqrt {-\tan ^2(e+f x)}\right )}{5 f \sqrt {-\tan ^2(e+f x)}} \]
(((3*I)/5)*a^2*(d*Sec[e + f*x])^(5/3)*(I*Hypergeometric2F1[-1/2, 5/6, 11/6 , Sec[e + f*x]^2]*Tan[e + f*x] + I*Hypergeometric2F1[1/2, 5/6, 11/6, Sec[e + f*x]^2]*Tan[e + f*x] + 2*Sqrt[-Tan[e + f*x]^2]))/(f*Sqrt[-Tan[e + f*x]^ 2])
Time = 0.45 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3986, 3042, 4006, 80, 27, 79}
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.
\(\displaystyle \int (a+i a \tan (e+f x))^2 (d \sec (e+f x))^{5/3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (e+f x))^2 (d \sec (e+f x))^{5/3}dx\) |
\(\Big \downarrow \) 3986 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \int (a-i a \tan (e+f x))^{5/6} (i \tan (e+f x) a+a)^{17/6}dx}{(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(d \sec (e+f x))^{5/3} \int (a-i a \tan (e+f x))^{5/6} (i \tan (e+f x) a+a)^{17/6}dx}{(a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}}\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a^2 (d \sec (e+f x))^{5/3} \int \frac {(i \tan (e+f x) a+a)^{11/6}}{\sqrt [6]{a-i a \tan (e+f x)}}d\tan (e+f x)}{f (a-i a \tan (e+f x))^{5/6} (a+i a \tan (e+f x))^{5/6}}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {2\ 2^{5/6} a^3 (d \sec (e+f x))^{5/3} \int \frac {(i \tan (e+f x)+1)^{11/6}}{2\ 2^{5/6} \sqrt [6]{a-i a \tan (e+f x)}}d\tan (e+f x)}{f (1+i \tan (e+f x))^{5/6} (a-i a \tan (e+f x))^{5/6}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^3 (d \sec (e+f x))^{5/3} \int \frac {(i \tan (e+f x)+1)^{11/6}}{\sqrt [6]{a-i a \tan (e+f x)}}d\tan (e+f x)}{f (1+i \tan (e+f x))^{5/6} (a-i a \tan (e+f x))^{5/6}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {12 i 2^{5/6} a^2 (d \sec (e+f x))^{5/3} \operatorname {Hypergeometric2F1}\left (-\frac {11}{6},\frac {5}{6},\frac {11}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{5 f (1+i \tan (e+f x))^{5/6}}\) |
(((12*I)/5)*2^(5/6)*a^2*Hypergeometric2F1[-11/6, 5/6, 11/6, (1 - I*Tan[e + f*x])/2]*(d*Sec[e + f*x])^(5/3))/(f*(1 + I*Tan[e + f*x])^(5/6))
3.3.67.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_.), x_Symbol] :> Simp[(d*Sec[e + f*x])^m/((a + b*Tan[e + f*x])^(m/ 2)*(a - b*Tan[e + f*x])^(m/2)) Int[(a + b*Tan[e + f*x])^(m/2 + n)*(a - b* Tan[e + f*x])^(m/2), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*( c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
\[\int \left (d \sec \left (f x +e \right )\right )^{\frac {5}{3}} \left (a +i a \tan \left (f x +e \right )\right )^{2}d x\]
\[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \,d x } \]
-1/80*(3*2^(2/3)*(55*I*a^2*d*e^(5*I*f*x + 5*I*e) + 26*I*a^2*d*e^(3*I*f*x + 3*I*e) + 11*I*a^2*d*e^(I*f*x + I*e))*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)* e^(2/3*I*f*x + 2/3*I*e) - 80*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I *e) + f)*integral(11/16*I*2^(2/3)*a^2*d*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3 )*e^(2/3*I*f*x + 2/3*I*e)/f, x))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)
Timed out. \[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\text {Timed out} \]
\[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \,d x } \]
\[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\int { \left (d \sec \left (f x + e\right )\right )^{\frac {5}{3}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} \,d x } \]
Timed out. \[ \int (d \sec (e+f x))^{5/3} (a+i a \tan (e+f x))^2 \, dx=\int {\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{5/3}\,{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \]