Integrand size = 21, antiderivative size = 67 \[ \int \sec ^6(e+f x) (d \tan (e+f x))^{5/2} \, dx=\frac {2 (d \tan (e+f x))^{7/2}}{7 d f}+\frac {4 (d \tan (e+f x))^{11/2}}{11 d^3 f}+\frac {2 (d \tan (e+f x))^{15/2}}{15 d^5 f} \]
2/7*(d*tan(f*x+e))^(7/2)/d/f+4/11*(d*tan(f*x+e))^(11/2)/d^3/f+2/15*(d*tan( f*x+e))^(15/2)/d^5/f
Time = 0.65 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.78 \[ \int \sec ^6(e+f x) (d \tan (e+f x))^{5/2} \, dx=\frac {2 (117+44 \cos (2 (e+f x))+4 \cos (4 (e+f x))) \sec ^4(e+f x) (d \tan (e+f x))^{7/2}}{1155 d f} \]
(2*(117 + 44*Cos[2*(e + f*x)] + 4*Cos[4*(e + f*x)])*Sec[e + f*x]^4*(d*Tan[ e + f*x])^(7/2))/(1155*d*f)
Time = 0.24 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 3087, 244, 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.
\(\displaystyle \int \sec ^6(e+f x) (d \tan (e+f x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sec (e+f x)^6 (d \tan (e+f x))^{5/2}dx\) |
\(\Big \downarrow \) 3087 |
\(\displaystyle \frac {\int (d \tan (e+f x))^{5/2} \left (\tan ^2(e+f x)+1\right )^2d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 244 |
\(\displaystyle \frac {\int \left (\frac {(d \tan (e+f x))^{13/2}}{d^4}+\frac {2 (d \tan (e+f x))^{9/2}}{d^2}+(d \tan (e+f x))^{5/2}\right )d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {2 (d \tan (e+f x))^{15/2}}{15 d^5}+\frac {4 (d \tan (e+f x))^{11/2}}{11 d^3}+\frac {2 (d \tan (e+f x))^{7/2}}{7 d}}{f}\) |
((2*(d*Tan[e + f*x])^(7/2))/(7*d) + (4*(d*Tan[e + f*x])^(11/2))/(11*d^3) + (2*(d*Tan[e + f*x])^(15/2))/(15*d^5))/f
3.3.47.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand Integrand[(c*x)^m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && IGtQ[p , 0]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_S ymbol] :> Simp[1/f Subst[Int[(b*x)^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] && !(IntegerQ[(n - 1) /2] && LtQ[0, n, m - 1])
Time = 0.21 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.78
\[\frac {\frac {2 \left (d \tan \left (f x +e \right )\right )^{\frac {15}{2}}}{15}+\frac {4 d^{2} \left (d \tan \left (f x +e \right )\right )^{\frac {11}{2}}}{11}+\frac {2 d^{4} \left (d \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}}{d^{5} f}\]
2/d^5/f*(1/15*(d*tan(f*x+e))^(15/2)+2/11*d^2*(d*tan(f*x+e))^(11/2)+1/7*d^4 *(d*tan(f*x+e))^(7/2))
Time = 0.30 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.22 \[ \int \sec ^6(e+f x) (d \tan (e+f x))^{5/2} \, dx=-\frac {2 \, {\left (32 \, d^{2} \cos \left (f x + e\right )^{6} + 24 \, d^{2} \cos \left (f x + e\right )^{4} + 21 \, d^{2} \cos \left (f x + e\right )^{2} - 77 \, d^{2}\right )} \sqrt {\frac {d \sin \left (f x + e\right )}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{1155 \, f \cos \left (f x + e\right )^{7}} \]
-2/1155*(32*d^2*cos(f*x + e)^6 + 24*d^2*cos(f*x + e)^4 + 21*d^2*cos(f*x + e)^2 - 77*d^2)*sqrt(d*sin(f*x + e)/cos(f*x + e))*sin(f*x + e)/(f*cos(f*x + e)^7)
Timed out. \[ \int \sec ^6(e+f x) (d \tan (e+f x))^{5/2} \, dx=\text {Timed out} \]
Time = 0.39 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.76 \[ \int \sec ^6(e+f x) (d \tan (e+f x))^{5/2} \, dx=\frac {2 \, {\left (77 \, \left (d \tan \left (f x + e\right )\right )^{\frac {15}{2}} + 210 \, \left (d \tan \left (f x + e\right )\right )^{\frac {11}{2}} d^{2} + 165 \, \left (d \tan \left (f x + e\right )\right )^{\frac {7}{2}} d^{4}\right )}}{1155 \, d^{5} f} \]
2/1155*(77*(d*tan(f*x + e))^(15/2) + 210*(d*tan(f*x + e))^(11/2)*d^2 + 165 *(d*tan(f*x + e))^(7/2)*d^4)/(d^5*f)
Time = 0.42 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.16 \[ \int \sec ^6(e+f x) (d \tan (e+f x))^{5/2} \, dx=\frac {2 \, {\left (77 \, \sqrt {d \tan \left (f x + e\right )} d^{7} \tan \left (f x + e\right )^{7} + 210 \, \sqrt {d \tan \left (f x + e\right )} d^{7} \tan \left (f x + e\right )^{5} + 165 \, \sqrt {d \tan \left (f x + e\right )} d^{7} \tan \left (f x + e\right )^{3}\right )}}{1155 \, d^{5} f} \]
2/1155*(77*sqrt(d*tan(f*x + e))*d^7*tan(f*x + e)^7 + 210*sqrt(d*tan(f*x + e))*d^7*tan(f*x + e)^5 + 165*sqrt(d*tan(f*x + e))*d^7*tan(f*x + e)^3)/(d^5 *f)
Time = 18.33 (sec) , antiderivative size = 474, normalized size of antiderivative = 7.07 \[ \int \sec ^6(e+f x) (d \tan (e+f x))^{5/2} \, dx=\frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{1155\,f}+\frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,64{}\mathrm {i}}{1155\,f\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}+\frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{385\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,2432{}\mathrm {i}}{231\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3}+\frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,1504{}\mathrm {i}}{33\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,4288{}\mathrm {i}}{55\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^5}+\frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,896{}\mathrm {i}}{15\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^6}-\frac {d^2\,\sqrt {-\frac {d\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,256{}\mathrm {i}}{15\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^7} \]
(d^2*(-(d*(exp(e*2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*64 i)/(1155*f) + (d^2*(-(d*(exp(e*2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*64i)/(1155*f*(exp(e*2i + f*x*2i) + 1)) + (d^2*(-(d*(exp(e*2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*32i)/(385*f*(exp(e*2i + f*x*2i) + 1)^2) - (d^2*(-(d*(exp(e*2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x *2i) + 1))^(1/2)*2432i)/(231*f*(exp(e*2i + f*x*2i) + 1)^3) + (d^2*(-(d*(ex p(e*2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*1504i)/(33*f*(e xp(e*2i + f*x*2i) + 1)^4) - (d^2*(-(d*(exp(e*2i + f*x*2i)*1i - 1i))/(exp(e *2i + f*x*2i) + 1))^(1/2)*4288i)/(55*f*(exp(e*2i + f*x*2i) + 1)^5) + (d^2* (-(d*(exp(e*2i + f*x*2i)*1i - 1i))/(exp(e*2i + f*x*2i) + 1))^(1/2)*896i)/( 15*f*(exp(e*2i + f*x*2i) + 1)^6) - (d^2*(-(d*(exp(e*2i + f*x*2i)*1i - 1i)) /(exp(e*2i + f*x*2i) + 1))^(1/2)*256i)/(15*f*(exp(e*2i + f*x*2i) + 1)^7)