Integrand size = 33, antiderivative size = 62 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=\frac {4 i a^2 (c-i c \tan (e+f x))^{5/2}}{5 f}-\frac {2 i a^2 (c-i c \tan (e+f x))^{7/2}}{7 c f} \]
Time = 2.10 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.87 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=\frac {2 a^2 c^2 (i+\tan (e+f x))^2 (-9 i+5 \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{35 f} \]
Time = 0.37 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3042, 4005, 3042, 3968, 53, 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.
\(\displaystyle \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2}dx\) |
\(\Big \downarrow \) 4005 |
\(\displaystyle a^2 c^2 \int \sec ^4(e+f x) \sqrt {c-i c \tan (e+f x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a^2 c^2 \int \sec (e+f x)^4 \sqrt {c-i c \tan (e+f x)}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle \frac {i a^2 \int (c-i c \tan (e+f x))^{3/2} (i \tan (e+f x) c+c)d(-i c \tan (e+f x))}{c f}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {i a^2 \int \left (2 c (c-i c \tan (e+f x))^{3/2}-(c-i c \tan (e+f x))^{5/2}\right )d(-i c \tan (e+f x))}{c f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {i a^2 \left (\frac {4}{5} c (c-i c \tan (e+f x))^{5/2}-\frac {2}{7} (c-i c \tan (e+f x))^{7/2}\right )}{c f}\) |
3.10.69.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[a^m*c^m Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[ b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] && !(IGtQ[n, 0] && (LtQ[ m, 0] || GtQ[m, n]))
Time = 0.63 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.76
method | result | size |
derivativedivides | \(\frac {2 i a^{2} \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {2 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}\right )}{f c}\) | \(47\) |
default | \(\frac {2 i a^{2} \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {2 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}\right )}{f c}\) | \(47\) |
parts | \(\frac {2 i a^{2} c \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 c \sqrt {c -i c \tan \left (f x +e \right )}+2 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}+\frac {2 i a^{2} \left (\frac {2 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}-4 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}-\frac {2 i a^{2} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}-2 c^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f c}\) | \(268\) |
Time = 0.24 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.40 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {16 \, \sqrt {2} {\left (-7 i \, a^{2} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, a^{2} c^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{35 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
-16/35*sqrt(2)*(-7*I*a^2*c^2*e^(2*I*f*x + 2*I*e) - 2*I*a^2*c^2)*sqrt(c/(e^ (2*I*f*x + 2*I*e) + 1))/(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)
\[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=- a^{2} \left (\int \left (- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}\right )\, dx + \int \left (- 2 c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}\right )\, dx\right ) \]
-a**2*(Integral(-c**2*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-2*c**2*s qrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2, x) + Integral(-c**2*sqrt(-I*c* tan(e + f*x) + c)*tan(e + f*x)**4, x))
Time = 0.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.74 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {2 i \, {\left (5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{2} - 14 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} c\right )}}{35 \, c f} \]
-2/35*I*(5*(-I*c*tan(f*x + e) + c)^(7/2)*a^2 - 14*(-I*c*tan(f*x + e) + c)^ (5/2)*a^2*c)/(c*f)
\[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]
Time = 11.61 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.34 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=\frac {16\,a^2\,c^2\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,7{}\mathrm {i}+2{}\mathrm {i}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}}{35\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3} \]