Integrand size = 33, antiderivative size = 94 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2} \, dx=\frac {8 i a^3 (c-i c \tan (e+f x))^{5/2}}{5 f}-\frac {8 i a^3 (c-i c \tan (e+f x))^{7/2}}{7 c f}+\frac {2 i a^3 (c-i c \tan (e+f x))^{9/2}}{9 c^2 f} \]
8/5*I*a^3*(c-I*c*tan(f*x+e))^(5/2)/f-8/7*I*a^3*(c-I*c*tan(f*x+e))^(7/2)/c/ f+2/9*I*a^3*(c-I*c*tan(f*x+e))^(9/2)/c^2/f
Time = 2.97 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.70 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2} \, dx=\frac {2 a^3 c^2 (i+\tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)} \left (-107 i+110 \tan (e+f x)+35 i \tan ^2(e+f x)\right )}{315 f} \]
(2*a^3*c^2*(I + Tan[e + f*x])^2*Sqrt[c - I*c*Tan[e + f*x]]*(-107*I + 110*T an[e + f*x] + (35*I)*Tan[e + f*x]^2))/(315*f)
Time = 0.41 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.86, 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.
| \(\displaystyle \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2}dx\) |
| \(\Big \downarrow \) 4005 |
| \(\displaystyle a^3 c^3 \int \frac {\sec ^6(e+f x)}{\sqrt {c-i c \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a^3 c^3 \int \frac {\sec (e+f x)^6}{\sqrt {c-i c \tan (e+f x)}}dx\) |
| \(\Big \downarrow \) 3968 |
| \(\displaystyle \frac {i a^3 \int (c-i c \tan (e+f x))^{3/2} (i \tan (e+f x) c+c)^2d(-i c \tan (e+f x))}{c^2 f}\) |
| \(\Big \downarrow \) 53 |
| \(\displaystyle \frac {i a^3 \int \left ((c-i c \tan (e+f x))^{7/2}-4 c (c-i c \tan (e+f x))^{5/2}+4 c^2 (c-i c \tan (e+f x))^{3/2}\right )d(-i c \tan (e+f x))}{c^2 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {i a^3 \left (\frac {8}{5} c^2 (c-i c \tan (e+f x))^{5/2}+\frac {2}{9} (c-i c \tan (e+f x))^{9/2}-\frac {8}{7} c (c-i c \tan (e+f x))^{7/2}\right )}{c^2 f}\) |
(I*a^3*((8*c^2*(c - I*c*Tan[e + f*x])^(5/2))/5 - (8*c*(c - I*c*Tan[e + f*x ])^(7/2))/7 + (2*(c - I*c*Tan[e + f*x])^(9/2))/9))/(c^2*f)
3.10.68.3.1 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 = 0.74 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.70
| method | result | size |
| derivativedivides | \(\frac {2 i a^{3} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}-\frac {4 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {4 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}\right )}{f \,c^{2}}\) | \(66\) |
| default | \(\frac {2 i a^{3} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}-\frac {4 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {4 c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}\right )}{f \,c^{2}}\) | \(66\) |
| parts | \(\frac {2 i a^{3} c \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 c \sqrt {c -i c \tan \left (f x +e \right )}+2 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}-\frac {2 i a^{3} \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {9}{2}}}{9}+\frac {c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}-\frac {c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}-\frac {c^{3} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 c^{4} \sqrt {c -i c \tan \left (f x +e \right )}+2 c^{\frac {9}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f \,c^{2}}+\frac {3 i a^{3} \left (\frac {2 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+4 c^{2} \sqrt {c -i c \tan \left (f x +e \right )}-4 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}-\frac {6 i a^{3} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}-2 c^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f c}\) | \(402\) |
2*I/f*a^3/c^2*(1/9*(c-I*c*tan(f*x+e))^(9/2)-4/7*c*(c-I*c*tan(f*x+e))^(7/2) +4/5*c^2*(c-I*c*tan(f*x+e))^(5/2))
Time = 0.26 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.23 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {32 \, \sqrt {2} {\left (-63 i \, a^{3} c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} - 36 i \, a^{3} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 8 i \, a^{3} c^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{315 \, {\left (f e^{\left (8 i \, f x + 8 i \, e\right )} + 4 \, f e^{\left (6 i \, f x + 6 i \, e\right )} + 6 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 4 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \]
-32/315*sqrt(2)*(-63*I*a^3*c^2*e^(4*I*f*x + 4*I*e) - 36*I*a^3*c^2*e^(2*I*f *x + 2*I*e) - 8*I*a^3*c^2)*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(8*I*f*x + 8*I*e) + 4*f*e^(6*I*f*x + 6*I*e) + 6*f*e^(4*I*f*x + 4*I*e) + 4*f*e^(2*I *f*x + 2*I*e) + f)
\[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2} \, dx=- i a^{3} \left (\int i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}\, dx + \int \left (- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan {\left (e + f x \right )}\right )\, dx + \int \left (- 2 c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{3}{\left (e + f x \right )}\right )\, dx + \int \left (- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{5}{\left (e + f x \right )}\right )\, dx + \int 2 i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\, dx + \int i c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}\, dx\right ) \]
-I*a**3*(Integral(I*c**2*sqrt(-I*c*tan(e + f*x) + c), x) + Integral(-c**2* sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x), x) + Integral(-2*c**2*sqrt(-I*c* tan(e + f*x) + c)*tan(e + f*x)**3, x) + Integral(-c**2*sqrt(-I*c*tan(e + f *x) + c)*tan(e + f*x)**5, x) + Integral(2*I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**2, x) + Integral(I*c**2*sqrt(-I*c*tan(e + f*x) + c)*tan(e + f*x)**4, x))
Time = 0.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.71 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2} \, dx=\frac {2 i \, {\left (35 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {9}{2}} a^{3} - 180 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{3} c + 252 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{3} c^{2}\right )}}{315 \, c^{2} f} \]
2/315*I*(35*(-I*c*tan(f*x + e) + c)^(9/2)*a^3 - 180*(-I*c*tan(f*x + e) + c )^(7/2)*a^3*c + 252*(-I*c*tan(f*x + e) + c)^(5/2)*a^3*c^2)/(c^2*f)
\[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2} \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{3} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \]
Time = 9.80 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.03 \[ \int (a+i a \tan (e+f x))^3 (c-i c \tan (e+f x))^{5/2} \, dx=\frac {32\,a^3\,c^2\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,36{}\mathrm {i}+{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,63{}\mathrm {i}+8{}\mathrm {i}\right )}{315\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4} \]