Integrand size = 33, antiderivative size = 62 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=\frac {4 i a^2 (c-i c \tan (e+f x))^{5/2}}{5 f}-\frac {2 i a^2 (c-i c \tan (e+f x))^{7/2}}{7 c f} \] Output:
4/5*I*a^2*(c-I*c*tan(f*x+e))^(5/2)/f-2/7*I*a^2*(c-I*c*tan(f*x+e))^(7/2)/c/ f
Time = 1.30 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.87 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=\frac {2 a^2 c^2 (i+\tan (e+f x))^2 (-9 i+5 \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{35 f} \] Input:
Integrate[(a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(5/2),x]
Output:
(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 |
\(\displaystyle \frac {i a^2 \left (\frac {4}{5} c (c-i c \tan (e+f x))^{5/2}-\frac {2}{7} (c-i c \tan (e+f x))^{7/2}\right )}{c f}\) |
Input:
Int[(a + I*a*Tan[e + f*x])^2*(c - I*c*Tan[e + f*x])^(5/2),x]
Output:
(I*a^2*((4*c*(c - I*c*Tan[e + f*x])^(5/2))/5 - (2*(c - I*c*Tan[e + f*x])^( 7/2))/7))/(c*f)
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 = 0.33 (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 | \(\frac {2 i a^{2} c \left (-\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}-2 \sqrt {c -i c \tan \left (f x +e \right )}\, c +2 c^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}+\frac {2 i a^{2} \left (\frac {2 \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5}+\frac {2 c \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+4 \sqrt {c -i c \tan \left (f x +e \right )}\, c^{2}-4 c^{\frac {5}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f}-\frac {2 i a^{2} \left (\frac {\left (c -i c \tan \left (f x +e \right )\right )^{\frac {7}{2}}}{7}+\frac {c^{2} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3}+2 c^{3} \sqrt {c -i c \tan \left (f x +e \right )}-2 c^{\frac {7}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c -i c \tan \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {c}}\right )\right )}{f c}\) | \(268\) |
Input:
int((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERBOSE)
Output:
2*I/f*a^2/c*(-1/7*(c-I*c*tan(f*x+e))^(7/2)+2/5*c*(c-I*c*tan(f*x+e))^(5/2))
Time = 0.08 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.40 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {16 \, \sqrt {2} {\left (-7 i \, a^{2} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} - 2 i \, a^{2} c^{2}\right )} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{35 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \] Input:
integrate((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fric as")
Output:
-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)
\[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=- a^{2} \left (\int \left (- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c}\right )\, dx + \int \left (- 2 c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{2}{\left (e + f x \right )}\right )\, dx + \int \left (- c^{2} \sqrt {- i c \tan {\left (e + f x \right )} + c} \tan ^{4}{\left (e + f x \right )}\right )\, dx\right ) \] Input:
integrate((a+I*a*tan(f*x+e))**2*(c-I*c*tan(f*x+e))**(5/2),x)
Output:
-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 = 0.04 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.74 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {2 i \, {\left (5 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {7}{2}} a^{2} - 14 \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a^{2} c\right )}}{35 \, c f} \] Input:
integrate((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxi ma")
Output:
-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)
Exception generated. \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac ")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeDone
Time = 7.02 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.34 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=\frac {16\,a^2\,c^2\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,7{}\mathrm {i}+2{}\mathrm {i}\right )\,\sqrt {c+\frac {c\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1}}}{35\,f\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3} \] Input:
int((a + a*tan(e + f*x)*1i)^2*(c - c*tan(e + f*x)*1i)^(5/2),x)
Output:
(16*a^2*c^2*(exp(e*2i + f*x*2i)*7i + 2i)*(c + (c*(exp(e*2i + f*x*2i)*1i - 1i)*1i)/(exp(e*2i + f*x*2i) + 1))^(1/2))/(35*f*(exp(e*2i + f*x*2i) + 1)^3)
Time = 0.19 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.92 \[ \int (a+i a \tan (e+f x))^2 (c-i c \tan (e+f x))^{5/2} \, dx=\frac {2 \sqrt {c}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, a^{2} c^{2} \left (5 \tan \left (f x +e \right )^{3}+\tan \left (f x +e \right )^{2} i +13 \tan \left (f x +e \right )+9 i \right )}{35 f} \] Input:
int((a+I*a*tan(f*x+e))^2*(c-I*c*tan(f*x+e))^(5/2),x)
Output:
(2*sqrt(c)*sqrt( - tan(e + f*x)*i + 1)*a**2*c**2*(5*tan(e + f*x)**3 + tan( e + f*x)**2*i + 13*tan(e + f*x) + 9*i))/(35*f)