Integrand size = 41, antiderivative size = 122 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=\frac {4 a^3 (A-i B)}{5 c^5 f (i+\tan (e+f x))^5}+\frac {a^3 (i A+2 B)}{c^5 f (i+\tan (e+f x))^4}-\frac {a^3 (A-5 i B)}{3 c^5 f (i+\tan (e+f x))^3}-\frac {a^3 B}{2 c^5 f (i+\tan (e+f x))^2} \] Output:
4/5*a^3*(A-I*B)/c^5/f/(I+tan(f*x+e))^5+a^3*(I*A+2*B)/c^5/f/(I+tan(f*x+e))^ 4-1/3*a^3*(A-5*I*B)/c^5/f/(I+tan(f*x+e))^3-1/2*a^3*B/c^5/f/(I+tan(f*x+e))^ 2
Time = 5.42 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.64 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=\frac {a^3 \left (4 A+i B+5 (2 i A+B) \tan (e+f x)+(-10 A+5 i B) \tan ^2(e+f x)-15 B \tan ^3(e+f x)\right )}{30 c^5 f (i+\tan (e+f x))^5} \] Input:
Integrate[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^5,x]
Output:
(a^3*(4*A + I*B + 5*((2*I)*A + B)*Tan[e + f*x] + (-10*A + (5*I)*B)*Tan[e + f*x]^2 - 15*B*Tan[e + f*x]^3))/(30*c^5*f*(I + Tan[e + f*x])^5)
Time = 0.57 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.79, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3042, 4071, 27, 86, 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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {a^2 (i \tan (e+f x)+1)^2 (A+B \tan (e+f x))}{c^6 (1-i \tan (e+f x))^6}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {a^3 \int \frac {(i \tan (e+f x)+1)^2 (A+B \tan (e+f x))}{(1-i \tan (e+f x))^6}d\tan (e+f x)}{c^5 f}\) |
| \(\Big \downarrow \) 86 |
| \(\displaystyle \frac {a^3 \int \left (-\frac {4 (A-i B)}{(\tan (e+f x)+i)^6}+\frac {B}{(\tan (e+f x)+i)^3}+\frac {A-5 i B}{(\tan (e+f x)+i)^4}-\frac {4 i (A-2 i B)}{(\tan (e+f x)+i)^5}\right )d\tan (e+f x)}{c^5 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {a^3 \left (\frac {4 (A-i B)}{5 (\tan (e+f x)+i)^5}-\frac {A-5 i B}{3 (\tan (e+f x)+i)^3}+\frac {2 B+i A}{(\tan (e+f x)+i)^4}-\frac {B}{2 (\tan (e+f x)+i)^2}\right )}{c^5 f}\) |
Input:
Int[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x]) ^5,x]
Output:
(a^3*((4*(A - I*B))/(5*(I + Tan[e + f*x])^5) + (I*A + 2*B)/(I + Tan[e + f* x])^4 - (A - (5*I)*B)/(3*(I + Tan[e + f*x])^3) - B/(2*(I + Tan[e + f*x])^2 )))/(c^5*f)
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_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_ .), x_] :> Int[ExpandIntegrand[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && ((ILtQ[n, 0] && ILtQ[p, 0]) || EqQ[p, 1 ] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x], x , Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
Time = 0.49 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.71
| method | result | size |
| derivativedivides | \(\frac {a^{3} \left (-\frac {B}{2 \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {-4 i A -8 B}{4 \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {-5 i B +A}{3 \left (i+\tan \left (f x +e \right )\right )^{3}}-\frac {4 i B -4 A}{5 \left (i+\tan \left (f x +e \right )\right )^{5}}\right )}{f \,c^{5}}\) | \(87\) |
| default | \(\frac {a^{3} \left (-\frac {B}{2 \left (i+\tan \left (f x +e \right )\right )^{2}}-\frac {-4 i A -8 B}{4 \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {-5 i B +A}{3 \left (i+\tan \left (f x +e \right )\right )^{3}}-\frac {4 i B -4 A}{5 \left (i+\tan \left (f x +e \right )\right )^{5}}\right )}{f \,c^{5}}\) | \(87\) |
| risch | \(-\frac {a^{3} {\mathrm e}^{10 i \left (f x +e \right )} B}{40 c^{5} f}-\frac {i a^{3} {\mathrm e}^{10 i \left (f x +e \right )} A}{40 c^{5} f}-\frac {i A \,a^{3} {\mathrm e}^{8 i \left (f x +e \right )}}{16 c^{5} f}+\frac {a^{3} {\mathrm e}^{6 i \left (f x +e \right )} B}{24 c^{5} f}-\frac {i a^{3} {\mathrm e}^{6 i \left (f x +e \right )} A}{24 c^{5} f}\) | \(110\) |
| norman | \(\frac {\frac {a^{3} A \tan \left (f x +e \right )}{f c}+\frac {-4 i A \,a^{3}+B \,a^{3}}{30 c f}-\frac {\left (-8 i B \,a^{3}+19 A \,a^{3}\right ) \tan \left (f x +e \right )^{3}}{3 c f}+\frac {7 \left (-16 i B \,a^{3}+11 A \,a^{3}\right ) \tan \left (f x +e \right )^{5}}{15 c f}-\frac {\left (-8 i B \,a^{3}+A \,a^{3}\right ) \tan \left (f x +e \right )^{7}}{3 c f}+\frac {2 \left (i A \,a^{3}+3 B \,a^{3}\right ) \tan \left (f x +e \right )^{6}}{c f}+\frac {\left (10 i A \,a^{3}+2 B \,a^{3}\right ) \tan \left (f x +e \right )^{2}}{3 c f}-\frac {\left (22 i A \,a^{3}+17 B \,a^{3}\right ) \tan \left (f x +e \right )^{4}}{3 c f}-\frac {B \,a^{3} \tan \left (f x +e \right )^{8}}{2 c f}}{\left (1+\tan \left (f x +e \right )^{2}\right )^{5} c^{4}}\) | \(255\) |
Input:
int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^5,x,method=_R ETURNVERBOSE)
Output:
1/f*a^3/c^5*(-1/2*B/(I+tan(f*x+e))^2-1/4*(-4*I*A-8*B)/(I+tan(f*x+e))^4-1/3 *(A-5*I*B)/(I+tan(f*x+e))^3-1/5*(-4*A+4*I*B)/(I+tan(f*x+e))^5)
Time = 0.08 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.52 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=-\frac {6 \, {\left (i \, A + B\right )} a^{3} e^{\left (10 i \, f x + 10 i \, e\right )} + 15 i \, A a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} + 10 \, {\left (i \, A - B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )}}{240 \, c^{5} f} \] Input:
integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^5,x, al gorithm="fricas")
Output:
-1/240*(6*(I*A + B)*a^3*e^(10*I*f*x + 10*I*e) + 15*I*A*a^3*e^(8*I*f*x + 8* I*e) + 10*(I*A - B)*a^3*e^(6*I*f*x + 6*I*e))/(c^5*f)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 218 vs. \(2 (100) = 200\).
Time = 0.47 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.79 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=\begin {cases} \frac {- 960 i A a^{3} c^{10} f^{2} e^{8 i e} e^{8 i f x} + \left (- 640 i A a^{3} c^{10} f^{2} e^{6 i e} + 640 B a^{3} c^{10} f^{2} e^{6 i e}\right ) e^{6 i f x} + \left (- 384 i A a^{3} c^{10} f^{2} e^{10 i e} - 384 B a^{3} c^{10} f^{2} e^{10 i e}\right ) e^{10 i f x}}{15360 c^{15} f^{3}} & \text {for}\: c^{15} f^{3} \neq 0 \\\frac {x \left (A a^{3} e^{10 i e} + 2 A a^{3} e^{8 i e} + A a^{3} e^{6 i e} - i B a^{3} e^{10 i e} + i B a^{3} e^{6 i e}\right )}{4 c^{5}} & \text {otherwise} \end {cases} \] Input:
integrate((a+I*a*tan(f*x+e))**3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**5,x)
Output:
Piecewise(((-960*I*A*a**3*c**10*f**2*exp(8*I*e)*exp(8*I*f*x) + (-640*I*A*a **3*c**10*f**2*exp(6*I*e) + 640*B*a**3*c**10*f**2*exp(6*I*e))*exp(6*I*f*x) + (-384*I*A*a**3*c**10*f**2*exp(10*I*e) - 384*B*a**3*c**10*f**2*exp(10*I* e))*exp(10*I*f*x))/(15360*c**15*f**3), Ne(c**15*f**3, 0)), (x*(A*a**3*exp( 10*I*e) + 2*A*a**3*exp(8*I*e) + A*a**3*exp(6*I*e) - I*B*a**3*exp(10*I*e) + I*B*a**3*exp(6*I*e))/(4*c**5), True))
Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^5,x, al gorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.65 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.80 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=-\frac {15 \, B a^{3} \tan \left (f x + e\right )^{3} + 10 \, A a^{3} \tan \left (f x + e\right )^{2} - 5 i \, B a^{3} \tan \left (f x + e\right )^{2} - 10 i \, A a^{3} \tan \left (f x + e\right ) - 5 \, B a^{3} \tan \left (f x + e\right ) - 4 \, A a^{3} - i \, B a^{3}}{30 \, c^{5} f {\left (\tan \left (f x + e\right ) + i\right )}^{5}} \] Input:
integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^5,x, al gorithm="giac")
Output:
-1/30*(15*B*a^3*tan(f*x + e)^3 + 10*A*a^3*tan(f*x + e)^2 - 5*I*B*a^3*tan(f *x + e)^2 - 10*I*A*a^3*tan(f*x + e) - 5*B*a^3*tan(f*x + e) - 4*A*a^3 - I*B *a^3)/(c^5*f*(tan(f*x + e) + I)^5)
Time = 5.66 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.05 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=\frac {\frac {a^3\,\left (4\,A+B\,1{}\mathrm {i}\right )}{30}+\frac {a^3\,\mathrm {tan}\left (e+f\,x\right )\,\left (5\,B+A\,10{}\mathrm {i}\right )}{30}-\frac {B\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^3}{2}-\frac {a^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (10\,A-B\,5{}\mathrm {i}\right )}{30}}{c^5\,f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^5+{\mathrm {tan}\left (e+f\,x\right )}^4\,5{}\mathrm {i}-10\,{\mathrm {tan}\left (e+f\,x\right )}^3-{\mathrm {tan}\left (e+f\,x\right )}^2\,10{}\mathrm {i}+5\,\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )} \] Input:
int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^3)/(c - c*tan(e + f*x)*1 i)^5,x)
Output:
((a^3*(4*A + B*1i))/30 + (a^3*tan(e + f*x)*(A*10i + 5*B))/30 - (B*a^3*tan( e + f*x)^3)/2 - (a^3*tan(e + f*x)^2*(10*A - B*5i))/30)/(c^5*f*(5*tan(e + f *x) - tan(e + f*x)^2*10i - 10*tan(e + f*x)^3 + tan(e + f*x)^4*5i + tan(e + f*x)^5 + 1i))
\[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^5} \, dx=\frac {a^{3} \left (\left (\int \frac {\tan \left (f x +e \right )^{4}}{\tan \left (f x +e \right )^{5} i -5 \tan \left (f x +e \right )^{4}-10 \tan \left (f x +e \right )^{3} i +10 \tan \left (f x +e \right )^{2}+5 \tan \left (f x +e \right ) i -1}d x \right ) b i +\left (\int \frac {\tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{5} i -5 \tan \left (f x +e \right )^{4}-10 \tan \left (f x +e \right )^{3} i +10 \tan \left (f x +e \right )^{2}+5 \tan \left (f x +e \right ) i -1}d x \right ) a i +3 \left (\int \frac {\tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{5} i -5 \tan \left (f x +e \right )^{4}-10 \tan \left (f x +e \right )^{3} i +10 \tan \left (f x +e \right )^{2}+5 \tan \left (f x +e \right ) i -1}d x \right ) b +3 \left (\int \frac {\tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{5} i -5 \tan \left (f x +e \right )^{4}-10 \tan \left (f x +e \right )^{3} i +10 \tan \left (f x +e \right )^{2}+5 \tan \left (f x +e \right ) i -1}d x \right ) a -3 \left (\int \frac {\tan \left (f x +e \right )^{2}}{\tan \left (f x +e \right )^{5} i -5 \tan \left (f x +e \right )^{4}-10 \tan \left (f x +e \right )^{3} i +10 \tan \left (f x +e \right )^{2}+5 \tan \left (f x +e \right ) i -1}d x \right ) b i -3 \left (\int \frac {\tan \left (f x +e \right )}{\tan \left (f x +e \right )^{5} i -5 \tan \left (f x +e \right )^{4}-10 \tan \left (f x +e \right )^{3} i +10 \tan \left (f x +e \right )^{2}+5 \tan \left (f x +e \right ) i -1}d x \right ) a i -\left (\int \frac {\tan \left (f x +e \right )}{\tan \left (f x +e \right )^{5} i -5 \tan \left (f x +e \right )^{4}-10 \tan \left (f x +e \right )^{3} i +10 \tan \left (f x +e \right )^{2}+5 \tan \left (f x +e \right ) i -1}d x \right ) b -\left (\int \frac {1}{\tan \left (f x +e \right )^{5} i -5 \tan \left (f x +e \right )^{4}-10 \tan \left (f x +e \right )^{3} i +10 \tan \left (f x +e \right )^{2}+5 \tan \left (f x +e \right ) i -1}d x \right ) a \right )}{c^{5}} \] Input:
int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^5,x)
Output:
(a**3*(int(tan(e + f*x)**4/(tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 10*tan (e + f*x)**3*i + 10*tan(e + f*x)**2 + 5*tan(e + f*x)*i - 1),x)*b*i + int(t an(e + f*x)**3/(tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 10*tan(e + f*x)**3 *i + 10*tan(e + f*x)**2 + 5*tan(e + f*x)*i - 1),x)*a*i + 3*int(tan(e + f*x )**3/(tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 10*tan(e + f*x)**3*i + 10*ta n(e + f*x)**2 + 5*tan(e + f*x)*i - 1),x)*b + 3*int(tan(e + f*x)**2/(tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 10*tan(e + f*x)**3*i + 10*tan(e + f*x)** 2 + 5*tan(e + f*x)*i - 1),x)*a - 3*int(tan(e + f*x)**2/(tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 10*tan(e + f*x)**3*i + 10*tan(e + f*x)**2 + 5*tan(e + f*x)*i - 1),x)*b*i - 3*int(tan(e + f*x)/(tan(e + f*x)**5*i - 5*tan(e + f *x)**4 - 10*tan(e + f*x)**3*i + 10*tan(e + f*x)**2 + 5*tan(e + f*x)*i - 1) ,x)*a*i - int(tan(e + f*x)/(tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 10*tan (e + f*x)**3*i + 10*tan(e + f*x)**2 + 5*tan(e + f*x)*i - 1),x)*b - int(1/( tan(e + f*x)**5*i - 5*tan(e + f*x)**4 - 10*tan(e + f*x)**3*i + 10*tan(e + f*x)**2 + 5*tan(e + f*x)*i - 1),x)*a))/c**5