Integrand size = 41, antiderivative size = 127 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^6} \, dx=\frac {2 a^3 (i A+B)}{3 c^6 f (i+\tan (e+f x))^6}-\frac {4 a^3 (A-2 i B)}{5 c^6 f (i+\tan (e+f x))^5}-\frac {a^3 (i A+5 B)}{4 c^6 f (i+\tan (e+f x))^4}-\frac {i a^3 B}{3 c^6 f (i+\tan (e+f x))^3} \] Output:
2/3*a^3*(I*A+B)/c^6/f/(I+tan(f*x+e))^6-4/5*a^3*(A-2*I*B)/c^6/f/(I+tan(f*x+ e))^5-1/4*a^3*(I*A+5*B)/c^6/f/(I+tan(f*x+e))^4-1/3*I*a^3*B/c^6/f/(I+tan(f* x+e))^3
Time = 5.43 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.63 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^6} \, dx=-\frac {i a^3 \left (-7 A-i B+(-18 i A-6 B) \tan (e+f x)+15 (A-i B) \tan ^2(e+f x)+20 B \tan ^3(e+f x)\right )}{60 c^6 f (i+\tan (e+f x))^6} \] Input:
Integrate[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x])^6,x]
Output:
((-1/60*I)*a^3*(-7*A - I*B + ((-18*I)*A - 6*B)*Tan[e + f*x] + 15*(A - I*B) *Tan[e + f*x]^2 + 20*B*Tan[e + f*x]^3))/(c^6*f*(I + Tan[e + f*x])^6)
Time = 0.57 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.80, 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.
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^6} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^6}dx\) |
\(\Big \downarrow \) 4071 |
\(\displaystyle \frac {a c \int \frac {a^2 (i \tan (e+f x)+1)^2 (A+B \tan (e+f x))}{c^7 (1-i \tan (e+f x))^7}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^3 \int \frac {(i \tan (e+f x)+1)^2 (A+B \tan (e+f x))}{(1-i \tan (e+f x))^7}d\tan (e+f x)}{c^6 f}\) |
\(\Big \downarrow \) 86 |
\(\displaystyle \frac {a^3 \int \left (-\frac {4 i (A-i B)}{(\tan (e+f x)+i)^7}+\frac {i B}{(\tan (e+f x)+i)^4}+\frac {i A+5 B}{(\tan (e+f x)+i)^5}+\frac {4 (A-2 i B)}{(\tan (e+f x)+i)^6}\right )d\tan (e+f x)}{c^6 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {a^3 \left (-\frac {4 (A-2 i B)}{5 (\tan (e+f x)+i)^5}-\frac {5 B+i A}{4 (\tan (e+f x)+i)^4}+\frac {2 (B+i A)}{3 (\tan (e+f x)+i)^6}-\frac {i B}{3 (\tan (e+f x)+i)^3}\right )}{c^6 f}\) |
Input:
Int[((a + I*a*Tan[e + f*x])^3*(A + B*Tan[e + f*x]))/(c - I*c*Tan[e + f*x]) ^6,x]
Output:
(a^3*((2*(I*A + B))/(3*(I + Tan[e + f*x])^6) - (4*(A - (2*I)*B))/(5*(I + T an[e + f*x])^5) - (I*A + 5*B)/(4*(I + Tan[e + f*x])^4) - ((I/3)*B)/(I + Ta n[e + f*x])^3))/(c^6*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 = 0.40 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.71
method | result | size |
derivativedivides | \(\frac {a^{3} \left (-\frac {i B}{3 \left (i+\tan \left (f x +e \right )\right )^{3}}-\frac {-8 i B +4 A}{5 \left (i+\tan \left (f x +e \right )\right )^{5}}-\frac {-4 i A -4 B}{6 \left (i+\tan \left (f x +e \right )\right )^{6}}-\frac {i A +5 B}{4 \left (i+\tan \left (f x +e \right )\right )^{4}}\right )}{f \,c^{6}}\) | \(90\) |
default | \(\frac {a^{3} \left (-\frac {i B}{3 \left (i+\tan \left (f x +e \right )\right )^{3}}-\frac {-8 i B +4 A}{5 \left (i+\tan \left (f x +e \right )\right )^{5}}-\frac {-4 i A -4 B}{6 \left (i+\tan \left (f x +e \right )\right )^{6}}-\frac {i A +5 B}{4 \left (i+\tan \left (f x +e \right )\right )^{4}}\right )}{f \,c^{6}}\) | \(90\) |
risch | \(-\frac {a^{3} {\mathrm e}^{12 i \left (f x +e \right )} B}{96 c^{6} f}-\frac {i a^{3} {\mathrm e}^{12 i \left (f x +e \right )} A}{96 c^{6} f}-\frac {{\mathrm e}^{10 i \left (f x +e \right )} B \,a^{3}}{80 c^{6} f}-\frac {3 i {\mathrm e}^{10 i \left (f x +e \right )} A \,a^{3}}{80 c^{6} f}+\frac {{\mathrm e}^{8 i \left (f x +e \right )} B \,a^{3}}{64 c^{6} f}-\frac {3 i {\mathrm e}^{8 i \left (f x +e \right )} A \,a^{3}}{64 c^{6} f}+\frac {a^{3} {\mathrm e}^{6 i \left (f x +e \right )} B}{48 c^{6} f}-\frac {i a^{3} {\mathrm e}^{6 i \left (f x +e \right )} A}{48 c^{6} f}\) | \(174\) |
Input:
int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^6,x,method=_R ETURNVERBOSE)
Output:
1/f*a^3/c^6*(-1/3*I*B/(I+tan(f*x+e))^3-1/5*(-8*I*B+4*A)/(I+tan(f*x+e))^5-1 /6*(-4*B-4*I*A)/(I+tan(f*x+e))^6-1/4*(I*A+5*B)/(I+tan(f*x+e))^4)
Time = 0.07 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.70 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^6} \, dx=-\frac {10 \, {\left (i \, A + B\right )} a^{3} e^{\left (12 i \, f x + 12 i \, e\right )} + 12 \, {\left (3 i \, A + B\right )} a^{3} e^{\left (10 i \, f x + 10 i \, e\right )} + 15 \, {\left (3 i \, A - B\right )} a^{3} e^{\left (8 i \, f x + 8 i \, e\right )} + 20 \, {\left (i \, A - B\right )} a^{3} e^{\left (6 i \, f x + 6 i \, e\right )}}{960 \, c^{6} f} \] Input:
integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^6,x, al gorithm="fricas")
Output:
-1/960*(10*(I*A + B)*a^3*e^(12*I*f*x + 12*I*e) + 12*(3*I*A + B)*a^3*e^(10* I*f*x + 10*I*e) + 15*(3*I*A - B)*a^3*e^(8*I*f*x + 8*I*e) + 20*(I*A - B)*a^ 3*e^(6*I*f*x + 6*I*e))/(c^6*f)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (105) = 210\).
Time = 0.54 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.61 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^6} \, dx=\begin {cases} \frac {\left (- 491520 i A a^{3} c^{18} f^{3} e^{6 i e} + 491520 B a^{3} c^{18} f^{3} e^{6 i e}\right ) e^{6 i f x} + \left (- 1105920 i A a^{3} c^{18} f^{3} e^{8 i e} + 368640 B a^{3} c^{18} f^{3} e^{8 i e}\right ) e^{8 i f x} + \left (- 884736 i A a^{3} c^{18} f^{3} e^{10 i e} - 294912 B a^{3} c^{18} f^{3} e^{10 i e}\right ) e^{10 i f x} + \left (- 245760 i A a^{3} c^{18} f^{3} e^{12 i e} - 245760 B a^{3} c^{18} f^{3} e^{12 i e}\right ) e^{12 i f x}}{23592960 c^{24} f^{4}} & \text {for}\: c^{24} f^{4} \neq 0 \\\frac {x \left (A a^{3} e^{12 i e} + 3 A a^{3} e^{10 i e} + 3 A a^{3} e^{8 i e} + A a^{3} e^{6 i e} - i B a^{3} e^{12 i e} - i B a^{3} e^{10 i e} + i B a^{3} e^{8 i e} + i B a^{3} e^{6 i e}\right )}{8 c^{6}} & \text {otherwise} \end {cases} \] Input:
integrate((a+I*a*tan(f*x+e))**3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))**6,x)
Output:
Piecewise((((-491520*I*A*a**3*c**18*f**3*exp(6*I*e) + 491520*B*a**3*c**18* f**3*exp(6*I*e))*exp(6*I*f*x) + (-1105920*I*A*a**3*c**18*f**3*exp(8*I*e) + 368640*B*a**3*c**18*f**3*exp(8*I*e))*exp(8*I*f*x) + (-884736*I*A*a**3*c** 18*f**3*exp(10*I*e) - 294912*B*a**3*c**18*f**3*exp(10*I*e))*exp(10*I*f*x) + (-245760*I*A*a**3*c**18*f**3*exp(12*I*e) - 245760*B*a**3*c**18*f**3*exp( 12*I*e))*exp(12*I*f*x))/(23592960*c**24*f**4), Ne(c**24*f**4, 0)), (x*(A*a **3*exp(12*I*e) + 3*A*a**3*exp(10*I*e) + 3*A*a**3*exp(8*I*e) + A*a**3*exp( 6*I*e) - I*B*a**3*exp(12*I*e) - I*B*a**3*exp(10*I*e) + I*B*a**3*exp(8*I*e) + I*B*a**3*exp(6*I*e))/(8*c**6), True))
Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^6} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^6,x, al gorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati ve exponent.
Time = 0.69 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.76 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^6} \, dx=-\frac {20 i \, B a^{3} \tan \left (f x + e\right )^{3} + 15 i \, A a^{3} \tan \left (f x + e\right )^{2} + 15 \, B a^{3} \tan \left (f x + e\right )^{2} + 18 \, A a^{3} \tan \left (f x + e\right ) - 6 i \, B a^{3} \tan \left (f x + e\right ) - 7 i \, A a^{3} + B a^{3}}{60 \, c^{6} f {\left (\tan \left (f x + e\right ) + i\right )}^{6}} \] Input:
integrate((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^6,x, al gorithm="giac")
Output:
-1/60*(20*I*B*a^3*tan(f*x + e)^3 + 15*I*A*a^3*tan(f*x + e)^2 + 15*B*a^3*ta n(f*x + e)^2 + 18*A*a^3*tan(f*x + e) - 6*I*B*a^3*tan(f*x + e) - 7*I*A*a^3 + B*a^3)/(c^6*f*(tan(f*x + e) + I)^6)
Time = 5.54 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.10 \[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^6} \, dx=-\frac {-\frac {a^3\,\left (-B+A\,7{}\mathrm {i}\right )}{60}+\frac {a^3\,\mathrm {tan}\left (e+f\,x\right )\,\left (18\,A-B\,6{}\mathrm {i}\right )}{60}+\frac {B\,a^3\,{\mathrm {tan}\left (e+f\,x\right )}^3\,1{}\mathrm {i}}{3}+\frac {a^3\,{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (15\,B+A\,15{}\mathrm {i}\right )}{60}}{c^6\,f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^6+{\mathrm {tan}\left (e+f\,x\right )}^5\,6{}\mathrm {i}-15\,{\mathrm {tan}\left (e+f\,x\right )}^4-{\mathrm {tan}\left (e+f\,x\right )}^3\,20{}\mathrm {i}+15\,{\mathrm {tan}\left (e+f\,x\right )}^2+\mathrm {tan}\left (e+f\,x\right )\,6{}\mathrm {i}-1\right )} \] Input:
int(((A + B*tan(e + f*x))*(a + a*tan(e + f*x)*1i)^3)/(c - c*tan(e + f*x)*1 i)^6,x)
Output:
-((a^3*tan(e + f*x)*(18*A - B*6i))/60 - (a^3*(A*7i - B))/60 + (B*a^3*tan(e + f*x)^3*1i)/3 + (a^3*tan(e + f*x)^2*(A*15i + 15*B))/60)/(c^6*f*(tan(e + f*x)*6i + 15*tan(e + f*x)^2 - tan(e + f*x)^3*20i - 15*tan(e + f*x)^4 + tan (e + f*x)^5*6i + tan(e + f*x)^6 - 1))
\[ \int \frac {(a+i a \tan (e+f x))^3 (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^6} \, dx =\text {Too large to display} \] Input:
int((a+I*a*tan(f*x+e))^3*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^6,x)
Output:
(a**3*( - int(tan(e + f*x)**4/(tan(e + f*x)**6*i - 6*tan(e + f*x)**5 - 15* tan(e + f*x)**4*i + 20*tan(e + f*x)**3 + 15*tan(e + f*x)**2*i - 6*tan(e + f*x) - i),x)*b - int(tan(e + f*x)**3/(tan(e + f*x)**6*i - 6*tan(e + f*x)** 5 - 15*tan(e + f*x)**4*i + 20*tan(e + f*x)**3 + 15*tan(e + f*x)**2*i - 6*t an(e + f*x) - i),x)*a + 3*int(tan(e + f*x)**3/(tan(e + f*x)**6*i - 6*tan(e + f*x)**5 - 15*tan(e + f*x)**4*i + 20*tan(e + f*x)**3 + 15*tan(e + f*x)** 2*i - 6*tan(e + f*x) - i),x)*b*i + 3*int(tan(e + f*x)**2/(tan(e + f*x)**6* i - 6*tan(e + f*x)**5 - 15*tan(e + f*x)**4*i + 20*tan(e + f*x)**3 + 15*tan (e + f*x)**2*i - 6*tan(e + f*x) - i),x)*a*i + 3*int(tan(e + f*x)**2/(tan(e + f*x)**6*i - 6*tan(e + f*x)**5 - 15*tan(e + f*x)**4*i + 20*tan(e + f*x)* *3 + 15*tan(e + f*x)**2*i - 6*tan(e + f*x) - i),x)*b + 3*int(tan(e + f*x)/ (tan(e + f*x)**6*i - 6*tan(e + f*x)**5 - 15*tan(e + f*x)**4*i + 20*tan(e + f*x)**3 + 15*tan(e + f*x)**2*i - 6*tan(e + f*x) - i),x)*a - int(tan(e + f *x)/(tan(e + f*x)**6*i - 6*tan(e + f*x)**5 - 15*tan(e + f*x)**4*i + 20*tan (e + f*x)**3 + 15*tan(e + f*x)**2*i - 6*tan(e + f*x) - i),x)*b*i - int(1/( tan(e + f*x)**6*i - 6*tan(e + f*x)**5 - 15*tan(e + f*x)**4*i + 20*tan(e + f*x)**3 + 15*tan(e + f*x)**2*i - 6*tan(e + f*x) - i),x)*a*i))/c**6