3.10.10 \(\int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx\) [910]

3.10.10.1 Optimal result

Integrand size = 31, antiderivative size = 83 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\frac {c^3 x}{a^2}+\frac {i c^3 \log (\cos (e+f x))}{a^2 f}+\frac {2 i c^3}{f (a+i a \tan (e+f x))^2}-\frac {4 i c^3}{f \left (a^2+i a^2 \tan (e+f x)\right )} \]

output

c^3*x/a^2+I*c^3*ln(cos(f*x+e))/a^2/f+2*I*c^3/f/(a+I*a*tan(f*x+e))^2-4*I*c^ 
3/f/(a^2+I*a^2*tan(f*x+e))
 

3.10.10.2 Mathematica [A] (verified)

Time = 5.39 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.63 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=-\frac {i c^3 \left (\log (i-\tan (e+f x))+\frac {-2-4 i \tan (e+f x)}{(-i+\tan (e+f x))^2}\right )}{a^2 f} \]

input

Integrate[(c - I*c*Tan[e + f*x])^3/(a + I*a*Tan[e + f*x])^2,x]
 

output

((-I)*c^3*(Log[I - Tan[e + f*x]] + (-2 - (4*I)*Tan[e + f*x])/(-I + Tan[e + 
 f*x])^2))/(a^2*f)
 

3.10.10.3 Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.80, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 4005, 3042, 3968, 49, 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.

\(\displaystyle \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2}dx\)

\(\Big \downarrow \) 4005

\(\displaystyle a^3 c^3 \int \frac {\sec ^6(e+f x)}{(i \tan (e+f x) a+a)^5}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle a^3 c^3 \int \frac {\sec (e+f x)^6}{(i \tan (e+f x) a+a)^5}dx\)

\(\Big \downarrow \) 3968

\(\displaystyle -\frac {i c^3 \int \frac {(a-i a \tan (e+f x))^2}{(i \tan (e+f x) a+a)^3}d(i a \tan (e+f x))}{a^2 f}\)

\(\Big \downarrow \) 49

\(\displaystyle -\frac {i c^3 \int \left (\frac {4 a^2}{(i \tan (e+f x) a+a)^3}-\frac {4 a}{(i \tan (e+f x) a+a)^2}+\frac {1}{i \tan (e+f x) a+a}\right )d(i a \tan (e+f x))}{a^2 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {i c^3 \left (-\frac {2 a^2}{(a+i a \tan (e+f x))^2}+\frac {4 a}{a+i a \tan (e+f x)}+\log (a+i a \tan (e+f x))\right )}{a^2 f}\)

input

Int[(c - I*c*Tan[e + f*x])^3/(a + I*a*Tan[e + f*x])^2,x]
 

output

((-I)*c^3*(Log[a + I*a*Tan[e + f*x]] - (2*a^2)/(a + I*a*Tan[e + f*x])^2 + 
(4*a)/(a + I*a*Tan[e + f*x])))/(a^2*f)
 

3.10.10.3.1 Defintions of rubi rules used

rule 49

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}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3968

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]
 

rule 4005

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]))
 

3.10.10.4 Maple [A] (verified)

Time = 0.25 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05

method	result	size
derivativedivides	\(-\frac {2 i c^{3}}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {4 c^{3}}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {i c^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \,a^{2}}+\frac {c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}\)	\(87\)
default	\(-\frac {2 i c^{3}}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )^{2}}-\frac {4 c^{3}}{f \,a^{2} \left (\tan \left (f x +e \right )-i\right )}-\frac {i c^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \,a^{2}}+\frac {c^{3} \arctan \left (\tan \left (f x +e \right )\right )}{f \,a^{2}}\)	\(87\)
risch	\(-\frac {i c^{3} {\mathrm e}^{-2 i \left (f x +e \right )}}{a^{2} f}+\frac {i c^{3} {\mathrm e}^{-4 i \left (f x +e \right )}}{2 a^{2} f}+\frac {2 c^{3} x}{a^{2}}+\frac {2 c^{3} e}{a^{2} f}+\frac {i c^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{a^{2} f}\)	\(89\)
norman	\(\frac {\frac {c^{3} x}{a}-\frac {2 i c^{3}}{a f}+\frac {c^{3} x \left (\tan ^{4}\left (f x +e \right )\right )}{a}-\frac {4 c^{3} \left (\tan ^{3}\left (f x +e \right )\right )}{a f}+\frac {2 c^{3} x \left (\tan ^{2}\left (f x +e \right )\right )}{a}-\frac {6 i c^{3} \left (\tan ^{2}\left (f x +e \right )\right )}{a f}}{a \left (1+\tan ^{2}\left (f x +e \right )\right )^{2}}-\frac {i c^{3} \ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \,a^{2}}\)	\(134\)

input

int((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
 

output

-2*I/f*c^3/a^2/(tan(f*x+e)-I)^2-4/f*c^3/a^2/(tan(f*x+e)-I)-1/2*I/f*c^3/a^2 
*ln(1+tan(f*x+e)^2)+1/f*c^3/a^2*arctan(tan(f*x+e))
 

3.10.10.5 Fricas [A] (verification not implemented)

Time = 0.28 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.95 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\frac {{\left (4 \, c^{3} f x e^{\left (4 i \, f x + 4 i \, e\right )} + 2 i \, c^{3} e^{\left (4 i \, f x + 4 i \, e\right )} \log \left (e^{\left (2 i \, f x + 2 i \, e\right )} + 1\right ) - 2 i \, c^{3} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{3}\right )} e^{\left (-4 i \, f x - 4 i \, e\right )}}{2 \, a^{2} f} \]

input

integrate((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^2,x, algorithm="fricas")
 

output

1/2*(4*c^3*f*x*e^(4*I*f*x + 4*I*e) + 2*I*c^3*e^(4*I*f*x + 4*I*e)*log(e^(2* 
I*f*x + 2*I*e) + 1) - 2*I*c^3*e^(2*I*f*x + 2*I*e) + I*c^3)*e^(-4*I*f*x - 4 
*I*e)/(a^2*f)
 

3.10.10.6 Sympy [A] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 167, normalized size of antiderivative = 2.01 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\begin {cases} \frac {\left (- 2 i a^{2} c^{3} f e^{4 i e} e^{- 2 i f x} + i a^{2} c^{3} f e^{2 i e} e^{- 4 i f x}\right ) e^{- 6 i e}}{2 a^{4} f^{2}} & \text {for}\: a^{4} f^{2} e^{6 i e} \neq 0 \\x \left (- \frac {2 c^{3}}{a^{2}} + \frac {\left (2 c^{3} e^{4 i e} - 2 c^{3} e^{2 i e} + 2 c^{3}\right ) e^{- 4 i e}}{a^{2}}\right ) & \text {otherwise} \end {cases} + \frac {2 c^{3} x}{a^{2}} + \frac {i c^{3} \log {\left (e^{2 i f x} + e^{- 2 i e} \right )}}{a^{2} f} \]

input

integrate((c-I*c*tan(f*x+e))**3/(a+I*a*tan(f*x+e))**2,x)
 

output

Piecewise(((-2*I*a**2*c**3*f*exp(4*I*e)*exp(-2*I*f*x) + I*a**2*c**3*f*exp( 
2*I*e)*exp(-4*I*f*x))*exp(-6*I*e)/(2*a**4*f**2), Ne(a**4*f**2*exp(6*I*e), 
0)), (x*(-2*c**3/a**2 + (2*c**3*exp(4*I*e) - 2*c**3*exp(2*I*e) + 2*c**3)*e 
xp(-4*I*e)/a**2), True)) + 2*c**3*x/a**2 + I*c**3*log(exp(2*I*f*x) + exp(- 
2*I*e))/(a**2*f)
 

3.10.10.7 Maxima [F(-2)]

Exception generated. \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=\text {Exception raised: RuntimeError} \]

input

integrate((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^2,x, algorithm="maxima")
 

output

Exception raised: RuntimeError >> ECL says: expt: undefined: 0 to a negati 
ve exponent.
 

3.10.10.8 Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 151 vs. \(2 (73) = 146\).

Time = 0.53 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.82 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=-\frac {-\frac {6 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{a^{2}} + \frac {12 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}{a^{2}} - \frac {6 i \, c^{3} \log \left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1\right )}{a^{2}} + \frac {-25 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 100 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 198 i \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 100 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 25 i \, c^{3}}{a^{2} {\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - i\right )}^{4}}}{6 \, f} \]

input

integrate((c-I*c*tan(f*x+e))^3/(a+I*a*tan(f*x+e))^2,x, algorithm="giac")
 

output

-1/6*(-6*I*c^3*log(tan(1/2*f*x + 1/2*e) + 1)/a^2 + 12*I*c^3*log(tan(1/2*f* 
x + 1/2*e) - I)/a^2 - 6*I*c^3*log(tan(1/2*f*x + 1/2*e) - 1)/a^2 + (-25*I*c 
^3*tan(1/2*f*x + 1/2*e)^4 - 100*c^3*tan(1/2*f*x + 1/2*e)^3 + 198*I*c^3*tan 
(1/2*f*x + 1/2*e)^2 + 100*c^3*tan(1/2*f*x + 1/2*e) - 25*I*c^3)/(a^2*(tan(1 
/2*f*x + 1/2*e) - I)^4))/f
 

3.10.10.9 Mupad [B] (verification not implemented)

Time = 5.61 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.92 \[ \int \frac {(c-i c \tan (e+f x))^3}{(a+i a \tan (e+f x))^2} \, dx=-\frac {\frac {2\,c^3}{a^2}+\frac {c^3\,\mathrm {tan}\left (e+f\,x\right )\,4{}\mathrm {i}}{a^2}}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^2\,1{}\mathrm {i}+2\,\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )}-\frac {c^3\,\ln \left (\mathrm {tan}\left (e+f\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{a^2\,f} \]

input

int((c - c*tan(e + f*x)*1i)^3/(a + a*tan(e + f*x)*1i)^2,x)
 

output

- ((2*c^3)/a^2 + (c^3*tan(e + f*x)*4i)/a^2)/(f*(2*tan(e + f*x) + tan(e + f 
*x)^2*1i - 1i)) - (c^3*log(tan(e + f*x) - 1i)*1i)/(a^2*f)