\(\int \frac {(c+d x)^2}{(a+b \tan (e+f x))^2} \, dx\) [60]

Optimal result

Integrand size = 20, antiderivative size = 654 \[ \int \frac {(c+d x)^2}{(a+b \tan (e+f x))^2} \, dx=-\frac {2 i b^2 (c+d x)^2}{\left (a^2+b^2\right )^2 f}+\frac {2 b^2 (c+d x)^2}{(a+i b) (i a+b)^2 \left (i a-b+(i a+b) e^{2 i e+2 i f x}\right ) f}+\frac {(c+d x)^3}{3 (a-i b)^2 d}+\frac {4 b (c+d x)^3}{3 (i a-b) (a-i b)^2 d}-\frac {4 b^2 (c+d x)^3}{3 \left (a^2+b^2\right )^2 d}+\frac {2 b^2 d (c+d x) \log \left (1+\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{\left (a^2+b^2\right )^2 f^2}+\frac {2 b (c+d x)^2 \log \left (1+\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{(a-i b)^2 (a+i b) f}-\frac {2 i b^2 (c+d x)^2 \log \left (1+\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{\left (a^2+b^2\right )^2 f}-\frac {i b^2 d^2 \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{\left (a^2+b^2\right )^2 f^3}+\frac {2 b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{(i a-b) (a-i b)^2 f^2}-\frac {2 b^2 d (c+d x) \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{\left (a^2+b^2\right )^2 f^2}+\frac {b d^2 \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{(a-i b)^2 (a+i b) f^3}-\frac {i b^2 d^2 \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{\left (a^2+b^2\right )^2 f^3} \] Output:

-2*I*b^2*(d*x+c)^2/(a^2+b^2)^2/f+2*b^2*(d*x+c)^2/(a+I*b)/(I*a+b)^2/(I*a-b+ 
(I*a+b)*exp(2*I*e+2*I*f*x))/f+1/3*(d*x+c)^3/(a-I*b)^2/d+4/3*b*(d*x+c)^3/(I 
*a-b)/(a-I*b)^2/d-4/3*b^2*(d*x+c)^3/(a^2+b^2)^2/d+2*b^2*d*(d*x+c)*ln(1+(a- 
I*b)*exp(2*I*e+2*I*f*x)/(a+I*b))/(a^2+b^2)^2/f^2+2*b*(d*x+c)^2*ln(1+(a-I*b 
)*exp(2*I*e+2*I*f*x)/(a+I*b))/(a-I*b)^2/(a+I*b)/f-2*I*b^2*(d*x+c)^2*ln(1+( 
a-I*b)*exp(2*I*e+2*I*f*x)/(a+I*b))/(a^2+b^2)^2/f-I*b^2*d^2*polylog(2,-(a-I 
*b)*exp(2*I*e+2*I*f*x)/(a+I*b))/(a^2+b^2)^2/f^3+2*b*d*(d*x+c)*polylog(2,-( 
a-I*b)*exp(2*I*e+2*I*f*x)/(a+I*b))/(I*a-b)/(a-I*b)^2/f^2-2*b^2*d*(d*x+c)*p 
olylog(2,-(a-I*b)*exp(2*I*e+2*I*f*x)/(a+I*b))/(a^2+b^2)^2/f^2+b*d^2*polylo 
g(3,-(a-I*b)*exp(2*I*e+2*I*f*x)/(a+I*b))/(a-I*b)^2/(a+I*b)/f^3-I*b^2*d^2*p 
olylog(3,-(a-I*b)*exp(2*I*e+2*I*f*x)/(a+I*b))/(a^2+b^2)^2/f^3
 

Mathematica [A] (warning: unable to verify)

Time = 6.11 (sec) , antiderivative size = 761, normalized size of antiderivative = 1.16 \[ \int \frac {(c+d x)^2}{(a+b \tan (e+f x))^2} \, dx=\frac {\frac {2 b \left (\frac {12 c (b d+a c f) x}{a-i b}-\frac {12 c \left (-i b \left (-1+e^{2 i e}\right )+a \left (1+e^{2 i e}\right )\right ) (b d+a c f) x}{a^2+b^2}+\frac {6 d (b d+2 a c f) x^2}{a-i b}+\frac {4 a d^2 f x^3}{a-i b}+\frac {6 d \left (-i b \left (-1+e^{2 i e}\right )+a \left (1+e^{2 i e}\right )\right ) (b d+2 a c f) x \log \left (1+\frac {(a+i b) e^{-2 i (e+f x)}}{a-i b}\right )}{(a+i b) (i a+b) f}+\frac {6 a d^2 \left (-i b \left (-1+e^{2 i e}\right )+a \left (1+e^{2 i e}\right )\right ) x^2 \log \left (1+\frac {(a+i b) e^{-2 i (e+f x)}}{a-i b}\right )}{(a+i b) (i a+b)}+\frac {6 c \left (-i b \left (-1+e^{2 i e}\right )+a \left (1+e^{2 i e}\right )\right ) (b d+a c f) \log \left (a+i b+(a-i b) e^{2 i (e+f x)}\right )}{(a+i b) (i a+b) f}+\frac {3 d \left (-i b \left (-1+e^{2 i e}\right )+a \left (1+e^{2 i e}\right )\right ) (b d+2 a c f) \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i (e+f x)}}{a-i b}\right )}{\left (a^2+b^2\right ) f^2}+\frac {3 a d^2 \left (-i b \left (-1+e^{2 i e}\right )+a \left (1+e^{2 i e}\right )\right ) \left (2 f x \operatorname {PolyLog}\left (2,\frac {(-a-i b) e^{-2 i (e+f x)}}{a-i b}\right )-i \operatorname {PolyLog}\left (3,\frac {(-a-i b) e^{-2 i (e+f x)}}{a-i b}\right )\right )}{\left (a^2+b^2\right ) f^2}\right )}{b-b e^{2 i e}-i a \left (1+e^{2 i e}\right )}+\frac {\left (a^2-b^2\right ) f x \left (3 c^2+3 c d x+d^2 x^2\right ) \cos (f x)+\left (a^2+b^2\right ) f x \left (3 c^2+3 c d x+d^2 x^2\right ) \cos (2 e+f x)+2 b \left (3 b (c+d x)^2+a f x \left (3 c^2+3 c d x+d^2 x^2\right )\right ) \sin (f x)}{(a \cos (e)+b \sin (e)) (a \cos (e+f x)+b \sin (e+f x))}}{6 \left (a^2+b^2\right ) f} \] Input:

Integrate[(c + d*x)^2/(a + b*Tan[e + f*x])^2,x]
 

Output:

((2*b*((12*c*(b*d + a*c*f)*x)/(a - I*b) - (12*c*((-I)*b*(-1 + E^((2*I)*e)) 
 + a*(1 + E^((2*I)*e)))*(b*d + a*c*f)*x)/(a^2 + b^2) + (6*d*(b*d + 2*a*c*f 
)*x^2)/(a - I*b) + (4*a*d^2*f*x^3)/(a - I*b) + (6*d*((-I)*b*(-1 + E^((2*I) 
*e)) + a*(1 + E^((2*I)*e)))*(b*d + 2*a*c*f)*x*Log[1 + (a + I*b)/((a - I*b) 
*E^((2*I)*(e + f*x)))])/((a + I*b)*(I*a + b)*f) + (6*a*d^2*((-I)*b*(-1 + E 
^((2*I)*e)) + a*(1 + E^((2*I)*e)))*x^2*Log[1 + (a + I*b)/((a - I*b)*E^((2* 
I)*(e + f*x)))])/((a + I*b)*(I*a + b)) + (6*c*((-I)*b*(-1 + E^((2*I)*e)) + 
 a*(1 + E^((2*I)*e)))*(b*d + a*c*f)*Log[a + I*b + (a - I*b)*E^((2*I)*(e + 
f*x))])/((a + I*b)*(I*a + b)*f) + (3*d*((-I)*b*(-1 + E^((2*I)*e)) + a*(1 + 
 E^((2*I)*e)))*(b*d + 2*a*c*f)*PolyLog[2, (-a - I*b)/((a - I*b)*E^((2*I)*( 
e + f*x)))])/((a^2 + b^2)*f^2) + (3*a*d^2*((-I)*b*(-1 + E^((2*I)*e)) + a*( 
1 + E^((2*I)*e)))*(2*f*x*PolyLog[2, (-a - I*b)/((a - I*b)*E^((2*I)*(e + f* 
x)))] - I*PolyLog[3, (-a - I*b)/((a - I*b)*E^((2*I)*(e + f*x)))]))/((a^2 + 
 b^2)*f^2)))/(b - b*E^((2*I)*e) - I*a*(1 + E^((2*I)*e))) + ((a^2 - b^2)*f* 
x*(3*c^2 + 3*c*d*x + d^2*x^2)*Cos[f*x] + (a^2 + b^2)*f*x*(3*c^2 + 3*c*d*x 
+ d^2*x^2)*Cos[2*e + f*x] + 2*b*(3*b*(c + d*x)^2 + a*f*x*(3*c^2 + 3*c*d*x 
+ d^2*x^2))*Sin[f*x])/((a*Cos[e] + b*Sin[e])*(a*Cos[e + f*x] + b*Sin[e + f 
*x])))/(6*(a^2 + b^2)*f)
 

Rubi [A] (verified)

Time = 1.79 (sec) , antiderivative size = 654, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {3042, 4217, 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.

\(\displaystyle \int \frac {(c+d x)^2}{(a+b \tan (e+f x))^2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c+d x)^2}{(a+b \tan (e+f x))^2}dx\)

\(\Big \downarrow \) 4217

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

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {2 b^2 d (c+d x) \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{f^2 \left (a^2+b^2\right )^2}+\frac {2 b^2 d (c+d x) \log \left (1+\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{f^2 \left (a^2+b^2\right )^2}-\frac {2 i b^2 (c+d x)^2 \log \left (1+\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{f \left (a^2+b^2\right )^2}-\frac {2 i b^2 (c+d x)^2}{f \left (a^2+b^2\right )^2}-\frac {4 b^2 (c+d x)^3}{3 d \left (a^2+b^2\right )^2}-\frac {i b^2 d^2 \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{f^3 \left (a^2+b^2\right )^2}-\frac {i b^2 d^2 \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{f^3 \left (a^2+b^2\right )^2}+\frac {2 b^2 (c+d x)^2}{f (a+i b) (b+i a)^2 \left ((b+i a) e^{2 i e+2 i f x}+i a-b\right )}+\frac {2 b d (c+d x) \operatorname {PolyLog}\left (2,-\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{f^2 (-b+i a) (a-i b)^2}+\frac {2 b (c+d x)^2 \log \left (1+\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{f (a-i b)^2 (a+i b)}+\frac {4 b (c+d x)^3}{3 d (-b+i a) (a-i b)^2}+\frac {(c+d x)^3}{3 d (a-i b)^2}+\frac {b d^2 \operatorname {PolyLog}\left (3,-\frac {(a-i b) e^{2 i e+2 i f x}}{a+i b}\right )}{f^3 (a-i b)^2 (a+i b)}\)

Input:

Int[(c + d*x)^2/(a + b*Tan[e + f*x])^2,x]
 

Output:

((-2*I)*b^2*(c + d*x)^2)/((a^2 + b^2)^2*f) + (2*b^2*(c + d*x)^2)/((a + I*b 
)*(I*a + b)^2*(I*a - b + (I*a + b)*E^((2*I)*e + (2*I)*f*x))*f) + (c + d*x) 
^3/(3*(a - I*b)^2*d) + (4*b*(c + d*x)^3)/(3*(I*a - b)*(a - I*b)^2*d) - (4* 
b^2*(c + d*x)^3)/(3*(a^2 + b^2)^2*d) + (2*b^2*d*(c + d*x)*Log[1 + ((a - I* 
b)*E^((2*I)*e + (2*I)*f*x))/(a + I*b)])/((a^2 + b^2)^2*f^2) + (2*b*(c + d* 
x)^2*Log[1 + ((a - I*b)*E^((2*I)*e + (2*I)*f*x))/(a + I*b)])/((a - I*b)^2* 
(a + I*b)*f) - ((2*I)*b^2*(c + d*x)^2*Log[1 + ((a - I*b)*E^((2*I)*e + (2*I 
)*f*x))/(a + I*b)])/((a^2 + b^2)^2*f) - (I*b^2*d^2*PolyLog[2, -(((a - I*b) 
*E^((2*I)*e + (2*I)*f*x))/(a + I*b))])/((a^2 + b^2)^2*f^3) + (2*b*d*(c + d 
*x)*PolyLog[2, -(((a - I*b)*E^((2*I)*e + (2*I)*f*x))/(a + I*b))])/((I*a - 
b)*(a - I*b)^2*f^2) - (2*b^2*d*(c + d*x)*PolyLog[2, -(((a - I*b)*E^((2*I)* 
e + (2*I)*f*x))/(a + I*b))])/((a^2 + b^2)^2*f^2) + (b*d^2*PolyLog[3, -(((a 
 - I*b)*E^((2*I)*e + (2*I)*f*x))/(a + I*b))])/((a - I*b)^2*(a + I*b)*f^3) 
- (I*b^2*d^2*PolyLog[3, -(((a - I*b)*E^((2*I)*e + (2*I)*f*x))/(a + I*b))]) 
/((a^2 + b^2)^2*f^3)
 

Defintions of rubi rules used

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 4217

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), 
 x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (1/(a - I*b) - 2*I*(b/(a^2 + 
 b^2 + (a - I*b)^2*E^(2*I*(e + f*x)))))^(-n), x], x] /; FreeQ[{a, b, c, d, 
e, f}, x] && NeQ[a^2 + b^2, 0] && ILtQ[n, 0] && IGtQ[m, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3900 vs. \(2 (588 ) = 1176\).

Time = 0.79 (sec) , antiderivative size = 3901, normalized size of antiderivative = 5.96

method	result	size
risch	\(\text {Expression too large to display}\)	\(3901\)

Input:

int((d*x+c)^2/(a+b*tan(f*x+e))^2,x,method=_RETURNVERBOSE)
 

Output:

-1/3*d^2/(2*I*a*b-a^2+b^2)*x^3-1/(2*I*a*b-a^2+b^2)*c^2*x-1/3/d/(2*I*a*b-a^ 
2+b^2)*c^3-8/3/(I*a+b)^2/f^3/(b-I*a)*b/(a+I*b)*e^3*a*d^2+4/(I*a+b)^2/f^2/( 
b-I*a)*b^2/(a+I*b)*d^2*e*x+4/(I*a+b)^2/(b-I*a)*b/(a+I*b)*d*a*c*x^2+4/3/(I* 
a+b)^2/(b-I*a)*b/(a+I*b)*d^2*a*x^3+2/(I*a+b)^2/f/(b-I*a)*b^2/(a+I*b)*d^2*x 
^2+2/(I*a+b)^2/f^3/(b-I*a)*b^2/(a+I*b)*d^2*e^2+1/(I*a+b)^2/f^3/(b-I*a)*b^2 
/(a+I*b)*d^2*polylog(2,(I*b-a)*exp(2*I*(f*x+e))/(a+I*b))-d/(2*I*a*b-a^2+b^ 
2)*c*x^2-1/(I*a+b)^2/f^3/(b-I*a)*b^2*e^2*a*d^2/(a+I*b)/(I*b-a)*ln(a^2*exp( 
4*I*(f*x+e))+exp(4*I*(f*x+e))*b^2+2*a^2*exp(2*I*(f*x+e))-2*b^2*exp(2*I*(f* 
x+e))+a^2+b^2)+2/(I*a+b)^2/f^3/(b-I*a)*b^2*e*d^2/(a+I*b)/(I*b-a)*arctan(1/ 
2/a*b*exp(2*I*(f*x+e))-1/2/a*b+1/2/b*a*exp(2*I*(f*x+e))+1/2/b*a)*a-2/(I*a+ 
b)^2/f^3/(b-I*a)*b^2*e*d^2/(a+I*b)/(I*b-a)*arctan(1/b*a)*a+2/(I*a+b)^2/f^2 
/(b-I*a)*b^2*c*d/(a+I*b)/(I*b-a)*arctan(1/b*a)*a-2/(I*a+b)^2/f^2/(b-I*a)*b 
^2*c*d/(a+I*b)/(I*b-a)*arctan(1/2/a*b*exp(2*I*(f*x+e))-1/2/a*b+1/2/b*a*exp 
(2*I*(f*x+e))+1/2/b*a)*a-2/(I*a+b)^2/f^3/(b-I*a)*b*e^2*a^2*d^2/(a+I*b)/(I* 
b-a)*arctan(1/2/a*b*exp(2*I*(f*x+e))-1/2/a*b+1/2/b*a*exp(2*I*(f*x+e))+1/2/ 
b*a)+2/(I*a+b)^2/f^3/(b-I*a)*b*e^2*a^2*d^2/(a+I*b)/(I*b-a)*arctan(1/b*a)+8 
/(I*a+b)^2/f/(b-I*a)*b/(a+I*b)*d*a*c*e*x-2*I/(I*a+b)^2/f^3/(b-I*a)*b^3*e*d 
^2/(a+I*b)/(I*b-a)*arctan(1/2/a*b*exp(2*I*(f*x+e))-1/2/a*b+1/2/b*a*exp(2*I 
*(f*x+e))+1/2/b*a)+2*I/(I*a+b)^2/f^3/(b-I*a)*b^3*e*d^2/(a+I*b)/(I*b-a)*arc 
tan(1/b*a)+2*I/(I*a+b)^2/f/(b-I*a)*b^2*a*c^2/(a+I*b)/(I*b-a)*arctan(1/2...
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1576 vs. \(2 (534) = 1068\).

Time = 0.12 (sec) , antiderivative size = 1576, normalized size of antiderivative = 2.41 \[ \int \frac {(c+d x)^2}{(a+b \tan (e+f x))^2} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^2/(a+b*tan(f*x+e))^2,x, algorithm="fricas")
 

Output:

1/6*(2*(a^3 - a*b^2)*d^2*f^3*x^3 - 6*b^3*c^2*f^2 - 6*(b^3*d^2*f^2 - (a^3 - 
 a*b^2)*c*d*f^3)*x^2 - 6*(2*b^3*c*d*f^2 - (a^3 - a*b^2)*c^2*f^3)*x - 3*(-2 
*I*a^2*b*d^2*f*x - 2*I*a^2*b*c*d*f - I*a*b^2*d^2 + (-2*I*a*b^2*d^2*f*x - 2 
*I*a*b^2*c*d*f - I*b^3*d^2)*tan(f*x + e))*dilog(2*((I*a*b - b^2)*tan(f*x + 
 e)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2)*tan(f*x + e))/((a^2 + b^2)*t 
an(f*x + e)^2 + a^2 + b^2) + 1) - 3*(2*I*a^2*b*d^2*f*x + 2*I*a^2*b*c*d*f + 
 I*a*b^2*d^2 + (2*I*a*b^2*d^2*f*x + 2*I*a*b^2*c*d*f + I*b^3*d^2)*tan(f*x + 
 e))*dilog(2*((-I*a*b - b^2)*tan(f*x + e)^2 - a^2 + I*a*b + (-I*a^2 - 2*a* 
b + I*b^2)*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^2) + 1) + 6 
*(a^2*b*d^2*f^2*x^2 - a^2*b*d^2*e^2 + 2*a^2*b*c*d*e*f + a*b^2*d^2*e + (2*a 
^2*b*c*d*f^2 + a*b^2*d^2*f)*x + (a*b^2*d^2*f^2*x^2 - a*b^2*d^2*e^2 + 2*a*b 
^2*c*d*e*f + b^3*d^2*e + (2*a*b^2*c*d*f^2 + b^3*d^2*f)*x)*tan(f*x + e))*lo 
g(-2*((I*a*b - b^2)*tan(f*x + e)^2 - a^2 - I*a*b + (I*a^2 - 2*a*b - I*b^2) 
*tan(f*x + e))/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^2)) + 6*(a^2*b*d^2*f^ 
2*x^2 - a^2*b*d^2*e^2 + 2*a^2*b*c*d*e*f + a*b^2*d^2*e + (2*a^2*b*c*d*f^2 + 
 a*b^2*d^2*f)*x + (a*b^2*d^2*f^2*x^2 - a*b^2*d^2*e^2 + 2*a*b^2*c*d*e*f + b 
^3*d^2*e + (2*a*b^2*c*d*f^2 + b^3*d^2*f)*x)*tan(f*x + e))*log(-2*((-I*a*b 
- b^2)*tan(f*x + e)^2 - a^2 + I*a*b + (-I*a^2 - 2*a*b + I*b^2)*tan(f*x + e 
))/((a^2 + b^2)*tan(f*x + e)^2 + a^2 + b^2)) + 6*(a^2*b*d^2*e^2 + a^2*b*c^ 
2*f^2 - a*b^2*d^2*e - (2*a^2*b*c*d*e - a*b^2*c*d)*f + (a*b^2*d^2*e^2 + ...
 

Sympy [F]

\[ \int \frac {(c+d x)^2}{(a+b \tan (e+f x))^2} \, dx=\int \frac {\left (c + d x\right )^{2}}{\left (a + b \tan {\left (e + f x \right )}\right )^{2}}\, dx \] Input:

integrate((d*x+c)**2/(a+b*tan(f*x+e))**2,x)
 

Output:

Integral((c + d*x)**2/(a + b*tan(e + f*x))**2, x)
 

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2530 vs. \(2 (534) = 1068\).

Time = 0.92 (sec) , antiderivative size = 2530, normalized size of antiderivative = 3.87 \[ \int \frac {(c+d x)^2}{(a+b \tan (e+f x))^2} \, dx=\text {Too large to display} \] Input:

integrate((d*x+c)^2/(a+b*tan(f*x+e))^2,x, algorithm="maxima")
 

Output:

-1/3*(6*c*d*e*(2*a*b*log(b*tan(f*x + e) + a)/((a^4 + 2*a^2*b^2 + b^4)*f) - 
 a*b*log(tan(f*x + e)^2 + 1)/((a^4 + 2*a^2*b^2 + b^4)*f) - b/((a^2*b + b^3 
)*f*tan(f*x + e) + (a^3 + a*b^2)*f) + (a^2 - b^2)*(f*x + e)/((a^4 + 2*a^2* 
b^2 + b^4)*f)) - 3*(2*a*b*log(b*tan(f*x + e) + a)/(a^4 + 2*a^2*b^2 + b^4) 
- a*b*log(tan(f*x + e)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) + (a^2 - b^2)*(f*x + 
 e)/(a^4 + 2*a^2*b^2 + b^4) - b/(a^3 + a*b^2 + (a^2*b + b^3)*tan(f*x + e)) 
)*c^2 - ((a^3 - I*a^2*b + a*b^2 - I*b^3)*(f*x + e)^3*d^2 + 3*(a^3 - I*a^2* 
b + a*b^2 - I*b^3)*(f*x + e)*d^2*e^2 - 6*(-I*a*b^2 + b^3)*d^2*e^2 - 3*((a^ 
3 - I*a^2*b + a*b^2 - I*b^3)*d^2*e - (a^3 - I*a^2*b + a*b^2 - I*b^3)*c*d*f 
)*(f*x + e)^2 - 6*((-I*a^2*b + a*b^2)*d^2*e^2 + (I*a*b^2 - b^3)*d^2*e + (- 
I*a*b^2 + b^3)*c*d*f + ((-I*a^2*b - a*b^2)*d^2*e^2 + (I*a*b^2 + b^3)*d^2*e 
 + (-I*a*b^2 - b^3)*c*d*f)*cos(2*f*x + 2*e) + ((a^2*b - I*a*b^2)*d^2*e^2 - 
 (a*b^2 - I*b^3)*d^2*e + (a*b^2 - I*b^3)*c*d*f)*sin(2*f*x + 2*e))*arctan2( 
-b*cos(2*f*x + 2*e) + a*sin(2*f*x + 2*e) + b, a*cos(2*f*x + 2*e) + b*sin(2 
*f*x + 2*e) + a) - 6*((I*a^2*b - a*b^2)*(f*x + e)^2*d^2 + (2*(-I*a^2*b + a 
*b^2)*d^2*e + 2*(I*a^2*b - a*b^2)*c*d*f + (I*a*b^2 - b^3)*d^2)*(f*x + e) + 
 ((I*a^2*b + a*b^2)*(f*x + e)^2*d^2 + (2*(-I*a^2*b - a*b^2)*d^2*e + 2*(I*a 
^2*b + a*b^2)*c*d*f + (I*a*b^2 + b^3)*d^2)*(f*x + e))*cos(2*f*x + 2*e) - ( 
(a^2*b - I*a*b^2)*(f*x + e)^2*d^2 - (2*(a^2*b - I*a*b^2)*d^2*e - 2*(a^2*b 
- I*a*b^2)*c*d*f - (a*b^2 - I*b^3)*d^2)*(f*x + e))*sin(2*f*x + 2*e))*ar...
 

Giac [F]

\[ \int \frac {(c+d x)^2}{(a+b \tan (e+f x))^2} \, dx=\int { \frac {{\left (d x + c\right )}^{2}}{{\left (b \tan \left (f x + e\right ) + a\right )}^{2}} \,d x } \] Input:

integrate((d*x+c)^2/(a+b*tan(f*x+e))^2,x, algorithm="giac")
 

Output:

integrate((d*x + c)^2/(b*tan(f*x + e) + a)^2, x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x)^2}{(a+b \tan (e+f x))^2} \, dx=\int \frac {{\left (c+d\,x\right )}^2}{{\left (a+b\,\mathrm {tan}\left (e+f\,x\right )\right )}^2} \,d x \] Input:

int((c + d*x)^2/(a + b*tan(e + f*x))^2,x)
 

Output:

int((c + d*x)^2/(a + b*tan(e + f*x))^2, x)
 

Reduce [F]

\[ \int \frac {(c+d x)^2}{(a+b \tan (e+f x))^2} \, dx =\text {Too large to display} \] Input:

int((d*x+c)^2/(a+b*tan(f*x+e))^2,x)
 

Output:

(int(x**2/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*tan(e + f* 
x)*a**5*b*d**2*f + 2*int(x**2/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + 
 a**2),x)*tan(e + f*x)*a**3*b**3*d**2*f + int(x**2/(tan(e + f*x)**2*b**2 + 
 2*tan(e + f*x)*a*b + a**2),x)*tan(e + f*x)*a*b**5*d**2*f + int(x**2/(tan( 
e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*a**6*d**2*f + 2*int(x**2/ 
(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*a**4*b**2*d**2*f + i 
nt(x**2/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*a**2*b**4*d* 
*2*f + 2*int(x/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*tan(e 
 + f*x)*a**5*b*c*d*f + 4*int(x/(tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b 
+ a**2),x)*tan(e + f*x)*a**3*b**3*c*d*f + 2*int(x/(tan(e + f*x)**2*b**2 + 
2*tan(e + f*x)*a*b + a**2),x)*tan(e + f*x)*a*b**5*c*d*f + 2*int(x/(tan(e + 
 f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*a**6*c*d*f + 4*int(x/(tan(e 
+ f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*a**4*b**2*c*d*f + 2*int(x/( 
tan(e + f*x)**2*b**2 + 2*tan(e + f*x)*a*b + a**2),x)*a**2*b**4*c*d*f - log 
(tan(e + f*x)**2 + 1)*tan(e + f*x)*a**2*b**2*c**2 - log(tan(e + f*x)**2 + 
1)*a**3*b*c**2 + 2*log(tan(e + f*x)*b + a)*tan(e + f*x)*a**2*b**2*c**2 + 2 
*log(tan(e + f*x)*b + a)*a**3*b*c**2 + tan(e + f*x)*a**3*b*c**2*f*x + tan( 
e + f*x)*a**2*b**2*c**2 - tan(e + f*x)*a*b**3*c**2*f*x + tan(e + f*x)*b**4 
*c**2 + a**4*c**2*f*x - a**2*b**2*c**2*f*x)/(a*f*(tan(e + f*x)*a**4*b + 2* 
tan(e + f*x)*a**2*b**3 + tan(e + f*x)*b**5 + a**5 + 2*a**3*b**2 + a*b**...