3.1.6 \(\int (c+d x)^3 (a+a \sec (e+f x))^2 \, dx\) [6]

3.1.6.1 Optimal result

Integrand size = 20, antiderivative size = 371 \[ \int (c+d x)^3 (a+a \sec (e+f x))^2 \, dx=-\frac {i a^2 (c+d x)^3}{f}+\frac {a^2 (c+d x)^4}{4 d}-\frac {4 i a^2 (c+d x)^3 \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {3 a^2 d (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f^2}+\frac {6 i a^2 d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {6 i a^2 d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {3 i a^2 d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^3}-\frac {12 a^2 d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac {12 a^2 d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac {3 a^2 d^3 \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^4}-\frac {12 i a^2 d^3 \operatorname {PolyLog}\left (4,-i e^{i (e+f x)}\right )}{f^4}+\frac {12 i a^2 d^3 \operatorname {PolyLog}\left (4,i e^{i (e+f x)}\right )}{f^4}+\frac {a^2 (c+d x)^3 \tan (e+f x)}{f} \]

output

-I*a^2*(d*x+c)^3/f+1/4*a^2*(d*x+c)^4/d-4*I*a^2*(d*x+c)^3*arctan(exp(I*(f*x 
+e)))/f+3*a^2*d*(d*x+c)^2*ln(1+exp(2*I*(f*x+e)))/f^2+6*I*a^2*d*(d*x+c)^2*p 
olylog(2,-I*exp(I*(f*x+e)))/f^2-6*I*a^2*d*(d*x+c)^2*polylog(2,I*exp(I*(f*x 
+e)))/f^2-3*I*a^2*d^2*(d*x+c)*polylog(2,-exp(2*I*(f*x+e)))/f^3-12*a^2*d^2* 
(d*x+c)*polylog(3,-I*exp(I*(f*x+e)))/f^3+12*a^2*d^2*(d*x+c)*polylog(3,I*ex 
p(I*(f*x+e)))/f^3+3/2*a^2*d^3*polylog(3,-exp(2*I*(f*x+e)))/f^4-12*I*a^2*d^ 
3*polylog(4,-I*exp(I*(f*x+e)))/f^4+12*I*a^2*d^3*polylog(4,I*exp(I*(f*x+e)) 
)/f^4+a^2*(d*x+c)^3*tan(f*x+e)/f
 

3.1.6.2 Mathematica [A] (verified)

Time = 2.05 (sec) , antiderivative size = 329, normalized size of antiderivative = 0.89 \[ \int (c+d x)^3 (a+a \sec (e+f x))^2 \, dx=\frac {1}{4} a^2 \left (-\frac {4 i (c+d x)^3}{f}+\frac {(c+d x)^4}{d}-\frac {16 i (c+d x)^3 \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {24 i d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {24 i d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}+\frac {6 d \left (2 f^2 (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )-2 i d f (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )+d^2 \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )\right )}{f^4}-\frac {48 d^2 \left (f (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )+i d \operatorname {PolyLog}\left (4,-i e^{i (e+f x)}\right )\right )}{f^4}+\frac {48 d^2 \left (f (c+d x) \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )+i d \operatorname {PolyLog}\left (4,i e^{i (e+f x)}\right )\right )}{f^4}+\frac {4 (c+d x)^3 \tan (e+f x)}{f}\right ) \]

input

Integrate[(c + d*x)^3*(a + a*Sec[e + f*x])^2,x]
 

output

(a^2*(((-4*I)*(c + d*x)^3)/f + (c + d*x)^4/d - ((16*I)*(c + d*x)^3*ArcTan[ 
E^(I*(e + f*x))])/f + ((24*I)*d*(c + d*x)^2*PolyLog[2, (-I)*E^(I*(e + f*x) 
)])/f^2 - ((24*I)*d*(c + d*x)^2*PolyLog[2, I*E^(I*(e + f*x))])/f^2 + (6*d* 
(2*f^2*(c + d*x)^2*Log[1 + E^((2*I)*(e + f*x))] - (2*I)*d*f*(c + d*x)*Poly 
Log[2, -E^((2*I)*(e + f*x))] + d^2*PolyLog[3, -E^((2*I)*(e + f*x))]))/f^4 
- (48*d^2*(f*(c + d*x)*PolyLog[3, (-I)*E^(I*(e + f*x))] + I*d*PolyLog[4, ( 
-I)*E^(I*(e + f*x))]))/f^4 + (48*d^2*(f*(c + d*x)*PolyLog[3, I*E^(I*(e + f 
*x))] + I*d*PolyLog[4, I*E^(I*(e + f*x))]))/f^4 + (4*(c + d*x)^3*Tan[e + f 
*x])/f))/4
 

3.1.6.3 Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 371, 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, 4678, 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.

\(\displaystyle \int (c+d x)^3 (a \sec (e+f x)+a)^2 \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d x)^3 \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^2dx\)

\(\Big \downarrow \) 4678

\(\displaystyle \int \left (a^2 (c+d x)^3 \sec ^2(e+f x)+2 a^2 (c+d x)^3 \sec (e+f x)+a^2 (c+d x)^3\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {4 i a^2 (c+d x)^3 \arctan \left (e^{i (e+f x)}\right )}{f}-\frac {3 i a^2 d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{2 i (e+f x)}\right )}{f^3}-\frac {12 a^2 d^2 (c+d x) \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac {12 a^2 d^2 (c+d x) \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3}+\frac {6 i a^2 d (c+d x)^2 \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {6 i a^2 d (c+d x)^2 \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}+\frac {3 a^2 d (c+d x)^2 \log \left (1+e^{2 i (e+f x)}\right )}{f^2}+\frac {a^2 (c+d x)^3 \tan (e+f x)}{f}-\frac {i a^2 (c+d x)^3}{f}+\frac {a^2 (c+d x)^4}{4 d}+\frac {3 a^2 d^3 \operatorname {PolyLog}\left (3,-e^{2 i (e+f x)}\right )}{2 f^4}-\frac {12 i a^2 d^3 \operatorname {PolyLog}\left (4,-i e^{i (e+f x)}\right )}{f^4}+\frac {12 i a^2 d^3 \operatorname {PolyLog}\left (4,i e^{i (e+f x)}\right )}{f^4}\)

input

Int[(c + d*x)^3*(a + a*Sec[e + f*x])^2,x]
 

output

((-I)*a^2*(c + d*x)^3)/f + (a^2*(c + d*x)^4)/(4*d) - ((4*I)*a^2*(c + d*x)^ 
3*ArcTan[E^(I*(e + f*x))])/f + (3*a^2*d*(c + d*x)^2*Log[1 + E^((2*I)*(e + 
f*x))])/f^2 + ((6*I)*a^2*d*(c + d*x)^2*PolyLog[2, (-I)*E^(I*(e + f*x))])/f 
^2 - ((6*I)*a^2*d*(c + d*x)^2*PolyLog[2, I*E^(I*(e + f*x))])/f^2 - ((3*I)* 
a^2*d^2*(c + d*x)*PolyLog[2, -E^((2*I)*(e + f*x))])/f^3 - (12*a^2*d^2*(c + 
 d*x)*PolyLog[3, (-I)*E^(I*(e + f*x))])/f^3 + (12*a^2*d^2*(c + d*x)*PolyLo 
g[3, I*E^(I*(e + f*x))])/f^3 + (3*a^2*d^3*PolyLog[3, -E^((2*I)*(e + f*x))] 
)/(2*f^4) - ((12*I)*a^2*d^3*PolyLog[4, (-I)*E^(I*(e + f*x))])/f^4 + ((12*I 
)*a^2*d^3*PolyLog[4, I*E^(I*(e + f*x))])/f^4 + (a^2*(c + d*x)^3*Tan[e + f* 
x])/f
 

3.1.6.3.1 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 4678

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

3.1.6.4 Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1516 vs. \(2 (334 ) = 668\).

Time = 1.70 (sec) , antiderivative size = 1517, normalized size of antiderivative = 4.09

method	result	size
risch	\(\text {Expression too large to display}\)	\(1517\)

input

int((d*x+c)^3*(a+a*sec(f*x+e))^2,x,method=_RETURNVERBOSE)
 

output

4*I/f^4*a^2*d^3*e^3*arctan(exp(I*(f*x+e)))+6*I/f^3*a^2*d^3*e^2*x-6*I/f^3*a 
^2*d^3*polylog(2,-I*exp(I*(f*x+e)))*x-6*I/f^4*a^2*d^3*polylog(2,-I*exp(I*( 
f*x+e)))*e-6*I/f^3*a^2*d^3*polylog(2,I*exp(I*(f*x+e)))*x-6*I/f^4*a^2*d^3*p 
olylog(2,I*exp(I*(f*x+e)))*e-6*I/f^2*a^2*c^2*d*polylog(2,I*exp(I*(f*x+e))) 
-6*I/f*a^2*c*d^2*x^2-3*I/f^3*a^2*c*d^2*polylog(2,-exp(2*I*(f*x+e)))+3*I/f^ 
4*a^2*e*d^3*polylog(2,-exp(2*I*(f*x+e)))+6*I/f^2*a^2*d^3*polylog(2,-I*exp( 
I*(f*x+e)))*x^2-6*I/f^2*a^2*d^3*polylog(2,I*exp(I*(f*x+e)))*x^2+6*I/f^2*a^ 
2*c^2*d*polylog(2,-I*exp(I*(f*x+e)))-6*I/f^3*a^2*e^2*c*d^2+6/f^4*a^2*d^3*p 
olylog(3,-I*exp(I*(f*x+e)))+6/f^4*a^2*d^3*polylog(3,I*exp(I*(f*x+e)))-6/f* 
a^2*d^2*c*ln(1+I*exp(I*(f*x+e)))*x^2+6/f*a^2*d^2*c*ln(1-I*exp(I*(f*x+e)))* 
x^2-6/f^3*a^2*e*d^3*ln(1+exp(2*I*(f*x+e)))*x+12/f^3*a^2*c*d^2*e*ln(exp(I*( 
f*x+e)))+6/f^3*a^2*d^3*ln(1+I*exp(I*(f*x+e)))*e*x+6/f^2*a^2*c^2*d*ln(1-I*e 
xp(I*(f*x+e)))*e-6/f^2*a^2*c^2*d*ln(1+I*exp(I*(f*x+e)))*e+6/f^2*a^2*c*d^2* 
ln(1+exp(2*I*(f*x+e)))*x+6/f*a^2*c^2*d*ln(1-I*exp(I*(f*x+e)))*x-6/f*a^2*c^ 
2*d*ln(1+I*exp(I*(f*x+e)))*x-6/f^3*a^2*e^2*c*d^2*ln(1-I*exp(I*(f*x+e)))+6/ 
f^3*a^2*e^2*c*d^2*ln(1+I*exp(I*(f*x+e)))+6/f^3*a^2*d^3*ln(1-I*exp(I*(f*x+e 
)))*e*x+1/4*a^2*d^3*x^4+1/4*a^2/d*c^4+a^2*d^2*c*x^3+3/2*a^2*d*c^2*x^2+a^2* 
c^3*x+2*I*a^2*(d^3*x^3+3*c*d^2*x^2+3*c^2*d*x+c^3)/f/(1+exp(2*I*(f*x+e)))+1 
2*I/f^2*a^2*c^2*d*e*arctan(exp(I*(f*x+e)))-12*I/f^2*a^2*d^2*c*polylog(2,I* 
exp(I*(f*x+e)))*x+12*I/f^2*a^2*d^2*c*polylog(2,-I*exp(I*(f*x+e)))*x-12*...
 

3.1.6.5 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. 1887 vs. \(2 (318) = 636\).

Time = 0.38 (sec) , antiderivative size = 1887, normalized size of antiderivative = 5.09 \[ \int (c+d x)^3 (a+a \sec (e+f x))^2 \, dx=\text {Too large to display} \]

input

integrate((d*x+c)^3*(a+a*sec(f*x+e))^2,x, algorithm="fricas")
 

output

1/4*(24*I*a^2*d^3*cos(f*x + e)*polylog(4, I*cos(f*x + e) + sin(f*x + e)) + 
 24*I*a^2*d^3*cos(f*x + e)*polylog(4, I*cos(f*x + e) - sin(f*x + e)) - 24* 
I*a^2*d^3*cos(f*x + e)*polylog(4, -I*cos(f*x + e) + sin(f*x + e)) - 24*I*a 
^2*d^3*cos(f*x + e)*polylog(4, -I*cos(f*x + e) - sin(f*x + e)) - 12*(I*a^2 
*d^3*f^2*x^2 + I*a^2*c^2*d*f^2 - I*a^2*c*d^2*f + I*(2*a^2*c*d^2*f^2 - a^2* 
d^3*f)*x)*cos(f*x + e)*dilog(I*cos(f*x + e) + sin(f*x + e)) - 12*(I*a^2*d^ 
3*f^2*x^2 + I*a^2*c^2*d*f^2 + I*a^2*c*d^2*f + I*(2*a^2*c*d^2*f^2 + a^2*d^3 
*f)*x)*cos(f*x + e)*dilog(I*cos(f*x + e) - sin(f*x + e)) - 12*(-I*a^2*d^3* 
f^2*x^2 - I*a^2*c^2*d*f^2 + I*a^2*c*d^2*f - I*(2*a^2*c*d^2*f^2 - a^2*d^3*f 
)*x)*cos(f*x + e)*dilog(-I*cos(f*x + e) + sin(f*x + e)) - 12*(-I*a^2*d^3*f 
^2*x^2 - I*a^2*c^2*d*f^2 - I*a^2*c*d^2*f - I*(2*a^2*c*d^2*f^2 + a^2*d^3*f) 
*x)*cos(f*x + e)*dilog(-I*cos(f*x + e) - sin(f*x + e)) - 2*(2*a^2*d^3*e^3 
- 2*a^2*c^3*f^3 - 3*a^2*d^3*e^2 + 3*(2*a^2*c^2*d*e - a^2*c^2*d)*f^2 - 6*(a 
^2*c*d^2*e^2 - a^2*c*d^2*e)*f)*cos(f*x + e)*log(cos(f*x + e) + I*sin(f*x + 
 e) + I) + 2*(2*a^2*d^3*e^3 - 2*a^2*c^3*f^3 + 3*a^2*d^3*e^2 + 3*(2*a^2*c^2 
*d*e + a^2*c^2*d)*f^2 - 6*(a^2*c*d^2*e^2 + a^2*c*d^2*e)*f)*cos(f*x + e)*lo 
g(cos(f*x + e) - I*sin(f*x + e) + I) + 2*(2*a^2*d^3*f^3*x^3 + 2*a^2*d^3*e^ 
3 + 6*a^2*c^2*d*e*f^2 - 3*a^2*d^3*e^2 + 3*(2*a^2*c*d^2*f^3 + a^2*d^3*f^2)* 
x^2 - 6*(a^2*c*d^2*e^2 - a^2*c*d^2*e)*f + 6*(a^2*c^2*d*f^3 + a^2*c*d^2*f^2 
)*x)*cos(f*x + e)*log(I*cos(f*x + e) + sin(f*x + e) + 1) - 2*(2*a^2*d^3...
 

3.1.6.6 Sympy [F]

\[ \int (c+d x)^3 (a+a \sec (e+f x))^2 \, dx=a^{2} \left (\int c^{3}\, dx + \int 2 c^{3} \sec {\left (e + f x \right )}\, dx + \int c^{3} \sec ^{2}{\left (e + f x \right )}\, dx + \int d^{3} x^{3}\, dx + \int 3 c d^{2} x^{2}\, dx + \int 3 c^{2} d x\, dx + \int 2 d^{3} x^{3} \sec {\left (e + f x \right )}\, dx + \int d^{3} x^{3} \sec ^{2}{\left (e + f x \right )}\, dx + \int 6 c d^{2} x^{2} \sec {\left (e + f x \right )}\, dx + \int 3 c d^{2} x^{2} \sec ^{2}{\left (e + f x \right )}\, dx + \int 6 c^{2} d x \sec {\left (e + f x \right )}\, dx + \int 3 c^{2} d x \sec ^{2}{\left (e + f x \right )}\, dx\right ) \]

input

integrate((d*x+c)**3*(a+a*sec(f*x+e))**2,x)
 

output

a**2*(Integral(c**3, x) + Integral(2*c**3*sec(e + f*x), x) + Integral(c**3 
*sec(e + f*x)**2, x) + Integral(d**3*x**3, x) + Integral(3*c*d**2*x**2, x) 
 + Integral(3*c**2*d*x, x) + Integral(2*d**3*x**3*sec(e + f*x), x) + Integ 
ral(d**3*x**3*sec(e + f*x)**2, x) + Integral(6*c*d**2*x**2*sec(e + f*x), x 
) + Integral(3*c*d**2*x**2*sec(e + f*x)**2, x) + Integral(6*c**2*d*x*sec(e 
 + f*x), x) + Integral(3*c**2*d*x*sec(e + f*x)**2, x))
 

3.1.6.7 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. 3403 vs. \(2 (318) = 636\).

Time = 0.71 (sec) , antiderivative size = 3403, normalized size of antiderivative = 9.17 \[ \int (c+d x)^3 (a+a \sec (e+f x))^2 \, dx=\text {Too large to display} \]

input

integrate((d*x+c)^3*(a+a*sec(f*x+e))^2,x, algorithm="maxima")
 

output

1/4*(4*(f*x + e)*a^2*c^3 + (f*x + e)^4*a^2*d^3/f^3 - 4*(f*x + e)^3*a^2*d^3 
*e/f^3 + 6*(f*x + e)^2*a^2*d^3*e^2/f^3 - 4*(f*x + e)*a^2*d^3*e^3/f^3 + 4*( 
f*x + e)^3*a^2*c*d^2/f^2 - 12*(f*x + e)^2*a^2*c*d^2*e/f^2 + 12*(f*x + e)*a 
^2*c*d^2*e^2/f^2 + 6*(f*x + e)^2*a^2*c^2*d/f - 12*(f*x + e)*a^2*c^2*d*e/f 
+ 8*a^2*c^3*log(sec(f*x + e) + tan(f*x + e)) - 8*a^2*d^3*e^3*log(sec(f*x + 
 e) + tan(f*x + e))/f^3 + 24*a^2*c*d^2*e^2*log(sec(f*x + e) + tan(f*x + e) 
)/f^2 - 24*a^2*c^2*d*e*log(sec(f*x + e) + tan(f*x + e))/f - 4*(4*a^2*d^3*e 
^3 - 12*a^2*c*d^2*e^2*f + 12*a^2*c^2*d*e*f^2 - 4*a^2*c^3*f^3 + 4*((f*x + e 
)^3*a^2*d^3 - 3*(a^2*d^3*e - a^2*c*d^2*f)*(f*x + e)^2 + 3*(a^2*d^3*e^2 - 2 
*a^2*c*d^2*e*f + a^2*c^2*d*f^2)*(f*x + e) + ((f*x + e)^3*a^2*d^3 - 3*(a^2* 
d^3*e - a^2*c*d^2*f)*(f*x + e)^2 + 3*(a^2*d^3*e^2 - 2*a^2*c*d^2*e*f + a^2* 
c^2*d*f^2)*(f*x + e))*cos(2*f*x + 2*e) + (I*(f*x + e)^3*a^2*d^3 + 3*(-I*a^ 
2*d^3*e + I*a^2*c*d^2*f)*(f*x + e)^2 + 3*(I*a^2*d^3*e^2 - 2*I*a^2*c*d^2*e* 
f + I*a^2*c^2*d*f^2)*(f*x + e))*sin(2*f*x + 2*e))*arctan2(cos(f*x + e), si 
n(f*x + e) + 1) + 4*((f*x + e)^3*a^2*d^3 - 3*(a^2*d^3*e - a^2*c*d^2*f)*(f* 
x + e)^2 + 3*(a^2*d^3*e^2 - 2*a^2*c*d^2*e*f + a^2*c^2*d*f^2)*(f*x + e) + ( 
(f*x + e)^3*a^2*d^3 - 3*(a^2*d^3*e - a^2*c*d^2*f)*(f*x + e)^2 + 3*(a^2*d^3 
*e^2 - 2*a^2*c*d^2*e*f + a^2*c^2*d*f^2)*(f*x + e))*cos(2*f*x + 2*e) + (I*( 
f*x + e)^3*a^2*d^3 + 3*(-I*a^2*d^3*e + I*a^2*c*d^2*f)*(f*x + e)^2 + 3*(I*a 
^2*d^3*e^2 - 2*I*a^2*c*d^2*e*f + I*a^2*c^2*d*f^2)*(f*x + e))*sin(2*f*x ...
 

3.1.6.8 Giac [F]

\[ \int (c+d x)^3 (a+a \sec (e+f x))^2 \, dx=\int { {\left (d x + c\right )}^{3} {\left (a \sec \left (f x + e\right ) + a\right )}^{2} \,d x } \]

input

integrate((d*x+c)^3*(a+a*sec(f*x+e))^2,x, algorithm="giac")
 

output

integrate((d*x + c)^3*(a*sec(f*x + e) + a)^2, x)
 

3.1.6.9 Mupad [F(-1)]

Timed out. \[ \int (c+d x)^3 (a+a \sec (e+f x))^2 \, dx=\int {\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^2\,{\left (c+d\,x\right )}^3 \,d x \]

input

int((a + a/cos(e + f*x))^2*(c + d*x)^3,x)
 

output

int((a + a/cos(e + f*x))^2*(c + d*x)^3, x)