\(\int (c+d x)^2 (a+b \sec (e+f x)) \, dx\) [25]

Optimal result

Integrand size = 18, antiderivative size = 157 \[ \int (c+d x)^2 (a+b \sec (e+f x)) \, dx=\frac {a (c+d x)^3}{3 d}-\frac {2 i b (c+d x)^2 \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {2 i b d (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {2 b d^2 \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac {2 b d^2 \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3} \] Output:

1/3*a*(d*x+c)^3/d-2*I*b*(d*x+c)^2*arctan(exp(I*(f*x+e)))/f+2*I*b*d*(d*x+c) 
*polylog(2,-I*exp(I*(f*x+e)))/f^2-2*I*b*d*(d*x+c)*polylog(2,I*exp(I*(f*x+e 
)))/f^2-2*b*d^2*polylog(3,-I*exp(I*(f*x+e)))/f^3+2*b*d^2*polylog(3,I*exp(I 
*(f*x+e)))/f^3
 

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.29 \[ \int (c+d x)^2 (a+b \sec (e+f x)) \, dx=a c^2 x+a c d x^2+\frac {1}{3} a d^2 x^3+\frac {b c^2 \coth ^{-1}(\sin (e+f x))}{f}-\frac {4 i b c d x \arctan \left (e^{i (e+f x)}\right )}{f}-\frac {2 i b d^2 x^2 \arctan \left (e^{i (e+f x)}\right )}{f}+\frac {2 i b d (c+d x) \operatorname {PolyLog}\left (2,-i e^{i (e+f x)}\right )}{f^2}-\frac {2 i b d (c+d x) \operatorname {PolyLog}\left (2,i e^{i (e+f x)}\right )}{f^2}-\frac {2 b d^2 \operatorname {PolyLog}\left (3,-i e^{i (e+f x)}\right )}{f^3}+\frac {2 b d^2 \operatorname {PolyLog}\left (3,i e^{i (e+f x)}\right )}{f^3} \] Input:

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

Output:

a*c^2*x + a*c*d*x^2 + (a*d^2*x^3)/3 + (b*c^2*ArcCoth[Sin[e + f*x]])/f - (( 
4*I)*b*c*d*x*ArcTan[E^(I*(e + f*x))])/f - ((2*I)*b*d^2*x^2*ArcTan[E^(I*(e 
+ f*x))])/f + ((2*I)*b*d*(c + d*x)*PolyLog[2, (-I)*E^(I*(e + f*x))])/f^2 - 
 ((2*I)*b*d*(c + d*x)*PolyLog[2, I*E^(I*(e + f*x))])/f^2 - (2*b*d^2*PolyLo 
g[3, (-I)*E^(I*(e + f*x))])/f^3 + (2*b*d^2*PolyLog[3, I*E^(I*(e + f*x))])/ 
f^3
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4678

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 439 vs. \(2 (138 ) = 276\).

Time = 0.18 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.80

method	result	size
risch	\(\frac {a \,d^{2} x^{3}}{3}+a d c \,x^{2}+a \,c^{2} x +\frac {a \,c^{3}}{3 d}+\frac {b \,d^{2} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{2}}{f}+\frac {2 b \,d^{2} \operatorname {polylog}\left (3, i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}+\frac {b \,e^{2} d^{2} \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {2 b c d \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}+\frac {2 i b \,d^{2} \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{2}}-\frac {2 b c d \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}-\frac {2 b \,d^{2} \operatorname {polylog}\left (3, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}+\frac {2 i b c d \operatorname {polylog}\left (2, -i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {2 i b \,d^{2} \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f^{2}}-\frac {2 i b \,d^{2} e^{2} \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}-\frac {2 i b \,c^{2} \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}+\frac {2 b c d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}-\frac {b \,d^{2} \ln \left (1+i {\mathrm e}^{i \left (f x +e \right )}\right ) x^{2}}{f}+\frac {4 i b c d e \arctan \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {b \,e^{2} d^{2} \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{3}}+\frac {2 b c d \ln \left (1-i {\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}-\frac {2 i b c d \operatorname {polylog}\left (2, i {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}\)	\(440\)

Input:

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

Output:

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

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. 675 vs. \(2 (129) = 258\).

Time = 0.12 (sec) , antiderivative size = 675, normalized size of antiderivative = 4.30 \[ \int (c+d x)^2 (a+b \sec (e+f x)) \, dx =\text {Too large to display} \] Input:

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

Output:

1/6*(2*a*d^2*f^3*x^3 + 6*a*c*d*f^3*x^2 + 6*a*c^2*f^3*x - 6*b*d^2*polylog(3 
, I*cos(f*x + e) + sin(f*x + e)) + 6*b*d^2*polylog(3, I*cos(f*x + e) - sin 
(f*x + e)) - 6*b*d^2*polylog(3, -I*cos(f*x + e) + sin(f*x + e)) + 6*b*d^2* 
polylog(3, -I*cos(f*x + e) - sin(f*x + e)) - 6*(I*b*d^2*f*x + I*b*c*d*f)*d 
ilog(I*cos(f*x + e) + sin(f*x + e)) - 6*(I*b*d^2*f*x + I*b*c*d*f)*dilog(I* 
cos(f*x + e) - sin(f*x + e)) - 6*(-I*b*d^2*f*x - I*b*c*d*f)*dilog(-I*cos(f 
*x + e) + sin(f*x + e)) - 6*(-I*b*d^2*f*x - I*b*c*d*f)*dilog(-I*cos(f*x + 
e) - sin(f*x + e)) + 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log(cos(f*x + 
 e) + I*sin(f*x + e) + I) - 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log(co 
s(f*x + e) - I*sin(f*x + e) + I) + 3*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x - b*d^ 
2*e^2 + 2*b*c*d*e*f)*log(I*cos(f*x + e) + sin(f*x + e) + 1) - 3*(b*d^2*f^2 
*x^2 + 2*b*c*d*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f)*log(I*cos(f*x + e) - sin(f 
*x + e) + 1) + 3*(b*d^2*f^2*x^2 + 2*b*c*d*f^2*x - b*d^2*e^2 + 2*b*c*d*e*f) 
*log(-I*cos(f*x + e) + sin(f*x + e) + 1) - 3*(b*d^2*f^2*x^2 + 2*b*c*d*f^2* 
x - b*d^2*e^2 + 2*b*c*d*e*f)*log(-I*cos(f*x + e) - sin(f*x + e) + 1) + 3*( 
b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log(-cos(f*x + e) + I*sin(f*x + e) + 
I) - 3*(b*d^2*e^2 - 2*b*c*d*e*f + b*c^2*f^2)*log(-cos(f*x + e) - I*sin(f*x 
 + e) + I))/f^3
 

Sympy [F]

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

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

Output:

Integral((a + b*sec(e + f*x))*(c + d*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. 516 vs. \(2 (129) = 258\).

Time = 0.18 (sec) , antiderivative size = 516, normalized size of antiderivative = 3.29 \[ \int (c+d x)^2 (a+b \sec (e+f x)) \, dx=\frac {6 \, {\left (f x + e\right )} a c^{2} + \frac {2 \, {\left (f x + e\right )}^{3} a d^{2}}{f^{2}} - \frac {6 \, {\left (f x + e\right )}^{2} a d^{2} e}{f^{2}} + \frac {6 \, {\left (f x + e\right )} a d^{2} e^{2}}{f^{2}} + \frac {6 \, {\left (f x + e\right )}^{2} a c d}{f} - \frac {12 \, {\left (f x + e\right )} a c d e}{f} + 6 \, b c^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right ) + \frac {6 \, b d^{2} e^{2} \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{f^{2}} - \frac {12 \, b c d e \log \left (\sec \left (f x + e\right ) + \tan \left (f x + e\right )\right )}{f} + \frac {3 \, {\left (4 \, b d^{2} {\rm Li}_{3}(i \, e^{\left (i \, f x + i \, e\right )}) - 4 \, b d^{2} {\rm Li}_{3}(-i \, e^{\left (i \, f x + i \, e\right )}) - 2 \, {\left (i \, {\left (f x + e\right )}^{2} b d^{2} + 2 \, {\left (-i \, b d^{2} e + i \, b c d f\right )} {\left (f x + e\right )}\right )} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (i \, {\left (f x + e\right )}^{2} b d^{2} + 2 \, {\left (-i \, b d^{2} e + i \, b c d f\right )} {\left (f x + e\right )}\right )} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) - 4 \, {\left (i \, {\left (f x + e\right )} b d^{2} - i \, b d^{2} e + i \, b c d f\right )} {\rm Li}_2\left (i \, e^{\left (i \, f x + i \, e\right )}\right ) - 4 \, {\left (-i \, {\left (f x + e\right )} b d^{2} + i \, b d^{2} e - i \, b c d f\right )} {\rm Li}_2\left (-i \, e^{\left (i \, f x + i \, e\right )}\right ) + {\left ({\left (f x + e\right )}^{2} b d^{2} - 2 \, {\left (b d^{2} e - b c d f\right )} {\left (f x + e\right )}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) - {\left ({\left (f x + e\right )}^{2} b d^{2} - 2 \, {\left (b d^{2} e - b c d f\right )} {\left (f x + e\right )}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right )\right )}}{f^{2}}}{6 \, f} \] Input:

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

Output:

1/6*(6*(f*x + e)*a*c^2 + 2*(f*x + e)^3*a*d^2/f^2 - 6*(f*x + e)^2*a*d^2*e/f 
^2 + 6*(f*x + e)*a*d^2*e^2/f^2 + 6*(f*x + e)^2*a*c*d/f - 12*(f*x + e)*a*c* 
d*e/f + 6*b*c^2*log(sec(f*x + e) + tan(f*x + e)) + 6*b*d^2*e^2*log(sec(f*x 
 + e) + tan(f*x + e))/f^2 - 12*b*c*d*e*log(sec(f*x + e) + tan(f*x + e))/f 
+ 3*(4*b*d^2*polylog(3, I*e^(I*f*x + I*e)) - 4*b*d^2*polylog(3, -I*e^(I*f* 
x + I*e)) - 2*(I*(f*x + e)^2*b*d^2 + 2*(-I*b*d^2*e + I*b*c*d*f)*(f*x + e)) 
*arctan2(cos(f*x + e), sin(f*x + e) + 1) - 2*(I*(f*x + e)^2*b*d^2 + 2*(-I* 
b*d^2*e + I*b*c*d*f)*(f*x + e))*arctan2(cos(f*x + e), -sin(f*x + e) + 1) - 
 4*(I*(f*x + e)*b*d^2 - I*b*d^2*e + I*b*c*d*f)*dilog(I*e^(I*f*x + I*e)) - 
4*(-I*(f*x + e)*b*d^2 + I*b*d^2*e - I*b*c*d*f)*dilog(-I*e^(I*f*x + I*e)) + 
 ((f*x + e)^2*b*d^2 - 2*(b*d^2*e - b*c*d*f)*(f*x + e))*log(cos(f*x + e)^2 
+ sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - ((f*x + e)^2*b*d^2 - 2*(b*d^2*e - 
 b*c*d*f)*(f*x + e))*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) 
+ 1))/f^2)/f
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (c+d x)^2 (a+b \sec (e+f x)) \, dx=\frac {-6 \left (\int \frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} x^{2}}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}d x \right ) b \,d^{2} f -12 \left (\int \frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} x}{\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1}d x \right ) b c d f -3 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) b \,c^{2}+3 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) b \,c^{2}+3 a \,c^{2} f x +3 a c d f \,x^{2}+a \,d^{2} f \,x^{3}+3 b c d f \,x^{2}+b \,d^{2} f \,x^{3}}{3 f} \] Input:

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

Output:

( - 6*int((tan((e + f*x)/2)**2*x**2)/(tan((e + f*x)/2)**2 - 1),x)*b*d**2*f 
 - 12*int((tan((e + f*x)/2)**2*x)/(tan((e + f*x)/2)**2 - 1),x)*b*c*d*f - 3 
*log(tan((e + f*x)/2) - 1)*b*c**2 + 3*log(tan((e + f*x)/2) + 1)*b*c**2 + 3 
*a*c**2*f*x + 3*a*c*d*f*x**2 + a*d**2*f*x**3 + 3*b*c*d*f*x**2 + b*d**2*f*x 
**3)/(3*f)