\(\int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx\) [11]

Optimal result

Integrand size = 20, antiderivative size = 152 \[ \int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx=\frac {i (c+d x)^3}{a f}+\frac {(c+d x)^4}{4 a d}-\frac {6 d (c+d x)^2 \log \left (1+e^{i (e+f x)}\right )}{a f^2}+\frac {12 i d^2 (c+d x) \operatorname {PolyLog}\left (2,-e^{i (e+f x)}\right )}{a f^3}-\frac {12 d^3 \operatorname {PolyLog}\left (3,-e^{i (e+f x)}\right )}{a f^4}-\frac {(c+d x)^3 \tan \left (\frac {e}{2}+\frac {f x}{2}\right )}{a f} \] Output:

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

Mathematica [A] (verified)

Time = 1.38 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.42 \[ \int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx=\frac {\cos \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \left (x \left (4 c^3+6 c^2 d x+4 c d^2 x^2+d^3 x^3\right ) \cos \left (\frac {1}{2} (e+f x)\right )+\frac {8 \cos \left (\frac {1}{2} (e+f x)\right ) \left (-\frac {i f^3 (c+d x)^3}{1+e^{i e}}-3 d f^2 (c+d x)^2 \log \left (1+e^{-i (e+f x)}\right )-6 i d^2 f (c+d x) \operatorname {PolyLog}\left (2,-e^{-i (e+f x)}\right )-6 d^3 \operatorname {PolyLog}\left (3,-e^{-i (e+f x)}\right )\right )}{f^4}-\frac {4 (c+d x)^3 \sec \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )}{f}\right )}{2 a (1+\sec (e+f x))} \] Input:

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

Output:

(Cos[(e + f*x)/2]*Sec[e + f*x]*(x*(4*c^3 + 6*c^2*d*x + 4*c*d^2*x^2 + d^3*x 
^3)*Cos[(e + f*x)/2] + (8*Cos[(e + f*x)/2]*(((-I)*f^3*(c + d*x)^3)/(1 + E^ 
(I*e)) - 3*d*f^2*(c + d*x)^2*Log[1 + E^((-I)*(e + f*x))] - (6*I)*d^2*f*(c 
+ d*x)*PolyLog[2, -E^((-I)*(e + f*x))] - 6*d^3*PolyLog[3, -E^((-I)*(e + f* 
x))]))/f^4 - (4*(c + d*x)^3*Sec[e/2]*Sin[(f*x)/2])/f))/(2*a*(1 + Sec[e + f 
*x]))
 

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 152, 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, 4679, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4679

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

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

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 4679

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(n_.)*((c_.) + (d_.)*(x_))^(m_.) 
, x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, 1/(Sin[e + f*x]^n/(b + a*Si 
n[e + f*x])^n), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && ILtQ[n, 0] && IGt 
Q[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. 416 vs. \(2 (138 ) = 276\).

Time = 0.18 (sec) , antiderivative size = 417, normalized size of antiderivative = 2.74

method	result	size
risch	\(\frac {d^{3} x^{4}}{4 a}+\frac {d^{2} c \,x^{3}}{a}+\frac {3 d \,c^{2} x^{2}}{2 a}+\frac {c^{3} x}{a}+\frac {c^{4}}{4 a d}+\frac {12 i d^{2} c e x}{a \,f^{2}}+\frac {12 i d^{3} \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (f x +e \right )}\right ) x}{a \,f^{3}}+\frac {2 i d^{3} x^{3}}{a f}-\frac {12 d^{2} e c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}+\frac {6 i d^{2} c \,x^{2}}{a f}+\frac {6 i d^{2} c \,e^{2}}{a \,f^{3}}+\frac {12 i d^{2} c \operatorname {polylog}\left (2, -{\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{3}}-\frac {6 i d^{3} e^{2} x}{a \,f^{3}}+\frac {6 d^{3} e^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{4}}-\frac {4 i d^{3} e^{3}}{a \,f^{4}}-\frac {12 d^{2} c \ln \left (1+{\mathrm e}^{i \left (f x +e \right )}\right ) x}{a \,f^{2}}-\frac {2 i \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )}{f a \left (1+{\mathrm e}^{i \left (f x +e \right )}\right )}-\frac {6 d \,c^{2} \ln \left (1+{\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{2}}+\frac {6 d \,c^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{2}}-\frac {6 d^{3} \ln \left (1+{\mathrm e}^{i \left (f x +e \right )}\right ) x^{2}}{a \,f^{2}}-\frac {12 d^{3} \operatorname {polylog}\left (3, -{\mathrm e}^{i \left (f x +e \right )}\right )}{a \,f^{4}}\)	\(417\)

Input:

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

Output:

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

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. 516 vs. \(2 (135) = 270\).

Time = 0.09 (sec) , antiderivative size = 516, normalized size of antiderivative = 3.39 \[ \int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx=\frac {d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2} + 4 \, c^{3} f^{4} x + {\left (d^{3} f^{4} x^{4} + 4 \, c d^{2} f^{4} x^{3} + 6 \, c^{2} d f^{4} x^{2} + 4 \, c^{3} f^{4} x\right )} \cos \left (f x + e\right ) - 24 \, {\left (i \, d^{3} f x + i \, c d^{2} f + {\left (i \, d^{3} f x + i \, c d^{2} f\right )} \cos \left (f x + e\right )\right )} {\rm Li}_2\left (-\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - 24 \, {\left (-i \, d^{3} f x - i \, c d^{2} f + {\left (-i \, d^{3} f x - i \, c d^{2} f\right )} \cos \left (f x + e\right )\right )} {\rm Li}_2\left (-\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 12 \, {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2} + {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right ) + 1\right ) - 12 \, {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2} + {\left (d^{3} f^{2} x^{2} + 2 \, c d^{2} f^{2} x + c^{2} d f^{2}\right )} \cos \left (f x + e\right )\right )} \log \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right ) + 1\right ) - 24 \, {\left (d^{3} \cos \left (f x + e\right ) + d^{3}\right )} {\rm polylog}\left (3, -\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right ) - 24 \, {\left (d^{3} \cos \left (f x + e\right ) + d^{3}\right )} {\rm polylog}\left (3, -\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right ) - 4 \, {\left (d^{3} f^{3} x^{3} + 3 \, c d^{2} f^{3} x^{2} + 3 \, c^{2} d f^{3} x + c^{3} f^{3}\right )} \sin \left (f x + e\right )}{4 \, {\left (a f^{4} \cos \left (f x + e\right ) + a f^{4}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

(Integral(c**3/(sec(e + f*x) + 1), x) + Integral(d**3*x**3/(sec(e + f*x) + 
 1), x) + Integral(3*c*d**2*x**2/(sec(e + f*x) + 1), x) + Integral(3*c**2* 
d*x/(sec(e + f*x) + 1), x))/a
 

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. 1285 vs. \(2 (135) = 270\).

Time = 0.25 (sec) , antiderivative size = 1285, normalized size of antiderivative = 8.45 \[ \int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/2*(6*c*d^2*e^2*(2*arctan(sin(f*x + e)/(cos(f*x + e) + 1))/(a*f^2) - sin( 
f*x + e)/(a*f^2*(cos(f*x + e) + 1))) - 6*c^2*d*e*(2*arctan(sin(f*x + e)/(c 
os(f*x + e) + 1))/(a*f) - sin(f*x + e)/(a*f*(cos(f*x + e) + 1))) - 6*((f*x 
 + e)^2*cos(f*x + e)^2 + (f*x + e)^2*sin(f*x + e)^2 + 2*(f*x + e)^2*cos(f* 
x + e) + (f*x + e)^2 - 2*(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*cos(f*x + e) 
 + 1)*log(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*cos(f*x + e) + 1) - 4*(f*x + 
 e)*sin(f*x + e))*c*d^2*e/(a*f^2*cos(f*x + e)^2 + a*f^2*sin(f*x + e)^2 + 2 
*a*f^2*cos(f*x + e) + a*f^2) + 2*c^3*(2*arctan(sin(f*x + e)/(cos(f*x + e) 
+ 1))/a - sin(f*x + e)/(a*(cos(f*x + e) + 1))) + 3*((f*x + e)^2*cos(f*x + 
e)^2 + (f*x + e)^2*sin(f*x + e)^2 + 2*(f*x + e)^2*cos(f*x + e) + (f*x + e) 
^2 - 2*(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*cos(f*x + e) + 1)*log(cos(f*x 
+ e)^2 + sin(f*x + e)^2 + 2*cos(f*x + e) + 1) - 4*(f*x + e)*sin(f*x + e))* 
c^2*d/(a*f*cos(f*x + e)^2 + a*f*sin(f*x + e)^2 + 2*a*f*cos(f*x + e) + a*f) 
 - 2*(I*(f*x + e)^4*d^3 + 6*I*(f*x + e)^2*d^3*e^2 - 4*I*(f*x + e)*d^3*e^3 
- 8*d^3*e^3 - 4*(I*d^3*e - I*c*d^2*f)*(f*x + e)^3 + 24*((f*x + e)^2*d^3 + 
d^3*e^2 - 2*(d^3*e - c*d^2*f)*(f*x + e) + ((f*x + e)^2*d^3 + d^3*e^2 - 2*( 
d^3*e - c*d^2*f)*(f*x + e))*cos(f*x + e) - (-I*(f*x + e)^2*d^3 - I*d^3*e^2 
 + 2*(I*d^3*e - I*c*d^2*f)*(f*x + e))*sin(f*x + e))*arctan2(sin(f*x + e), 
cos(f*x + e) + 1) + (I*(f*x + e)^4*d^3 - 4*(I*d^3*e - I*c*d^2*f + 2*d^3)*( 
f*x + e)^3 - 6*(-I*d^3*e^2 - 4*d^3*e + 4*c*d^2*f)*(f*x + e)^2 - 4*(I*d^...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(c+d x)^3}{a+a \sec (e+f x)} \, dx=\frac {12 \left (\int \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) x^{2}d x \right ) d^{3} f +24 \left (\int \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) x d x \right ) c \,d^{2} f +12 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) c^{2} d -4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c^{3} f -12 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c^{2} d f x -12 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c \,d^{2} f \,x^{2}-4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) d^{3} f \,x^{3}+4 c^{3} f^{2} x +6 c^{2} d \,f^{2} x^{2}+4 c \,d^{2} f^{2} x^{3}+d^{3} f^{2} x^{4}}{4 a \,f^{2}} \] Input:

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

Output:

(12*int(tan((e + f*x)/2)*x**2,x)*d**3*f + 24*int(tan((e + f*x)/2)*x,x)*c*d 
**2*f + 12*log(tan((e + f*x)/2)**2 + 1)*c**2*d - 4*tan((e + f*x)/2)*c**3*f 
 - 12*tan((e + f*x)/2)*c**2*d*f*x - 12*tan((e + f*x)/2)*c*d**2*f*x**2 - 4* 
tan((e + f*x)/2)*d**3*f*x**3 + 4*c**3*f**2*x + 6*c**2*d*f**2*x**2 + 4*c*d* 
*2*f**2*x**3 + d**3*f**2*x**4)/(4*a*f**2)