\(\int \frac {(c+d \sec (e+f x))^3}{a+b \cos (e+f x)} \, dx\) [10]

Optimal result

Integrand size = 25, antiderivative size = 170 \[ \int \frac {(c+d \sec (e+f x))^3}{a+b \cos (e+f x)} \, dx=\frac {2 (a c-b d)^3 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} f}+\frac {d^3 \text {arctanh}(\sin (e+f x))}{2 a f}+\frac {d \left (3 a^2 c^2-3 a b c d+b^2 d^2\right ) \text {arctanh}(\sin (e+f x))}{a^3 f}+\frac {d^2 (3 a c-b d) \tan (e+f x)}{a^2 f}+\frac {d^3 \sec (e+f x) \tan (e+f x)}{2 a f} \] Output:

2*(a*c-b*d)^3*arctan((a-b)^(1/2)*tan(1/2*f*x+1/2*e)/(a+b)^(1/2))/a^3/(a-b) 
^(1/2)/(a+b)^(1/2)/f+1/2*d^3*arctanh(sin(f*x+e))/a/f+d*(3*a^2*c^2-3*a*b*c* 
d+b^2*d^2)*arctanh(sin(f*x+e))/a^3/f+d^2*(3*a*c-b*d)*tan(f*x+e)/a^2/f+1/2* 
d^3*sec(f*x+e)*tan(f*x+e)/a/f
 

Mathematica [A] (verified)

Time = 3.54 (sec) , antiderivative size = 335, normalized size of antiderivative = 1.97 \[ \int \frac {(c+d \sec (e+f x))^3}{a+b \cos (e+f x)} \, dx=\frac {-\frac {8 (a c-b d)^3 \text {arctanh}\left (\frac {(a-b) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {-a^2+b^2}}\right )}{\sqrt {-a^2+b^2}}-2 d \left (-6 a b c d+2 b^2 d^2+a^2 \left (6 c^2+d^2\right )\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 d \left (-6 a b c d+2 b^2 d^2+a^2 \left (6 c^2+d^2\right )\right ) \log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+\frac {a^2 d^3}{\left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {4 a d^2 (3 a c-b d) \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )}-\frac {a^2 d^3}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2}+\frac {4 a d^2 (3 a c-b d) \sin \left (\frac {1}{2} (e+f x)\right )}{\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )}}{4 a^3 f} \] Input:

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

Output:

((-8*(a*c - b*d)^3*ArcTanh[((a - b)*Tan[(e + f*x)/2])/Sqrt[-a^2 + b^2]])/S 
qrt[-a^2 + b^2] - 2*d*(-6*a*b*c*d + 2*b^2*d^2 + a^2*(6*c^2 + d^2))*Log[Cos 
[(e + f*x)/2] - Sin[(e + f*x)/2]] + 2*d*(-6*a*b*c*d + 2*b^2*d^2 + a^2*(6*c 
^2 + d^2))*Log[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + (a^2*d^3)/(Cos[(e + 
f*x)/2] - Sin[(e + f*x)/2])^2 + (4*a*d^2*(3*a*c - b*d)*Sin[(e + f*x)/2])/( 
Cos[(e + f*x)/2] - Sin[(e + f*x)/2]) - (a^2*d^3)/(Cos[(e + f*x)/2] + Sin[( 
e + f*x)/2])^2 + (4*a*d^2*(3*a*c - b*d)*Sin[(e + f*x)/2])/(Cos[(e + f*x)/2 
] + Sin[(e + f*x)/2]))/(4*a^3*f)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3307, 3042, 3431, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3307

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3431

\(\displaystyle \int \left (\frac {(a c-b d)^3}{a^3 (a+b \cos (e+f x))}+\frac {d^2 (3 a c-b d) \sec ^2(e+f x)}{a^2}+\frac {d \left (3 a^2 c^2-3 a b c d+b^2 d^2\right ) \sec (e+f x)}{a^3}+\frac {d^3 \sec ^3(e+f x)}{a}\right )dx\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {2 (a c-b d)^3 \arctan \left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {a+b}}\right )}{a^3 f \sqrt {a-b} \sqrt {a+b}}+\frac {d^2 (3 a c-b d) \tan (e+f x)}{a^2 f}+\frac {d \left (3 a^2 c^2-3 a b c d+b^2 d^2\right ) \text {arctanh}(\sin (e+f x))}{a^3 f}+\frac {d^3 \text {arctanh}(\sin (e+f x))}{2 a f}+\frac {d^3 \tan (e+f x) \sec (e+f x)}{2 a f}\)

Input:

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

Output:

(2*(a*c - b*d)^3*ArcTan[(Sqrt[a - b]*Tan[(e + f*x)/2])/Sqrt[a + b]])/(a^3* 
Sqrt[a - b]*Sqrt[a + b]*f) + (d^3*ArcTanh[Sin[e + f*x]])/(2*a*f) + (d*(3*a 
^2*c^2 - 3*a*b*c*d + b^2*d^2)*ArcTanh[Sin[e + f*x]])/(a^3*f) + (d^2*(3*a*c 
 - b*d)*Tan[e + f*x])/(a^2*f) + (d^3*Sec[e + f*x]*Tan[e + 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 3307

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

rule 3431

Int[((g_.)*sin[(e_.) + (f_.)*(x_)])^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x 
_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Int[Exp 
andTrig[(g*sin[e + f*x])^p*(a + b*sin[e + f*x])^m*(c + d*sin[e + f*x])^n, x 
], x] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && NeQ[b*c - a*d, 0] && (Int 
egersQ[m, n] || IntegersQ[m, p] || IntegersQ[n, p]) && NeQ[p, 2]
 

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 291, normalized size of antiderivative = 1.71

method	result	size
derivativedivides	\(\frac {\frac {2 \left (a^{3} c^{3}-3 a^{2} b \,c^{2} d +3 a \,b^{2} c \,d^{2}-b^{3} d^{3}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {d^{3}}{2 a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {d \left (6 a^{2} c^{2}+d^{2} a^{2}-6 a b c d +2 b^{2} d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 a^{3}}-\frac {d^{2} \left (6 a c -a d -2 b d \right )}{2 a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {d^{3}}{2 a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {d \left (6 a^{2} c^{2}+d^{2} a^{2}-6 a b c d +2 b^{2} d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 a^{3}}-\frac {d^{2} \left (6 a c -a d -2 b d \right )}{2 a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}}{f}\)	\(291\)
default	\(\frac {\frac {2 \left (a^{3} c^{3}-3 a^{2} b \,c^{2} d +3 a \,b^{2} c \,d^{2}-b^{3} d^{3}\right ) \arctan \left (\frac {\left (a -b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {d^{3}}{2 a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )^{2}}+\frac {d \left (6 a^{2} c^{2}+d^{2} a^{2}-6 a b c d +2 b^{2} d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{2 a^{3}}-\frac {d^{2} \left (6 a c -a d -2 b d \right )}{2 a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}+\frac {d^{3}}{2 a \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )^{2}}-\frac {d \left (6 a^{2} c^{2}+d^{2} a^{2}-6 a b c d +2 b^{2} d^{2}\right ) \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{2 a^{3}}-\frac {d^{2} \left (6 a c -a d -2 b d \right )}{2 a^{2} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}}{f}\)	\(291\)
risch	\(-\frac {i d^{2} \left (a d \,{\mathrm e}^{3 i \left (f x +e \right )}-6 a c \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b d \,{\mathrm e}^{2 i \left (f x +e \right )}-d \,{\mathrm e}^{i \left (f x +e \right )} a -6 a c +2 b d \right )}{a^{2} f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}-\frac {3 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) c^{2}}{a f}-\frac {d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{2 a f}+\frac {3 d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) b c}{a^{2} f}-\frac {d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right ) b^{2}}{a^{3} f}+\frac {3 d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) c^{2}}{a f}+\frac {d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{2 a f}-\frac {3 d^{2} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) b c}{a^{2} f}+\frac {d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right ) b^{2}}{a^{3} f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) c^{3}}{\sqrt {-a^{2}+b^{2}}\, f}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b \,c^{2} d}{\sqrt {-a^{2}+b^{2}}\, f a}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2} c \,d^{2}}{\sqrt {-a^{2}+b^{2}}\, f \,a^{2}}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i a^{2}-i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{3} d^{3}}{\sqrt {-a^{2}+b^{2}}\, f \,a^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) c^{3}}{\sqrt {-a^{2}+b^{2}}\, f}-\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b \,c^{2} d}{\sqrt {-a^{2}+b^{2}}\, f a}+\frac {3 \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{2} c \,d^{2}}{\sqrt {-a^{2}+b^{2}}\, f \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {-i a^{2}+i b^{2}+a \sqrt {-a^{2}+b^{2}}}{b \sqrt {-a^{2}+b^{2}}}\right ) b^{3} d^{3}}{\sqrt {-a^{2}+b^{2}}\, f \,a^{3}}\)	\(892\)

Input:

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

Output:

1/f*(2*(a^3*c^3-3*a^2*b*c^2*d+3*a*b^2*c*d^2-b^3*d^3)/a^3/((a-b)*(a+b))^(1/ 
2)*arctan((a-b)*tan(1/2*f*x+1/2*e)/((a-b)*(a+b))^(1/2))-1/2*d^3/a/(tan(1/2 
*f*x+1/2*e)+1)^2+1/2*d*(6*a^2*c^2+a^2*d^2-6*a*b*c*d+2*b^2*d^2)/a^3*ln(tan( 
1/2*f*x+1/2*e)+1)-1/2*d^2*(6*a*c-a*d-2*b*d)/a^2/(tan(1/2*f*x+1/2*e)+1)+1/2 
*d^3/a/(tan(1/2*f*x+1/2*e)-1)^2-1/2*d*(6*a^2*c^2+a^2*d^2-6*a*b*c*d+2*b^2*d 
^2)/a^3*ln(tan(1/2*f*x+1/2*e)-1)-1/2*d^2*(6*a*c-a*d-2*b*d)/a^2/(tan(1/2*f* 
x+1/2*e)-1))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 339 vs. \(2 (157) = 314\).

Time = 21.48 (sec) , antiderivative size = 747, normalized size of antiderivative = 4.39 \[ \int \frac {(c+d \sec (e+f x))^3}{a+b \cos (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/4*(2*(a^3*c^3 - 3*a^2*b*c^2*d + 3*a*b^2*c*d^2 - b^3*d^3)*sqrt(-a^2 + b^ 
2)*cos(f*x + e)^2*log((2*a*b*cos(f*x + e) + (2*a^2 - b^2)*cos(f*x + e)^2 - 
 2*sqrt(-a^2 + b^2)*(a*cos(f*x + e) + b)*sin(f*x + e) - a^2 + 2*b^2)/(b^2* 
cos(f*x + e)^2 + 2*a*b*cos(f*x + e) + a^2)) + (6*(a^4 - a^2*b^2)*c^2*d - 6 
*(a^3*b - a*b^3)*c*d^2 + (a^4 + a^2*b^2 - 2*b^4)*d^3)*cos(f*x + e)^2*log(s 
in(f*x + e) + 1) - (6*(a^4 - a^2*b^2)*c^2*d - 6*(a^3*b - a*b^3)*c*d^2 + (a 
^4 + a^2*b^2 - 2*b^4)*d^3)*cos(f*x + e)^2*log(-sin(f*x + e) + 1) + 2*((a^4 
 - a^2*b^2)*d^3 + 2*(3*(a^4 - a^2*b^2)*c*d^2 - (a^3*b - a*b^3)*d^3)*cos(f* 
x + e))*sin(f*x + e))/((a^5 - a^3*b^2)*f*cos(f*x + e)^2), 1/4*(4*(a^3*c^3 
- 3*a^2*b*c^2*d + 3*a*b^2*c*d^2 - b^3*d^3)*sqrt(a^2 - b^2)*arctan(-(a*cos( 
f*x + e) + b)/(sqrt(a^2 - b^2)*sin(f*x + e)))*cos(f*x + e)^2 + (6*(a^4 - a 
^2*b^2)*c^2*d - 6*(a^3*b - a*b^3)*c*d^2 + (a^4 + a^2*b^2 - 2*b^4)*d^3)*cos 
(f*x + e)^2*log(sin(f*x + e) + 1) - (6*(a^4 - a^2*b^2)*c^2*d - 6*(a^3*b - 
a*b^3)*c*d^2 + (a^4 + a^2*b^2 - 2*b^4)*d^3)*cos(f*x + e)^2*log(-sin(f*x + 
e) + 1) + 2*((a^4 - a^2*b^2)*d^3 + 2*(3*(a^4 - a^2*b^2)*c*d^2 - (a^3*b - a 
*b^3)*d^3)*cos(f*x + e))*sin(f*x + e))/((a^5 - a^3*b^2)*f*cos(f*x + e)^2)]
 

Sympy [F]

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

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

Output:

Integral((c + d*sec(e + f*x))**3/(a + b*cos(e + f*x)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {(c+d \sec (e+f x))^3}{a+b \cos (e+f x)} \, dx=\text {Exception raised: ValueError} \] Input:

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?` f 
or more de
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 340 vs. \(2 (157) = 314\).

Time = 0.19 (sec) , antiderivative size = 340, normalized size of antiderivative = 2.00 \[ \int \frac {(c+d \sec (e+f x))^3}{a+b \cos (e+f x)} \, dx=\frac {\frac {{\left (6 \, a^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3} + 2 \, b^{2} d^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a^{3}} - \frac {{\left (6 \, a^{2} c^{2} d - 6 \, a b c d^{2} + a^{2} d^{3} + 2 \, b^{2} d^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a^{3}} - \frac {4 \, {\left (a^{3} c^{3} - 3 \, a^{2} b c^{2} d + 3 \, a b^{2} c d^{2} - b^{3} d^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{3}} - \frac {2 \, {\left (6 \, a c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - a d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, b d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 6 \, a c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - a d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, b d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )}^{2} a^{2}}}{2 \, f} \] Input:

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

Output:

1/2*((6*a^2*c^2*d - 6*a*b*c*d^2 + a^2*d^3 + 2*b^2*d^3)*log(abs(tan(1/2*f*x 
 + 1/2*e) + 1))/a^3 - (6*a^2*c^2*d - 6*a*b*c*d^2 + a^2*d^3 + 2*b^2*d^3)*lo 
g(abs(tan(1/2*f*x + 1/2*e) - 1))/a^3 - 4*(a^3*c^3 - 3*a^2*b*c^2*d + 3*a*b^ 
2*c*d^2 - b^3*d^3)*(pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(-2*a + 2*b) + arc 
tan(-(a*tan(1/2*f*x + 1/2*e) - b*tan(1/2*f*x + 1/2*e))/sqrt(a^2 - b^2)))/( 
sqrt(a^2 - b^2)*a^3) - 2*(6*a*c*d^2*tan(1/2*f*x + 1/2*e)^3 - a*d^3*tan(1/2 
*f*x + 1/2*e)^3 - 2*b*d^3*tan(1/2*f*x + 1/2*e)^3 - 6*a*c*d^2*tan(1/2*f*x + 
 1/2*e) - a*d^3*tan(1/2*f*x + 1/2*e) + 2*b*d^3*tan(1/2*f*x + 1/2*e))/((tan 
(1/2*f*x + 1/2*e)^2 - 1)^2*a^2))/f
 

Mupad [B] (verification not implemented)

Time = 5.70 (sec) , antiderivative size = 6735, normalized size of antiderivative = 39.62 \[ \int \frac {(c+d \sec (e+f x))^3}{a+b \cos (e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

((tan(e/2 + (f*x)/2)*(a*d^3 - 2*b*d^3 + 6*a*c*d^2))/a^2 + (tan(e/2 + (f*x) 
/2)^3*(a*d^3 + 2*b*d^3 - 6*a*c*d^2))/a^2)/(f*(tan(e/2 + (f*x)/2)^4 - 2*tan 
(e/2 + (f*x)/2)^2 + 1)) + (atan(((((8*tan(e/2 + (f*x)/2)*(4*a^7*c^6 + a^7* 
d^6 - 8*b^7*d^6 - 4*a^6*b*c^6 + 16*a*b^6*d^6 - 3*a^6*b*d^6 - 16*a^2*b^5*d^ 
6 + 16*a^3*b^4*d^6 - 13*a^4*b^3*d^6 + 7*a^5*b^2*d^6 + 12*a^7*c^2*d^4 + 36* 
a^7*c^4*d^2 - 96*a^2*b^5*c*d^5 + 84*a^3*b^4*c*d^5 - 60*a^4*b^3*c*d^5 + 36* 
a^5*b^2*c*d^5 + 24*a^5*b^2*c^5*d - 36*a^6*b*c^2*d^4 - 72*a^6*b*c^3*d^3 - 1 
08*a^6*b*c^4*d^2 - 120*a^2*b^5*c^2*d^4 + 240*a^3*b^4*c^2*d^4 + 152*a^3*b^4 
*c^3*d^3 - 192*a^4*b^3*c^2*d^4 - 296*a^4*b^3*c^3*d^3 - 96*a^4*b^3*c^4*d^2 
+ 96*a^5*b^2*c^2*d^4 + 216*a^5*b^2*c^3*d^3 + 168*a^5*b^2*c^4*d^2 + 48*a*b^ 
6*c*d^5 - 12*a^6*b*c*d^5 - 24*a^6*b*c^5*d))/a^4 + (((8*(4*a^10*c^3 + 2*a^1 
0*d^3 - 8*a^9*b*c^3 - 2*a^9*b*d^3 + 12*a^10*c^2*d + 4*a^8*b^2*c^3 + 4*a^6* 
b^4*d^3 - 6*a^7*b^3*d^3 + 2*a^8*b^2*d^3 - 12*a^7*b^3*c*d^2 + 24*a^8*b^2*c* 
d^2 + 12*a^8*b^2*c^2*d - 12*a^9*b*c*d^2 - 24*a^9*b*c^2*d))/a^6 + (8*tan(e/ 
2 + (f*x)/2)*(8*a^8*b + 8*a^6*b^3 - 16*a^7*b^2)*(a^2*(3*c^2*d + d^3/2) + b 
^2*d^3 - 3*a*b*c*d^2))/a^7)*(a^2*(3*c^2*d + d^3/2) + b^2*d^3 - 3*a*b*c*d^2 
))/a^3)*(a^2*(3*c^2*d + d^3/2) + b^2*d^3 - 3*a*b*c*d^2)*1i)/a^3 + (((8*tan 
(e/2 + (f*x)/2)*(4*a^7*c^6 + a^7*d^6 - 8*b^7*d^6 - 4*a^6*b*c^6 + 16*a*b^6* 
d^6 - 3*a^6*b*d^6 - 16*a^2*b^5*d^6 + 16*a^3*b^4*d^6 - 13*a^4*b^3*d^6 + 7*a 
^5*b^2*d^6 + 12*a^7*c^2*d^4 + 36*a^7*c^4*d^2 - 96*a^2*b^5*c*d^5 + 84*a^...
 

Reduce [B] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 1631, normalized size of antiderivative = 9.59 \[ \int \frac {(c+d \sec (e+f x))^3}{a+b \cos (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

(4*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt(a 
**2 - b**2))*cos(e + f*x)*sin(e + f*x)**2*a**3*c**3 - 12*sqrt(a**2 - b**2) 
*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt(a**2 - b**2))*cos(e + 
 f*x)*sin(e + f*x)**2*a**2*b*c**2*d + 12*sqrt(a**2 - b**2)*atan((tan((e + 
f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt(a**2 - b**2))*cos(e + f*x)*sin(e + f* 
x)**2*a*b**2*c*d**2 - 4*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a - tan(( 
e + f*x)/2)*b)/sqrt(a**2 - b**2))*cos(e + f*x)*sin(e + f*x)**2*b**3*d**3 - 
 4*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt(a 
**2 - b**2))*cos(e + f*x)*a**3*c**3 + 12*sqrt(a**2 - b**2)*atan((tan((e + 
f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt(a**2 - b**2))*cos(e + f*x)*a**2*b*c** 
2*d - 12*sqrt(a**2 - b**2)*atan((tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/ 
sqrt(a**2 - b**2))*cos(e + f*x)*a*b**2*c*d**2 + 4*sqrt(a**2 - b**2)*atan(( 
tan((e + f*x)/2)*a - tan((e + f*x)/2)*b)/sqrt(a**2 - b**2))*cos(e + f*x)*b 
**3*d**3 + 2*cos(e + f*x)**2*sin(e + f*x)*a**3*b*d**3 - 2*cos(e + f*x)**2* 
sin(e + f*x)*a*b**3*d**3 - 6*cos(e + f*x)*log(tan((e + f*x)/2) - 1)*sin(e 
+ f*x)**2*a**4*c**2*d - cos(e + f*x)*log(tan((e + f*x)/2) - 1)*sin(e + f*x 
)**2*a**4*d**3 + 6*cos(e + f*x)*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**2* 
a**3*b*c*d**2 + 6*cos(e + f*x)*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**2*a 
**2*b**2*c**2*d - cos(e + f*x)*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**2*a 
**2*b**2*d**3 - 6*cos(e + f*x)*log(tan((e + f*x)/2) - 1)*sin(e + f*x)**...