\(\int \frac {\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [122]

Optimal result

Integrand size = 33, antiderivative size = 133 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {(2 A+3 C) \text {arctanh}(\sin (c+d x))}{2 a d}+\frac {(3 A+4 C) \tan (c+d x)}{a d}-\frac {(2 A+3 C) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{d (a+a \sec (c+d x))}+\frac {(3 A+4 C) \tan ^3(c+d x)}{3 a d} \] Output:

-1/2*(2*A+3*C)*arctanh(sin(d*x+c))/a/d+(3*A+4*C)*tan(d*x+c)/a/d-1/2*(2*A+3 
*C)*sec(d*x+c)*tan(d*x+c)/a/d-(A+C)*sec(d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x 
+c))+1/3*(3*A+4*C)*tan(d*x+c)^3/a/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(1090\) vs. \(2(133)=266\).

Time = 7.44 (sec) , antiderivative size = 1090, normalized size of antiderivative = 8.20 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx =\text {Too large to display} \] Input:

Integrate[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]
 

Output:

(2*(2*A + 3*C)*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*Log[Cos[c/2 + (d*x)/2] - 
Sin[c/2 + (d*x)/2]]*(A + C*Sec[c + d*x]^2))/(d*(A + 2*C + A*Cos[2*c + 2*d* 
x])*(a + a*Sec[c + d*x])) - (2*(2*A + 3*C)*Cos[c/2 + (d*x)/2]^2*Cos[c + d* 
x]*Log[Cos[c/2 + (d*x)/2] + Sin[c/2 + (d*x)/2]]*(A + C*Sec[c + d*x]^2))/(d 
*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (4*Cos[c/2 + (d*x) 
/2]*Cos[c + d*x]*Sec[c/2]*(A + C*Sec[c + d*x]^2)*(A*Sin[(d*x)/2] + C*Sin[( 
d*x)/2]))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])) + (2*C*C 
os[c/2 + (d*x)/2]^2*Cos[c + d*x]*(A + C*Sec[c + d*x]^2)*Sin[(d*x)/2])/(3*d 
*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] - Sin[c/2]) 
*(Cos[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^3) - (2*Cos[c/2 + (d*x)/2]^2*Co 
s[c + d*x]*(A + C*Sec[c + d*x]^2)*(C*Cos[c/2] - 2*C*Sin[c/2]))/(3*d*(A + 2 
*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] - Sin[c/2])*(Cos[c 
/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])^2) + (4*Cos[c/2 + (d*x)/2]^2*Cos[c + d 
*x]*(A + C*Sec[c + d*x]^2)*(3*A*Sin[(d*x)/2] + 5*C*Sin[(d*x)/2]))/(3*d*(A 
+ 2*C + A*Cos[2*c + 2*d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] - Sin[c/2])*(Co 
s[c/2 + (d*x)/2] - Sin[c/2 + (d*x)/2])) + (2*C*Cos[c/2 + (d*x)/2]^2*Cos[c 
+ d*x]*(A + C*Sec[c + d*x]^2)*Sin[(d*x)/2])/(3*d*(A + 2*C + A*Cos[2*c + 2* 
d*x])*(a + a*Sec[c + d*x])*(Cos[c/2] + Sin[c/2])*(Cos[c/2 + (d*x)/2] + Sin 
[c/2 + (d*x)/2])^3) + (2*Cos[c/2 + (d*x)/2]^2*Cos[c + d*x]*(A + C*Sec[c + 
d*x]^2)*(C*Cos[c/2] + 2*C*Sin[c/2]))/(3*d*(A + 2*C + A*Cos[2*c + 2*d*x]...
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.88, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.303, Rules used = {3042, 4573, 3042, 4274, 3042, 4254, 2009, 4255, 3042, 4257}

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.

\(\displaystyle \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a \sec (c+d x)+a} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4573

\(\displaystyle -\frac {\int \sec ^3(c+d x) (a (2 A+3 C)-a (3 A+4 C) \sec (c+d x))dx}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (a (2 A+3 C)-a (3 A+4 C) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 4274

\(\displaystyle -\frac {a (2 A+3 C) \int \sec ^3(c+d x)dx-a (3 A+4 C) \int \sec ^4(c+d x)dx}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a (2 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx-a (3 A+4 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 4254

\(\displaystyle -\frac {\frac {a (3 A+4 C) \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}+a (2 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {a (2 A+3 C) \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {a (3 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 4255

\(\displaystyle -\frac {a (2 A+3 C) \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a (3 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 3042

\(\displaystyle -\frac {a (2 A+3 C) \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a (3 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\)

\(\Big \downarrow \) 4257

\(\displaystyle -\frac {a (2 A+3 C) \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {a (3 A+4 C) \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}}{a^2}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{d (a \sec (c+d x)+a)}\)

Input:

Int[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]
 

Output:

-(((A + C)*Sec[c + d*x]^3*Tan[c + d*x])/(d*(a + a*Sec[c + d*x]))) - (a*(2* 
A + 3*C)*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x])/(2*d)) 
 + (a*(3*A + 4*C)*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d)/a^2
 

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 4254

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1)   Subst[Int[Exp 
andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, 
 d}, x] && IGtQ[n/2, 0]
 

rule 4255

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) 
  Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] 
&& IntegerQ[2*n]
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4274

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[a   Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d   In 
t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
 

rule 4573

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_. 
))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a) 
*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*(2*m 
+ 1))), x] + Simp[1/(a*b*(2*m + 1))   Int[(a + b*Csc[e + f*x])^(m + 1)*(d*C 
sc[e + f*x])^n*Simp[b*C*n + A*b*(2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - 
n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x] && EqQ[ 
a^2 - b^2, 0] && LtQ[m, -2^(-1)]
 

Maple [A] (verified)

Time = 0.42 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.24

method	result	size
parallelrisch	\(\frac {3 \left (A +\frac {3 C}{2}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-3 \left (A +\frac {3 C}{2}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+6 \left (\frac {\left (A +\frac {4 C}{3}\right ) \cos \left (3 d x +3 c \right )}{3}+\frac {\left (A +\frac {7 C}{6}\right ) \cos \left (2 d x +2 c \right )}{3}+\left (A +\frac {11 C}{9}\right ) \cos \left (d x +c \right )+\frac {A}{3}+\frac {11 C}{18}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(165\)
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (A +\frac {3 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {\frac {5 C}{2}+A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-\frac {3 C}{2}-A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\frac {5 C}{2}+A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d a}\)	\(172\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A +\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}+\left (A +\frac {3 C}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\frac {\frac {5 C}{2}+A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}-\frac {C}{3 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {C}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\left (-\frac {3 C}{2}-A \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-\frac {\frac {5 C}{2}+A}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}}{d a}\)	\(172\)
norman	\(\frac {\frac {\left (A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a d}+\frac {\left (3 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {3 \left (2 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{a d}-\frac {\left (30 A +37 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}+\frac {\left (36 A +49 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}+\frac {\left (2 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}-\frac {\left (2 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\)	\(195\)
risch	\(\frac {i \left (6 A \,{\mathrm e}^{6 i \left (d x +c \right )}+9 C \,{\mathrm e}^{6 i \left (d x +c \right )}+6 A \,{\mathrm e}^{5 i \left (d x +c \right )}+9 C \,{\mathrm e}^{5 i \left (d x +c \right )}+24 A \,{\mathrm e}^{4 i \left (d x +c \right )}+24 C \,{\mathrm e}^{4 i \left (d x +c \right )}+12 A \,{\mathrm e}^{3 i \left (d x +c \right )}+24 C \,{\mathrm e}^{3 i \left (d x +c \right )}+30 A \,{\mathrm e}^{2 i \left (d x +c \right )}+39 C \,{\mathrm e}^{2 i \left (d x +c \right )}+6 A \,{\mathrm e}^{i \left (d x +c \right )}+7 C \,{\mathrm e}^{i \left (d x +c \right )}+12 A +16 C \right )}{3 d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{a d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{a d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 a d}\)	\(275\)

Input:

int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBO 
SE)
 

Output:

3*((A+3/2*C)*(1/3*cos(3*d*x+3*c)+cos(d*x+c))*ln(tan(1/2*d*x+1/2*c)-1)-(A+3 
/2*C)*(1/3*cos(3*d*x+3*c)+cos(d*x+c))*ln(tan(1/2*d*x+1/2*c)+1)+2*(1/3*(A+4 
/3*C)*cos(3*d*x+3*c)+1/3*(A+7/6*C)*cos(2*d*x+2*c)+(A+11/9*C)*cos(d*x+c)+1/ 
3*A+11/18*C)*tan(1/2*d*x+1/2*c))/a/d/(cos(3*d*x+3*c)+3*cos(d*x+c))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.29 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {3 \, {\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left ({\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{4} + {\left (2 \, A + 3 \, C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (6 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} - C \cos \left (d x + c\right ) + 2 \, C\right )} \sin \left (d x + c\right )}{12 \, {\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \] Input:

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="f 
ricas")
 

Output:

-1/12*(3*((2*A + 3*C)*cos(d*x + c)^4 + (2*A + 3*C)*cos(d*x + c)^3)*log(sin 
(d*x + c) + 1) - 3*((2*A + 3*C)*cos(d*x + c)^4 + (2*A + 3*C)*cos(d*x + c)^ 
3)*log(-sin(d*x + c) + 1) - 2*(4*(3*A + 4*C)*cos(d*x + c)^3 + (6*A + 7*C)* 
cos(d*x + c)^2 - C*cos(d*x + c) + 2*C)*sin(d*x + c))/(a*d*cos(d*x + c)^4 + 
 a*d*cos(d*x + c)^3)
 

Sympy [F]

\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \sec ^{5}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \] Input:

integrate(sec(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)
 

Output:

(Integral(A*sec(c + d*x)**3/(sec(c + d*x) + 1), x) + Integral(C*sec(c + d* 
x)**5/(sec(c + d*x) + 1), x))/a
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (127) = 254\).

Time = 0.04 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.44 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {C {\left (\frac {2 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a - \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {9 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {6 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 6 \, A {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a - \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \] Input:

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="m 
axima")
 

Output:

1/6*(C*(2*(9*sin(d*x + c)/(cos(d*x + c) + 1) - 16*sin(d*x + c)^3/(cos(d*x 
+ c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a - 3*a*sin(d*x + c 
)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - a*sin 
(d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) 
+ 1)/a + 9*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a + 6*sin(d*x + c)/(a* 
(cos(d*x + c) + 1))) - 6*A*(log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a - l 
og(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a - 2*sin(d*x + c)/((a - a*sin(d*x 
 + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) - sin(d*x + c)/(a*(cos(d 
*x + c) + 1))))/d
 

Giac [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.39 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {\frac {3 \, {\left (2 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {3 \, {\left (2 \, A + 3 \, C\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {6 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 16 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} a}}{6 \, d} \] Input:

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="g 
iac")
 

Output:

-1/6*(3*(2*A + 3*C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 3*(2*A + 3*C)*l 
og(abs(tan(1/2*d*x + 1/2*c) - 1))/a - 6*(A*tan(1/2*d*x + 1/2*c) + C*tan(1/ 
2*d*x + 1/2*c))/a + 2*(6*A*tan(1/2*d*x + 1/2*c)^5 + 15*C*tan(1/2*d*x + 1/2 
*c)^5 - 12*A*tan(1/2*d*x + 1/2*c)^3 - 16*C*tan(1/2*d*x + 1/2*c)^3 + 6*A*ta 
n(1/2*d*x + 1/2*c) + 9*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 - 
1)^3*a))/d
 

Mupad [B] (verification not implemented)

Time = 12.98 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\left (2\,A+5\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-4\,A-\frac {16\,C}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A+3\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (-a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+\frac {3\,C}{2}\right )}{a\,d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A+C\right )}{a\,d} \] Input:

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + a/cos(c + d*x))),x)
 

Output:

(tan(c/2 + (d*x)/2)^5*(2*A + 5*C) - tan(c/2 + (d*x)/2)^3*(4*A + (16*C)/3) 
+ tan(c/2 + (d*x)/2)*(2*A + 3*C))/(d*(a - 3*a*tan(c/2 + (d*x)/2)^2 + 3*a*t 
an(c/2 + (d*x)/2)^4 - a*tan(c/2 + (d*x)/2)^6)) - (2*atanh(tan(c/2 + (d*x)/ 
2))*(A + (3*C)/2))/(a*d) + (tan(c/2 + (d*x)/2)*(A + C))/(a*d)
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.73 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} a +9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{3} c -6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) a -9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right ) c -6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} a -9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{3} c +6 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) a +9 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right ) c +6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a +9 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} c -6 \cos \left (d x +c \right ) a -6 \cos \left (d x +c \right ) c +12 \sin \left (d x +c \right )^{4} a +16 \sin \left (d x +c \right )^{4} c -18 \sin \left (d x +c \right )^{2} a -24 \sin \left (d x +c \right )^{2} c +6 a +6 c}{6 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x)
 

Output:

(6*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*a + 9*cos(c + d* 
x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**3*c - 6*cos(c + d*x)*log(tan((c 
 + d*x)/2) - 1)*sin(c + d*x)*a - 9*cos(c + d*x)*log(tan((c + d*x)/2) - 1)* 
sin(c + d*x)*c - 6*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3* 
a - 9*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**3*c + 6*cos(c + 
 d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)*a + 9*cos(c + d*x)*log(tan((c 
 + d*x)/2) + 1)*sin(c + d*x)*c + 6*cos(c + d*x)*sin(c + d*x)**2*a + 9*cos( 
c + d*x)*sin(c + d*x)**2*c - 6*cos(c + d*x)*a - 6*cos(c + d*x)*c + 12*sin( 
c + d*x)**4*a + 16*sin(c + d*x)**4*c - 18*sin(c + d*x)**2*a - 24*sin(c + d 
*x)**2*c + 6*a + 6*c)/(6*cos(c + d*x)*sin(c + d*x)*a*d*(sin(c + d*x)**2 - 
1))