\(\int \sec (c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\) [94]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [A] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [A] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [B] (verification not implemented)
Reduce [B] (verification not implemented)

Optimal result

Integrand size = 31, antiderivative size = 132 \[ \int \sec (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (12 A+7 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 (12 A+7 C) \tan (c+d x)}{6 d}+\frac {a^2 (12 A+7 C) \sec (c+d x) \tan (c+d x)}{24 d}-\frac {C (a+a \sec (c+d x))^2 \tan (c+d x)}{12 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{4 a d} \] Output:

1/8*a^2*(12*A+7*C)*arctanh(sin(d*x+c))/d+1/6*a^2*(12*A+7*C)*tan(d*x+c)/d+1 
/24*a^2*(12*A+7*C)*sec(d*x+c)*tan(d*x+c)/d-1/12*C*(a+a*sec(d*x+c))^2*tan(d 
*x+c)/d+1/4*C*(a+a*sec(d*x+c))^3*tan(d*x+c)/a/d
 

Mathematica [A] (verified)

Time = 1.07 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.66 \[ \int \sec (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \left (24 A \coth ^{-1}(\sin (c+d x))+3 (4 A+7 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (48 (A+C)+3 (4 A+7 C) \sec (c+d x)+6 C \sec ^3(c+d x)+16 C \tan ^2(c+d x)\right )\right )}{24 d} \] Input:

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

Output:

(a^2*(24*A*ArcCoth[Sin[c + d*x]] + 3*(4*A + 7*C)*ArcTanh[Sin[c + d*x]] + T 
an[c + d*x]*(48*(A + C) + 3*(4*A + 7*C)*Sec[c + d*x] + 6*C*Sec[c + d*x]^3 
+ 16*C*Tan[c + d*x]^2)))/(24*d)
 

Rubi [A] (verified)

Time = 0.85 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.99, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4571, 3042, 4489, 3042, 4275, 3042, 4254, 24, 4534, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4571

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4489

\(\displaystyle \frac {\frac {1}{3} a (12 A+7 C) \int \sec (c+d x) (\sec (c+d x) a+a)^2dx-\frac {a C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} a (12 A+7 C) \int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2dx-\frac {a C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d}\)

\(\Big \downarrow \) 4275

\(\displaystyle \frac {\frac {1}{3} a (12 A+7 C) \left (2 a^2 \int \sec ^2(c+d x)dx+\int \sec (c+d x) \left (\sec ^2(c+d x) a^2+a^2\right )dx\right )-\frac {a C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} a (12 A+7 C) \left (2 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx\right )-\frac {a C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {\frac {1}{3} a (12 A+7 C) \left (\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx-\frac {2 a^2 \int 1d(-\tan (c+d x))}{d}\right )-\frac {a C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {\frac {1}{3} a (12 A+7 C) \left (\int \csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )^2 a^2+a^2\right )dx+\frac {2 a^2 \tan (c+d x)}{d}\right )-\frac {a C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d}\)

\(\Big \downarrow \) 4534

\(\displaystyle \frac {\frac {1}{3} a (12 A+7 C) \left (\frac {3}{2} a^2 \int \sec (c+d x)dx+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {a C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} a (12 A+7 C) \left (\frac {3}{2} a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {a C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\frac {1}{3} a (12 A+7 C) \left (\frac {3 a^2 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {2 a^2 \tan (c+d x)}{d}+\frac {a^2 \tan (c+d x) \sec (c+d x)}{2 d}\right )-\frac {a C \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}}{4 a}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{4 a d}\)

Input:

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

Output:

(C*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(4*a*d) + (-1/3*(a*C*(a + a*Sec[c 
+ d*x])^2*Tan[c + d*x])/d + (a*(12*A + 7*C)*((3*a^2*ArcTanh[Sin[c + d*x]]) 
/(2*d) + (2*a^2*Tan[c + d*x])/d + (a^2*Sec[c + d*x]*Tan[c + d*x])/(2*d)))/ 
3)/(4*a)
 

Defintions of rubi rules used

rule 24
Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]
 

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 4257
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4275
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_))^2, x_Symbol] :> Simp[2*a*(b/d)   Int[(d*Csc[e + f*x])^(n + 1), x], x] 
 + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b, d, 
 e, f, n}, x]
 

rule 4489
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs 
c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*Cot[e + f*x]*(( 
a + b*Csc[e + f*x])^m/(f*(m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(b*(m + 
 1))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m, x], x] /; FreeQ[{a, b, A, B 
, e, f, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b 
*(m + 1), 0] &&  !LtQ[m, -2^(-1)]
 

rule 4534
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) 
+ (A_)), x_Symbol] :> Simp[(-C)*Cot[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1) 
)), x] + Simp[(C*m + A*(m + 1))/(m + 1)   Int[(b*Csc[e + f*x])^m, x], x] /; 
 FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]
 

rule 4571
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[ 
(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Cot[e + f*x] 
*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Simp[1/(b*(m + 2))   In 
t[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) - a*C* 
Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 
Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.24

method result size
parts \(\frac {\left (a^{2} A +C \,a^{2}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {2 C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {2 a^{2} A \tan \left (d x +c \right )}{d}\) \(164\)
norman \(\frac {\frac {5 a^{2} \left (4 A +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {11 a^{2} \left (12 A +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {a^{2} \left (12 A +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}-\frac {a^{2} \left (156 A +83 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {a^{2} \left (12 A +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{2} \left (12 A +7 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) \(175\)
parallelrisch \(\frac {4 a^{2} \left (-\frac {3 \left (A +\frac {7 C}{12}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {3 \left (A +\frac {7 C}{12}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\left (A +\frac {4 C}{3}\right ) \sin \left (2 d x +2 c \right )+\frac {\left (A +\frac {7 C}{4}\right ) \sin \left (3 d x +3 c \right )}{4}+\left (\frac {A}{2}+\frac {C}{3}\right ) \sin \left (4 d x +4 c \right )+\frac {\sin \left (d x +c \right ) \left (A +\frac {15 C}{4}\right )}{4}\right )}{d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) \(177\)
derivativedivides \(\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a^{2} A \tan \left (d x +c \right )-2 C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{2} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) \(182\)
default \(\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a^{2} A \tan \left (d x +c \right )-2 C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{2} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) \(182\)
risch \(-\frac {i a^{2} \left (12 A \,{\mathrm e}^{7 i \left (d x +c \right )}+21 C \,{\mathrm e}^{7 i \left (d x +c \right )}-48 A \,{\mathrm e}^{6 i \left (d x +c \right )}+12 A \,{\mathrm e}^{5 i \left (d x +c \right )}+45 C \,{\mathrm e}^{5 i \left (d x +c \right )}-144 A \,{\mathrm e}^{4 i \left (d x +c \right )}-96 C \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{3 i \left (d x +c \right )}-45 C \,{\mathrm e}^{3 i \left (d x +c \right )}-144 A \,{\mathrm e}^{2 i \left (d x +c \right )}-128 C \,{\mathrm e}^{2 i \left (d x +c \right )}-12 A \,{\mathrm e}^{i \left (d x +c \right )}-21 C \,{\mathrm e}^{i \left (d x +c \right )}-48 A -32 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\) \(275\)

Input:

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

Output:

(A*a^2+C*a^2)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+ 
C*a^2/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+ 
tan(d*x+c)))+a^2*A/d*ln(sec(d*x+c)+tan(d*x+c))-2*C*a^2/d*(-2/3-1/3*sec(d*x 
+c)^2)*tan(d*x+c)+2*a^2*A/d*tan(d*x+c)
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.07 \[ \int \sec (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (12 \, A + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (12 \, A + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (3 \, A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 7 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 16 \, C a^{2} \cos \left (d x + c\right ) + 6 \, C a^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:

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

Output:

1/48*(3*(12*A + 7*C)*a^2*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(12*A + 
7*C)*a^2*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(16*(3*A + 2*C)*a^2*cos 
(d*x + c)^3 + 3*(4*A + 7*C)*a^2*cos(d*x + c)^2 + 16*C*a^2*cos(d*x + c) + 6 
*C*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4)
 

Sympy [F]

\[ \int \sec (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=a^{2} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 2 A \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sec ^{3}{\left (c + d x \right )}\, dx + \int C \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 C \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx\right ) \] Input:

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

Output:

a**2*(Integral(A*sec(c + d*x), x) + Integral(2*A*sec(c + d*x)**2, x) + Int 
egral(A*sec(c + d*x)**3, x) + Integral(C*sec(c + d*x)**3, x) + Integral(2* 
C*sec(c + d*x)**4, x) + Integral(C*sec(c + d*x)**5, x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.72 \[ \int \sec (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 3 \, C a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{2} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 96 \, A a^{2} \tan \left (d x + c\right )}{48 \, d} \] Input:

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

Output:

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 3*C*a^2*(2*(3*sin(d*x + 
 c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(si 
n(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 12*A*a^2*(2*sin(d*x + c)/(sin 
(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*C*a 
^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin( 
d*x + c) - 1)) + 48*A*a^2*log(sec(d*x + c) + tan(d*x + c)) + 96*A*a^2*tan( 
d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.35 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.61 \[ \int \sec (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (12 \, A a^{2} + 7 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (12 \, A a^{2} + 7 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (36 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 21 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 132 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 77 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 156 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 83 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 75 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \] Input:

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

Output:

1/24*(3*(12*A*a^2 + 7*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(12*A* 
a^2 + 7*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(36*A*a^2*tan(1/2*d* 
x + 1/2*c)^7 + 21*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 132*A*a^2*tan(1/2*d*x + 1 
/2*c)^5 - 77*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 156*A*a^2*tan(1/2*d*x + 1/2*c) 
^3 + 83*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 60*A*a^2*tan(1/2*d*x + 1/2*c) - 75* 
C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
 

Mupad [B] (verification not implemented)

Time = 14.61 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.40 \[ \int \sec (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (-3\,A\,a^2-\frac {7\,C\,a^2}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (11\,A\,a^2+\frac {77\,C\,a^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-13\,A\,a^2-\frac {83\,C\,a^2}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (5\,A\,a^2+\frac {25\,C\,a^2}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (12\,A+7\,C\right )}{4\,d} \] Input:

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

Output:

(tan(c/2 + (d*x)/2)*(5*A*a^2 + (25*C*a^2)/4) - tan(c/2 + (d*x)/2)^7*(3*A*a 
^2 + (7*C*a^2)/4) + tan(c/2 + (d*x)/2)^5*(11*A*a^2 + (77*C*a^2)/12) - tan( 
c/2 + (d*x)/2)^3*(13*A*a^2 + (83*C*a^2)/12))/(d*(6*tan(c/2 + (d*x)/2)^4 - 
4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1 
)) + (a^2*atanh(tan(c/2 + (d*x)/2))*(12*A + 7*C))/(4*d)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 379, normalized size of antiderivative = 2.87 \[ \int \sec (c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^{2} \left (-48 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a -32 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} c +48 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a +48 \cos \left (d x +c \right ) \sin \left (d x +c \right ) c -36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} a -21 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} c +72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a +42 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} c -36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a -21 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) c +36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} a +21 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} c -72 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a -42 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} c +36 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a +21 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) c -12 \sin \left (d x +c \right )^{3} a -21 \sin \left (d x +c \right )^{3} c +12 \sin \left (d x +c \right ) a +27 \sin \left (d x +c \right ) c \right )}{24 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:

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

Output:

(a**2*( - 48*cos(c + d*x)*sin(c + d*x)**3*a - 32*cos(c + d*x)*sin(c + d*x) 
**3*c + 48*cos(c + d*x)*sin(c + d*x)*a + 48*cos(c + d*x)*sin(c + d*x)*c - 
36*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*a - 21*log(tan((c + d*x)/2) - 
 1)*sin(c + d*x)**4*c + 72*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a + 4 
2*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*c - 36*log(tan((c + d*x)/2) - 
1)*a - 21*log(tan((c + d*x)/2) - 1)*c + 36*log(tan((c + d*x)/2) + 1)*sin(c 
 + d*x)**4*a + 21*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4*c - 72*log(tan 
((c + d*x)/2) + 1)*sin(c + d*x)**2*a - 42*log(tan((c + d*x)/2) + 1)*sin(c 
+ d*x)**2*c + 36*log(tan((c + d*x)/2) + 1)*a + 21*log(tan((c + d*x)/2) + 1 
)*c - 12*sin(c + d*x)**3*a - 21*sin(c + d*x)**3*c + 12*sin(c + d*x)*a + 27 
*sin(c + d*x)*c))/(24*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))