\(\int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\) [65]

Optimal result

Integrand size = 23, antiderivative size = 111 \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=a^3 A x+\frac {a^3 (7 A+5 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^3 (A+B) \tan (c+d x)}{2 d}+\frac {a B (a+a \sec (c+d x))^2 \tan (c+d x)}{3 d}+\frac {(3 A+5 B) \left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{6 d} \] Output:

a^3*A*x+1/2*a^3*(7*A+5*B)*arctanh(sin(d*x+c))/d+5/2*a^3*(A+B)*tan(d*x+c)/d 
+1/3*a*B*(a+a*sec(d*x+c))^2*tan(d*x+c)/d+1/6*(3*A+5*B)*(a^3+a^3*sec(d*x+c) 
)*tan(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.04 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.93 \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {a^3 \left (6 A d x+6 (3 A+B) \coth ^{-1}(\sin (c+d x))+3 (A+3 B) \text {arctanh}(\sin (c+d x))+18 A \tan (c+d x)+24 B \tan (c+d x)+3 A \sec (c+d x) \tan (c+d x)+9 B \sec (c+d x) \tan (c+d x)+2 B \tan ^3(c+d x)\right )}{6 d} \] Input:

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

Output:

(a^3*(6*A*d*x + 6*(3*A + B)*ArcCoth[Sin[c + d*x]] + 3*(A + 3*B)*ArcTanh[Si 
n[c + d*x]] + 18*A*Tan[c + d*x] + 24*B*Tan[c + d*x] + 3*A*Sec[c + d*x]*Tan 
[c + d*x] + 9*B*Sec[c + d*x]*Tan[c + d*x] + 2*B*Tan[c + d*x]^3))/(6*d)
 

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.05, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {3042, 4405, 3042, 4405, 27, 3042, 4402, 3042, 4254, 24, 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.

\(\displaystyle \int (a \sec (c+d x)+a)^3 (A+B \sec (c+d x)) \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4405

\(\displaystyle \frac {1}{3} \int (\sec (c+d x) a+a)^2 (3 a A+a (3 A+5 B) \sec (c+d x))dx+\frac {a B \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2 \left (3 a A+a (3 A+5 B) \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {a B \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\)

\(\Big \downarrow \) 4405

\(\displaystyle \frac {1}{3} \left (\frac {1}{2} \int 3 (\sec (c+d x) a+a) \left (2 A a^2+5 (A+B) \sec (c+d x) a^2\right )dx+\frac {(3 A+5 B) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\right )+\frac {a B \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \int (\sec (c+d x) a+a) \left (2 A a^2+5 (A+B) \sec (c+d x) a^2\right )dx+\frac {(3 A+5 B) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\right )+\frac {a B \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (2 A a^2+5 (A+B) \csc \left (c+d x+\frac {\pi }{2}\right ) a^2\right )dx+\frac {(3 A+5 B) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\right )+\frac {a B \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\)

\(\Big \downarrow \) 4402

\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \left (5 a^3 (A+B) \int \sec ^2(c+d x)dx+a^3 (7 A+5 B) \int \sec (c+d x)dx+2 a^3 A x\right )+\frac {(3 A+5 B) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\right )+\frac {a B \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \left (a^3 (7 A+5 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+5 a^3 (A+B) \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+2 a^3 A x\right )+\frac {(3 A+5 B) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\right )+\frac {a B \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\)

\(\Big \downarrow \) 4254

\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \left (-\frac {5 a^3 (A+B) \int 1d(-\tan (c+d x))}{d}+a^3 (7 A+5 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 a^3 A x\right )+\frac {(3 A+5 B) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\right )+\frac {a B \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\)

\(\Big \downarrow \) 24

\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \left (a^3 (7 A+5 B) \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {5 a^3 (A+B) \tan (c+d x)}{d}+2 a^3 A x\right )+\frac {(3 A+5 B) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\right )+\frac {a B \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {1}{3} \left (\frac {3}{2} \left (\frac {a^3 (7 A+5 B) \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^3 (A+B) \tan (c+d x)}{d}+2 a^3 A x\right )+\frac {(3 A+5 B) \tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\right )+\frac {a B \tan (c+d x) (a \sec (c+d x)+a)^2}{3 d}\)

Input:

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

Output:

(a*B*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(3*d) + (((3*A + 5*B)*(a^3 + a^3 
*Sec[c + d*x])*Tan[c + d*x])/(2*d) + (3*(2*a^3*A*x + (a^3*(7*A + 5*B)*ArcT 
anh[Sin[c + d*x]])/d + (5*a^3*(A + B)*Tan[c + d*x])/d))/2)/3
 

Defintions of rubi rules used

rule 24

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

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, 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 4402

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + 
 (c_)), x_Symbol] :> Simp[a*c*x, x] + (Simp[b*d   Int[Csc[e + f*x]^2, x], x 
] + Simp[(b*c + a*d)   Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, f 
}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]
 

rule 4405

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d 
_.) + (c_)), x_Symbol] :> Simp[(-b)*d*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m 
 - 1)/(f*m)), x] + Simp[1/m   Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m + 
 (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, 
f}, x] && NeQ[b*c - a*d, 0] && GtQ[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2 
*m]
 

Maple [A] (verified)

Time = 0.78 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.22

method	result	size
parts	\(a^{3} A x +\frac {\left (a^{3} A +3 B \,a^{3}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (3 a^{3} A +B \,a^{3}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (3 a^{3} A +3 B \,a^{3}\right ) \tan \left (d x +c \right )}{d}-\frac {B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(135\)
parallelrisch	\(\frac {a^{3} \left (-\frac {21 \left (A +\frac {5 B}{7}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {21 \left (A +\frac {5 B}{7}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+A x d \cos \left (3 d x +3 c \right )+\left (A +3 B \right ) \sin \left (2 d x +2 c \right )+\left (3 A +\frac {11 B}{3}\right ) \sin \left (3 d x +3 c \right )+3 A x d \cos \left (d x +c \right )+3 \sin \left (d x +c \right ) \left (A +\frac {5 B}{3}\right )\right )}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(172\)
derivativedivides	\(\frac {a^{3} A \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \tan \left (d x +c \right )+3 a^{3} A \tan \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(176\)
default	\(\frac {a^{3} A \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 B \,a^{3} \tan \left (d x +c \right )+3 a^{3} A \tan \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a^{3} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(176\)
norman	\(\frac {a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}-a^{3} A x +3 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-3 a^{3} A x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\frac {5 a^{3} \left (A +B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{d}-\frac {a^{3} \left (7 A +11 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {4 a^{3} \left (9 A +10 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{3}}-\frac {a^{3} \left (7 A +5 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{3} \left (7 A +5 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(205\)
risch	\(a^{3} A x -\frac {i a^{3} \left (3 A \,{\mathrm e}^{5 i \left (d x +c \right )}+9 B \,{\mathrm e}^{5 i \left (d x +c \right )}-18 A \,{\mathrm e}^{4 i \left (d x +c \right )}-18 B \,{\mathrm e}^{4 i \left (d x +c \right )}-36 A \,{\mathrm e}^{2 i \left (d x +c \right )}-48 B \,{\mathrm e}^{2 i \left (d x +c \right )}-3 \,{\mathrm e}^{i \left (d x +c \right )} A -9 B \,{\mathrm e}^{i \left (d x +c \right )}-18 A -22 B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}+\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}\)	\(221\)

Input:

int((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.27 \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {12 \, A a^{3} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (7 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (9 \, A + 11 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right ) + 2 \, B a^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \] Input:

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="fricas")
 

Output:

1/12*(12*A*a^3*d*x*cos(d*x + c)^3 + 3*(7*A + 5*B)*a^3*cos(d*x + c)^3*log(s 
in(d*x + c) + 1) - 3*(7*A + 5*B)*a^3*cos(d*x + c)^3*log(-sin(d*x + c) + 1) 
 + 2*(2*(9*A + 11*B)*a^3*cos(d*x + c)^2 + 3*(A + 3*B)*a^3*cos(d*x + c) + 2 
*B*a^3)*sin(d*x + c))/(d*cos(d*x + c)^3)
 

Sympy [F]

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

integrate((a+a*sec(d*x+c))**3*(A+B*sec(d*x+c)),x)
 

Output:

a**3*(Integral(A, x) + Integral(3*A*sec(c + d*x), x) + Integral(3*A*sec(c 
+ d*x)**2, x) + Integral(A*sec(c + d*x)**3, x) + Integral(B*sec(c + d*x), 
x) + Integral(3*B*sec(c + d*x)**2, x) + Integral(3*B*sec(c + d*x)**3, x) + 
 Integral(B*sec(c + d*x)**4, x))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.78 \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {12 \, {\left (d x + c\right )} A a^{3} + 4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} - 3 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 9 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 12 \, B a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 36 \, A a^{3} \tan \left (d x + c\right ) + 36 \, B a^{3} \tan \left (d x + c\right )}{12 \, d} \] Input:

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="maxima")
 

Output:

1/12*(12*(d*x + c)*A*a^3 + 4*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 - 3*A 
*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(si 
n(d*x + c) - 1)) - 9*B*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin( 
d*x + c) + 1) + log(sin(d*x + c) - 1)) + 36*A*a^3*log(sec(d*x + c) + tan(d 
*x + c)) + 12*B*a^3*log(sec(d*x + c) + tan(d*x + c)) + 36*A*a^3*tan(d*x + 
c) + 36*B*a^3*tan(d*x + c))/d
 

Giac [A] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.70 \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {6 \, {\left (d x + c\right )} A a^{3} + 3 \, {\left (7 \, A a^{3} + 5 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (7 \, A a^{3} + 5 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 21 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 33 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \] Input:

integrate((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="giac")
 

Output:

1/6*(6*(d*x + c)*A*a^3 + 3*(7*A*a^3 + 5*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c 
) + 1)) - 3*(7*A*a^3 + 5*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15 
*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 36*A*a^3 
*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 21*A*a^3*tan(1 
/2*d*x + 1/2*c) + 33*B*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 
 1)^3)/d
 

Mupad [B] (verification not implemented)

Time = 10.97 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.88 \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {2\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {5\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {11\,B\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {3\,B\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3} \] Input:

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

Output:

(2*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (7*A*a^3*atanh(s 
in(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (5*B*a^3*atanh(sin(c/2 + (d*x)/ 
2)/cos(c/2 + (d*x)/2)))/d + (3*A*a^3*sin(c + d*x))/(d*cos(c + d*x)) + (A*a 
^3*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (11*B*a^3*sin(c + d*x))/(3*d*cos(c 
 + d*x)) + (3*B*a^3*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (B*a^3*sin(c + d* 
x))/(3*d*cos(c + d*x)^3)
 

Reduce [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 329, normalized size of antiderivative = 2.96 \[ \int (a+a \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {a^{3} \left (-21 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} a -15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b +21 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) a +15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +21 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} a +15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b -21 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) a -15 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b +6 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} a d x -3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -9 \cos \left (d x +c \right ) \sin \left (d x +c \right ) b -6 \cos \left (d x +c \right ) a d x +18 \sin \left (d x +c \right )^{3} a +22 \sin \left (d x +c \right )^{3} b -18 \sin \left (d x +c \right ) a -24 \sin \left (d x +c \right ) b \right )}{6 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{2}-1\right )} \] Input:

int((a+a*sec(d*x+c))^3*(A+B*sec(d*x+c)),x)
 

Output:

(a**3*( - 21*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*a - 15 
*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2*b + 21*cos(c + d*x 
)*log(tan((c + d*x)/2) - 1)*a + 15*cos(c + d*x)*log(tan((c + d*x)/2) - 1)* 
b + 21*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*a + 15*cos(c 
 + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b - 21*cos(c + d*x)*log( 
tan((c + d*x)/2) + 1)*a - 15*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*b + 6* 
cos(c + d*x)*sin(c + d*x)**2*a*d*x - 3*cos(c + d*x)*sin(c + d*x)*a - 9*cos 
(c + d*x)*sin(c + d*x)*b - 6*cos(c + d*x)*a*d*x + 18*sin(c + d*x)**3*a + 2 
2*sin(c + d*x)**3*b - 18*sin(c + d*x)*a - 24*sin(c + d*x)*b))/(6*cos(c + d 
*x)*d*(sin(c + d*x)**2 - 1))