\(\int \sec ^3(c+d x) (a+a \sec (c+d x)) (A+C \sec ^2(c+d x)) \, dx\) [84]

Optimal result

Integrand size = 31, antiderivative size = 140 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (5 A+4 C) \tan (c+d x)}{5 d}+\frac {a (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a (5 A+4 C) \tan ^3(c+d x)}{15 d} \]

[Out]

1/8*a*(4*A+3*C)*arctanh(sin(d*x+c))/d+1/5*a*(5*A+4*C)*tan(d*x+c)/d+1/8*a*(4*A+3*C)*sec(d*x+c)*tan(d*x+c)/d+1/4
*a*C*sec(d*x+c)^3*tan(d*x+c)/d+1/5*a*C*sec(d*x+c)^4*tan(d*x+c)/d+1/15*a*(5*A+4*C)*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4162, 4132, 3852, 4131, 3853, 3855} \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac {a (5 A+4 C) \tan (c+d x)}{5 d}+\frac {a (4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a C \tan (c+d x) \sec ^4(c+d x)}{5 d}+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

[In]

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

[Out]

(a*(4*A + 3*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(5*A + 4*C)*Tan[c + d*x])/(5*d) + (a*(4*A + 3*C)*Sec[c + d*x]
*Tan[c + d*x])/(8*d) + (a*C*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a*C*Sec[c + d*x]^4*Tan[c + d*x])/(5*d) + (a*
(5*A + 4*C)*Tan[c + d*x]^3)/(15*d)

Rule 3852

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

Rule 3853

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] + Dist[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 3855

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

Rule 4131

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] + Dist[(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 4132

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4162

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

Rubi steps \begin{align*} \text {integral}& = \frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) \left (5 a A+a (5 A+4 C) \sec (c+d x)+5 a C \sec ^2(c+d x)\right ) \, dx \\ & = \frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) \left (5 a A+5 a C \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} (a (5 A+4 C)) \int \sec ^4(c+d x) \, dx \\ & = \frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (a (4 A+3 C)) \int \sec ^3(c+d x) \, dx-\frac {(a (5 A+4 C)) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d} \\ & = \frac {a (5 A+4 C) \tan (c+d x)}{5 d}+\frac {a (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a (5 A+4 C) \tan ^3(c+d x)}{15 d}+\frac {1}{8} (a (4 A+3 C)) \int \sec (c+d x) \, dx \\ & = \frac {a (4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (5 A+4 C) \tan (c+d x)}{5 d}+\frac {a (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a C \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {a (5 A+4 C) \tan ^3(c+d x)}{15 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.66 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a \left (15 (4 A+3 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 (4 A+3 C) \sec (c+d x)+30 C \sec ^3(c+d x)+8 \left (15 (A+C)+5 (A+2 C) \tan ^2(c+d x)+3 C \tan ^4(c+d x)\right )\right )\right )}{120 d} \]

[In]

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

[Out]

(a*(15*(4*A + 3*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(4*A + 3*C)*Sec[c + d*x] + 30*C*Sec[c + d*x]^3 + 8
*(15*(A + C) + 5*(A + 2*C)*Tan[c + d*x]^2 + 3*C*Tan[c + d*x]^4))))/(120*d)

Maple [A] (verified)

Time = 0.58 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(141\)
default	\(\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(141\)
parts	\(\frac {C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {C a \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {a A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {a A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(149\)
norman	\(\frac {-\frac {a \left (4 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {a \left (12 A +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (20 A +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}-\frac {4 a \left (25 A +29 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {a \left (44 A +19 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(187\)
parallelrisch	\(\frac {8 a \left (-\frac {15 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8}+\frac {15 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8}+\frac {3 \left (A +\frac {7 C}{4}\right ) \sin \left (2 d x +2 c \right )}{4}+\left (\frac {5 A}{4}+C \right ) \sin \left (3 d x +3 c \right )+\frac {3 \left (A +\frac {3 C}{4}\right ) \sin \left (4 d x +4 c \right )}{8}+\left (\frac {A}{4}+\frac {C}{5}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +2 C \right )\right )}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(211\)
risch	\(-\frac {i a \left (60 A \,{\mathrm e}^{9 i \left (d x +c \right )}+45 C \,{\mathrm e}^{9 i \left (d x +c \right )}+120 A \,{\mathrm e}^{7 i \left (d x +c \right )}+210 C \,{\mathrm e}^{7 i \left (d x +c \right )}-240 A \,{\mathrm e}^{6 i \left (d x +c \right )}-560 A \,{\mathrm e}^{4 i \left (d x +c \right )}-640 C \,{\mathrm e}^{4 i \left (d x +c \right )}-120 A \,{\mathrm e}^{3 i \left (d x +c \right )}-210 C \,{\mathrm e}^{3 i \left (d x +c \right )}-400 A \,{\mathrm e}^{2 i \left (d x +c \right )}-320 C \,{\mathrm e}^{2 i \left (d x +c \right )}-60 A \,{\mathrm e}^{i \left (d x +c \right )}-45 C \,{\mathrm e}^{i \left (d x +c \right )}-80 A -64 C \right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\)	\(265\)

[In]

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

[Out]

1/d*(a*A*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+C*a*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*ta
n(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))-a*A*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)-C*a*(-8/15-1/5*sec(d*x+c)^4-4/1
5*sec(d*x+c)^2)*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.05 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, A + 4 \, C\right )} a \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, A + 3 \, C\right )} a \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 4 \, C\right )} a \cos \left (d x + c\right )^{2} + 30 \, C a \cos \left (d x + c\right ) + 24 \, C a\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

[In]

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

[Out]

1/240*(15*(4*A + 3*C)*a*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*A + 3*C)*a*cos(d*x + c)^5*log(-sin(d*x +
c) + 1) + 2*(16*(5*A + 4*C)*a*cos(d*x + c)^4 + 15*(4*A + 3*C)*a*cos(d*x + c)^3 + 8*(5*A + 4*C)*a*cos(d*x + c)^
2 + 30*C*a*cos(d*x + c) + 24*C*a)*sin(d*x + c))/(d*cos(d*x + c)^5)

Sympy [F]

\[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=a \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sec ^{4}{\left (c + d x \right )}\, dx + \int C \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a*(Integral(A*sec(c + d*x)**3, x) + Integral(A*sec(c + d*x)**4, x) + Integral(C*sec(c + d*x)**5, x) + Integral
(C*sec(c + d*x)**6, x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.25 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a - 15 \, C a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

[In]

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

[Out]

1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*
C*a - 15*C*a*(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(sin(d*x +
c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*A*a*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log
(sin(d*x + c) - 1)))/d

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.56 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, A a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, A a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 200 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 130 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 400 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 464 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 440 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 190 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 180 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 195 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

[In]

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

[Out]

1/120*(15*(4*A*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*A*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c)
- 1)) - 2*(60*A*a*tan(1/2*d*x + 1/2*c)^9 + 45*C*a*tan(1/2*d*x + 1/2*c)^9 - 200*A*a*tan(1/2*d*x + 1/2*c)^7 - 1
30*C*a*tan(1/2*d*x + 1/2*c)^7 + 400*A*a*tan(1/2*d*x + 1/2*c)^5 + 464*C*a*tan(1/2*d*x + 1/2*c)^5 - 440*A*a*tan(
1/2*d*x + 1/2*c)^3 - 190*C*a*tan(1/2*d*x + 1/2*c)^3 + 180*A*a*tan(1/2*d*x + 1/2*c) + 195*C*a*tan(1/2*d*x + 1/2
*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

Mupad [B] (verification not implemented)

Time = 18.82 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.44 \[ \int \sec ^3(c+d x) (a+a \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+3\,C\right )}{4\,d}-\frac {\left (A\,a+\frac {3\,C\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {10\,A\,a}{3}-\frac {13\,C\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,A\,a}{3}+\frac {116\,C\,a}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {22\,A\,a}{3}-\frac {19\,C\,a}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,A\,a+\frac {13\,C\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

(a*atanh(tan(c/2 + (d*x)/2))*(4*A + 3*C))/(4*d) - (tan(c/2 + (d*x)/2)*(3*A*a + (13*C*a)/4) + tan(c/2 + (d*x)/2
)^9*(A*a + (3*C*a)/4) - tan(c/2 + (d*x)/2)^7*((10*A*a)/3 + (13*C*a)/6) - tan(c/2 + (d*x)/2)^3*((22*A*a)/3 + (1
9*C*a)/6) + tan(c/2 + (d*x)/2)^5*((20*A*a)/3 + (116*C*a)/15))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/
2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))