\(\int \sec ^2(c+d x) (a+b \sec (c+d x)) (A+C \sec ^2(c+d x)) \, dx\) [638]

Optimal result

Integrand size = 31, antiderivative size = 117 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {b (4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (3 A+2 C) \tan (c+d x)}{3 d}+\frac {b (4 A+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {b C \sec ^3(c+d x) \tan (c+d x)}{4 d} \]

[Out]

1/8*b*(4*A+3*C)*arctanh(sin(d*x+c))/d+1/3*a*(3*A+2*C)*tan(d*x+c)/d+1/8*b*(4*A+3*C)*sec(d*x+c)*tan(d*x+c)/d+1/3
*a*C*sec(d*x+c)^2*tan(d*x+c)/d+1/4*b*C*sec(d*x+c)^3*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4162, 4132, 3853, 3855, 4131, 3852, 8} \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a (3 A+2 C) \tan (c+d x)}{3 d}+\frac {a C \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {b (4 A+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {b (4 A+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {b C \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

[In]

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

[Out]

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

Rule 8

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

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

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.68 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 b (4 A+3 C) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (3 b (4 A+3 C) \sec (c+d x)+6 b C \sec ^3(c+d x)+8 a \left (3 (A+C)+C \tan ^2(c+d x)\right )\right )}{24 d} \]

[In]

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

[Out]

(3*b*(4*A + 3*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(3*b*(4*A + 3*C)*Sec[c + d*x] + 6*b*C*Sec[c + d*x]^3 + 8
*a*(3*(A + C) + C*Tan[c + d*x]^2)))/(24*d)

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.01

method	result	size
derivativedivides	\(\frac {a A \tan \left (d x +c \right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(118\)
default	\(\frac {a A \tan \left (d x +c \right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+C b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(118\)
parts	\(\frac {A b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {a A \tan \left (d x +c \right )}{d}\)	\(126\)
parallelrisch	\(\frac {-2 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) b \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) b \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+2 \left (A +\frac {4 C}{3}\right ) a \sin \left (2 d x +2 c \right )+b \left (A +\frac {3 C}{4}\right ) \sin \left (3 d x +3 c \right )+a \left (A +\frac {2 C}{3}\right ) \sin \left (4 d x +4 c \right )+b \left (A +\frac {11 C}{4}\right ) \sin \left (d x +c \right )}{d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(176\)
norman	\(\frac {-\frac {\left (8 a A -4 A b +8 C a -5 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {\left (8 a A +4 A b +8 C a +5 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (72 a A -12 A b +40 C a +9 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {\left (72 a A +12 A b +40 C a -9 C b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {b \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {b \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(199\)
risch	\(-\frac {i \left (12 A b \,{\mathrm e}^{7 i \left (d x +c \right )}+9 C b \,{\mathrm e}^{7 i \left (d x +c \right )}-24 A a \,{\mathrm e}^{6 i \left (d x +c \right )}+12 A b \,{\mathrm e}^{5 i \left (d x +c \right )}+33 C b \,{\mathrm e}^{5 i \left (d x +c \right )}-72 A a \,{\mathrm e}^{4 i \left (d x +c \right )}-48 C a \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A b \,{\mathrm e}^{3 i \left (d x +c \right )}-33 C b \,{\mathrm e}^{3 i \left (d x +c \right )}-72 a A \,{\mathrm e}^{2 i \left (d x +c \right )}-64 C a \,{\mathrm e}^{2 i \left (d x +c \right )}-12 A b \,{\mathrm e}^{i \left (d x +c \right )}-9 C b \,{\mathrm e}^{i \left (d x +c \right )}-24 a A -16 C a \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}\)	\(279\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.10 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (4 \, A + 3 \, C\right )} b \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, A + 3 \, C\right )} b \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (3 \, A + 2 \, C\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 3 \, C\right )} b \cos \left (d x + c\right )^{2} + 8 \, C a \cos \left (d x + c\right ) + 6 \, C b\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

1/48*(3*(4*A + 3*C)*b*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*A + 3*C)*b*cos(d*x + c)^4*log(-sin(d*x + c)
+ 1) + 2*(8*(3*A + 2*C)*a*cos(d*x + c)^3 + 3*(4*A + 3*C)*b*cos(d*x + c)^2 + 8*C*a*cos(d*x + c) + 6*C*b)*sin(d*
x + c))/(d*cos(d*x + c)^4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.30 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a - 3 \, C b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a \tan \left (d x + c\right )}{48 \, d} \]

[In]

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

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a - 3*C*b*(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)) - 12*A*b*(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*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (107) = 214\).

Time = 0.34 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.60 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (4 \, A b + 3 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, A b + 3 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (24 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

[In]

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

[Out]

1/24*(3*(4*A*b + 3*C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*A*b + 3*C*b)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) - 2*(24*A*a*tan(1/2*d*x + 1/2*c)^7 + 24*C*a*tan(1/2*d*x + 1/2*c)^7 - 12*A*b*tan(1/2*d*x + 1/2*c)^7 - 15*C*
b*tan(1/2*d*x + 1/2*c)^7 - 72*A*a*tan(1/2*d*x + 1/2*c)^5 - 40*C*a*tan(1/2*d*x + 1/2*c)^5 + 12*A*b*tan(1/2*d*x
+ 1/2*c)^5 - 9*C*b*tan(1/2*d*x + 1/2*c)^5 + 72*A*a*tan(1/2*d*x + 1/2*c)^3 + 40*C*a*tan(1/2*d*x + 1/2*c)^3 + 12
*A*b*tan(1/2*d*x + 1/2*c)^3 - 9*C*b*tan(1/2*d*x + 1/2*c)^3 - 24*A*a*tan(1/2*d*x + 1/2*c) - 24*C*a*tan(1/2*d*x
+ 1/2*c) - 12*A*b*tan(1/2*d*x + 1/2*c) - 15*C*b*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 = 20.19 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.68 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x)) \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (A\,b-2\,A\,a-2\,C\,a+\frac {5\,C\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (6\,A\,a-A\,b+\frac {10\,C\,a}{3}+\frac {3\,C\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,C\,b}{4}-A\,b-\frac {10\,C\,a}{3}-6\,A\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a+A\,b+2\,C\,a+\frac {5\,C\,b}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,A+3\,C\right )}{4\,d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(2*A*a + A*b + 2*C*a + (5*C*b)/4) - tan(c/2 + (d*x)/2)^7*(2*A*a - A*b + 2*C*a - (5*C*b)/4)
- tan(c/2 + (d*x)/2)^3*(6*A*a + A*b + (10*C*a)/3 - (3*C*b)/4) + tan(c/2 + (d*x)/2)^5*(6*A*a - A*b + (10*C*a)/
3 + (3*C*b)/4))/(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)) + (b*atanh(tan(c/2 + (d*x)/2))*(4*A + 3*C))/(4*d)