\(\int \sec ^2(c+d x) (a+a \sec (c+d x)) (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [306]

Optimal result

Integrand size = 38, antiderivative size = 106 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a (4 B+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (B+C) \tan (c+d x)}{d}+\frac {a (4 B+3 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a (B+C) \tan ^3(c+d x)}{3 d} \]

[Out]

1/8*a*(4*B+3*C)*arctanh(sin(d*x+c))/d+a*(B+C)*tan(d*x+c)/d+1/8*a*(4*B+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+1/3*a*(B+C)*tan(d*x+c)^3/d

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4157, 4082, 3872, 3853, 3855, 3852} \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a (4 B+3 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (B+C) \tan ^3(c+d x)}{3 d}+\frac {a (B+C) \tan (c+d x)}{d}+\frac {a (4 B+3 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a C \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

[In]

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

[Out]

(a*(4*B + 3*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(B + C)*Tan[c + d*x])/d + (a*(4*B + 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*(B + C)*Tan[c + d*x]^3)/(3*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 3872

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

Rule 4082

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.
) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[e + f*x]*((d*Csc[e + f*x])^n/(f*(n + 1))), x] + Dist[1/(n + 1), Int[(d
*Csc[e + f*x])^n*Simp[A*a*(n + 1) + B*b*n + (A*b + B*a)*(n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e,
f, A, B}, x] && NeQ[A*b - a*B, 0] &&  !LeQ[n, -1]

Rule 4157

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

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

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(337\) vs. \(2(106)=212\).

Time = 1.16 (sec) , antiderivative size = 337, normalized size of antiderivative = 3.18 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {a \sec ^4(c+d x) \left (36 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+27 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 (4 B+3 C) \cos (2 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+3 (4 B+3 C) \cos (4 (c+d x)) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )-36 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-27 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-24 B \sin (c+d x)-66 C \sin (c+d x)-64 B \sin (2 (c+d x))-64 C \sin (2 (c+d x))-24 B \sin (3 (c+d x))-18 C \sin (3 (c+d x))-16 B \sin (4 (c+d x))-16 C \sin (4 (c+d x))\right )}{192 d} \]

[In]

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

[Out]

-1/192*(a*Sec[c + d*x]^4*(36*B*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 27*C*Log[Cos[(c + d*x)/2] - Sin[(c +
d*x)/2]] + 12*(4*B + 3*C)*Cos[2*(c + d*x)]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]]) + 3*(4*B + 3*C)*Cos[4*(c + d*x)]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c +
d*x)/2] + Sin[(c + d*x)/2]]) - 36*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 27*C*Log[Cos[(c + d*x)/2] + Sin
[(c + d*x)/2]] - 24*B*Sin[c + d*x] - 66*C*Sin[c + d*x] - 64*B*Sin[2*(c + d*x)] - 64*C*Sin[2*(c + d*x)] - 24*B*
Sin[3*(c + d*x)] - 18*C*Sin[3*(c + d*x)] - 16*B*Sin[4*(c + d*x)] - 16*C*Sin[4*(c + d*x)]))/d

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.12

method	result	size
parts	\(-\frac {\left (a B +C a \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {a B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(119\)
derivativedivides	\(\frac {-a B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(131\)
default	\(\frac {-a B \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C a \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a B \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(131\)
norman	\(\frac {-\frac {a \left (4 B +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {7 a \left (4 B +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}+\frac {a \left (12 B +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {a \left (52 B +31 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {a \left (4 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a \left (4 B +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(163\)
parallelrisch	\(\frac {8 a \left (-\frac {3 \left (B +\frac {3 C}{4}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4}+\frac {3 \left (B +\frac {3 C}{4}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}+\left (B +C \right ) \sin \left (2 d x +2 c \right )+\frac {3 \left (B +\frac {3 C}{4}\right ) \sin \left (3 d x +3 c \right )}{8}+\frac {\left (B +C \right ) \sin \left (4 d x +4 c \right )}{4}+\frac {3 \sin \left (d x +c \right ) \left (B +\frac {11 C}{4}\right )}{8}\right )}{3 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(170\)
risch	\(-\frac {i a \left (12 B \,{\mathrm e}^{7 i \left (d x +c \right )}+9 C \,{\mathrm e}^{7 i \left (d x +c \right )}+12 B \,{\mathrm e}^{5 i \left (d x +c \right )}+33 C \,{\mathrm e}^{5 i \left (d x +c \right )}-48 B \,{\mathrm e}^{4 i \left (d x +c \right )}-48 C \,{\mathrm e}^{4 i \left (d x +c \right )}-12 B \,{\mathrm e}^{3 i \left (d x +c \right )}-33 C \,{\mathrm e}^{3 i \left (d x +c \right )}-64 B \,{\mathrm e}^{2 i \left (d x +c \right )}-64 C \,{\mathrm e}^{2 i \left (d x +c \right )}-12 B \,{\mathrm e}^{i \left (d x +c \right )}-9 C \,{\mathrm e}^{i \left (d x +c \right )}-16 B -16 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {3 a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}\)	\(253\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.20 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (4 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (B + C\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, B + 3 \, C\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (B + C\right )} a \cos \left (d x + c\right ) + 6 \, C a\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

1/48*(3*(4*B + 3*C)*a*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*B + 3*C)*a*cos(d*x + c)^4*log(-sin(d*x + c)
+ 1) + 2*(16*(B + C)*a*cos(d*x + c)^3 + 3*(4*B + 3*C)*a*cos(d*x + c)^2 + 8*(B + C)*a*cos(d*x + c) + 6*C*a)*sin
(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.54 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a - 3 \, C a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

[In]

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

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a - 3*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)) - 12*B*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 = 188, normalized size of antiderivative = 1.77 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (4 \, B a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, B a + 3 \, C a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 28 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 49 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 52 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 31 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 39 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

[In]

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

[Out]

1/24*(3*(4*B*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*B*a + 3*C*a)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) - 2*(12*B*a*tan(1/2*d*x + 1/2*c)^7 + 9*C*a*tan(1/2*d*x + 1/2*c)^7 - 28*B*a*tan(1/2*d*x + 1/2*c)^5 - 49*C*a
*tan(1/2*d*x + 1/2*c)^5 + 52*B*a*tan(1/2*d*x + 1/2*c)^3 + 31*C*a*tan(1/2*d*x + 1/2*c)^3 - 36*B*a*tan(1/2*d*x +
1/2*c) - 39*C*a*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 = 18.76 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.57 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x)) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (-B\,a-\frac {3\,C\,a}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {7\,B\,a}{3}+\frac {49\,C\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {13\,B\,a}{3}-\frac {31\,C\,a}{12}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,B\,a+\frac {13\,C\,a}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (4\,B+3\,C\right )}{4\,d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(3*B*a + (13*C*a)/4) - tan(c/2 + (d*x)/2)^7*(B*a + (3*C*a)/4) - tan(c/2 + (d*x)/2)^3*((13*
B*a)/3 + (31*C*a)/12) + tan(c/2 + (d*x)/2)^5*((7*B*a)/3 + (49*C*a)/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*atanh(tan(c/2 + (d*x)/2))*(4*B + 3*C)
)/(4*d)