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

Optimal. Leaf size=169 \[ \frac{a^2 (10 B+9 C) \tan ^3(c+d x)}{15 d}+\frac{a^2 (10 B+9 C) \tan (c+d x)}{5 d}+\frac{a^2 (7 B+6 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (5 B+6 C) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{a^2 (7 B+6 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d} \]

[Out]

(a^2*(7*B + 6*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(10*B + 9*C)*Tan[c + d*x])/(5*d) + (a^2*(7*B + 6*C)*Sec[c
+ d*x]*Tan[c + d*x])/(8*d) + (a^2*(5*B + 6*C)*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (C*Sec[c + d*x]^3*(a^2 +
a^2*Sec[c + d*x])*Tan[c + d*x])/(5*d) + (a^2*(10*B + 9*C)*Tan[c + d*x]^3)/(15*d)

________________________________________________________________________________________

Rubi [A]  time = 0.323242, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.175, Rules used = {4072, 4018, 3997, 3787, 3768, 3770, 3767} \[ \frac{a^2 (10 B+9 C) \tan ^3(c+d x)}{15 d}+\frac{a^2 (10 B+9 C) \tan (c+d x)}{5 d}+\frac{a^2 (7 B+6 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (5 B+6 C) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac{a^2 (7 B+6 C) \tan (c+d x) \sec (c+d x)}{8 d}+\frac{C \tan (c+d x) \sec ^3(c+d x) \left (a^2 \sec (c+d x)+a^2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(7*B + 6*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(10*B + 9*C)*Tan[c + d*x])/(5*d) + (a^2*(7*B + 6*C)*Sec[c
+ d*x]*Tan[c + d*x])/(8*d) + (a^2*(5*B + 6*C)*Sec[c + d*x]^3*Tan[c + d*x])/(20*d) + (C*Sec[c + d*x]^3*(a^2 +
a^2*Sec[c + d*x])*Tan[c + d*x])/(5*d) + (a^2*(10*B + 9*C)*Tan[c + d*x]^3)/(15*d)

Rule 4072

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]

Rule 4018

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

Rule 3997

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*C
sc[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 3787

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 3768

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 3770

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

Rule 3767

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]

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^3(c+d x) (a+a \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx\\ &=\frac{C \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac{1}{5} \int \sec ^3(c+d x) (a+a \sec (c+d x)) (a (5 B+3 C)+a (5 B+6 C) \sec (c+d x)) \, dx\\ &=\frac{a^2 (5 B+6 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{C \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac{1}{20} \int \sec ^3(c+d x) \left (5 a^2 (7 B+6 C)+4 a^2 (10 B+9 C) \sec (c+d x)\right ) \, dx\\ &=\frac{a^2 (5 B+6 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{C \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac{1}{4} \left (a^2 (7 B+6 C)\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{5} \left (a^2 (10 B+9 C)\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac{a^2 (7 B+6 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^2 (5 B+6 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{C \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac{1}{8} \left (a^2 (7 B+6 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^2 (10 B+9 C)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac{a^2 (7 B+6 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (10 B+9 C) \tan (c+d x)}{5 d}+\frac{a^2 (7 B+6 C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac{a^2 (5 B+6 C) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac{C \sec ^3(c+d x) \left (a^2+a^2 \sec (c+d x)\right ) \tan (c+d x)}{5 d}+\frac{a^2 (10 B+9 C) \tan ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [B]  time = 0.768951, size = 391, normalized size = 2.31 \[ -\frac{a^2 \sec ^5(c+d x) \left (150 (7 B+6 C) \cos (c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+75 (7 B+6 C) \cos (3 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-640 B \sin (c+d x)-660 B \sin (2 (c+d x))-800 B \sin (3 (c+d x))-210 B \sin (4 (c+d x))-160 B \sin (5 (c+d x))+105 B \cos (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-105 B \cos (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-960 C \sin (c+d x)-840 C \sin (2 (c+d x))-720 C \sin (3 (c+d x))-180 C \sin (4 (c+d x))-144 C \sin (5 (c+d x))+90 C \cos (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-90 C \cos (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{1920 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*Sec[c + d*x]^5*(105*B*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 90*C*Cos[5*(c + d*x)]*
Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 150*(7*B + 6*C)*Cos[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]]) + 75*(7*B + 6*C)*Cos[3*(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]]) - 105*B*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[
(c + d*x)/2]] - 90*C*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 640*B*Sin[c + d*x] - 960*C*Si
n[c + d*x] - 660*B*Sin[2*(c + d*x)] - 840*C*Sin[2*(c + d*x)] - 800*B*Sin[3*(c + d*x)] - 720*C*Sin[3*(c + d*x)]
- 210*B*Sin[4*(c + d*x)] - 180*C*Sin[4*(c + d*x)] - 160*B*Sin[5*(c + d*x)] - 144*C*Sin[5*(c + d*x)]))/(1920*d
)

________________________________________________________________________________________

Maple [A]  time = 0.05, size = 235, normalized size = 1.4 \begin{align*}{\frac{7\,B{a}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{7\,B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{6\,{a}^{2}C\tan \left ( dx+c \right ) }{5\,d}}+{\frac{3\,{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{4\,B{a}^{2}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{2\,B{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,{a}^{2}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{3\,{a}^{2}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{B{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{{a}^{2}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/8/d*B*a^2*sec(d*x+c)*tan(d*x+c)+7/8/d*B*a^2*ln(sec(d*x+c)+tan(d*x+c))+6/5/d*a^2*C*tan(d*x+c)+3/5/d*a^2*C*tan
(d*x+c)*sec(d*x+c)^2+4/3/d*B*a^2*tan(d*x+c)+2/3/d*B*a^2*tan(d*x+c)*sec(d*x+c)^2+1/2/d*a^2*C*tan(d*x+c)*sec(d*x
+c)^3+3/4/d*a^2*C*sec(d*x+c)*tan(d*x+c)+3/4/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*B*a^2*tan(d*x+c)*sec(d*x+c
)^3+1/5/d*a^2*C*tan(d*x+c)*sec(d*x+c)^4

________________________________________________________________________________________

Maxima [A]  time = 0.952453, size = 375, normalized size = 2.22 \begin{align*} \frac{160 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 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^{2} + 80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 15 \, B 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 )} - 30 \, 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 )} - 60 \, B 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 )}}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(160*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*C*a^2 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 15*B*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(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 30*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(sin(d*x + c) + 1) + 3*log(sin(d*x +
c) - 1)) - 60*B*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)))/d

________________________________________________________________________________________

Fricas [A]  time = 0.532336, size = 421, normalized size = 2.49 \begin{align*} \frac{15 \,{\left (7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (10 \, B + 9 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \,{\left (7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \,{\left (10 \, B + 9 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \,{\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 24 \, C a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(7*B + 6*C)*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(7*B + 6*C)*a^2*cos(d*x + c)^5*log(-sin(d*
x + c) + 1) + 2*(16*(10*B + 9*C)*a^2*cos(d*x + c)^4 + 15*(7*B + 6*C)*a^2*cos(d*x + c)^3 + 8*(10*B + 9*C)*a^2*c
os(d*x + c)^2 + 30*(B + 2*C)*a^2*cos(d*x + c) + 24*C*a^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int B \sec ^{3}{\left (c + d x \right )}\, dx + \int 2 B \sec ^{4}{\left (c + d x \right )}\, dx + \int B \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx + \int 2 C \sec ^{5}{\left (c + d x \right )}\, dx + \int C \sec ^{6}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(B*sec(c + d*x)**3, x) + Integral(2*B*sec(c + d*x)**4, x) + Integral(B*sec(c + d*x)**5, x) + Int
egral(C*sec(c + d*x)**4, x) + Integral(2*C*sec(c + d*x)**5, x) + Integral(C*sec(c + d*x)**6, x))

________________________________________________________________________________________

Giac [A]  time = 1.17684, size = 332, normalized size = 1.96 \begin{align*} \frac{15 \,{\left (7 \, B a^{2} + 6 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (7 \, B a^{2} + 6 \, 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 (105 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 90 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 490 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 420 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 800 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 864 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 790 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 540 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 375 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 390 \, 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 )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(7*B*a^2 + 6*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(7*B*a^2 + 6*C*a^2)*log(abs(tan(1/2*d*x
+ 1/2*c) - 1)) - 2*(105*B*a^2*tan(1/2*d*x + 1/2*c)^9 + 90*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 490*B*a^2*tan(1/2*d*x
+ 1/2*c)^7 - 420*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 800*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 864*C*a^2*tan(1/2*d*x + 1/
2*c)^5 - 790*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 540*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 375*B*a^2*tan(1/2*d*x + 1/2*c)
+ 390*C*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d