3.323 \(\int \sec (c+d x) (a+a \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=163 \[ \frac{a^3 (15 B+13 C) \tan ^3(c+d x)}{60 d}+\frac{a^3 (15 B+13 C) \tan (c+d x)}{5 d}+\frac{a^3 (15 B+13 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 a^3 (15 B+13 C) \tan (c+d x) \sec (c+d x)}{40 d}+\frac{(5 B-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{20 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 a d} \]

[Out]

(a^3*(15*B + 13*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(15*B + 13*C)*Tan[c + d*x])/(5*d) + (3*a^3*(15*B + 13*C
)*Sec[c + d*x]*Tan[c + d*x])/(40*d) + ((5*B - C)*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(20*d) + (C*(a + a*Sec[c
+ d*x])^4*Tan[c + d*x])/(5*a*d) + (a^3*(15*B + 13*C)*Tan[c + d*x]^3)/(60*d)

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Rubi [A]  time = 0.312872, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 8, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {4072, 4010, 4001, 3791, 3770, 3767, 8, 3768} \[ \frac{a^3 (15 B+13 C) \tan ^3(c+d x)}{60 d}+\frac{a^3 (15 B+13 C) \tan (c+d x)}{5 d}+\frac{a^3 (15 B+13 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{3 a^3 (15 B+13 C) \tan (c+d x) \sec (c+d x)}{40 d}+\frac{(5 B-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{20 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^4}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(15*B + 13*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(15*B + 13*C)*Tan[c + d*x])/(5*d) + (3*a^3*(15*B + 13*C
)*Sec[c + d*x]*Tan[c + d*x])/(40*d) + ((5*B - C)*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(20*d) + (C*(a + a*Sec[c
+ d*x])^4*Tan[c + d*x])/(5*a*d) + (a^3*(15*B + 13*C)*Tan[c + d*x]^3)/(60*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 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4001

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

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[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]

Rule 8

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

Rubi steps

\begin{align*} \int \sec (c+d x) (a+a \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac{\int \sec (c+d x) (a+a \sec (c+d x))^3 (4 a C+a (5 B-C) \sec (c+d x)) \, dx}{5 a}\\ &=\frac{(5 B-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac{1}{20} (15 B+13 C) \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx\\ &=\frac{(5 B-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac{1}{20} (15 B+13 C) \int \left (a^3 \sec (c+d x)+3 a^3 \sec ^2(c+d x)+3 a^3 \sec ^3(c+d x)+a^3 \sec ^4(c+d x)\right ) \, dx\\ &=\frac{(5 B-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac{1}{20} \left (a^3 (15 B+13 C)\right ) \int \sec (c+d x) \, dx+\frac{1}{20} \left (a^3 (15 B+13 C)\right ) \int \sec ^4(c+d x) \, dx+\frac{1}{20} \left (3 a^3 (15 B+13 C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{20} \left (3 a^3 (15 B+13 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{a^3 (15 B+13 C) \tanh ^{-1}(\sin (c+d x))}{20 d}+\frac{3 a^3 (15 B+13 C) \sec (c+d x) \tan (c+d x)}{40 d}+\frac{(5 B-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac{1}{40} \left (3 a^3 (15 B+13 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^3 (15 B+13 C)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{20 d}-\frac{\left (3 a^3 (15 B+13 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{20 d}\\ &=\frac{a^3 (15 B+13 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3 (15 B+13 C) \tan (c+d x)}{5 d}+\frac{3 a^3 (15 B+13 C) \sec (c+d x) \tan (c+d x)}{40 d}+\frac{(5 B-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{5 a d}+\frac{a^3 (15 B+13 C) \tan ^3(c+d x)}{60 d}\\ \end{align*}

Mathematica [B]  time = 0.80796, size = 391, normalized size = 2.4 \[ -\frac{a^3 \sec ^5(c+d x) \left (150 (15 B+13 C) \cos (c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )+75 (15 B+13 C) \cos (3 (c+d x)) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-1200 B \sin (c+d x)-1140 B \sin (2 (c+d x))-1560 B \sin (3 (c+d x))-450 B \sin (4 (c+d x))-360 B \sin (5 (c+d x))+225 B \cos (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-225 B \cos (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-1600 C \sin (c+d x)-1500 C \sin (2 (c+d x))-1520 C \sin (3 (c+d x))-390 C \sin (4 (c+d x))-304 C \sin (5 (c+d x))+195 C \cos (5 (c+d x)) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-195 C \cos (5 (c+d x)) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{1920 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*Sec[c + d*x]^5*(225*B*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 195*C*Cos[5*(c + d*x)]
*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 150*(15*B + 13*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*(15*B + 13*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]]) - 225*B*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] +
Sin[(c + d*x)/2]] - 195*C*Cos[5*(c + d*x)]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 1200*B*Sin[c + d*x] - 1
600*C*Sin[c + d*x] - 1140*B*Sin[2*(c + d*x)] - 1500*C*Sin[2*(c + d*x)] - 1560*B*Sin[3*(c + d*x)] - 1520*C*Sin[
3*(c + d*x)] - 450*B*Sin[4*(c + d*x)] - 390*C*Sin[4*(c + d*x)] - 360*B*Sin[5*(c + d*x)] - 304*C*Sin[5*(c + d*x
)]))/(1920*d)

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Maple [A]  time = 0.049, size = 234, normalized size = 1.4 \begin{align*} 3\,{\frac{B{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{13\,{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{13\,{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{15\,B{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{15\,B{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{38\,{a}^{3}C\tan \left ( dx+c \right ) }{15\,d}}+{\frac{19\,{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{B{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{3\,{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{B{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{{a}^{3}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)*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

3/d*B*a^3*tan(d*x+c)+13/8/d*a^3*C*sec(d*x+c)*tan(d*x+c)+13/8/d*a^3*C*ln(sec(d*x+c)+tan(d*x+c))+15/8/d*B*a^3*se
c(d*x+c)*tan(d*x+c)+15/8/d*B*a^3*ln(sec(d*x+c)+tan(d*x+c))+38/15*a^3*C*tan(d*x+c)/d+19/15/d*a^3*C*tan(d*x+c)*s
ec(d*x+c)^2+1/d*B*a^3*tan(d*x+c)*sec(d*x+c)^2+3/4/d*a^3*C*tan(d*x+c)*sec(d*x+c)^3+1/4/d*B*a^3*tan(d*x+c)*sec(d
*x+c)^3+1/5/d*a^3*C*tan(d*x+c)*sec(d*x+c)^4

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Maxima [B]  time = 0.958117, size = 455, normalized size = 2.79 \begin{align*} \frac{240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} + 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^{3} + 240 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 15 \, B a^{3}{\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 )} - 45 \, C a^{3}{\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 )} - 180 \, 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 )} - 60 \, C 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 )} + 240 \, B a^{3} \tan \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(240*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 + 16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*C*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 15*B*a^3*(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)) - 45*C*a^3*(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)) - 180*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)) -
60*C*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)) + 240*B*a^3*ta
n(d*x + c))/d

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Fricas [A]  time = 0.522507, size = 431, normalized size = 2.64 \begin{align*} \frac{15 \,{\left (15 \, B + 13 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (15 \, B + 13 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (8 \,{\left (45 \, B + 38 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 15 \,{\left (15 \, B + 13 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 8 \,{\left (15 \, B + 19 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 30 \,{\left (B + 3 \, C\right )} a^{3} \cos \left (d x + c\right ) + 24 \, C a^{3}\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)*(a+a*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*(15*B + 13*C)*a^3*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(15*B + 13*C)*a^3*cos(d*x + c)^5*log(-si
n(d*x + c) + 1) + 2*(8*(45*B + 38*C)*a^3*cos(d*x + c)^4 + 15*(15*B + 13*C)*a^3*cos(d*x + c)^3 + 8*(15*B + 19*C
)*a^3*cos(d*x + c)^2 + 30*(B + 3*C)*a^3*cos(d*x + c) + 24*C*a^3)*sin(d*x + c))/(d*cos(d*x + c)^5)

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

[Out]

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

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Giac [A]  time = 1.19213, size = 332, normalized size = 2.04 \begin{align*} \frac{15 \,{\left (15 \, B a^{3} + 13 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (15 \, B a^{3} + 13 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (225 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 195 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 1050 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 910 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 1920 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 1664 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 1830 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 1330 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 735 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 765 \, C 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 )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(15*B*a^3 + 13*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(15*B*a^3 + 13*C*a^3)*log(abs(tan(1/2*
d*x + 1/2*c) - 1)) - 2*(225*B*a^3*tan(1/2*d*x + 1/2*c)^9 + 195*C*a^3*tan(1/2*d*x + 1/2*c)^9 - 1050*B*a^3*tan(1
/2*d*x + 1/2*c)^7 - 910*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 1920*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 1664*C*a^3*tan(1/2*
d*x + 1/2*c)^5 - 1830*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 1330*C*a^3*tan(1/2*d*x + 1/2*c)^3 + 735*B*a^3*tan(1/2*d*x
+ 1/2*c) + 765*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d