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

Optimal. Leaf size=190 \[ \frac{a^2 (8 A+7 B+6 C) \tan (c+d x)}{6 d}+\frac{a^2 (8 A+7 B+6 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (8 A+7 B+6 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{(20 A-5 B+6 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{60 d}+\frac{(5 B+2 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{20 a d}+\frac{C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^2}{5 d} \]

[Out]

(a^2*(8*A + 7*B + 6*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(8*A + 7*B + 6*C)*Tan[c + d*x])/(6*d) + (a^2*(8*A +
7*B + 6*C)*Sec[c + d*x]*Tan[c + d*x])/(24*d) + ((20*A - 5*B + 6*C)*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(60*d
) + (C*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(5*d) + ((5*B + 2*C)*(a + a*Sec[c + d*x])^3*Tan[c +
d*x])/(20*a*d)

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Rubi [A]  time = 0.41012, antiderivative size = 190, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.195, Rules used = {4088, 4010, 4001, 3788, 3767, 8, 4046, 3770} \[ \frac{a^2 (8 A+7 B+6 C) \tan (c+d x)}{6 d}+\frac{a^2 (8 A+7 B+6 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (8 A+7 B+6 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac{(20 A-5 B+6 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{60 d}+\frac{(5 B+2 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{20 a d}+\frac{C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(8*A + 7*B + 6*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(8*A + 7*B + 6*C)*Tan[c + d*x])/(6*d) + (a^2*(8*A +
7*B + 6*C)*Sec[c + d*x]*Tan[c + d*x])/(24*d) + ((20*A - 5*B + 6*C)*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(60*d
) + (C*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(5*d) + ((5*B + 2*C)*(a + a*Sec[c + d*x])^3*Tan[c +
d*x])/(20*a*d)

Rule 4088

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d
*Csc[e + f*x])^n)/(f*(m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n*
Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b*B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A,
B, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] &&  !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 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 3788

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[(2*a*b)/
d, Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, 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 4046

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 3770

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

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac{\int \sec ^2(c+d x) (a+a \sec (c+d x))^2 (a (5 A+2 C)+a (5 B+2 C) \sec (c+d x)) \, dx}{5 a}\\ &=\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac{(5 B+2 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 a d}+\frac{\int \sec (c+d x) (a+a \sec (c+d x))^2 \left (3 a^2 (5 B+2 C)+a^2 (20 A-5 B+6 C) \sec (c+d x)\right ) \, dx}{20 a^2}\\ &=\frac{(20 A-5 B+6 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac{(5 B+2 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 a d}+\frac{1}{12} (8 A+7 B+6 C) \int \sec (c+d x) (a+a \sec (c+d x))^2 \, dx\\ &=\frac{(20 A-5 B+6 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac{(5 B+2 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 a d}+\frac{1}{12} (8 A+7 B+6 C) \int \sec (c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx+\frac{1}{6} \left (a^2 (8 A+7 B+6 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{a^2 (8 A+7 B+6 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(20 A-5 B+6 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac{(5 B+2 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 a d}+\frac{1}{8} \left (a^2 (8 A+7 B+6 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^2 (8 A+7 B+6 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=\frac{a^2 (8 A+7 B+6 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^2 (8 A+7 B+6 C) \tan (c+d x)}{6 d}+\frac{a^2 (8 A+7 B+6 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac{(20 A-5 B+6 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac{(5 B+2 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 a d}\\ \end{align*}

Mathematica [B]  time = 3.06567, size = 417, normalized size = 2.19 \[ -\frac{a^2 (\cos (c+d x)+1)^2 \sec ^4\left (\frac{1}{2} (c+d x)\right ) \sec ^5(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (240 (8 A+7 B+6 C) \cos ^5(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-240 (3 A+2 B+C) \sin (2 c+d x)+80 (16 A+14 B+15 C) \sin (d x)+240 A \sin (c+2 d x)+240 A \sin (3 c+2 d x)+880 A \sin (2 c+3 d x)-120 A \sin (4 c+3 d x)+120 A \sin (3 c+4 d x)+120 A \sin (5 c+4 d x)+200 A \sin (4 c+5 d x)+330 B \sin (c+2 d x)+330 B \sin (3 c+2 d x)+800 B \sin (2 c+3 d x)+105 B \sin (3 c+4 d x)+105 B \sin (5 c+4 d x)+160 B \sin (4 c+5 d x)+420 C \sin (c+2 d x)+420 C \sin (3 c+2 d x)+720 C \sin (2 c+3 d x)+90 C \sin (3 c+4 d x)+90 C \sin (5 c+4 d x)+144 C \sin (4 c+5 d x))\right )}{3840 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(1 + Cos[c + d*x])^2*(C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^4*Sec[c + d*x]^5*(240*(8*A
+ 7*B + 6*C)*Cos[c + d*x]^5*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/
2]]) - Sec[c]*(80*(16*A + 14*B + 15*C)*Sin[d*x] - 240*(3*A + 2*B + C)*Sin[2*c + d*x] + 240*A*Sin[c + 2*d*x] +
330*B*Sin[c + 2*d*x] + 420*C*Sin[c + 2*d*x] + 240*A*Sin[3*c + 2*d*x] + 330*B*Sin[3*c + 2*d*x] + 420*C*Sin[3*c
+ 2*d*x] + 880*A*Sin[2*c + 3*d*x] + 800*B*Sin[2*c + 3*d*x] + 720*C*Sin[2*c + 3*d*x] - 120*A*Sin[4*c + 3*d*x] +
120*A*Sin[3*c + 4*d*x] + 105*B*Sin[3*c + 4*d*x] + 90*C*Sin[3*c + 4*d*x] + 120*A*Sin[5*c + 4*d*x] + 105*B*Sin[
5*c + 4*d*x] + 90*C*Sin[5*c + 4*d*x] + 200*A*Sin[4*c + 5*d*x] + 160*B*Sin[4*c + 5*d*x] + 144*C*Sin[4*c + 5*d*x
])))/(3840*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.061, size = 315, normalized size = 1.7 \begin{align*}{\frac{5\,{a}^{2}A\tan \left ( dx+c \right ) }{3\,d}}+{\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{{a}^{2}A\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{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{{a}^{2}A\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,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*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

5/3/d*a^2*A*tan(d*x+c)+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+1/d*a^2*A*sec(d*x+c)*tan(d*x+c)+1/d*a^2*A*ln(sec(d*x+c)+tan(d*x+c)
)+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*s
ec(d*x+c)*tan(d*x+c)+3/4/d*a^2*C*ln(sec(d*x+c)+tan(d*x+c))+1/3/d*a^2*A*tan(d*x+c)*sec(d*x+c)^2+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

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Maxima [B]  time = 0.967308, size = 486, normalized size = 2.56 \begin{align*} \frac{80 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 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 )} - 120 \, A 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 )} - 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 \, A a^{2} \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)^2*(a+a*sec(d*x+c))^2*(A+B*sec(d*x+c)+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^2 + 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*l
og(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)) - 120*A*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)) - 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)) + 240*A*a^2*tan(d*x + c))/d

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Fricas [A]  time = 0.536121, size = 463, normalized size = 2.44 \begin{align*} \frac{15 \,{\left (8 \, A + 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 (8 \, A + 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 (8 \,{\left (25 \, A + 20 \, B + 18 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \,{\left (8 \, A + 7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \,{\left (5 \, A + 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*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/240*(15*(8*A + 7*B + 6*C)*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(8*A + 7*B + 6*C)*a^2*cos(d*x + c)^5
*log(-sin(d*x + c) + 1) + 2*(8*(25*A + 20*B + 18*C)*a^2*cos(d*x + c)^4 + 15*(8*A + 7*B + 6*C)*a^2*cos(d*x + c)
^3 + 8*(5*A + 10*B + 9*C)*a^2*cos(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)

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

[Out]

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

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Giac [A]  time = 1.31246, size = 460, normalized size = 2.42 \begin{align*} \frac{15 \,{\left (8 \, A a^{2} + 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 (8 \, A a^{2} + 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 (120 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 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} - 560 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 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} + 1120 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 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} - 1040 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 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} + 360 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 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*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/120*(15*(8*A*a^2 + 7*B*a^2 + 6*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(8*A*a^2 + 7*B*a^2 + 6*C*a^2)*
log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 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 - 560*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 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 + 1120*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 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 - 1040*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 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 + 360*A*a^2*tan(1/2*d*x + 1/2*c) + 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