3.111 \(\int \sec ^2(c+d x) (a+a \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=228 \[ \frac{8 a^4 (14 A+11 C) \tan ^3(c+d x)}{105 d}+\frac{16 a^4 (14 A+11 C) \tan (c+d x)}{35 d}+\frac{a^4 (14 A+11 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a^4 (14 A+11 C) \tan (c+d x) \sec ^3(c+d x)}{70 d}+\frac{27 a^4 (14 A+11 C) \tan (c+d x) \sec (c+d x)}{140 d}+\frac{(21 A+4 C) \tan (c+d x) (a \sec (c+d x)+a)^4}{105 d}+\frac{C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^4}{7 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^5}{21 a d} \]

[Out]

(a^4*(14*A + 11*C)*ArcTanh[Sin[c + d*x]])/(4*d) + (16*a^4*(14*A + 11*C)*Tan[c + d*x])/(35*d) + (27*a^4*(14*A +
11*C)*Sec[c + d*x]*Tan[c + d*x])/(140*d) + (a^4*(14*A + 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(70*d) + ((21*A +
4*C)*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(105*d) + (C*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(7*
d) + (2*C*(a + a*Sec[c + d*x])^5*Tan[c + d*x])/(21*a*d) + (8*a^4*(14*A + 11*C)*Tan[c + d*x]^3)/(105*d)

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Rubi [A]  time = 0.481239, antiderivative size = 228, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 8, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.242, Rules used = {4089, 4010, 4001, 3791, 3770, 3767, 8, 3768} \[ \frac{8 a^4 (14 A+11 C) \tan ^3(c+d x)}{105 d}+\frac{16 a^4 (14 A+11 C) \tan (c+d x)}{35 d}+\frac{a^4 (14 A+11 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{a^4 (14 A+11 C) \tan (c+d x) \sec ^3(c+d x)}{70 d}+\frac{27 a^4 (14 A+11 C) \tan (c+d x) \sec (c+d x)}{140 d}+\frac{(21 A+4 C) \tan (c+d x) (a \sec (c+d x)+a)^4}{105 d}+\frac{C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^4}{7 d}+\frac{2 C \tan (c+d x) (a \sec (c+d x)+a)^5}{21 a d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^4*(14*A + 11*C)*ArcTanh[Sin[c + d*x]])/(4*d) + (16*a^4*(14*A + 11*C)*Tan[c + d*x])/(35*d) + (27*a^4*(14*A +
11*C)*Sec[c + d*x]*Tan[c + d*x])/(140*d) + (a^4*(14*A + 11*C)*Sec[c + d*x]^3*Tan[c + d*x])/(70*d) + ((21*A +
4*C)*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(105*d) + (C*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(7*
d) + (2*C*(a + a*Sec[c + d*x])^5*Tan[c + d*x])/(21*a*d) + (8*a^4*(14*A + 11*C)*Tan[c + d*x]^3)/(105*d)

Rule 4089

Int[((A_.) + 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*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 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 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 ^2(c+d x) (a+a \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{\int \sec ^2(c+d x) (a+a \sec (c+d x))^4 (a (7 A+2 C)+4 a C \sec (c+d x)) \, dx}{7 a}\\ &=\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{2 C (a+a \sec (c+d x))^5 \tan (c+d x)}{21 a d}+\frac{\int \sec (c+d x) (a+a \sec (c+d x))^4 \left (20 a^2 C+2 a^2 (21 A+4 C) \sec (c+d x)\right ) \, dx}{42 a^2}\\ &=\frac{(21 A+4 C) (a+a \sec (c+d x))^4 \tan (c+d x)}{105 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{2 C (a+a \sec (c+d x))^5 \tan (c+d x)}{21 a d}+\frac{1}{35} (2 (14 A+11 C)) \int \sec (c+d x) (a+a \sec (c+d x))^4 \, dx\\ &=\frac{(21 A+4 C) (a+a \sec (c+d x))^4 \tan (c+d x)}{105 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{2 C (a+a \sec (c+d x))^5 \tan (c+d x)}{21 a d}+\frac{1}{35} (2 (14 A+11 C)) \int \left (a^4 \sec (c+d x)+4 a^4 \sec ^2(c+d x)+6 a^4 \sec ^3(c+d x)+4 a^4 \sec ^4(c+d x)+a^4 \sec ^5(c+d x)\right ) \, dx\\ &=\frac{(21 A+4 C) (a+a \sec (c+d x))^4 \tan (c+d x)}{105 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{2 C (a+a \sec (c+d x))^5 \tan (c+d x)}{21 a d}+\frac{1}{35} \left (2 a^4 (14 A+11 C)\right ) \int \sec (c+d x) \, dx+\frac{1}{35} \left (2 a^4 (14 A+11 C)\right ) \int \sec ^5(c+d x) \, dx+\frac{1}{35} \left (8 a^4 (14 A+11 C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{35} \left (8 a^4 (14 A+11 C)\right ) \int \sec ^4(c+d x) \, dx+\frac{1}{35} \left (12 a^4 (14 A+11 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{2 a^4 (14 A+11 C) \tanh ^{-1}(\sin (c+d x))}{35 d}+\frac{6 a^4 (14 A+11 C) \sec (c+d x) \tan (c+d x)}{35 d}+\frac{a^4 (14 A+11 C) \sec ^3(c+d x) \tan (c+d x)}{70 d}+\frac{(21 A+4 C) (a+a \sec (c+d x))^4 \tan (c+d x)}{105 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{2 C (a+a \sec (c+d x))^5 \tan (c+d x)}{21 a d}+\frac{1}{70} \left (3 a^4 (14 A+11 C)\right ) \int \sec ^3(c+d x) \, dx+\frac{1}{35} \left (6 a^4 (14 A+11 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (8 a^4 (14 A+11 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{35 d}-\frac{\left (8 a^4 (14 A+11 C)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{35 d}\\ &=\frac{8 a^4 (14 A+11 C) \tanh ^{-1}(\sin (c+d x))}{35 d}+\frac{16 a^4 (14 A+11 C) \tan (c+d x)}{35 d}+\frac{27 a^4 (14 A+11 C) \sec (c+d x) \tan (c+d x)}{140 d}+\frac{a^4 (14 A+11 C) \sec ^3(c+d x) \tan (c+d x)}{70 d}+\frac{(21 A+4 C) (a+a \sec (c+d x))^4 \tan (c+d x)}{105 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{2 C (a+a \sec (c+d x))^5 \tan (c+d x)}{21 a d}+\frac{8 a^4 (14 A+11 C) \tan ^3(c+d x)}{105 d}+\frac{1}{140} \left (3 a^4 (14 A+11 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{a^4 (14 A+11 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac{16 a^4 (14 A+11 C) \tan (c+d x)}{35 d}+\frac{27 a^4 (14 A+11 C) \sec (c+d x) \tan (c+d x)}{140 d}+\frac{a^4 (14 A+11 C) \sec ^3(c+d x) \tan (c+d x)}{70 d}+\frac{(21 A+4 C) (a+a \sec (c+d x))^4 \tan (c+d x)}{105 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^4 \tan (c+d x)}{7 d}+\frac{2 C (a+a \sec (c+d x))^5 \tan (c+d x)}{21 a d}+\frac{8 a^4 (14 A+11 C) \tan ^3(c+d x)}{105 d}\\ \end{align*}

Mathematica [A]  time = 4.9479, size = 419, normalized size = 1.84 \[ -\frac{a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \sec ^7(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (6720 (14 A+11 C) \cos ^7(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-140 (217 A+122 C) \sin (2 c+d x)+10710 A \sin (c+2 d x)+10710 A \sin (3 c+2 d x)+41244 A \sin (2 c+3 d x)-7560 A \sin (4 c+3 d x)+7560 A \sin (3 c+4 d x)+7560 A \sin (5 c+4 d x)+15848 A \sin (4 c+5 d x)-420 A \sin (6 c+5 d x)+1470 A \sin (5 c+6 d x)+1470 A \sin (7 c+6 d x)+2324 A \sin (6 c+7 d x)+560 (91 A+83 C) \sin (d x)+16415 C \sin (c+2 d x)+16415 C \sin (3 c+2 d x)+37296 C \sin (2 c+3 d x)-840 C \sin (4 c+3 d x)+7700 C \sin (3 c+4 d x)+7700 C \sin (5 c+4 d x)+12712 C \sin (4 c+5 d x)+1155 C \sin (5 c+6 d x)+1155 C \sin (7 c+6 d x)+1816 C \sin (6 c+7 d x))\right )}{215040 d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2),x]

[Out]

-(a^4*(1 + Cos[c + d*x])^4*(C + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^8*Sec[c + d*x]^7*(6720*(14*A + 11*C)*Cos[c
+ d*x]^7*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(560*(
91*A + 83*C)*Sin[d*x] - 140*(217*A + 122*C)*Sin[2*c + d*x] + 10710*A*Sin[c + 2*d*x] + 16415*C*Sin[c + 2*d*x] +
10710*A*Sin[3*c + 2*d*x] + 16415*C*Sin[3*c + 2*d*x] + 41244*A*Sin[2*c + 3*d*x] + 37296*C*Sin[2*c + 3*d*x] - 7
560*A*Sin[4*c + 3*d*x] - 840*C*Sin[4*c + 3*d*x] + 7560*A*Sin[3*c + 4*d*x] + 7700*C*Sin[3*c + 4*d*x] + 7560*A*S
in[5*c + 4*d*x] + 7700*C*Sin[5*c + 4*d*x] + 15848*A*Sin[4*c + 5*d*x] + 12712*C*Sin[4*c + 5*d*x] - 420*A*Sin[6*
c + 5*d*x] + 1470*A*Sin[5*c + 6*d*x] + 1155*C*Sin[5*c + 6*d*x] + 1470*A*Sin[7*c + 6*d*x] + 1155*C*Sin[7*c + 6*
d*x] + 2324*A*Sin[6*c + 7*d*x] + 1816*C*Sin[6*c + 7*d*x])))/(215040*d*(A + 2*C + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.065, size = 303, normalized size = 1.3 \begin{align*}{\frac{83\,A{a}^{4}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{454\,{a}^{4}C\tan \left ( dx+c \right ) }{105\,d}}+{\frac{227\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{105\,d}}+{\frac{7\,A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{7\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{11\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{6\,d}}+{\frac{11\,{a}^{4}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{4\,d}}+{\frac{11\,{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{34\,A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{48\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{35\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{2\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{3\,d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{6}}{7\,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))^4*(A+C*sec(d*x+c)^2),x)

[Out]

83/15/d*A*a^4*tan(d*x+c)+454/105/d*a^4*C*tan(d*x+c)+227/105/d*a^4*C*tan(d*x+c)*sec(d*x+c)^2+7/2/d*A*a^4*sec(d*
x+c)*tan(d*x+c)+7/2/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+11/6/d*a^4*C*tan(d*x+c)*sec(d*x+c)^3+11/4/d*a^4*C*sec(d*
x+c)*tan(d*x+c)+11/4/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+34/15/d*A*a^4*tan(d*x+c)*sec(d*x+c)^2+48/35/d*a^4*C*tan
(d*x+c)*sec(d*x+c)^4+1/d*A*a^4*tan(d*x+c)*sec(d*x+c)^3+2/3/d*a^4*C*tan(d*x+c)*sec(d*x+c)^5+1/5/d*A*a^4*tan(d*x
+c)*sec(d*x+c)^4+1/7/d*a^4*C*tan(d*x+c)*sec(d*x+c)^6

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Maxima [B]  time = 0.965796, size = 624, normalized size = 2.74 \begin{align*} \frac{56 \,{\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 )} A a^{4} + 1680 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{4} + 24 \,{\left (5 \, \tan \left (d x + c\right )^{7} + 21 \, \tan \left (d x + c\right )^{5} + 35 \, \tan \left (d x + c\right )^{3} + 35 \, \tan \left (d x + c\right )\right )} C a^{4} + 336 \,{\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^{4} + 280 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} - 35 \, C a^{4}{\left (\frac{2 \,{\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 210 \, A a^{4}{\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 )} - 210 \, C a^{4}{\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 )} - 840 \, A a^{4}{\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 )} + 840 \, A a^{4} \tan \left (d x + c\right )}{840 \, 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))^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/840*(56*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^4 + 1680*(tan(d*x + c)^3 + 3*tan(d*x +
c))*A*a^4 + 24*(5*tan(d*x + c)^7 + 21*tan(d*x + c)^5 + 35*tan(d*x + c)^3 + 35*tan(d*x + c))*C*a^4 + 336*(3*tan
(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a^4 + 280*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^4 - 35*C*
a^4*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*sin(d*x + c)^4 + 3*sin(d*
x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 210*A*a^4*(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)) - 210*C
*a^4*(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)) - 840*A*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(
d*x + c) - 1)) + 840*A*a^4*tan(d*x + c))/d

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Fricas [A]  time = 0.542161, size = 531, normalized size = 2.33 \begin{align*} \frac{105 \,{\left (14 \, A + 11 \, C\right )} a^{4} \cos \left (d x + c\right )^{7} \log \left (\sin \left (d x + c\right ) + 1\right ) - 105 \,{\left (14 \, A + 11 \, C\right )} a^{4} \cos \left (d x + c\right )^{7} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (4 \,{\left (581 \, A + 454 \, C\right )} a^{4} \cos \left (d x + c\right )^{6} + 105 \,{\left (14 \, A + 11 \, C\right )} a^{4} \cos \left (d x + c\right )^{5} + 4 \,{\left (238 \, A + 227 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} + 70 \,{\left (6 \, A + 11 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 12 \,{\left (7 \, A + 48 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 280 \, C a^{4} \cos \left (d x + c\right ) + 60 \, C a^{4}\right )} \sin \left (d x + c\right )}{840 \, d \cos \left (d x + c\right )^{7}} \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))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/840*(105*(14*A + 11*C)*a^4*cos(d*x + c)^7*log(sin(d*x + c) + 1) - 105*(14*A + 11*C)*a^4*cos(d*x + c)^7*log(-
sin(d*x + c) + 1) + 2*(4*(581*A + 454*C)*a^4*cos(d*x + c)^6 + 105*(14*A + 11*C)*a^4*cos(d*x + c)^5 + 4*(238*A
+ 227*C)*a^4*cos(d*x + c)^4 + 70*(6*A + 11*C)*a^4*cos(d*x + c)^3 + 12*(7*A + 48*C)*a^4*cos(d*x + c)^2 + 280*C*
a^4*cos(d*x + c) + 60*C*a^4)*sin(d*x + c))/(d*cos(d*x + c)^7)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{4} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 A \sec ^{3}{\left (c + d x \right )}\, dx + \int 6 A \sec ^{4}{\left (c + d x \right )}\, dx + \int 4 A \sec ^{5}{\left (c + d x \right )}\, dx + \int A \sec ^{6}{\left (c + d x \right )}\, dx + \int C \sec ^{4}{\left (c + d x \right )}\, dx + \int 4 C \sec ^{5}{\left (c + d x \right )}\, dx + \int 6 C \sec ^{6}{\left (c + d x \right )}\, dx + \int 4 C \sec ^{7}{\left (c + d x \right )}\, dx + \int C \sec ^{8}{\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))**4*(A+C*sec(d*x+c)**2),x)

[Out]

a**4*(Integral(A*sec(c + d*x)**2, x) + Integral(4*A*sec(c + d*x)**3, x) + Integral(6*A*sec(c + d*x)**4, x) + I
ntegral(4*A*sec(c + d*x)**5, x) + Integral(A*sec(c + d*x)**6, x) + Integral(C*sec(c + d*x)**4, x) + Integral(4
*C*sec(c + d*x)**5, x) + Integral(6*C*sec(c + d*x)**6, x) + Integral(4*C*sec(c + d*x)**7, x) + Integral(C*sec(
c + d*x)**8, x))

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Giac [A]  time = 1.26657, size = 424, normalized size = 1.86 \begin{align*} \frac{105 \,{\left (14 \, A a^{4} + 11 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 105 \,{\left (14 \, A a^{4} + 11 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (1470 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} + 1155 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{13} - 9800 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 7700 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 27734 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 21791 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 43008 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 33792 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 39914 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 31521 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 21560 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 14700 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5250 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5565 \, C a^{4} \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 )}^{7}}}{420 \, 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))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/420*(105*(14*A*a^4 + 11*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 105*(14*A*a^4 + 11*C*a^4)*log(abs(tan(1/
2*d*x + 1/2*c) - 1)) - 2*(1470*A*a^4*tan(1/2*d*x + 1/2*c)^13 + 1155*C*a^4*tan(1/2*d*x + 1/2*c)^13 - 9800*A*a^4
*tan(1/2*d*x + 1/2*c)^11 - 7700*C*a^4*tan(1/2*d*x + 1/2*c)^11 + 27734*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 21791*C*a
^4*tan(1/2*d*x + 1/2*c)^9 - 43008*A*a^4*tan(1/2*d*x + 1/2*c)^7 - 33792*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 39914*A*
a^4*tan(1/2*d*x + 1/2*c)^5 + 31521*C*a^4*tan(1/2*d*x + 1/2*c)^5 - 21560*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 14700*C
*a^4*tan(1/2*d*x + 1/2*c)^3 + 5250*A*a^4*tan(1/2*d*x + 1/2*c) + 5565*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x
+ 1/2*c)^2 - 1)^7)/d