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

Optimal. Leaf size=197 \[ \frac{a^3 (30 A+23 C) \tan ^3(c+d x)}{120 d}+\frac{a^3 (30 A+23 C) \tan (c+d x)}{10 d}+\frac{a^3 (30 A+23 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{3 a^3 (30 A+23 C) \tan (c+d x) \sec (c+d x)}{80 d}+\frac{(30 A+7 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{120 d}+\frac{C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^3}{6 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^4}{10 a d} \]

[Out]

(a^3*(30*A + 23*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^3*(30*A + 23*C)*Tan[c + d*x])/(10*d) + (3*a^3*(30*A + 23
*C)*Sec[c + d*x]*Tan[c + d*x])/(80*d) + ((30*A + 7*C)*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(120*d) + (C*Sec[c
+ d*x]^2*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(6*d) + (C*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(10*a*d) + (a^3*
(30*A + 23*C)*Tan[c + d*x]^3)/(120*d)

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Rubi [A]  time = 0.416659, antiderivative size = 197, normalized size of antiderivative = 1., number of steps used = 12, 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} \[ \frac{a^3 (30 A+23 C) \tan ^3(c+d x)}{120 d}+\frac{a^3 (30 A+23 C) \tan (c+d x)}{10 d}+\frac{a^3 (30 A+23 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{3 a^3 (30 A+23 C) \tan (c+d x) \sec (c+d x)}{80 d}+\frac{(30 A+7 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{120 d}+\frac{C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^3}{6 d}+\frac{C \tan (c+d x) (a \sec (c+d x)+a)^4}{10 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(30*A + 23*C)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^3*(30*A + 23*C)*Tan[c + d*x])/(10*d) + (3*a^3*(30*A + 23
*C)*Sec[c + d*x]*Tan[c + d*x])/(80*d) + ((30*A + 7*C)*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(120*d) + (C*Sec[c
+ d*x]^2*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(6*d) + (C*(a + a*Sec[c + d*x])^4*Tan[c + d*x])/(10*a*d) + (a^3*
(30*A + 23*C)*Tan[c + d*x]^3)/(120*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))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac{\int \sec ^2(c+d x) (a+a \sec (c+d x))^3 (2 a (3 A+C)+3 a C \sec (c+d x)) \, dx}{6 a}\\ &=\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{10 a d}+\frac{\int \sec (c+d x) (a+a \sec (c+d x))^3 \left (12 a^2 C+a^2 (30 A+7 C) \sec (c+d x)\right ) \, dx}{30 a^2}\\ &=\frac{(30 A+7 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{120 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{10 a d}+\frac{1}{40} (30 A+23 C) \int \sec (c+d x) (a+a \sec (c+d x))^3 \, dx\\ &=\frac{(30 A+7 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{120 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{10 a d}+\frac{1}{40} (30 A+23 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{(30 A+7 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{120 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{10 a d}+\frac{1}{40} \left (a^3 (30 A+23 C)\right ) \int \sec (c+d x) \, dx+\frac{1}{40} \left (a^3 (30 A+23 C)\right ) \int \sec ^4(c+d x) \, dx+\frac{1}{40} \left (3 a^3 (30 A+23 C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{40} \left (3 a^3 (30 A+23 C)\right ) \int \sec ^3(c+d x) \, dx\\ &=\frac{a^3 (30 A+23 C) \tanh ^{-1}(\sin (c+d x))}{40 d}+\frac{3 a^3 (30 A+23 C) \sec (c+d x) \tan (c+d x)}{80 d}+\frac{(30 A+7 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{120 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{10 a d}+\frac{1}{80} \left (3 a^3 (30 A+23 C)\right ) \int \sec (c+d x) \, dx-\frac{\left (a^3 (30 A+23 C)\right ) \operatorname{Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{40 d}-\frac{\left (3 a^3 (30 A+23 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{40 d}\\ &=\frac{a^3 (30 A+23 C) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac{a^3 (30 A+23 C) \tan (c+d x)}{10 d}+\frac{3 a^3 (30 A+23 C) \sec (c+d x) \tan (c+d x)}{80 d}+\frac{(30 A+7 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{120 d}+\frac{C \sec ^2(c+d x) (a+a \sec (c+d x))^3 \tan (c+d x)}{6 d}+\frac{C (a+a \sec (c+d x))^4 \tan (c+d x)}{10 a d}+\frac{a^3 (30 A+23 C) \tan ^3(c+d x)}{120 d}\\ \end{align*}

Mathematica [A]  time = 2.78739, size = 387, normalized size = 1.96 \[ -\frac{a^3 (\cos (c+d x)+1)^3 \sec ^6\left (\frac{1}{2} (c+d x)\right ) \sec ^6(c+d x) \left (A \cos ^2(c+d x)+C\right ) \left (480 (30 A+23 C) \cos ^6(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )-\sec (c) (-160 (45 A+34 C) \sin (c)+1140 A \sin (2 c+d x)+8160 A \sin (c+2 d x)-2640 A \sin (3 c+2 d x)+1590 A \sin (2 c+3 d x)+1590 A \sin (4 c+3 d x)+4080 A \sin (3 c+4 d x)-240 A \sin (5 c+4 d x)+450 A \sin (4 c+5 d x)+450 A \sin (6 c+5 d x)+720 A \sin (5 c+6 d x)+30 (38 A+75 C) \sin (d x)+2250 C \sin (2 c+d x)+7680 C \sin (c+2 d x)-480 C \sin (3 c+2 d x)+1955 C \sin (2 c+3 d x)+1955 C \sin (4 c+3 d x)+3264 C \sin (3 c+4 d x)+345 C \sin (4 c+5 d x)+345 C \sin (6 c+5 d x)+544 C \sin (5 c+6 d x))\right )}{30720 d (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^3*(1 + Cos[c + d*x])^3*(C + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^6*Sec[c + d*x]^6*(480*(30*A + 23*C)*Cos[c +
d*x]^6*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) - Sec[c]*(-160*(
45*A + 34*C)*Sin[c] + 30*(38*A + 75*C)*Sin[d*x] + 1140*A*Sin[2*c + d*x] + 2250*C*Sin[2*c + d*x] + 8160*A*Sin[c
+ 2*d*x] + 7680*C*Sin[c + 2*d*x] - 2640*A*Sin[3*c + 2*d*x] - 480*C*Sin[3*c + 2*d*x] + 1590*A*Sin[2*c + 3*d*x]
+ 1955*C*Sin[2*c + 3*d*x] + 1590*A*Sin[4*c + 3*d*x] + 1955*C*Sin[4*c + 3*d*x] + 4080*A*Sin[3*c + 4*d*x] + 326
4*C*Sin[3*c + 4*d*x] - 240*A*Sin[5*c + 4*d*x] + 450*A*Sin[4*c + 5*d*x] + 345*C*Sin[4*c + 5*d*x] + 450*A*Sin[6*
c + 5*d*x] + 345*C*Sin[6*c + 5*d*x] + 720*A*Sin[5*c + 6*d*x] + 544*C*Sin[5*c + 6*d*x])))/(30720*d*(A + 2*C + A
*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.06, size = 257, normalized size = 1.3 \begin{align*} 3\,{\frac{A{a}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{34\,{a}^{3}C\tan \left ( dx+c \right ) }{15\,d}}+{\frac{17\,{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{15\,A{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{15\,A{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{23\,{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{23\,{a}^{3}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{23\,{a}^{3}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}}+{\frac{A{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 ) ^{4}}{5\,d}}+{\frac{A{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 ) ^{5}}{6\,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))^3*(A+C*sec(d*x+c)^2),x)

[Out]

3/d*A*a^3*tan(d*x+c)+34/15*a^3*C*tan(d*x+c)/d+17/15/d*a^3*C*tan(d*x+c)*sec(d*x+c)^2+15/8/d*A*a^3*sec(d*x+c)*ta
n(d*x+c)+15/8/d*A*a^3*ln(sec(d*x+c)+tan(d*x+c))+23/24/d*a^3*C*tan(d*x+c)*sec(d*x+c)^3+23/16/d*a^3*C*sec(d*x+c)
*tan(d*x+c)+23/16/d*a^3*C*ln(sec(d*x+c)+tan(d*x+c))+1/d*A*a^3*tan(d*x+c)*sec(d*x+c)^2+3/5/d*a^3*C*tan(d*x+c)*s
ec(d*x+c)^4+1/4/d*A*a^3*tan(d*x+c)*sec(d*x+c)^3+1/6/d*a^3*C*tan(d*x+c)*sec(d*x+c)^5

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Maxima [B]  time = 0.960638, size = 516, normalized size = 2.62 \begin{align*} \frac{480 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 96 \,{\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} + 160 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} - 5 \, C a^{3}{\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 )} - 30 \, A 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 )} - 90 \, 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 )} - 360 \, A 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 )} + 480 \, A a^{3} \tan \left (d x + c\right )}{480 \, 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))^3*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/480*(480*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 96*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c
))*C*a^3 + 160*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 - 5*C*a^3*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 3
3*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)) - 30*A*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)) - 90*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)) - 360*A*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)) + 480*A*a^3*tan(d*x + c))/d

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Fricas [A]  time = 0.531258, size = 474, normalized size = 2.41 \begin{align*} \frac{15 \,{\left (30 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \,{\left (30 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (16 \,{\left (45 \, A + 34 \, C\right )} a^{3} \cos \left (d x + c\right )^{5} + 15 \,{\left (30 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{4} + 16 \,{\left (15 \, A + 17 \, C\right )} a^{3} \cos \left (d x + c\right )^{3} + 10 \,{\left (6 \, A + 23 \, C\right )} a^{3} \cos \left (d x + c\right )^{2} + 144 \, C a^{3} \cos \left (d x + c\right ) + 40 \, C a^{3}\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \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))^3*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/480*(15*(30*A + 23*C)*a^3*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(30*A + 23*C)*a^3*cos(d*x + c)^6*log(-si
n(d*x + c) + 1) + 2*(16*(45*A + 34*C)*a^3*cos(d*x + c)^5 + 15*(30*A + 23*C)*a^3*cos(d*x + c)^4 + 16*(15*A + 17
*C)*a^3*cos(d*x + c)^3 + 10*(6*A + 23*C)*a^3*cos(d*x + c)^2 + 144*C*a^3*cos(d*x + c) + 40*C*a^3)*sin(d*x + c))
/(d*cos(d*x + c)^6)

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

[Out]

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

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Giac [A]  time = 1.24728, size = 378, normalized size = 1.92 \begin{align*} \frac{15 \,{\left (30 \, A a^{3} + 23 \, C a^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 15 \,{\left (30 \, A a^{3} + 23 \, 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 (450 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} + 345 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{11} - 2550 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} - 1955 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{9} + 5940 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 4554 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 7500 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 5814 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 5130 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3165 \, C a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 1470 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1575 \, 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 )}^{6}}}{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))^3*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/240*(15*(30*A*a^3 + 23*C*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(30*A*a^3 + 23*C*a^3)*log(abs(tan(1/2*
d*x + 1/2*c) - 1)) - 2*(450*A*a^3*tan(1/2*d*x + 1/2*c)^11 + 345*C*a^3*tan(1/2*d*x + 1/2*c)^11 - 2550*A*a^3*tan
(1/2*d*x + 1/2*c)^9 - 1955*C*a^3*tan(1/2*d*x + 1/2*c)^9 + 5940*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 4554*C*a^3*tan(1
/2*d*x + 1/2*c)^7 - 7500*A*a^3*tan(1/2*d*x + 1/2*c)^5 - 5814*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 5130*A*a^3*tan(1/2
*d*x + 1/2*c)^3 + 3165*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 1470*A*a^3*tan(1/2*d*x + 1/2*c) - 1575*C*a^3*tan(1/2*d*x
+ 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d