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

Optimal. Leaf size=209 \[ \frac{5 a^4 (A-B-2 C) \sin (c+d x)}{2 d}+\frac{a^4 (8 A+13 B+12 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(A-B-2 C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac{(3 A+18 B+22 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac{1}{2} a^4 x (13 A+8 B+2 C)-\frac{a (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{6 d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d} \]

[Out]

(a^4*(13*A + 8*B + 2*C)*x)/2 + (a^4*(8*A + 13*B + 12*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^4*(A - B - 2*C)*Si
n[c + d*x])/(2*d) - (a*(3*A - 2*C)*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]*(a + a*Sec[c +
d*x])^4*Sin[c + d*x])/(2*d) - ((A - B - 2*C)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + ((3*A + 18*B +
22*C)*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(6*d)

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Rubi [A]  time = 0.565657, antiderivative size = 209, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.098, Rules used = {4086, 4018, 3996, 3770} \[ \frac{5 a^4 (A-B-2 C) \sin (c+d x)}{2 d}+\frac{a^4 (8 A+13 B+12 C) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac{(A-B-2 C) \sin (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{2 d}+\frac{(3 A+18 B+22 C) \sin (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{6 d}+\frac{1}{2} a^4 x (13 A+8 B+2 C)-\frac{a (3 A-2 C) \sin (c+d x) (a \sec (c+d x)+a)^3}{6 d}+\frac{A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^4}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(13*A + 8*B + 2*C)*x)/2 + (a^4*(8*A + 13*B + 12*C)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^4*(A - B - 2*C)*Si
n[c + d*x])/(2*d) - (a*(3*A - 2*C)*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]*(a + a*Sec[c +
d*x])^4*Sin[c + d*x])/(2*d) - ((A - B - 2*C)*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x])/(2*d) + ((3*A + 18*B +
22*C)*(a^4 + a^4*Sec[c + d*x])*Sin[c + d*x])/(6*d)

Rule 4086

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[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*
Csc[e + f*x])^n)/(f*n), x] - Dist[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m -
b*B*n - b*(A*(m + n + 1) + C*n)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, m}, x] && EqQ[a^2 -
b^2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -2^(-1)] || EqQ[m + n + 1, 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 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*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 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 \cos ^2(c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^4 (2 a (2 A+B)-a (3 A-2 C) \sec (c+d x)) \, dx}{2 a}\\ &=-\frac{a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^3 \left (a^2 (15 A+6 B-2 C)-6 a^2 (A-B-2 C) \sec (c+d x)\right ) \, dx}{6 a}\\ &=-\frac{a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{(A-B-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x))^2 \left (2 a^3 (18 A+3 B-8 C)+2 a^3 (3 A+18 B+22 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=-\frac{a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{(A-B-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{(3 A+18 B+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{\int \cos (c+d x) (a+a \sec (c+d x)) \left (30 a^4 (A-B-2 C)+6 a^4 (8 A+13 B+12 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac{5 a^4 (A-B-2 C) \sin (c+d x)}{2 d}-\frac{a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{(A-B-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{(3 A+18 B+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}-\frac{\int \left (-6 a^5 (13 A+8 B+2 C)-6 a^5 (8 A+13 B+12 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac{1}{2} a^4 (13 A+8 B+2 C) x+\frac{5 a^4 (A-B-2 C) \sin (c+d x)}{2 d}-\frac{a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{(A-B-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{(3 A+18 B+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}+\frac{1}{2} \left (a^4 (8 A+13 B+12 C)\right ) \int \sec (c+d x) \, dx\\ &=\frac{1}{2} a^4 (13 A+8 B+2 C) x+\frac{a^4 (8 A+13 B+12 C) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{5 a^4 (A-B-2 C) \sin (c+d x)}{2 d}-\frac{a (3 A-2 C) (a+a \sec (c+d x))^3 \sin (c+d x)}{6 d}+\frac{A \cos (c+d x) (a+a \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{(A-B-2 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{2 d}+\frac{(3 A+18 B+22 C) \left (a^4+a^4 \sec (c+d x)\right ) \sin (c+d x)}{6 d}\\ \end{align*}

Mathematica [B]  time = 4.62698, size = 524, normalized size = 2.51 \[ \frac{a^4 (\cos (c+d x)+1)^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (\sec (c) (36 d x (13 A+8 B+2 C) \cos (2 c+d x)+36 d x (13 A+8 B+2 C) \cos (d x)-42 A \sin (2 c+d x)+96 A \sin (c+2 d x)+96 A \sin (3 c+2 d x)+57 A \sin (2 c+3 d x)+9 A \sin (4 c+3 d x)+48 A \sin (3 c+4 d x)+48 A \sin (5 c+4 d x)+3 A \sin (4 c+5 d x)+3 A \sin (6 c+5 d x)+156 A d x \cos (2 c+3 d x)+156 A d x \cos (4 c+3 d x)+102 A \sin (d x)-192 B \sin (2 c+d x)+48 B \sin (c+2 d x)+48 B \sin (3 c+2 d x)+192 B \sin (2 c+3 d x)+12 B \sin (3 c+4 d x)+12 B \sin (5 c+4 d x)+96 B d x \cos (2 c+3 d x)+96 B d x \cos (4 c+3 d x)+384 B \sin (d x)-288 C \sin (2 c+d x)+96 C \sin (c+2 d x)+96 C \sin (3 c+2 d x)+320 C \sin (2 c+3 d x)+24 C d x \cos (2 c+3 d x)+24 C d x \cos (4 c+3 d x)+672 C \sin (d x))-96 (8 A+13 B+12 C) \cos ^3(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{1536 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(1 + Cos[c + d*x])^4*(C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^8*Sec[c + d*x]^3*(-96*(8*A
+ 13*B + 12*C)*Cos[c + d*x]^3*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)
/2]]) + Sec[c]*(36*(13*A + 8*B + 2*C)*d*x*Cos[d*x] + 36*(13*A + 8*B + 2*C)*d*x*Cos[2*c + d*x] + 156*A*d*x*Cos[
2*c + 3*d*x] + 96*B*d*x*Cos[2*c + 3*d*x] + 24*C*d*x*Cos[2*c + 3*d*x] + 156*A*d*x*Cos[4*c + 3*d*x] + 96*B*d*x*C
os[4*c + 3*d*x] + 24*C*d*x*Cos[4*c + 3*d*x] + 102*A*Sin[d*x] + 384*B*Sin[d*x] + 672*C*Sin[d*x] - 42*A*Sin[2*c
+ d*x] - 192*B*Sin[2*c + d*x] - 288*C*Sin[2*c + d*x] + 96*A*Sin[c + 2*d*x] + 48*B*Sin[c + 2*d*x] + 96*C*Sin[c
+ 2*d*x] + 96*A*Sin[3*c + 2*d*x] + 48*B*Sin[3*c + 2*d*x] + 96*C*Sin[3*c + 2*d*x] + 57*A*Sin[2*c + 3*d*x] + 192
*B*Sin[2*c + 3*d*x] + 320*C*Sin[2*c + 3*d*x] + 9*A*Sin[4*c + 3*d*x] + 48*A*Sin[3*c + 4*d*x] + 12*B*Sin[3*c + 4
*d*x] + 48*A*Sin[5*c + 4*d*x] + 12*B*Sin[5*c + 4*d*x] + 3*A*Sin[4*c + 5*d*x] + 3*A*Sin[6*c + 5*d*x])))/(1536*d
*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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Maple [A]  time = 0.125, size = 279, normalized size = 1.3 \begin{align*}{\frac{A{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{13\,{a}^{4}Ax}{2}}+{\frac{13\,A{a}^{4}c}{2\,d}}+{\frac{B{a}^{4}\sin \left ( dx+c \right ) }{d}}+{a}^{4}Cx+{\frac{C{a}^{4}c}{d}}+4\,{\frac{A{a}^{4}\sin \left ( dx+c \right ) }{d}}+4\,B{a}^{4}x+4\,{\frac{B{a}^{4}c}{d}}+6\,{\frac{{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{13\,B{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{20\,{a}^{4}C\tan \left ( dx+c \right ) }{3\,d}}+4\,{\frac{A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{B{a}^{4}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{4}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{A{a}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{B{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/2/d*A*a^4*cos(d*x+c)*sin(d*x+c)+13/2*a^4*A*x+13/2/d*A*a^4*c+1/d*B*a^4*sin(d*x+c)+a^4*C*x+1/d*C*a^4*c+4/d*A*a
^4*sin(d*x+c)+4*B*a^4*x+4/d*B*a^4*c+6/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+13/2/d*B*a^4*ln(sec(d*x+c)+tan(d*x+c))
+20/3/d*a^4*C*tan(d*x+c)+4/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+4/d*B*a^4*tan(d*x+c)+2/d*a^4*C*sec(d*x+c)*tan(d*x
+c)+1/d*A*a^4*tan(d*x+c)+1/2/d*B*a^4*sec(d*x+c)*tan(d*x+c)+1/3/d*a^4*C*tan(d*x+c)*sec(d*x+c)^2

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Maxima [A]  time = 0.966271, size = 432, normalized size = 2.07 \begin{align*} \frac{3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 72 \,{\left (d x + c\right )} A a^{4} + 48 \,{\left (d x + c\right )} B a^{4} + 4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} + 12 \,{\left (d x + c\right )} C a^{4} - 3 \, B 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 )} - 12 \, C 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 )} + 24 \, A a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, B a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{4}{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{4} \sin \left (d x + c\right ) + 12 \, B a^{4} \sin \left (d x + c\right ) + 12 \, A a^{4} \tan \left (d x + c\right ) + 48 \, B a^{4} \tan \left (d x + c\right ) + 72 \, C a^{4} \tan \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 + 72*(d*x + c)*A*a^4 + 48*(d*x + c)*B*a^4 + 4*(tan(d*x + c)^3 +
3*tan(d*x + c))*C*a^4 + 12*(d*x + c)*C*a^4 - 3*B*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)) - 12*C*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(si
n(d*x + c) - 1)) + 24*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*B*a^4*(log(sin(d*x + c) + 1)
- log(sin(d*x + c) - 1)) + 24*C*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*A*a^4*sin(d*x + c) +
12*B*a^4*sin(d*x + c) + 12*A*a^4*tan(d*x + c) + 48*B*a^4*tan(d*x + c) + 72*C*a^4*tan(d*x + c))/d

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Fricas [A]  time = 0.560103, size = 487, normalized size = 2.33 \begin{align*} \frac{6 \,{\left (13 \, A + 8 \, B + 2 \, C\right )} a^{4} d x \cos \left (d x + c\right )^{3} + 3 \,{\left (8 \, A + 13 \, B + 12 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (8 \, A + 13 \, B + 12 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 6 \,{\left (4 \, A + B\right )} a^{4} \cos \left (d x + c\right )^{3} + 2 \,{\left (3 \, A + 12 \, B + 20 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 3 \,{\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 2 \, C a^{4}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/12*(6*(13*A + 8*B + 2*C)*a^4*d*x*cos(d*x + c)^3 + 3*(8*A + 13*B + 12*C)*a^4*cos(d*x + c)^3*log(sin(d*x + c)
+ 1) - 3*(8*A + 13*B + 12*C)*a^4*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(3*A*a^4*cos(d*x + c)^4 + 6*(4*A +
B)*a^4*cos(d*x + c)^3 + 2*(3*A + 12*B + 20*C)*a^4*cos(d*x + c)^2 + 3*(B + 4*C)*a^4*cos(d*x + c) + 2*C*a^4)*sin
(d*x + c))/(d*cos(d*x + c)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+a*sec(d*x+c))**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.31841, size = 468, normalized size = 2.24 \begin{align*} \frac{3 \,{\left (13 \, A a^{4} + 8 \, B a^{4} + 2 \, C a^{4}\right )}{\left (d x + c\right )} + 3 \,{\left (8 \, A a^{4} + 13 \, B a^{4} + 12 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (8 \, A a^{4} + 13 \, B a^{4} + 12 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) + \frac{6 \,{\left (7 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 9 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B 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 )}^{2}} - \frac{2 \,{\left (6 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 21 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 30 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 12 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 48 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 76 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 27 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 54 \, 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 )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*sec(d*x+c))^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(13*A*a^4 + 8*B*a^4 + 2*C*a^4)*(d*x + c) + 3*(8*A*a^4 + 13*B*a^4 + 12*C*a^4)*log(abs(tan(1/2*d*x + 1/2*
c) + 1)) - 3*(8*A*a^4 + 13*B*a^4 + 12*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 6*(7*A*a^4*tan(1/2*d*x + 1/2
*c)^3 + 2*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 9*A*a^4*tan(1/2*d*x + 1/2*c) + 2*B*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2
*d*x + 1/2*c)^2 + 1)^2 - 2*(6*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 21*B*a^4*tan(1/2*d*x + 1/2*c)^5 + 30*C*a^4*tan(1/
2*d*x + 1/2*c)^5 - 12*A*a^4*tan(1/2*d*x + 1/2*c)^3 - 48*B*a^4*tan(1/2*d*x + 1/2*c)^3 - 76*C*a^4*tan(1/2*d*x +
1/2*c)^3 + 6*A*a^4*tan(1/2*d*x + 1/2*c) + 27*B*a^4*tan(1/2*d*x + 1/2*c) + 54*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(
1/2*d*x + 1/2*c)^2 - 1)^3)/d