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

Optimal. Leaf size=219 \[ -\frac{b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \tan (c+d x)}{6 d}+\frac{2 a b \left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{1}{2} a^2 x \left (a^2 (A+2 C)+12 A b^2\right )-\frac{b^2 (15 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^2}{6 d}-\frac{a b^3 (9 A-4 C) \tan (c+d x) \sec (c+d x)}{3 d}+\frac{2 A b \sin (c+d x) (a+b \sec (c+d x))^3}{d}+\frac{A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^4}{2 d} \]

[Out]

(a^2*(12*A*b^2 + a^2*(A + 2*C))*x)/2 + (2*a*b*(2*A*b^2 + (2*a^2 + b^2)*C)*ArcTanh[Sin[c + d*x]])/d + (2*A*b*(a
+ b*Sec[c + d*x])^3*Sin[c + d*x])/d + (A*Cos[c + d*x]*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(2*d) - (b^2*(a^2*
(39*A - 34*C) - 2*b^2*(3*A + 2*C))*Tan[c + d*x])/(6*d) - (a*b^3*(9*A - 4*C)*Sec[c + d*x]*Tan[c + d*x])/(3*d) -
(b^2*(15*A - 2*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(6*d)

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Rubi [A]  time = 0.60505, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {4095, 4094, 4056, 4048, 3770, 3767, 8} \[ -\frac{b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \tan (c+d x)}{6 d}+\frac{2 a b \left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{1}{2} a^2 x \left (a^2 (A+2 C)+12 A b^2\right )-\frac{b^2 (15 A-2 C) \tan (c+d x) (a+b \sec (c+d x))^2}{6 d}-\frac{a b^3 (9 A-4 C) \tan (c+d x) \sec (c+d x)}{3 d}+\frac{2 A b \sin (c+d x) (a+b \sec (c+d x))^3}{d}+\frac{A \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^4}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(12*A*b^2 + a^2*(A + 2*C))*x)/2 + (2*a*b*(2*A*b^2 + (2*a^2 + b^2)*C)*ArcTanh[Sin[c + d*x]])/d + (2*A*b*(a
+ b*Sec[c + d*x])^3*Sin[c + d*x])/d + (A*Cos[c + d*x]*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(2*d) - (b^2*(a^2*
(39*A - 34*C) - 2*b^2*(3*A + 2*C))*Tan[c + d*x])/(6*d) - (a*b^3*(9*A - 4*C)*Sec[c + d*x]*Tan[c + d*x])/(3*d) -
(b^2*(15*A - 2*C)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(6*d)

Rule 4095

Int[((A_.) + 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] - Dis
t[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*(C*n + A*(n + 1))*Csc[e +
f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C}, x] && NeQ[a^2 - b^2,
0] && GtQ[m, 0] && LeQ[n, -1]

Rule 4094

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/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*
m - a*B*n - (b*B*n + a*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 1))*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && LeQ[n, -1]

Rule 4056

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

Rule 4048

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

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]

Rubi steps

\begin{align*} \int \cos ^2(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac{1}{2} \int \cos (c+d x) (a+b \sec (c+d x))^3 \left (4 A b+a (A+2 C) \sec (c+d x)-b (3 A-2 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{2 A b (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}+\frac{1}{2} \int (a+b \sec (c+d x))^2 \left (12 A b^2+a^2 (A+2 C)-2 a b (A-2 C) \sec (c+d x)-b^2 (15 A-2 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{2 A b (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{b^2 (15 A-2 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac{1}{6} \int (a+b \sec (c+d x)) \left (3 a \left (12 A b^2+a^2 (A+2 C)\right )-b \left (3 a^2 (A-6 C)-2 b^2 (3 A+2 C)\right ) \sec (c+d x)-4 a b^2 (9 A-4 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{2 A b (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{a b^3 (9 A-4 C) \sec (c+d x) \tan (c+d x)}{3 d}-\frac{b^2 (15 A-2 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac{1}{12} \int \left (6 a^2 \left (12 A b^2+a^2 (A+2 C)\right )+24 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \sec (c+d x)-2 b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{1}{2} a^2 \left (12 A b^2+a^2 (A+2 C)\right ) x+\frac{2 A b (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{a b^3 (9 A-4 C) \sec (c+d x) \tan (c+d x)}{3 d}-\frac{b^2 (15 A-2 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\left (2 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right )\right ) \int \sec (c+d x) \, dx-\frac{1}{6} \left (b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac{1}{2} a^2 \left (12 A b^2+a^2 (A+2 C)\right ) x+\frac{2 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 A b (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{a b^3 (9 A-4 C) \sec (c+d x) \tan (c+d x)}{3 d}-\frac{b^2 (15 A-2 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac{\left (b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=\frac{1}{2} a^2 \left (12 A b^2+a^2 (A+2 C)\right ) x+\frac{2 a b \left (2 A b^2+\left (2 a^2+b^2\right ) C\right ) \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 A b (a+b \sec (c+d x))^3 \sin (c+d x)}{d}+\frac{A \cos (c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{2 d}-\frac{b^2 \left (a^2 (39 A-34 C)-2 b^2 (3 A+2 C)\right ) \tan (c+d x)}{6 d}-\frac{a b^3 (9 A-4 C) \sec (c+d x) \tan (c+d x)}{3 d}-\frac{b^2 (15 A-2 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}\\ \end{align*}

Mathematica [A]  time = 6.11514, size = 416, normalized size = 1.9 \[ \frac{6 a^2 (c+d x) \left (a^2 (A+2 C)+12 A b^2\right )+\frac{4 b^2 \left (2 C \left (9 a^2+b^2\right )+3 A b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{4 b^2 \left (2 C \left (9 a^2+b^2\right )+3 A b^2\right ) \sin \left (\frac{1}{2} (c+d x)\right )}{\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )}-24 a b \left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+24 a b \left (C \left (2 a^2+b^2\right )+2 A b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+48 a^3 A b \sin (c+d x)+3 a^4 A \sin (2 (c+d x))+\frac{b^3 C (12 a+b)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}-\frac{b^3 C (12 a+b)}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{2 b^4 C \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2 b^4 C \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3}}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*a^2*(12*A*b^2 + a^2*(A + 2*C))*(c + d*x) - 24*a*b*(2*A*b^2 + (2*a^2 + b^2)*C)*Log[Cos[(c + d*x)/2] - Sin[(c
+ d*x)/2]] + 24*a*b*(2*A*b^2 + (2*a^2 + b^2)*C)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (b^3*(12*a + b)*C)
/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*b^4*C*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3
+ (4*b^2*(3*A*b^2 + 2*(9*a^2 + b^2)*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (2*b^4*C*Sin[
(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - (b^3*(12*a + b)*C)/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2
])^2 + (4*b^2*(3*A*b^2 + 2*(9*a^2 + b^2)*C)*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 48*a^3*A
*b*Sin[c + d*x] + 3*a^4*A*Sin[2*(c + d*x)])/(12*d)

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Maple [A]  time = 0.085, size = 258, normalized size = 1.2 \begin{align*}{\frac{A{a}^{4}\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2\,d}}+{\frac{{a}^{4}Ax}{2}}+{\frac{A{a}^{4}c}{2\,d}}+{a}^{4}Cx+{\frac{C{a}^{4}c}{d}}+4\,{\frac{A{a}^{3}b\sin \left ( dx+c \right ) }{d}}+4\,{\frac{{a}^{3}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+6\,A{a}^{2}{b}^{2}x+6\,{\frac{A{a}^{2}{b}^{2}c}{d}}+6\,{\frac{C{a}^{2}{b}^{2}\tan \left ( dx+c \right ) }{d}}+4\,{\frac{Aa{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{Ca{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+2\,{\frac{Ca{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{A{b}^{4}\tan \left ( dx+c \right ) }{d}}+{\frac{2\,C{b}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{C{b}^{4}\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+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 0.987792, size = 298, normalized size = 1.36 \begin{align*} \frac{3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} + 12 \,{\left (d x + c\right )} C a^{4} + 72 \,{\left (d x + c\right )} A a^{2} b^{2} + 4 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C b^{4} - 12 \, C a b^{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 )} + 24 \, C a^{3} b{\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, A a b^{3}{\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^{3} b \sin \left (d x + c\right ) + 72 \, C a^{2} b^{2} \tan \left (d x + c\right ) + 12 \, A b^{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+b*sec(d*x+c))^4*(A+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 + 12*(d*x + c)*C*a^4 + 72*(d*x + c)*A*a^2*b^2 + 4*(tan(d*x + c)
^3 + 3*tan(d*x + c))*C*b^4 - 12*C*a*b^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin
(d*x + c) - 1)) + 24*C*a^3*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 24*A*a*b^3*(log(sin(d*x + c) +
1) - log(sin(d*x + c) - 1)) + 48*A*a^3*b*sin(d*x + c) + 72*C*a^2*b^2*tan(d*x + c) + 12*A*b^4*tan(d*x + c))/d

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Fricas [A]  time = 0.584596, size = 509, normalized size = 2.32 \begin{align*} \frac{3 \,{\left ({\left (A + 2 \, C\right )} a^{4} + 12 \, A a^{2} b^{2}\right )} d x \cos \left (d x + c\right )^{3} + 6 \,{\left (2 \, C a^{3} b +{\left (2 \, A + C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 6 \,{\left (2 \, C a^{3} b +{\left (2 \, A + C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) +{\left (3 \, A a^{4} \cos \left (d x + c\right )^{4} + 24 \, A a^{3} b \cos \left (d x + c\right )^{3} + 12 \, C a b^{3} \cos \left (d x + c\right ) + 2 \, C b^{4} + 2 \,{\left (18 \, C a^{2} b^{2} +{\left (3 \, A + 2 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{6 \, 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+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/6*(3*((A + 2*C)*a^4 + 12*A*a^2*b^2)*d*x*cos(d*x + c)^3 + 6*(2*C*a^3*b + (2*A + C)*a*b^3)*cos(d*x + c)^3*log(
sin(d*x + c) + 1) - 6*(2*C*a^3*b + (2*A + C)*a*b^3)*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + (3*A*a^4*cos(d*x +
c)^4 + 24*A*a^3*b*cos(d*x + c)^3 + 12*C*a*b^3*cos(d*x + c) + 2*C*b^4 + 2*(18*C*a^2*b^2 + (3*A + 2*C)*b^4)*cos
(d*x + c)^2)*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+b*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [A]  time = 1.21575, size = 536, normalized size = 2.45 \begin{align*} \frac{3 \,{\left (A a^{4} + 2 \, C a^{4} + 12 \, A a^{2} b^{2}\right )}{\left (d x + c\right )} + 12 \,{\left (2 \, C a^{3} b + 2 \, A a b^{3} + C a b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 12 \,{\left (2 \, C a^{3} b + 2 \, A a b^{3} + C a b^{3}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{6 \,{\left (A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 8 \, A a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 8 \, A a^{3} b \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{4 \,{\left (18 \, C a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, C a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, A b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 3 \, C b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 36 \, C a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 6 \, A b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 18 \, C a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, A b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, C b^{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+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(A*a^4 + 2*C*a^4 + 12*A*a^2*b^2)*(d*x + c) + 12*(2*C*a^3*b + 2*A*a*b^3 + C*a*b^3)*log(abs(tan(1/2*d*x +
1/2*c) + 1)) - 12*(2*C*a^3*b + 2*A*a*b^3 + C*a*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 6*(A*a^4*tan(1/2*d*x
+ 1/2*c)^3 - 8*A*a^3*b*tan(1/2*d*x + 1/2*c)^3 - A*a^4*tan(1/2*d*x + 1/2*c) - 8*A*a^3*b*tan(1/2*d*x + 1/2*c))/
(tan(1/2*d*x + 1/2*c)^2 + 1)^2 - 4*(18*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^5 - 6*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 3
*A*b^4*tan(1/2*d*x + 1/2*c)^5 + 3*C*b^4*tan(1/2*d*x + 1/2*c)^5 - 36*C*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 6*A*b^4
*tan(1/2*d*x + 1/2*c)^3 - 2*C*b^4*tan(1/2*d*x + 1/2*c)^3 + 18*C*a^2*b^2*tan(1/2*d*x + 1/2*c) + 6*C*a*b^3*tan(1
/2*d*x + 1/2*c) + 3*A*b^4*tan(1/2*d*x + 1/2*c) + 3*C*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^3)
/d