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

Optimal. Leaf size=196 \[ \frac{5 a^4 (4 A+8 B+7 C) \tan (c+d x)}{8 d}+\frac{a^4 (52 A+48 B+35 C) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{(12 A-4 B-7 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{12 d}-\frac{(12 A-32 B-35 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+a^4 x (4 A+B)-\frac{a (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}+\frac{A \sin (c+d x) (a \sec (c+d x)+a)^4}{d} \]

[Out]

a^4*(4*A + B)*x + (a^4*(52*A + 48*B + 35*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (A*(a + a*Sec[c + d*x])^4*Sin[c + d
*x])/d + (5*a^4*(4*A + 8*B + 7*C)*Tan[c + d*x])/(8*d) - (a*(4*A - C)*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(4*d
) - ((12*A - 4*B - 7*C)*(a^2 + a^2*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) - ((12*A - 32*B - 35*C)*(a^4 + a^4*Sec
[c + d*x])*Tan[c + d*x])/(24*d)

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Rubi [A]  time = 0.382038, antiderivative size = 196, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4086, 3917, 3914, 3767, 8, 3770} \[ \frac{5 a^4 (4 A+8 B+7 C) \tan (c+d x)}{8 d}+\frac{a^4 (52 A+48 B+35 C) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{(12 A-4 B-7 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{12 d}-\frac{(12 A-32 B-35 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+a^4 x (4 A+B)-\frac{a (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}+\frac{A \sin (c+d x) (a \sec (c+d x)+a)^4}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^4*(4*A + B)*x + (a^4*(52*A + 48*B + 35*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (A*(a + a*Sec[c + d*x])^4*Sin[c + d
*x])/d + (5*a^4*(4*A + 8*B + 7*C)*Tan[c + d*x])/(8*d) - (a*(4*A - C)*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(4*d
) - ((12*A - 4*B - 7*C)*(a^2 + a^2*Sec[c + d*x])^2*Tan[c + d*x])/(12*d) - ((12*A - 32*B - 35*C)*(a^4 + a^4*Sec
[c + d*x])*Tan[c + d*x])/(24*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 3917

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> -Simp[(b*
d*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1))/(f*m), x] + Dist[1/m, Int[(a + b*Csc[e + f*x])^(m - 1)*Simp[a*c*m
+ (b*c*m + a*d*(2*m - 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Gt
Q[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 3914

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x]
+ (Dist[b*d, Int[Csc[e + f*x]^2, x], x] + Dist[b*c + a*d, Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]

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 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 (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 (a+a \sec (c+d x))^4 \sin (c+d x)}{d}+\frac{\int (a+a \sec (c+d x))^4 (a (4 A+B)-a (4 A-C) \sec (c+d x)) \, dx}{a}\\ &=\frac{A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}-\frac{a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{\int (a+a \sec (c+d x))^3 \left (4 a^2 (4 A+B)-a^2 (12 A-4 B-7 C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac{A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}-\frac{a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{(12 A-4 B-7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}+\frac{\int (a+a \sec (c+d x))^2 \left (12 a^3 (4 A+B)-a^3 (12 A-32 B-35 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac{A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}-\frac{a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{(12 A-4 B-7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}-\frac{(12 A-32 B-35 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}+\frac{\int (a+a \sec (c+d x)) \left (24 a^4 (4 A+B)+15 a^4 (4 A+8 B+7 C) \sec (c+d x)\right ) \, dx}{24 a}\\ &=a^4 (4 A+B) x+\frac{A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}-\frac{a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{(12 A-4 B-7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}-\frac{(12 A-32 B-35 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}+\frac{1}{8} \left (5 a^4 (4 A+8 B+7 C)\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (a^4 (52 A+48 B+35 C)\right ) \int \sec (c+d x) \, dx\\ &=a^4 (4 A+B) x+\frac{a^4 (52 A+48 B+35 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}-\frac{a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{(12 A-4 B-7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}-\frac{(12 A-32 B-35 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}-\frac{\left (5 a^4 (4 A+8 B+7 C)\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{8 d}\\ &=a^4 (4 A+B) x+\frac{a^4 (52 A+48 B+35 C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}+\frac{5 a^4 (4 A+8 B+7 C) \tan (c+d x)}{8 d}-\frac{a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{(12 A-4 B-7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}-\frac{(12 A-32 B-35 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d}\\ \end{align*}

Mathematica [B]  time = 4.65642, size = 530, normalized size = 2.7 \[ \frac{a^4 \sec ^8\left (\frac{1}{2} (c+d x)\right ) (\sec (c+d x)+1)^4 \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (\sec (c) (72 d x (4 A+B) \cos (c)+48 d x (4 A+B) \cos (c+2 d x)+24 A \sin (2 c+d x)+288 A \sin (c+2 d x)-96 A \sin (3 c+2 d x)+30 A \sin (2 c+3 d x)+30 A \sin (4 c+3 d x)+96 A \sin (3 c+4 d x)+6 A \sin (4 c+5 d x)+6 A \sin (6 c+5 d x)+192 A d x \cos (3 c+2 d x)+48 A d x \cos (3 c+4 d x)+48 A d x \cos (5 c+4 d x)-288 A \sin (c)+24 A \sin (d x)+48 B \sin (2 c+d x)+496 B \sin (c+2 d x)-144 B \sin (3 c+2 d x)+48 B \sin (2 c+3 d x)+48 B \sin (4 c+3 d x)+160 B \sin (3 c+4 d x)+48 B d x \cos (3 c+2 d x)+12 B d x \cos (3 c+4 d x)+12 B d x \cos (5 c+4 d x)-480 B \sin (c)+48 B \sin (d x)+105 C \sin (2 c+d x)+544 C \sin (c+2 d x)-96 C \sin (3 c+2 d x)+81 C \sin (2 c+3 d x)+81 C \sin (4 c+3 d x)+160 C \sin (3 c+4 d x)-480 C \sin (c)+105 C \sin (d x))-24 (52 A+48 B+35 C) \cos ^4(c+d x) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{1536 d (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*Sec[(c + d*x)/2]^8*(1 + Sec[c + d*x])^4*(-24*(52*A + 48*B + 35*C)
*Cos[c + d*x]^4*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c]
*(72*(4*A + B)*d*x*Cos[c] + 48*(4*A + B)*d*x*Cos[c + 2*d*x] + 192*A*d*x*Cos[3*c + 2*d*x] + 48*B*d*x*Cos[3*c +
2*d*x] + 48*A*d*x*Cos[3*c + 4*d*x] + 12*B*d*x*Cos[3*c + 4*d*x] + 48*A*d*x*Cos[5*c + 4*d*x] + 12*B*d*x*Cos[5*c
+ 4*d*x] - 288*A*Sin[c] - 480*B*Sin[c] - 480*C*Sin[c] + 24*A*Sin[d*x] + 48*B*Sin[d*x] + 105*C*Sin[d*x] + 24*A*
Sin[2*c + d*x] + 48*B*Sin[2*c + d*x] + 105*C*Sin[2*c + d*x] + 288*A*Sin[c + 2*d*x] + 496*B*Sin[c + 2*d*x] + 54
4*C*Sin[c + 2*d*x] - 96*A*Sin[3*c + 2*d*x] - 144*B*Sin[3*c + 2*d*x] - 96*C*Sin[3*c + 2*d*x] + 30*A*Sin[2*c + 3
*d*x] + 48*B*Sin[2*c + 3*d*x] + 81*C*Sin[2*c + 3*d*x] + 30*A*Sin[4*c + 3*d*x] + 48*B*Sin[4*c + 3*d*x] + 81*C*S
in[4*c + 3*d*x] + 96*A*Sin[3*c + 4*d*x] + 160*B*Sin[3*c + 4*d*x] + 160*C*Sin[3*c + 4*d*x] + 6*A*Sin[4*c + 5*d*
x] + 6*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 = 0.117, size = 294, normalized size = 1.5 \begin{align*}{\frac{A{a}^{4}\sin \left ( dx+c \right ) }{d}}+B{a}^{4}x+{\frac{B{a}^{4}c}{d}}+{\frac{35\,{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+4\,{a}^{4}Ax+4\,{\frac{A{a}^{4}c}{d}}+6\,{\frac{B{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{20\,{a}^{4}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{13\,A{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{20\,B{a}^{4}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{27\,{a}^{4}C\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+4\,{\frac{A{a}^{4}\tan \left ( dx+c \right ) }{d}}+2\,{\frac{B{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+{\frac{4\,{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{A{a}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{{a}^{4}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*A*a^4*sin(d*x+c)+B*a^4*x+1/d*B*a^4*c+35/8/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+4*a^4*A*x+4/d*A*a^4*c+6/d*B*a^
4*ln(sec(d*x+c)+tan(d*x+c))+20/3/d*a^4*C*tan(d*x+c)+13/2/d*A*a^4*ln(sec(d*x+c)+tan(d*x+c))+20/3/d*B*a^4*tan(d*
x+c)+27/8/d*a^4*C*sec(d*x+c)*tan(d*x+c)+4/d*A*a^4*tan(d*x+c)+2/d*B*a^4*sec(d*x+c)*tan(d*x+c)+4/3/d*a^4*C*tan(d
*x+c)*sec(d*x+c)^2+1/2/d*A*a^4*sec(d*x+c)*tan(d*x+c)+1/3/d*B*a^4*tan(d*x+c)*sec(d*x+c)^2+1/4/d*a^4*C*tan(d*x+c
)*sec(d*x+c)^3

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Maxima [B]  time = 0.978593, size = 562, normalized size = 2.87 \begin{align*} \frac{192 \,{\left (d x + c\right )} A a^{4} + 16 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 48 \,{\left (d x + c\right )} B a^{4} + 64 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} - 3 \, 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 )} - 12 \, 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 )} - 48 \, 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 )} - 72 \, 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 )} + 144 \, 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 )} + 96 \, 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 ) + 192 \, A a^{4} \tan \left (d x + c\right ) + 288 \, B a^{4} \tan \left (d x + c\right ) + 192 \, C a^{4} \tan \left (d x + c\right )}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(192*(d*x + c)*A*a^4 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^4 + 48*(d*x + c)*B*a^4 + 64*(tan(d*x + c)
^3 + 3*tan(d*x + c))*C*a^4 - 3*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)) - 12*A*a^4*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - l
og(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 48*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)) - 72*C*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)) + 144*A*a^4*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 96*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
) + 192*A*a^4*tan(d*x + c) + 288*B*a^4*tan(d*x + c) + 192*C*a^4*tan(d*x + c))/d

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Fricas [A]  time = 0.565522, size = 493, normalized size = 2.52 \begin{align*} \frac{48 \,{\left (4 \, A + B\right )} a^{4} d x \cos \left (d x + c\right )^{4} + 3 \,{\left (52 \, A + 48 \, B + 35 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (52 \, A + 48 \, B + 35 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \,{\left (24 \, A a^{4} \cos \left (d x + c\right )^{4} + 32 \,{\left (3 \, A + 5 \, B + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \,{\left (4 \, A + 16 \, B + 27 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 8 \,{\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 6 \, C a^{4}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(48*(4*A + B)*a^4*d*x*cos(d*x + c)^4 + 3*(52*A + 48*B + 35*C)*a^4*cos(d*x + c)^4*log(sin(d*x + c) + 1) -
3*(52*A + 48*B + 35*C)*a^4*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(24*A*a^4*cos(d*x + c)^4 + 32*(3*A + 5*B
+ 5*C)*a^4*cos(d*x + c)^3 + 3*(4*A + 16*B + 27*C)*a^4*cos(d*x + c)^2 + 8*(B + 4*C)*a^4*cos(d*x + c) + 6*C*a^4)
*sin(d*x + c))/(d*cos(d*x + c)^4)

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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)*(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 = 1.32555, size = 458, normalized size = 2.34 \begin{align*} \frac{\frac{48 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + 24 \,{\left (4 \, A a^{4} + B a^{4}\right )}{\left (d x + c\right )} + 3 \,{\left (52 \, A a^{4} + 48 \, B a^{4} + 35 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right ) - 3 \,{\left (52 \, A a^{4} + 48 \, B a^{4} + 35 \, 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 (84 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 120 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 105 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 276 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 424 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 385 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 300 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 520 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 511 \, C a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 108 \, A a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 216 \, B a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 279 \, 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 )}^{4}}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(48*A*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 24*(4*A*a^4 + B*a^4)*(d*x + c) + 3*(52*A*a^
4 + 48*B*a^4 + 35*C*a^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(52*A*a^4 + 48*B*a^4 + 35*C*a^4)*log(abs(tan(1
/2*d*x + 1/2*c) - 1)) - 2*(84*A*a^4*tan(1/2*d*x + 1/2*c)^7 + 120*B*a^4*tan(1/2*d*x + 1/2*c)^7 + 105*C*a^4*tan(
1/2*d*x + 1/2*c)^7 - 276*A*a^4*tan(1/2*d*x + 1/2*c)^5 - 424*B*a^4*tan(1/2*d*x + 1/2*c)^5 - 385*C*a^4*tan(1/2*d
*x + 1/2*c)^5 + 300*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 520*B*a^4*tan(1/2*d*x + 1/2*c)^3 + 511*C*a^4*tan(1/2*d*x +
1/2*c)^3 - 108*A*a^4*tan(1/2*d*x + 1/2*c) - 216*B*a^4*tan(1/2*d*x + 1/2*c) - 279*C*a^4*tan(1/2*d*x + 1/2*c))/(
tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d