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

Optimal. Leaf size=229 \[ -\frac{a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \tan (c+d x)}{6 d}+\frac{\left (24 a^2 b^2 (2 A+C)+8 a^4 C+b^4 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \tan (c+d x) \sec (c+d x)}{24 d}+4 a^3 A b x-\frac{b (4 A-C) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}-\frac{a b (12 A-7 C) \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{A \sin (c+d x) (a+b \sec (c+d x))^4}{d} \]

[Out]

4*a^3*A*b*x + ((8*a^4*C + 24*a^2*b^2*(2*A + C) + b^4*(4*A + 3*C))*ArcTanh[Sin[c + d*x]])/(8*d) + (A*(a + b*Sec
[c + d*x])^4*Sin[c + d*x])/d - (a*b*(a^2*(12*A - 19*C) - 8*b^2*(3*A + 2*C))*Tan[c + d*x])/(6*d) - (b^2*(a^2*(2
4*A - 26*C) - 3*b^2*(4*A + 3*C))*Sec[c + d*x]*Tan[c + d*x])/(24*d) - (a*b*(12*A - 7*C)*(a + b*Sec[c + d*x])^2*
Tan[c + d*x])/(12*d) - (b*(4*A - C)*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*d)

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Rubi [A]  time = 0.492718, antiderivative size = 229, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4095, 4056, 4048, 3770, 3767, 8} \[ -\frac{a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \tan (c+d x)}{6 d}+\frac{\left (24 a^2 b^2 (2 A+C)+8 a^4 C+b^4 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}-\frac{b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \tan (c+d x) \sec (c+d x)}{24 d}+4 a^3 A b x-\frac{b (4 A-C) \tan (c+d x) (a+b \sec (c+d x))^3}{4 d}-\frac{a b (12 A-7 C) \tan (c+d x) (a+b \sec (c+d x))^2}{12 d}+\frac{A \sin (c+d x) (a+b \sec (c+d x))^4}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

4*a^3*A*b*x + ((8*a^4*C + 24*a^2*b^2*(2*A + C) + b^4*(4*A + 3*C))*ArcTanh[Sin[c + d*x]])/(8*d) + (A*(a + b*Sec
[c + d*x])^4*Sin[c + d*x])/d - (a*b*(a^2*(12*A - 19*C) - 8*b^2*(3*A + 2*C))*Tan[c + d*x])/(6*d) - (b^2*(a^2*(2
4*A - 26*C) - 3*b^2*(4*A + 3*C))*Sec[c + d*x]*Tan[c + d*x])/(24*d) - (a*b*(12*A - 7*C)*(a + b*Sec[c + d*x])^2*
Tan[c + d*x])/(12*d) - (b*(4*A - C)*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*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 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 (c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}+\int (a+b \sec (c+d x))^3 \left (4 A b+a C \sec (c+d x)-b (4 A-C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}-\frac{b (4 A-C) (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{4} \int (a+b \sec (c+d x))^2 \left (16 a A b+\left (4 A b^2+4 a^2 C+3 b^2 C\right ) \sec (c+d x)-a b (12 A-7 C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}-\frac{a b (12 A-7 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{12} \int (a+b \sec (c+d x)) \left (48 a^2 A b+a \left (36 A b^2+12 a^2 C+23 b^2 C\right ) \sec (c+d x)-b \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}-\frac{b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{a b (12 A-7 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{1}{24} \int \left (96 a^3 A b+3 \left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) \sec (c+d x)-4 a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \sec ^2(c+d x)\right ) \, dx\\ &=4 a^3 A b x+\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}-\frac{b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{a b (12 A-7 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac{1}{6} \left (a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right )\right ) \int \sec ^2(c+d x) \, dx+\frac{1}{8} \left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) \int \sec (c+d x) \, dx\\ &=4 a^3 A b x+\frac{\left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}-\frac{b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{a b (12 A-7 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}+\frac{\left (a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right )\right ) \operatorname{Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d}\\ &=4 a^3 A b x+\frac{\left (8 a^4 C+24 a^2 b^2 (2 A+C)+b^4 (4 A+3 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{A (a+b \sec (c+d x))^4 \sin (c+d x)}{d}-\frac{a b \left (a^2 (12 A-19 C)-8 b^2 (3 A+2 C)\right ) \tan (c+d x)}{6 d}-\frac{b^2 \left (a^2 (24 A-26 C)-3 b^2 (4 A+3 C)\right ) \sec (c+d x) \tan (c+d x)}{24 d}-\frac{a b (12 A-7 C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 d}-\frac{b (4 A-C) (a+b \sec (c+d x))^3 \tan (c+d x)}{4 d}\\ \end{align*}

Mathematica [B]  time = 6.5741, size = 1357, normalized size = 5.93 \[ \text{result too large to display} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^3*A*b*(c + d*x)*Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(d*(b + a*Cos[c + d*x])^4*(
A + 2*C + A*Cos[2*c + 2*d*x])) + ((-48*a^2*A*b^2 - 4*A*b^4 - 8*a^4*C - 24*a^2*b^2*C - 3*b^4*C)*Cos[c + d*x]^6*
Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(4*d*(b + a*Cos[c + d*
x])^4*(A + 2*C + A*Cos[2*c + 2*d*x])) + ((48*a^2*A*b^2 + 4*A*b^4 + 8*a^4*C + 24*a^2*b^2*C + 3*b^4*C)*Cos[c + d
*x]^6*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(4*d*(b + a*Cos[
c + d*x])^4*(A + 2*C + A*Cos[2*c + 2*d*x])) + (b^4*C*Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]
^2))/(8*d*(b + a*Cos[c + d*x])^4*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) + ((1
2*A*b^4 + 72*a^2*b^2*C + 16*a*b^3*C + 9*b^4*C)*Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/(
24*d*(b + a*Cos[c + d*x])^4*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2) + (4*a*b^3
*C*Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[(c + d*x)/2])/(3*d*(b + a*Cos[c + d*x])^4*
(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3) - (b^4*C*Cos[c + d*x]^6*(a + b*Sec[c +
d*x])^4*(A + C*Sec[c + d*x]^2))/(8*d*(b + a*Cos[c + d*x])^4*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])^4) + (4*a*b^3*C*Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*Sin[(c + d*x)
/2])/(3*d*(b + a*Cos[c + d*x])^4*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3) + ((-
12*A*b^4 - 72*a^2*b^2*C - 16*a*b^3*C - 9*b^4*C)*Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2))/
(24*d*(b + a*Cos[c + d*x])^4*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2) + (8*Cos[
c + d*x]^6*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*(3*a*A*b^3*Sin[(c + d*x)/2] + 3*a^3*b*C*Sin[(c + d*x)
/2] + 2*a*b^3*C*Sin[(c + d*x)/2]))/(3*d*(b + a*Cos[c + d*x])^4*(A + 2*C + A*Cos[2*c + 2*d*x])*(Cos[(c + d*x)/2
] - Sin[(c + d*x)/2])) + (8*Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4*(A + C*Sec[c + d*x]^2)*(3*a*A*b^3*Sin[(c + d
*x)/2] + 3*a^3*b*C*Sin[(c + d*x)/2] + 2*a*b^3*C*Sin[(c + d*x)/2]))/(3*d*(b + a*Cos[c + d*x])^4*(A + 2*C + A*Co
s[2*c + 2*d*x])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])) + (2*a^4*A*Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4*(A + C
*Sec[c + d*x]^2)*Sin[c + d*x])/(d*(b + a*Cos[c + d*x])^4*(A + 2*C + A*Cos[2*c + 2*d*x]))

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Maple [A]  time = 0.082, size = 316, normalized size = 1.4 \begin{align*}{\frac{A{a}^{4}\sin \left ( dx+c \right ) }{d}}+{\frac{{a}^{4}C\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{a}^{3}Abx+4\,{\frac{A{a}^{3}bc}{d}}+4\,{\frac{{a}^{3}bC\tan \left ( dx+c \right ) }{d}}+6\,{\frac{A{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+3\,{\frac{C{a}^{2}{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{d}}+3\,{\frac{C{a}^{2}{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+4\,{\frac{Aa{b}^{3}\tan \left ( dx+c \right ) }{d}}+{\frac{8\,Ca{b}^{3}\tan \left ( dx+c \right ) }{3\,d}}+{\frac{4\,Ca{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+{\frac{A{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{A{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{C{b}^{4}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,C{b}^{4}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,C{b}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*A*a^4*sin(d*x+c)+1/d*a^4*C*ln(sec(d*x+c)+tan(d*x+c))+4*a^3*A*b*x+4/d*A*a^3*b*c+4/d*a^3*b*C*tan(d*x+c)+6/d*
A*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c))+3/d*C*a^2*b^2*sec(d*x+c)*tan(d*x+c)+3/d*C*a^2*b^2*ln(sec(d*x+c)+tan(d*x+c)
)+4/d*A*a*b^3*tan(d*x+c)+8/3/d*C*a*b^3*tan(d*x+c)+4/3/d*C*a*b^3*tan(d*x+c)*sec(d*x+c)^2+1/2/d*A*b^4*sec(d*x+c)
*tan(d*x+c)+1/2/d*A*b^4*ln(sec(d*x+c)+tan(d*x+c))+1/4/d*C*b^4*tan(d*x+c)*sec(d*x+c)^3+3/8/d*C*b^4*sec(d*x+c)*t
an(d*x+c)+3/8/d*C*b^4*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.992264, size = 413, normalized size = 1.8 \begin{align*} \frac{192 \,{\left (d x + c\right )} A a^{3} b + 64 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b^{3} - 3 \, C b^{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 )} - 72 \, C a^{2} b^{2}{\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 \, A b^{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 \, 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 )} + 144 \, A a^{2} b^{2}{\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 \, C a^{3} b \tan \left (d x + c\right ) + 192 \, A a b^{3} \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+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

1/48*(192*(d*x + c)*A*a^3*b + 64*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a*b^3 - 3*C*b^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)) - 7
2*C*a^2*b^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 12*A*b^4*(
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^4*(log(sin(d*x +
c) + 1) - log(sin(d*x + c) - 1)) + 144*A*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*A*a^4*s
in(d*x + c) + 192*C*a^3*b*tan(d*x + c) + 192*A*a*b^3*tan(d*x + c))/d

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Fricas [A]  time = 0.59003, size = 575, normalized size = 2.51 \begin{align*} \frac{192 \, A a^{3} b d x \cos \left (d x + c\right )^{4} + 3 \,{\left (8 \, C a^{4} + 24 \,{\left (2 \, A + C\right )} a^{2} b^{2} +{\left (4 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (8 \, C a^{4} + 24 \,{\left (2 \, A + C\right )} a^{2} b^{2} +{\left (4 \, A + 3 \, C\right )} b^{4}\right )} \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 \, C a b^{3} \cos \left (d x + c\right ) + 6 \, C b^{4} + 32 \,{\left (3 \, C a^{3} b +{\left (3 \, A + 2 \, C\right )} a b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (24 \, C a^{2} b^{2} +{\left (4 \, A + 3 \, C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2}\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+b*sec(d*x+c))^4*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/48*(192*A*a^3*b*d*x*cos(d*x + c)^4 + 3*(8*C*a^4 + 24*(2*A + C)*a^2*b^2 + (4*A + 3*C)*b^4)*cos(d*x + c)^4*log
(sin(d*x + c) + 1) - 3*(8*C*a^4 + 24*(2*A + C)*a^2*b^2 + (4*A + 3*C)*b^4)*cos(d*x + c)^4*log(-sin(d*x + c) + 1
) + 2*(24*A*a^4*cos(d*x + c)^4 + 32*C*a*b^3*cos(d*x + c) + 6*C*b^4 + 32*(3*C*a^3*b + (3*A + 2*C)*a*b^3)*cos(d*
x + c)^3 + 3*(24*C*a^2*b^2 + (4*A + 3*C)*b^4)*cos(d*x + c)^2)*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+b*sec(d*x+c))**4*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.26404, size = 797, normalized size = 3.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(96*(d*x + c)*A*a^3*b + 48*A*a^4*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 + 1) + 3*(8*C*a^4 + 48*A*a^
2*b^2 + 24*C*a^2*b^2 + 4*A*b^4 + 3*C*b^4)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(8*C*a^4 + 48*A*a^2*b^2 + 24*
C*a^2*b^2 + 4*A*b^4 + 3*C*b^4)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(96*C*a^3*b*tan(1/2*d*x + 1/2*c)^7 - 72*
C*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 + 96*A*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 96*C*a*b^3*tan(1/2*d*x + 1/2*c)^7 - 12*
A*b^4*tan(1/2*d*x + 1/2*c)^7 - 15*C*b^4*tan(1/2*d*x + 1/2*c)^7 - 288*C*a^3*b*tan(1/2*d*x + 1/2*c)^5 + 72*C*a^2
*b^2*tan(1/2*d*x + 1/2*c)^5 - 288*A*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 160*C*a*b^3*tan(1/2*d*x + 1/2*c)^5 + 12*A*b
^4*tan(1/2*d*x + 1/2*c)^5 - 9*C*b^4*tan(1/2*d*x + 1/2*c)^5 + 288*C*a^3*b*tan(1/2*d*x + 1/2*c)^3 + 72*C*a^2*b^2
*tan(1/2*d*x + 1/2*c)^3 + 288*A*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 160*C*a*b^3*tan(1/2*d*x + 1/2*c)^3 + 12*A*b^4*t
an(1/2*d*x + 1/2*c)^3 - 9*C*b^4*tan(1/2*d*x + 1/2*c)^3 - 96*C*a^3*b*tan(1/2*d*x + 1/2*c) - 72*C*a^2*b^2*tan(1/
2*d*x + 1/2*c) - 96*A*a*b^3*tan(1/2*d*x + 1/2*c) - 96*C*a*b^3*tan(1/2*d*x + 1/2*c) - 12*A*b^4*tan(1/2*d*x + 1/
2*c) - 15*C*b^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d