3.2.8 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^3 (A+C \sec ^2(c+d x)) \, dx\) [108]

Optimal. Leaf size=169 \[ \frac {1}{8} a^3 (15 A+28 C) x+\frac {a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac {(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d} \]

[Out]

1/8*a^3*(15*A+28*C)*x+a^3*C*arctanh(sin(d*x+c))/d+5/8*a^3*(3*A+4*C)*sin(d*x+c)/d+1/4*A*cos(d*x+c)^3*(a+a*sec(d
*x+c))^3*sin(d*x+c)/d+1/4*A*cos(d*x+c)^2*(a^2+a^2*sec(d*x+c))^2*sin(d*x+c)/a/d+1/8*(5*A+4*C)*cos(d*x+c)*(a^3+a
^3*sec(d*x+c))*sin(d*x+c)/d

________________________________________________________________________________________

Rubi [A]
time = 0.28, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {4172, 4102, 4081, 3855} \begin {gather*} \frac {5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac {(5 A+4 C) \sin (c+d x) \cos (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{8 d}+\frac {1}{8} a^3 x (15 A+28 C)+\frac {a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {A \sin (c+d x) \cos ^2(c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{4 a d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^3}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(15*A + 28*C)*x)/8 + (a^3*C*ArcTanh[Sin[c + d*x]])/d + (5*a^3*(3*A + 4*C)*Sin[c + d*x])/(8*d) + (A*Cos[c
+ d*x]^3*(a + a*Sec[c + d*x])^3*Sin[c + d*x])/(4*d) + (A*Cos[c + d*x]^2*(a^2 + a^2*Sec[c + d*x])^2*Sin[c + d*x
])/(4*a*d) + ((5*A + 4*C)*Cos[c + d*x]*(a^3 + a^3*Sec[c + d*x])*Sin[c + d*x])/(8*d)

Rule 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4081

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 4102

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

Rule 4172

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

Rubi steps

\begin {align*} \int \cos ^4(c+d x) (a+a \sec (c+d x))^3 \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x))^3 (3 a A+4 a C \sec (c+d x)) \, dx}{4 a}\\ &=\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac {\int \cos ^2(c+d x) (a+a \sec (c+d x))^2 \left (3 a^2 (5 A+4 C)+12 a^2 C \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac {(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x)) \left (15 a^3 (3 A+4 C)+24 a^3 C \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac {5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac {(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d}-\frac {\int \left (-3 a^4 (15 A+28 C)-24 a^4 C \sec (c+d x)\right ) \, dx}{24 a}\\ &=\frac {1}{8} a^3 (15 A+28 C) x+\frac {5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac {(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d}+\left (a^3 C\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{8} a^3 (15 A+28 C) x+\frac {a^3 C \tanh ^{-1}(\sin (c+d x))}{d}+\frac {5 a^3 (3 A+4 C) \sin (c+d x)}{8 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^3 \sin (c+d x)}{4 d}+\frac {A \cos ^2(c+d x) \left (a^2+a^2 \sec (c+d x)\right )^2 \sin (c+d x)}{4 a d}+\frac {(5 A+4 C) \cos (c+d x) \left (a^3+a^3 \sec (c+d x)\right ) \sin (c+d x)}{8 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.32, size = 124, normalized size = 0.73 \begin {gather*} \frac {a^3 \left (60 A d x+112 C d x-32 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+32 C \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 (13 A+12 C) \sin (c+d x)+8 (4 A+C) \sin (2 (c+d x))+8 A \sin (3 (c+d x))+A \sin (4 (c+d x))\right )}{32 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^3*(A + C*Sec[c + d*x]^2),x]

[Out]

(a^3*(60*A*d*x + 112*C*d*x - 32*C*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 32*C*Log[Cos[(c + d*x)/2] + Sin[(
c + d*x)/2]] + 8*(13*A + 12*C)*Sin[c + d*x] + 8*(4*A + C)*Sin[2*(c + d*x)] + 8*A*Sin[3*(c + d*x)] + A*Sin[4*(c
+ d*x)]))/(32*d)

________________________________________________________________________________________

Maple [A]
time = 0.66, size = 173, normalized size = 1.02

method	result	size
derivativedivides	\(\frac {A \,a^{3} \sin \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \left (d x +c \right )+A \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 a^{3} C \sin \left (d x +c \right )+A \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(173\)
default	\(\frac {A \,a^{3} \sin \left (d x +c \right )+a^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+3 a^{3} C \left (d x +c \right )+A \,a^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 a^{3} C \sin \left (d x +c \right )+A \,a^{3} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{3} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(173\)
risch	\(\frac {15 a^{3} A x}{8}+\frac {7 a^{3} x C}{2}-\frac {13 i A \,a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} a^{3} C}{2 d}+\frac {13 i A \,a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} a^{3} C}{2 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}+\frac {A \,a^{3} \sin \left (4 d x +4 c \right )}{32 d}+\frac {A \,a^{3} \sin \left (3 d x +3 c \right )}{4 d}+\frac {\sin \left (2 d x +2 c \right ) A \,a^{3}}{d}+\frac {\sin \left (2 d x +2 c \right ) a^{3} C}{4 d}\)	\(206\)
norman	\(\frac {\left (\frac {15}{8} A \,a^{3}+\frac {7}{2} a^{3} C \right ) x +\left (-\frac {15}{2} A \,a^{3}-14 a^{3} C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-\frac {15}{2} A \,a^{3}-14 a^{3} C \right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {15}{8} A \,a^{3}+\frac {7}{2} a^{3} C \right ) x \left (\tan ^{16}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (\frac {45}{4} A \,a^{3}+21 a^{3} C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {19 a^{3} \left (3 A +4 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {5 a^{3} \left (3 A +4 C \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {a^{3} \left (5 A +12 C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {7 a^{3} \left (7 A +4 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{3} \left (27 A +52 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a^{3} \left (37 A +92 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {3 a^{3} \left (41 A +12 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a^{3} \left (-68 C +57 A \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {a^{3} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{3} C \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(400\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.31, size = 171, normalized size = 1.01 \begin {gather*} -\frac {32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{3} - {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} - 8 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{3} - 96 \, {\left (d x + c\right )} C a^{3} - 16 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 32 \, A a^{3} \sin \left (d x + c\right ) - 96 \, C a^{3} \sin \left (d x + c\right )}{32 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/32*(32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^3 - (12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*
a^3 - 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3 - 8*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^3 - 96*(d*x + c)*C*a^
3 - 16*C*a^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 32*A*a^3*sin(d*x + c) - 96*C*a^3*sin(d*x + c))/
d

________________________________________________________________________________________

Fricas [A]
time = 3.81, size = 112, normalized size = 0.66 \begin {gather*} \frac {{\left (15 \, A + 28 \, C\right )} a^{3} d x + 4 \, C a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 4 \, C a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, A a^{3} \cos \left (d x + c\right )^{3} + 8 \, A a^{3} \cos \left (d x + c\right )^{2} + {\left (15 \, A + 4 \, C\right )} a^{3} \cos \left (d x + c\right ) + 24 \, {\left (A + C\right )} a^{3}\right )} \sin \left (d x + c\right )}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/8*((15*A + 28*C)*a^3*d*x + 4*C*a^3*log(sin(d*x + c) + 1) - 4*C*a^3*log(-sin(d*x + c) + 1) + (2*A*a^3*cos(d*x
+ c)^3 + 8*A*a^3*cos(d*x + c)^2 + (15*A + 4*C)*a^3*cos(d*x + c) + 24*(A + C)*a^3)*sin(d*x + c))/d

________________________________________________________________________________________

Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+a*sec(d*x+c))**3*(A+C*sec(d*x+c)**2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

________________________________________________________________________________________

Giac [A]
time = 0.50, size = 213, normalized size = 1.26 \begin {gather*} \frac {8 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 8 \, C a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + {\left (15 \, A a^{3} + 28 \, C a^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 20 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 55 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 68 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 73 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 76 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 49 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 28 \, 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 )}^{4}}}{8 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^3*(A+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/8*(8*C*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 8*C*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + (15*A*a^3 + 28*
C*a^3)*(d*x + c) + 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 20*C*a^3*tan(1/2*d*x + 1/2*c)^7 + 55*A*a^3*tan(1/2*d*x
+ 1/2*c)^5 + 68*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 73*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 76*C*a^3*tan(1/2*d*x + 1/2*c
)^3 + 49*A*a^3*tan(1/2*d*x + 1/2*c) + 28*C*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d

________________________________________________________________________________________

Mupad [B]
time = 2.85, size = 195, normalized size = 1.15 \begin {gather*} \frac {13\,A\,a^3\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,C\,a^3\,\sin \left (c+d\,x\right )}{d}+\frac {15\,A\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{4\,d}+\frac {7\,C\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {A\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{d}+\frac {A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {A\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {C\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^3,x)

[Out]

(13*A*a^3*sin(c + d*x))/(4*d) + (3*C*a^3*sin(c + d*x))/d + (15*A*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2
)))/(4*d) + (7*C*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/
2 + (d*x)/2)))/d + (A*a^3*sin(2*c + 2*d*x))/d + (A*a^3*sin(3*c + 3*d*x))/(4*d) + (A*a^3*sin(4*c + 4*d*x))/(32*
d) + (C*a^3*sin(2*c + 2*d*x))/(4*d)

________________________________________________________________________________________