\(\int \cos ^3(c+d x) (a+b \sec (c+d x))^4 (A+C \sec ^2(c+d x)) \, dx\) [668]

Optimal result

Integrand size = 33, antiderivative size = 251 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=2 a b \left (2 A b^2+a^2 (A+2 C)\right ) x+\frac {b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\left (6 A b^2+a^2 (2 A+3 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {2 A b \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {2 a b \left (b^2 (11 A-6 C)+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}-\frac {b^2 \left (3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d} \]

[Out]

2*a*b*(2*A*b^2+a^2*(A+2*C))*x+1/2*b^2*(2*A*b^2+(12*a^2+b^2)*C)*arctanh(sin(d*x+c))/d+1/3*(6*A*b^2+a^2*(2*A+3*C
))*(a+b*sec(d*x+c))^2*sin(d*x+c)/d+2/3*A*b*cos(d*x+c)*(a+b*sec(d*x+c))^3*sin(d*x+c)/d+1/3*A*cos(d*x+c)^2*(a+b*
sec(d*x+c))^4*sin(d*x+c)/d-2/3*a*b*(b^2*(11*A-6*C)+a^2*(2*A+3*C))*tan(d*x+c)/d-1/6*b^2*(3*b^2*(6*A-C)+a^2*(4*A
+6*C))*sec(d*x+c)*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4180, 4179, 4133, 3855, 3852, 8} \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {b^2 \left (C \left (12 a^2+b^2\right )+2 A b^2\right ) \text {arctanh}(\sin (c+d x))}{2 d}-\frac {2 a b \left (a^2 (2 A+3 C)+b^2 (11 A-6 C)\right ) \tan (c+d x)}{3 d}+\frac {\left (a^2 (2 A+3 C)+6 A b^2\right ) \sin (c+d x) (a+b \sec (c+d x))^2}{3 d}-\frac {b^2 \left (a^2 (4 A+6 C)+3 b^2 (6 A-C)\right ) \tan (c+d x) \sec (c+d x)}{6 d}+2 a b x \left (a^2 (A+2 C)+2 A b^2\right )+\frac {A \sin (c+d x) \cos ^2(c+d x) (a+b \sec (c+d x))^4}{3 d}+\frac {2 A b \sin (c+d x) \cos (c+d x) (a+b \sec (c+d x))^3}{3 d} \]

[In]

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

[Out]

2*a*b*(2*A*b^2 + a^2*(A + 2*C))*x + (b^2*(2*A*b^2 + (12*a^2 + b^2)*C)*ArcTanh[Sin[c + d*x]])/(2*d) + ((6*A*b^2
+ a^2*(2*A + 3*C))*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(3*d) + (2*A*b*Cos[c + d*x]*(a + b*Sec[c + d*x])^3*Si
n[c + d*x])/(3*d) + (A*Cos[c + d*x]^2*(a + b*Sec[c + d*x])^4*Sin[c + d*x])/(3*d) - (2*a*b*(b^2*(11*A - 6*C) +
a^2*(2*A + 3*C))*Tan[c + d*x])/(3*d) - (b^2*(3*b^2*(6*A - C) + a^2*(4*A + 6*C))*Sec[c + d*x]*Tan[c + d*x])/(6*
d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 3852

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 3855

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

Rule 4133

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 4179

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 4180

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]

Rubi steps \begin{align*} \text {integral}& = \frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac {1}{3} \int \cos ^2(c+d x) (a+b \sec (c+d x))^3 \left (4 A b+a (2 A+3 C) \sec (c+d x)-b (2 A-3 C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {2 A b \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac {1}{6} \int \cos (c+d x) (a+b \sec (c+d x))^2 \left (2 \left (6 A b^2+\frac {1}{2} a^2 (4 A+6 C)\right )+4 a b (A+3 C) \sec (c+d x)-6 b^2 (2 A-C) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {\left (6 A b^2+a^2 (2 A+3 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {2 A b \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}+\frac {1}{6} \int (a+b \sec (c+d x)) \left (12 b \left (2 A b^2+a^2 (A+2 C)\right )-2 a b^2 (4 A-9 C) \sec (c+d x)-2 b \left (3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec ^2(c+d x)\right ) \, dx \\ & = \frac {\left (6 A b^2+a^2 (2 A+3 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {2 A b \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {b^2 \left (3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {1}{12} \int \left (24 a b \left (2 A b^2+a^2 (A+2 C)\right )+6 b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \sec (c+d x)-8 a b \left (b^2 (11 A-6 C)+a^2 (2 A+3 C)\right ) \sec ^2(c+d x)\right ) \, dx \\ & = 2 a b \left (2 A b^2+a^2 (A+2 C)\right ) x+\frac {\left (6 A b^2+a^2 (2 A+3 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {2 A b \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {b^2 \left (3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {1}{2} \left (b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right )\right ) \int \sec (c+d x) \, dx-\frac {1}{3} \left (2 a b \left (b^2 (11 A-6 C)+a^2 (2 A+3 C)\right )\right ) \int \sec ^2(c+d x) \, dx \\ & = 2 a b \left (2 A b^2+a^2 (A+2 C)\right ) x+\frac {b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\left (6 A b^2+a^2 (2 A+3 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {2 A b \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {b^2 \left (3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {\left (2 a b \left (b^2 (11 A-6 C)+a^2 (2 A+3 C)\right )\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d} \\ & = 2 a b \left (2 A b^2+a^2 (A+2 C)\right ) x+\frac {b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {\left (6 A b^2+a^2 (2 A+3 C)\right ) (a+b \sec (c+d x))^2 \sin (c+d x)}{3 d}+\frac {2 A b \cos (c+d x) (a+b \sec (c+d x))^3 \sin (c+d x)}{3 d}+\frac {A \cos ^2(c+d x) (a+b \sec (c+d x))^4 \sin (c+d x)}{3 d}-\frac {2 a b \left (b^2 (11 A-6 C)+a^2 (2 A+3 C)\right ) \tan (c+d x)}{3 d}-\frac {b^2 \left (3 b^2 (6 A-C)+a^2 (4 A+6 C)\right ) \sec (c+d x) \tan (c+d x)}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.83 (sec) , antiderivative size = 324, normalized size of antiderivative = 1.29 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {24 a b \left (2 A b^2+a^2 (A+2 C)\right ) (c+d x)-6 b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 b^2 \left (2 A b^2+\left (12 a^2+b^2\right ) C\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {3 b^4 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {48 a b^3 C \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}-\frac {3 b^4 C}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {48 a b^3 C \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )}+3 a^2 \left (24 A b^2+a^2 (3 A+4 C)\right ) \sin (c+d x)+12 a^3 A b \sin (2 (c+d x))+a^4 A \sin (3 (c+d x))}{12 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.00 (sec) , antiderivative size = 198, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(\frac {\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{4} C \sin \left (d x +c \right )+4 A \,a^{3} b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b C \left (d x +c \right )+6 A \,a^{2} b^{2} \sin \left (d x +c \right )+6 C \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a A \,b^{3} \left (d x +c \right )+4 C \tan \left (d x +c \right ) a \,b^{3}+A \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,b^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(198\)
default	\(\frac {\frac {a^{4} A \left (2+\cos \left (d x +c \right )^{2}\right ) \sin \left (d x +c \right )}{3}+a^{4} C \sin \left (d x +c \right )+4 A \,a^{3} b \left (\frac {\sin \left (d x +c \right ) \cos \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 a^{3} b C \left (d x +c \right )+6 A \,a^{2} b^{2} \sin \left (d x +c \right )+6 C \,a^{2} b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a A \,b^{3} \left (d x +c \right )+4 C \tan \left (d x +c \right ) a \,b^{3}+A \,b^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,b^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(198\)
parallelrisch	\(\frac {-24 b^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \left (\left (A +\frac {C}{2}\right ) b^{2}+6 C \,a^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+24 b^{2} \left (1+\cos \left (2 d x +2 c \right )\right ) \left (\left (A +\frac {C}{2}\right ) b^{2}+6 C \,a^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+48 \left (2 A \,b^{2}+a^{2} \left (A +2 C \right )\right ) a x b d \cos \left (2 d x +2 c \right )+\left (72 A \,a^{2} b^{2}+11 \left (A +\frac {12 C}{11}\right ) a^{4}\right ) \sin \left (3 d x +3 c \right )+\left (24 A \,a^{3} b +96 C a \,b^{3}\right ) \sin \left (2 d x +2 c \right )+12 A \,a^{3} b \sin \left (4 d x +4 c \right )+a^{4} A \sin \left (5 d x +5 c \right )+\left (24 C \,b^{4}+72 A \,a^{2} b^{2}+10 \left (A +\frac {6 C}{5}\right ) a^{4}\right ) \sin \left (d x +c \right )+48 \left (2 A \,b^{2}+a^{2} \left (A +2 C \right )\right ) a x b d}{24 d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(279\)
risch	\(2 a^{3} A b x +4 A a \,b^{3} x +4 C \,a^{3} b x -\frac {i a^{4} A \,{\mathrm e}^{3 i \left (d x +c \right )}}{24 d}+\frac {3 i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{8 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A \,a^{2} b^{2}}{d}-\frac {i A \,a^{3} b \,{\mathrm e}^{2 i \left (d x +c \right )}}{2 d}-\frac {3 i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{8 d}-\frac {i C \,b^{3} \left (b \,{\mathrm e}^{3 i \left (d x +c \right )}-8 a \,{\mathrm e}^{2 i \left (d x +c \right )}-b \,{\mathrm e}^{i \left (d x +c \right )}-8 a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {i A \,a^{3} b \,{\mathrm e}^{-2 i \left (d x +c \right )}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{4} C}{2 d}+\frac {i a^{4} A \,{\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{4} C}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A \,a^{2} b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{4}}{d}+\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,a^{2} b^{2}}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,b^{4}}{2 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{4}}{d}-\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,a^{2} b^{2}}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,b^{4}}{2 d}\)	\(423\)
norman	\(\frac {\left (-2 A \,a^{3} b -4 a A \,b^{3}-4 a^{3} b C \right ) x +\left (-12 A \,a^{3} b -24 a A \,b^{3}-24 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}+\left (-4 A \,a^{3} b -8 a A \,b^{3}-8 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (-4 A \,a^{3} b -8 a A \,b^{3}-8 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{14}+\left (2 A \,a^{3} b +4 a A \,b^{3}+4 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{16}+\left (4 A \,a^{3} b +8 a A \,b^{3}+8 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (4 A \,a^{3} b +8 a A \,b^{3}+8 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (12 A \,a^{3} b +24 a A \,b^{3}+24 a^{3} b C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\frac {\left (2 a^{4} A -4 A \,a^{3} b +12 A \,a^{2} b^{2}+2 a^{4} C -8 C a \,b^{3}+C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{15}}{d}-\frac {\left (2 a^{4} A +4 A \,a^{3} b +12 A \,a^{2} b^{2}+2 a^{4} C +8 C a \,b^{3}+C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {\left (26 a^{4} A -60 A \,a^{3} b +108 A \,a^{2} b^{2}+18 a^{4} C -24 C a \,b^{3}-3 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{13}}{3 d}+\frac {\left (26 a^{4} A +60 A \,a^{3} b +108 A \,a^{2} b^{2}+18 a^{4} C +24 C a \,b^{3}-3 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}+\frac {\left (46 a^{4} A -108 A \,a^{3} b +36 A \,a^{2} b^{2}+6 a^{4} C +72 C a \,b^{3}-9 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{3 d}-\frac {\left (46 a^{4} A +108 A \,a^{3} b +36 A \,a^{2} b^{2}+6 a^{4} C -72 C a \,b^{3}-9 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{3 d}-\frac {\left (50 a^{4} A -60 A \,a^{3} b -180 A \,a^{2} b^{2}-30 a^{4} C +72 C a \,b^{3}+9 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{3 d}+\frac {\left (50 a^{4} A +60 A \,a^{3} b -180 A \,a^{2} b^{2}-30 a^{4} C -72 C a \,b^{3}+9 C \,b^{4}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {b^{2} \left (2 A \,b^{2}+12 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {b^{2} \left (2 A \,b^{2}+12 C \,a^{2}+C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(838\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.84 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {24 \, {\left ({\left (A + 2 \, C\right )} a^{3} b + 2 \, A a b^{3}\right )} d x \cos \left (d x + c\right )^{2} + 3 \, {\left (12 \, C a^{2} b^{2} + {\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (12 \, C a^{2} b^{2} + {\left (2 \, A + C\right )} b^{4}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, A a^{4} \cos \left (d x + c\right )^{4} + 12 \, A a^{3} b \cos \left (d x + c\right )^{3} + 24 \, C a b^{3} \cos \left (d x + c\right ) + 3 \, C b^{4} + 2 \, {\left ({\left (2 \, A + 3 \, C\right )} a^{4} + 18 \, A a^{2} b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.88 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=-\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 12 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{3} b - 48 \, {\left (d x + c\right )} C a^{3} b - 48 \, {\left (d x + c\right )} A a b^{3} + 3 \, C b^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{2} b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 6 \, A b^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{4} \sin \left (d x + c\right ) - 72 \, A a^{2} b^{2} \sin \left (d x + c\right ) - 48 \, C a b^{3} \tan \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 12*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^3*b - 48*(d*x + c)*
C*a^3*b - 48*(d*x + c)*A*a*b^3 + 3*C*b^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)) - 36*C*a^2*b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) - 6*A*b^4*(log(sin(d*x + c) +
1) - log(sin(d*x + c) - 1)) - 12*C*a^4*sin(d*x + c) - 72*A*a^2*b^2*sin(d*x + c) - 48*C*a*b^3*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 398, normalized size of antiderivative = 1.59 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {12 \, {\left (A a^{3} b + 2 \, C a^{3} b + 2 \, A a b^{3}\right )} {\left (d x + c\right )} + 3 \, {\left (12 \, C a^{2} b^{2} + 2 \, A b^{4} + C b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (12 \, C a^{2} b^{2} + 2 \, A b^{4} + C b^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (8 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 8 \, C a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} + \frac {4 \, {\left (3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, A a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 18.05 (sec) , antiderivative size = 2660, normalized size of antiderivative = 10.60 \[ \int \cos ^3(c+d x) (a+b \sec (c+d x))^4 \left (A+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^5*((4*A*a^4)/3 - 4*C*a^4 + 6*C*b^4 - 24*A*a^2*b^2) - tan(c/2 + (d*x)/2)^3*((8*A*a^4)/3 - 4
*C*b^4 + 8*A*a^3*b - 16*C*a*b^3) - tan(c/2 + (d*x)/2)^7*((8*A*a^4)/3 - 4*C*b^4 - 8*A*a^3*b + 16*C*a*b^3) + tan
(c/2 + (d*x)/2)*(2*A*a^4 + 2*C*a^4 + C*b^4 + 12*A*a^2*b^2 + 4*A*a^3*b + 8*C*a*b^3) + tan(c/2 + (d*x)/2)^9*(2*A
*a^4 + 2*C*a^4 + C*b^4 + 12*A*a^2*b^2 - 4*A*a^3*b - 8*C*a*b^3))/(d*(tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2
)^4 - 2*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 + 1)) - (atan((((A*b^4 + (C*b^4)/2
+ 6*C*a^2*b^2)*(32*A*b^4 + 16*C*b^4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*C*a^3*b) + tan(c/2 + (d*
x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C
^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2))*(A*b^4 + (C
*b^4)/2 + 6*C*a^2*b^2)*1i - ((A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2)*(32*A*b^4 + 16*C*b^4 + 192*C*a^2*b^2 + 128*A*a*
b^3 + 64*A*a^3*b + 128*C*a^3*b) - tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b
^4 + 128*A^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C*a^2*b^6 + 1
024*A*C*a^4*b^4 + 512*A*C*a^6*b^2))*(A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2)*1i)/(((A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2)*
(32*A*b^4 + 16*C*b^4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*C*a^3*b) + tan(c/2 + (d*x)/2)*(32*A^2*b^
8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512
*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2))*(A*b^4 + (C*b^4)/2 + 6*C*a^
2*b^2) + ((A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2)*(32*A*b^4 + 16*C*b^4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b +
128*C*a^3*b) - tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^
2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 5
12*A*C*a^6*b^2))*(A*b^4 + (C*b^4)/2 + 6*C*a^2*b^2) - 256*A^3*a*b^11 + 1024*A^3*a^2*b^10 - 128*A^3*a^3*b^9 + 10
24*A^3*a^4*b^8 + 256*A^3*a^6*b^6 - 64*C^3*a^3*b^9 - 1536*C^3*a^5*b^7 + 512*C^3*a^6*b^6 - 9216*C^3*a^7*b^5 + 61
44*C^3*a^8*b^4 - 64*A*C^2*a*b^11 - 256*A^2*C*a*b^11 - 1824*A*C^2*a^3*b^9 + 1024*A*C^2*a^4*b^8 - 13056*A*C^2*a^
5*b^7 + 13824*A*C^2*a^6*b^6 - 4608*A*C^2*a^7*b^5 + 6144*A*C^2*a^8*b^4 + 512*A^2*C*a^2*b^10 - 3456*A^2*C*a^3*b^
9 + 8704*A^2*C*a^4*b^8 - 1536*A^2*C*a^5*b^7 + 7296*A^2*C*a^6*b^6 + 1536*A^2*C*a^8*b^4))*(A*b^4*2i + C*b^4*1i +
C*a^2*b^2*12i))/d + (4*a*b*atan((2*a*b*(tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^
2*a^4*b^4 + 128*A^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C*a^2*
b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2) - a*b*(A*a^2 + 2*A*b^2 + 2*C*a^2)*(32*A*b^4 + 16*C*b^4 + 192*C*a^2*b
^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*C*a^3*b)*2i)*(A*a^2 + 2*A*b^2 + 2*C*a^2) + 2*a*b*(tan(c/2 + (d*x)/2)*(32*A
^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4
+ 512*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2) + a*b*(A*a^2 + 2*A*b^2
+ 2*C*a^2)*(32*A*b^4 + 16*C*b^4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*C*a^3*b)*2i)*(A*a^2 + 2*A*b^2
+ 2*C*a^2))/(256*A^3*a*b^11 - 1024*A^3*a^2*b^10 + 128*A^3*a^3*b^9 - 1024*A^3*a^4*b^8 - 256*A^3*a^6*b^6 + 64*C
^3*a^3*b^9 + 1536*C^3*a^5*b^7 - 512*C^3*a^6*b^6 + 9216*C^3*a^7*b^5 - 6144*C^3*a^8*b^4 + a*b*(tan(c/2 + (d*x)/2
)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 + 128*A^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a
^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C*a^2*b^6 + 1024*A*C*a^4*b^4 + 512*A*C*a^6*b^2) - a*b*(A*a^2 + 2
*A*b^2 + 2*C*a^2)*(32*A*b^4 + 16*C*b^4 + 192*C*a^2*b^2 + 128*A*a*b^3 + 64*A*a^3*b + 128*C*a^3*b)*2i)*(A*a^2 +
2*A*b^2 + 2*C*a^2)*2i - a*b*(tan(c/2 + (d*x)/2)*(32*A^2*b^8 + 8*C^2*b^8 + 512*A^2*a^2*b^6 + 512*A^2*a^4*b^4 +
128*A^2*a^6*b^2 + 192*C^2*a^2*b^6 + 1152*C^2*a^4*b^4 + 512*C^2*a^6*b^2 + 32*A*C*b^8 + 384*A*C*a^2*b^6 + 1024*A
*C*a^4*b^4 + 512*A*C*a^6*b^2) + a*b*(A*a^2 + 2*A*b^2 + 2*C*a^2)*(32*A*b^4 + 16*C*b^4 + 192*C*a^2*b^2 + 128*A*a
*b^3 + 64*A*a^3*b + 128*C*a^3*b)*2i)*(A*a^2 + 2*A*b^2 + 2*C*a^2)*2i + 64*A*C^2*a*b^11 + 256*A^2*C*a*b^11 + 182
4*A*C^2*a^3*b^9 - 1024*A*C^2*a^4*b^8 + 13056*A*C^2*a^5*b^7 - 13824*A*C^2*a^6*b^6 + 4608*A*C^2*a^7*b^5 - 6144*A
*C^2*a^8*b^4 - 512*A^2*C*a^2*b^10 + 3456*A^2*C*a^3*b^9 - 8704*A^2*C*a^4*b^8 + 1536*A^2*C*a^5*b^7 - 7296*A^2*C*
a^6*b^6 - 1536*A^2*C*a^8*b^4))*(A*a^2 + 2*A*b^2 + 2*C*a^2))/d