\(\int \cos (c+d x) (a+a \sec (c+d x))^4 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [441]

Optimal result

Integrand size = 39, antiderivative size = 196 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^4 (4 A+B) x+\frac {a^4 (52 A+48 B+35 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {A (a+a \sec (c+d x))^4 \sin (c+d x)}{d}+\frac {5 a^4 (4 A+8 B+7 C) \tan (c+d x)}{8 d}-\frac {a (4 A-C) (a+a \sec (c+d x))^3 \tan (c+d x)}{4 d}-\frac {(12 A-4 B-7 C) \left (a^2+a^2 \sec (c+d x)\right )^2 \tan (c+d x)}{12 d}-\frac {(12 A-32 B-35 C) \left (a^4+a^4 \sec (c+d x)\right ) \tan (c+d x)}{24 d} \]

[Out]

a^4*(4*A+B)*x+1/8*a^4*(52*A+48*B+35*C)*arctanh(sin(d*x+c))/d+A*(a+a*sec(d*x+c))^4*sin(d*x+c)/d+5/8*a^4*(4*A+8*
B+7*C)*tan(d*x+c)/d-1/4*a*(4*A-C)*(a+a*sec(d*x+c))^3*tan(d*x+c)/d-1/12*(12*A-4*B-7*C)*(a^2+a^2*sec(d*x+c))^2*t
an(d*x+c)/d-1/24*(12*A-32*B-35*C)*(a^4+a^4*sec(d*x+c))*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {4171, 4002, 3999, 3852, 8, 3855} \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^4 (52 A+48 B+35 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {5 a^4 (4 A+8 B+7 C) \tan (c+d x)}{8 d}-\frac {(12 A-32 B-35 C) \tan (c+d x) \left (a^4 \sec (c+d x)+a^4\right )}{24 d}+a^4 x (4 A+B)-\frac {(12 A-4 B-7 C) \tan (c+d x) \left (a^2 \sec (c+d x)+a^2\right )^2}{12 d}-\frac {a (4 A-C) \tan (c+d x) (a \sec (c+d x)+a)^3}{4 d}+\frac {A \sin (c+d x) (a \sec (c+d x)+a)^4}{d} \]

[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 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 3999

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 4002

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] &&
GtQ[m, 1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]

Rule 4171

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])

Rubi steps \begin{align*} \text {integral}& = \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) \text {arctanh}(\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 ) \text {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) \text {arctanh}(\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 [A] (verified)

Time = 5.71 (sec) , antiderivative size = 179, normalized size of antiderivative = 0.91 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^4 \left (96 A d x+24 B d x+3 (52 A+48 B+35 C) \text {arctanh}(\sin (c+d x))+24 A \sin (c+d x)+168 B \tan (c+d x)+192 C \tan (c+d x)+12 A \sec (c+d x) \tan (c+d x)+48 B \sec (c+d x) \tan (c+d x)+81 C \sec (c+d x) \tan (c+d x)+96 A \sec ^2(c+d x) \tan (c+d x)+6 C \sec ^3(c+d x) \tan (c+d x)-96 A \tan ^3(c+d x)+8 B \tan ^3(c+d x)+32 C \tan ^3(c+d x)\right )}{24 d} \]

[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*(96*A*d*x + 24*B*d*x + 3*(52*A + 48*B + 35*C)*ArcTanh[Sin[c + d*x]] + 24*A*Sin[c + d*x] + 168*B*Tan[c + d
*x] + 192*C*Tan[c + d*x] + 12*A*Sec[c + d*x]*Tan[c + d*x] + 48*B*Sec[c + d*x]*Tan[c + d*x] + 81*C*Sec[c + d*x]
*Tan[c + d*x] + 96*A*Sec[c + d*x]^2*Tan[c + d*x] + 6*C*Sec[c + d*x]^3*Tan[c + d*x] - 96*A*Tan[c + d*x]^3 + 8*B
*Tan[c + d*x]^3 + 32*C*Tan[c + d*x]^3))/(24*d)

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 254, normalized size of antiderivative = 1.30

method	result	size
parallelrisch	\(\frac {2 \left (-13 \left (A +\frac {12 B}{13}+\frac {35 C}{52}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+13 \left (A +\frac {12 B}{13}+\frac {35 C}{52}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 x d \left (A +\frac {B}{4}\right ) \cos \left (2 d x +2 c \right )+2 x d \left (A +\frac {B}{4}\right ) \cos \left (4 d x +4 c \right )+\left (4 A +\frac {22 B}{3}+\frac {28 C}{3}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {5 A}{4}+2 B +\frac {27 C}{8}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {10 C}{3}+2 A +\frac {10 B}{3}\right ) \sin \left (4 d x +4 c \right )+\frac {A \sin \left (5 d x +5 c \right )}{4}+\left (2 B +\frac {35 C}{8}+A \right ) \sin \left (d x +c \right )+6 x d \left (A +\frac {B}{4}\right )\right ) a^{4}}{d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(254\)
derivativedivides	\(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{4} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-4 a^{4} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+6 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 B \,a^{4} \tan \left (d x +c \right )+6 a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} A \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} C \tan \left (d x +c \right )+a^{4} A \sin \left (d x +c \right )+B \,a^{4} \left (d x +c \right )+a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(340\)
default	\(\frac {a^{4} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-B \,a^{4} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+a^{4} C \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+4 a^{4} A \tan \left (d x +c \right )+4 B \,a^{4} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-4 a^{4} C \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+6 a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+6 B \,a^{4} \tan \left (d x +c \right )+6 a^{4} C \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+4 a^{4} A \left (d x +c \right )+4 B \,a^{4} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+4 a^{4} C \tan \left (d x +c \right )+a^{4} A \sin \left (d x +c \right )+B \,a^{4} \left (d x +c \right )+a^{4} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(340\)
norman	\(\frac {\left (-4 a^{4} A -B \,a^{4}\right ) x +\left (-20 a^{4} A -5 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\left (-16 a^{4} A -4 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}+\left (4 a^{4} A +B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{12}+\left (16 a^{4} A +4 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\left (20 a^{4} A +5 B \,a^{4}\right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}-\frac {5 a^{4} \left (4 A +8 B +7 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{4 d}-\frac {a^{4} \left (44 A +72 B +93 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{4} \left (420 A +520 B +511 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {a^{4} \left (-200 B -203 C +12 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}-\frac {a^{4} \left (104 B +53 C +204 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 d}+\frac {a^{4} \left (424 B +385 C +156 A \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{12 d}}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a^{4} \left (52 A +48 B +35 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{4} \left (52 A +48 B +35 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(410\)
risch	\(4 a^{4} A x +a^{4} x B -\frac {i a^{4} A \,{\mathrm e}^{i \left (d x +c \right )}}{2 d}+\frac {i a^{4} A \,{\mathrm e}^{-i \left (d x +c \right )}}{2 d}-\frac {i a^{4} \left (12 A \,{\mathrm e}^{7 i \left (d x +c \right )}+48 B \,{\mathrm e}^{7 i \left (d x +c \right )}+81 C \,{\mathrm e}^{7 i \left (d x +c \right )}-96 A \,{\mathrm e}^{6 i \left (d x +c \right )}-144 B \,{\mathrm e}^{6 i \left (d x +c \right )}-96 C \,{\mathrm e}^{6 i \left (d x +c \right )}+12 A \,{\mathrm e}^{5 i \left (d x +c \right )}+48 B \,{\mathrm e}^{5 i \left (d x +c \right )}+105 C \,{\mathrm e}^{5 i \left (d x +c \right )}-288 A \,{\mathrm e}^{4 i \left (d x +c \right )}-480 B \,{\mathrm e}^{4 i \left (d x +c \right )}-480 C \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{3 i \left (d x +c \right )}-48 B \,{\mathrm e}^{3 i \left (d x +c \right )}-105 C \,{\mathrm e}^{3 i \left (d x +c \right )}-288 A \,{\mathrm e}^{2 i \left (d x +c \right )}-496 B \,{\mathrm e}^{2 i \left (d x +c \right )}-544 C \,{\mathrm e}^{2 i \left (d x +c \right )}-12 A \,{\mathrm e}^{i \left (d x +c \right )}-48 B \,{\mathrm e}^{i \left (d x +c \right )}-81 C \,{\mathrm e}^{i \left (d x +c \right )}-96 A -160 B -160 C \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{2 d}+\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}+\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{8 d}-\frac {13 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{2 d}-\frac {6 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}-\frac {35 a^{4} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{8 d}\)	\(469\)

[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,method=_RETURNVERBOSE)

[Out]

2*(-13*(A+12/13*B+35/52*C)*(3/4+1/4*cos(4*d*x+4*c)+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)-1)+13*(A+12/13*B+35/5
2*C)*(3/4+1/4*cos(4*d*x+4*c)+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)+1)+8*x*d*(A+1/4*B)*cos(2*d*x+2*c)+2*x*d*(A+
1/4*B)*cos(4*d*x+4*c)+(4*A+22/3*B+28/3*C)*sin(2*d*x+2*c)+(5/4*A+2*B+27/8*C)*sin(3*d*x+3*c)+(10/3*C+2*A+10/3*B)
*sin(4*d*x+4*c)+1/4*A*sin(5*d*x+5*c)+(2*B+35/8*C+A)*sin(d*x+c)+6*x*d*(A+1/4*B))*a^4/d/(cos(4*d*x+4*c)+4*cos(2*
d*x+2*c)+3)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.97 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {48 \, {\left (4 \, A + B\right )} a^{4} d x \cos \left (d x + c\right )^{4} + 3 \, {\left (52 \, A + 48 \, B + 35 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (52 \, A + 48 \, B + 35 \, C\right )} a^{4} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (24 \, A a^{4} \cos \left (d x + c\right )^{4} + 32 \, {\left (3 \, A + 5 \, B + 5 \, C\right )} a^{4} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, A + 16 \, B + 27 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 8 \, {\left (B + 4 \, C\right )} a^{4} \cos \left (d x + c\right ) + 6 \, C a^{4}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[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)

Sympy [F]

\[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=a^{4} \left (\int A \cos {\left (c + d x \right )}\, dx + \int 4 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 A \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 A \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int A \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 6 B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 4 B \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 4 C \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int 6 C \cos {\left (c + d x \right )} \sec ^{4}{\left (c + d x \right )}\, dx + \int 4 C \cos {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int C \cos {\left (c + d x \right )} \sec ^{6}{\left (c + d x \right )}\, dx\right ) \]

[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]

a**4*(Integral(A*cos(c + d*x), x) + Integral(4*A*cos(c + d*x)*sec(c + d*x), x) + Integral(6*A*cos(c + d*x)*sec
(c + d*x)**2, x) + Integral(4*A*cos(c + d*x)*sec(c + d*x)**3, x) + Integral(A*cos(c + d*x)*sec(c + d*x)**4, x)
+ Integral(B*cos(c + d*x)*sec(c + d*x), x) + Integral(4*B*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(6*B*cos
(c + d*x)*sec(c + d*x)**3, x) + Integral(4*B*cos(c + d*x)*sec(c + d*x)**4, x) + Integral(B*cos(c + d*x)*sec(c
+ d*x)**5, x) + Integral(C*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(4*C*cos(c + d*x)*sec(c + d*x)**3, x) +
Integral(6*C*cos(c + d*x)*sec(c + d*x)**4, x) + Integral(4*C*cos(c + d*x)*sec(c + d*x)**5, x) + Integral(C*cos
(c + d*x)*sec(c + d*x)**6, x))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 416 vs. \(2 (186) = 372\).

Time = 0.23 (sec) , antiderivative size = 416, normalized size of antiderivative = 2.12 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {192 \, {\left (d x + c\right )} A a^{4} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{4} + 48 \, {\left (d x + c\right )} B a^{4} + 64 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{4} - 3 \, C a^{4} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 48 \, B a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 72 \, C a^{4} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 144 \, A a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 96 \, B a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C a^{4} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{4} \sin \left (d x + c\right ) + 192 \, A a^{4} \tan \left (d x + c\right ) + 288 \, B a^{4} \tan \left (d x + c\right ) + 192 \, C a^{4} \tan \left (d x + c\right )}{48 \, d} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.38 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.73 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\frac {48 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + 24 \, {\left (4 \, A a^{4} + B a^{4}\right )} {\left (d x + c\right )} + 3 \, {\left (52 \, A a^{4} + 48 \, B a^{4} + 35 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (52 \, A a^{4} + 48 \, B a^{4} + 35 \, C a^{4}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (84 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 105 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 276 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 424 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 385 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 300 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 520 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 511 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 108 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 216 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 279 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 17.97 (sec) , antiderivative size = 1346, normalized size of antiderivative = 6.87 \[ \int \cos (c+d x) (a+a \sec (c+d x))^4 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Too large to display} \]

[In]

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

[Out]

((117*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/2 + 54*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x
)/2)) + (315*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/8 + 12*A*a^4*sin(2*c + 2*d*x) + (15*A*a^4*sin
(3*c + 3*d*x))/4 + 6*A*a^4*sin(4*c + 4*d*x) + (3*A*a^4*sin(5*c + 5*d*x))/4 + 22*B*a^4*sin(2*c + 2*d*x) + 6*B*a
^4*sin(3*c + 3*d*x) + 10*B*a^4*sin(4*c + 4*d*x) + 28*C*a^4*sin(2*c + 2*d*x) + (81*C*a^4*sin(3*c + 3*d*x))/8 +
10*C*a^4*sin(4*c + 4*d*x) + 36*A*a^4*atan((3728*A^2*sin(c/2 + (d*x)/2) + 2368*B^2*sin(c/2 + (d*x)/2) + 1225*C^
2*sin(c/2 + (d*x)/2) + 5504*A*B*sin(c/2 + (d*x)/2) + 3640*A*C*sin(c/2 + (d*x)/2) + 3360*B*C*sin(c/2 + (d*x)/2)
)/(cos(c/2 + (d*x)/2)*(3728*A^2 + 2368*B^2 + 1225*C^2 + 5504*A*B + 3640*A*C + 3360*B*C))) + 9*B*a^4*atan((3728
*A^2*sin(c/2 + (d*x)/2) + 2368*B^2*sin(c/2 + (d*x)/2) + 1225*C^2*sin(c/2 + (d*x)/2) + 5504*A*B*sin(c/2 + (d*x)
/2) + 3640*A*C*sin(c/2 + (d*x)/2) + 3360*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(3728*A^2 + 2368*B^2 + 12
25*C^2 + 5504*A*B + 3640*A*C + 3360*B*C))) + 3*A*a^4*sin(c + d*x) + 6*B*a^4*sin(c + d*x) + (105*C*a^4*sin(c +
d*x))/8 + 48*A*a^4*atan((3728*A^2*sin(c/2 + (d*x)/2) + 2368*B^2*sin(c/2 + (d*x)/2) + 1225*C^2*sin(c/2 + (d*x)/
2) + 5504*A*B*sin(c/2 + (d*x)/2) + 3640*A*C*sin(c/2 + (d*x)/2) + 3360*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)
/2)*(3728*A^2 + 2368*B^2 + 1225*C^2 + 5504*A*B + 3640*A*C + 3360*B*C)))*cos(2*c + 2*d*x) + 12*A*a^4*atan((3728
*A^2*sin(c/2 + (d*x)/2) + 2368*B^2*sin(c/2 + (d*x)/2) + 1225*C^2*sin(c/2 + (d*x)/2) + 5504*A*B*sin(c/2 + (d*x)
/2) + 3640*A*C*sin(c/2 + (d*x)/2) + 3360*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(3728*A^2 + 2368*B^2 + 12
25*C^2 + 5504*A*B + 3640*A*C + 3360*B*C)))*cos(4*c + 4*d*x) + 12*B*a^4*atan((3728*A^2*sin(c/2 + (d*x)/2) + 236
8*B^2*sin(c/2 + (d*x)/2) + 1225*C^2*sin(c/2 + (d*x)/2) + 5504*A*B*sin(c/2 + (d*x)/2) + 3640*A*C*sin(c/2 + (d*x
)/2) + 3360*B*C*sin(c/2 + (d*x)/2))/(cos(c/2 + (d*x)/2)*(3728*A^2 + 2368*B^2 + 1225*C^2 + 5504*A*B + 3640*A*C
+ 3360*B*C)))*cos(2*c + 2*d*x) + 3*B*a^4*atan((3728*A^2*sin(c/2 + (d*x)/2) + 2368*B^2*sin(c/2 + (d*x)/2) + 122
5*C^2*sin(c/2 + (d*x)/2) + 5504*A*B*sin(c/2 + (d*x)/2) + 3640*A*C*sin(c/2 + (d*x)/2) + 3360*B*C*sin(c/2 + (d*x
)/2))/(cos(c/2 + (d*x)/2)*(3728*A^2 + 2368*B^2 + 1225*C^2 + 5504*A*B + 3640*A*C + 3360*B*C)))*cos(4*c + 4*d*x)
+ 78*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x) + (39*A*a^4*atanh(sin(c/2 + (d*x)/2)
/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x))/2 + 72*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d
*x) + 18*B*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c + 4*d*x) + (105*C*a^4*atanh(sin(c/2 + (d*x
)/2)/cos(c/2 + (d*x)/2))*cos(2*c + 2*d*x))/2 + (105*C*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*cos(4*c
+ 4*d*x))/8)/(12*d*(cos(2*c + 2*d*x)/2 + cos(4*c + 4*d*x)/8 + 3/8))