\(\int \sec ^2(c+d x) (a+a \sec (c+d x))^2 (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [418]

Optimal result

Integrand size = 41, antiderivative size = 190 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (8 A+7 B+6 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 (8 A+7 B+6 C) \tan (c+d x)}{6 d}+\frac {a^2 (8 A+7 B+6 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(20 A-5 B+6 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac {(5 B+2 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 a d} \]

[Out]

1/8*a^2*(8*A+7*B+6*C)*arctanh(sin(d*x+c))/d+1/6*a^2*(8*A+7*B+6*C)*tan(d*x+c)/d+1/24*a^2*(8*A+7*B+6*C)*sec(d*x+
c)*tan(d*x+c)/d+1/60*(20*A-5*B+6*C)*(a+a*sec(d*x+c))^2*tan(d*x+c)/d+1/5*C*sec(d*x+c)^2*(a+a*sec(d*x+c))^2*tan(
d*x+c)/d+1/20*(5*B+2*C)*(a+a*sec(d*x+c))^3*tan(d*x+c)/a/d

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.195, Rules used = {4173, 4095, 4086, 3873, 3852, 8, 4131, 3855} \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (8 A+7 B+6 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 (8 A+7 B+6 C) \tan (c+d x)}{6 d}+\frac {a^2 (8 A+7 B+6 C) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {(20 A-5 B+6 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{60 d}+\frac {(5 B+2 C) \tan (c+d x) (a \sec (c+d x)+a)^3}{20 a d}+\frac {C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^2}{5 d} \]

[In]

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

[Out]

(a^2*(8*A + 7*B + 6*C)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(8*A + 7*B + 6*C)*Tan[c + d*x])/(6*d) + (a^2*(8*A +
7*B + 6*C)*Sec[c + d*x]*Tan[c + d*x])/(24*d) + ((20*A - 5*B + 6*C)*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(60*d
) + (C*Sec[c + d*x]^2*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(5*d) + ((5*B + 2*C)*(a + a*Sec[c + d*x])^3*Tan[c +
d*x])/(20*a*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 3873

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2, x_Symbol] :> Dist[2*a*(b/d
), Int[(d*Csc[e + f*x])^(n + 1), x], x] + Int[(d*Csc[e + f*x])^n*(a^2 + b^2*Csc[e + f*x]^2), x] /; FreeQ[{a, b
, d, e, f, n}, x]

Rule 4086

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

Rule 4095

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

Rule 4131

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(-C)*Cot
[e + f*x]*((b*Csc[e + f*x])^m/(f*(m + 1))), x] + Dist[(C*m + A*(m + 1))/(m + 1), Int[(b*Csc[e + f*x])^m, x], x
] /; FreeQ[{b, e, f, A, C, m}, x] && NeQ[C*m + A*(m + 1), 0] &&  !LeQ[m, -1]

Rule 4173

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

Rubi steps \begin{align*} \text {integral}& = \frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac {\int \sec ^2(c+d x) (a+a \sec (c+d x))^2 (a (5 A+2 C)+a (5 B+2 C) \sec (c+d x)) \, dx}{5 a} \\ & = \frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac {(5 B+2 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 a d}+\frac {\int \sec (c+d x) (a+a \sec (c+d x))^2 \left (3 a^2 (5 B+2 C)+a^2 (20 A-5 B+6 C) \sec (c+d x)\right ) \, dx}{20 a^2} \\ & = \frac {(20 A-5 B+6 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac {(5 B+2 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 a d}+\frac {1}{12} (8 A+7 B+6 C) \int \sec (c+d x) (a+a \sec (c+d x))^2 \, dx \\ & = \frac {(20 A-5 B+6 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac {(5 B+2 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 a d}+\frac {1}{12} (8 A+7 B+6 C) \int \sec (c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx+\frac {1}{6} \left (a^2 (8 A+7 B+6 C)\right ) \int \sec ^2(c+d x) \, dx \\ & = \frac {a^2 (8 A+7 B+6 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(20 A-5 B+6 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac {(5 B+2 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 a d}+\frac {1}{8} \left (a^2 (8 A+7 B+6 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^2 (8 A+7 B+6 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 d} \\ & = \frac {a^2 (8 A+7 B+6 C) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 (8 A+7 B+6 C) \tan (c+d x)}{6 d}+\frac {a^2 (8 A+7 B+6 C) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(20 A-5 B+6 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{60 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac {(5 B+2 C) (a+a \sec (c+d x))^3 \tan (c+d x)}{20 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.73 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.60 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \left (15 (8 A+7 B+6 C) \text {arctanh}(\sin (c+d x))+\left (8 (25 A+20 B+18 C)+15 (8 A+7 B+6 C) \sec (c+d x)+8 (5 A+10 B+9 C) \sec ^2(c+d x)+30 (B+2 C) \sec ^3(c+d x)+24 C \sec ^4(c+d x)\right ) \tan (c+d x)\right )}{120 d} \]

[In]

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

[Out]

(a^2*(15*(8*A + 7*B + 6*C)*ArcTanh[Sin[c + d*x]] + (8*(25*A + 20*B + 18*C) + 15*(8*A + 7*B + 6*C)*Sec[c + d*x]
+ 8*(5*A + 10*B + 9*C)*Sec[c + d*x]^2 + 30*(B + 2*C)*Sec[c + d*x]^3 + 24*C*Sec[c + d*x]^4)*Tan[c + d*x]))/(12
0*d)

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.05

method	result	size
parts	\(\frac {\left (2 a^{2} A +B \,a^{2}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (B \,a^{2}+2 C \,a^{2}\right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (a^{2} A +2 B \,a^{2}+C \,a^{2}\right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}-\frac {C \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} A \tan \left (d x +c \right )}{d}\)	\(200\)
norman	\(\frac {\frac {7 a^{2} \left (8 A +7 B +6 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}-\frac {a^{2} \left (8 A +7 B +6 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {a^{2} \left (24 A +25 B +26 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {8 a^{2} \left (35 A +25 B +27 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {a^{2} \left (104 A +79 B +54 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a^{2} \left (8 A +7 B +6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{2} \left (8 A +7 B +6 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(222\)
parallelrisch	\(-\frac {\left (\left (A +\frac {7 B}{8}+\frac {3 C}{4}\right ) \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\left (A +\frac {7 B}{8}+\frac {3 C}{4}\right ) \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-4 A -\frac {11 B}{2}-7 C \right ) \sin \left (2 d x +2 c \right )+\left (-6 C -\frac {19 A}{3}-\frac {20 B}{3}\right ) \sin \left (3 d x +3 c \right )+\left (-\frac {7 B}{4}-\frac {3 C}{2}-2 A \right ) \sin \left (4 d x +4 c \right )+\left (-\frac {6 C}{5}-\frac {5 A}{3}-\frac {4 B}{3}\right ) \sin \left (5 d x +5 c \right )-\frac {14 \sin \left (d x +c \right ) \left (A +\frac {8 B}{7}+\frac {12 C}{7}\right )}{3}\right ) a^{2}}{d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(238\)
derivativedivides	\(\frac {-a^{2} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+2 a^{2} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 B \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} A \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(295\)
default	\(\frac {-a^{2} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+B \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-C \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )+2 a^{2} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 B \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} A \tan \left (d x +c \right )+B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(295\)
risch	\(-\frac {i a^{2} \left (-144 C -200 A -160 B -120 A \,{\mathrm e}^{8 i \left (d x +c \right )}+240 A \,{\mathrm e}^{7 i \left (d x +c \right )}-240 A \,{\mathrm e}^{3 i \left (d x +c \right )}-120 A \,{\mathrm e}^{i \left (d x +c \right )}-1120 B \,{\mathrm e}^{4 i \left (d x +c \right )}-720 A \,{\mathrm e}^{6 i \left (d x +c \right )}-240 C \,{\mathrm e}^{6 i \left (d x +c \right )}-1280 A \,{\mathrm e}^{4 i \left (d x +c \right )}-1200 C \,{\mathrm e}^{4 i \left (d x +c \right )}-880 A \,{\mathrm e}^{2 i \left (d x +c \right )}-720 C \,{\mathrm e}^{2 i \left (d x +c \right )}+420 C \,{\mathrm e}^{7 i \left (d x +c \right )}-420 C \,{\mathrm e}^{3 i \left (d x +c \right )}-800 B \,{\mathrm e}^{2 i \left (d x +c \right )}-90 C \,{\mathrm e}^{i \left (d x +c \right )}+120 A \,{\mathrm e}^{9 i \left (d x +c \right )}+90 C \,{\mathrm e}^{9 i \left (d x +c \right )}-105 B \,{\mathrm e}^{i \left (d x +c \right )}+330 B \,{\mathrm e}^{7 i \left (d x +c \right )}-330 B \,{\mathrm e}^{3 i \left (d x +c \right )}+105 B \,{\mathrm e}^{9 i \left (d x +c \right )}-480 B \,{\mathrm e}^{6 i \left (d x +c \right )}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{8 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{4 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {7 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{8 d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{4 d}\)	\(429\)

[In]

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

[Out]

(2*A*a^2+B*a^2)/d*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+(B*a^2+2*C*a^2)/d*(-(-1/4*sec(d*x+
c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))-(A*a^2+2*B*a^2+C*a^2)/d*(-2/3-1/3*sec(d*x+c)^2)
*tan(d*x+c)-C*a^2/d*(-8/15-1/5*sec(d*x+c)^4-4/15*sec(d*x+c)^2)*tan(d*x+c)+a^2*A/d*tan(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.95 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (8 \, A + 7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (8 \, A + 7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (25 \, A + 20 \, B + 18 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \, {\left (8 \, A + 7 \, B + 6 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 8 \, {\left (5 \, A + 10 \, B + 9 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 24 \, C a^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

[In]

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

[Out]

1/240*(15*(8*A + 7*B + 6*C)*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(8*A + 7*B + 6*C)*a^2*cos(d*x + c)^5
*log(-sin(d*x + c) + 1) + 2*(8*(25*A + 20*B + 18*C)*a^2*cos(d*x + c)^4 + 15*(8*A + 7*B + 6*C)*a^2*cos(d*x + c)
^3 + 8*(5*A + 10*B + 9*C)*a^2*cos(d*x + c)^2 + 30*(B + 2*C)*a^2*cos(d*x + c) + 24*C*a^2)*sin(d*x + c))/(d*cos(
d*x + c)^5)

Sympy [F]

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

[In]

integrate(sec(d*x+c)**2*(a+a*sec(d*x+c))**2*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

a**2*(Integral(A*sec(c + d*x)**2, x) + Integral(2*A*sec(c + d*x)**3, x) + Integral(A*sec(c + d*x)**4, x) + Int
egral(B*sec(c + d*x)**3, x) + Integral(2*B*sec(c + d*x)**4, x) + Integral(B*sec(c + d*x)**5, x) + Integral(C*s
ec(c + d*x)**4, x) + Integral(2*C*sec(c + d*x)**5, x) + Integral(C*sec(c + d*x)**6, x))

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (178) = 356\).

Time = 0.23 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.89 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a^{2} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 15 \, B a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 30 \, C a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{2} \tan \left (d x + c\right )}{240 \, d} \]

[In]

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

[Out]

1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 160*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 + 16*(3*tan(d*
x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a^2 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 15*B*a^2*
(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*l
og(sin(d*x + c) - 1)) - 30*C*a^2*(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)) - 120*A*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(
sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 60*B*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c)
+ 1) + log(sin(d*x + c) - 1)) + 240*A*a^2*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.79 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (8 \, A a^{2} + 7 \, B a^{2} + 6 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (8 \, A a^{2} + 7 \, B a^{2} + 6 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 105 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 90 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 560 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 490 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 420 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1120 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 800 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 864 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 1040 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 790 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 540 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 360 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 375 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 390 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{120 \, d} \]

[In]

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

[Out]

1/120*(15*(8*A*a^2 + 7*B*a^2 + 6*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(8*A*a^2 + 7*B*a^2 + 6*C*a^2)*
log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 105*B*a^2*tan(1/2*d*x + 1/2*c)^9 +
90*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 560*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 490*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 420*C*
a^2*tan(1/2*d*x + 1/2*c)^7 + 1120*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 800*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 864*C*a^2*
tan(1/2*d*x + 1/2*c)^5 - 1040*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 790*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 540*C*a^2*tan(
1/2*d*x + 1/2*c)^3 + 360*A*a^2*tan(1/2*d*x + 1/2*c) + 375*B*a^2*tan(1/2*d*x + 1/2*c) + 390*C*a^2*tan(1/2*d*x +
1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^5)/d

Mupad [B] (verification not implemented)

Time = 16.64 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.51 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (A+\frac {7\,B}{8}+\frac {3\,C}{4}\right )}{d}-\frac {\left (2\,A\,a^2+\frac {7\,B\,a^2}{4}+\frac {3\,C\,a^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {28\,A\,a^2}{3}-\frac {49\,B\,a^2}{6}-7\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {56\,A\,a^2}{3}+\frac {40\,B\,a^2}{3}+\frac {72\,C\,a^2}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {52\,A\,a^2}{3}-\frac {79\,B\,a^2}{6}-9\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,A\,a^2+\frac {25\,B\,a^2}{4}+\frac {13\,C\,a^2}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )}-\frac {a^2\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (8\,A+7\,B+6\,C\right )}{8\,d} \]

[In]

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

[Out]

(a^2*log(tan(c/2 + (d*x)/2) + 1)*(A + (7*B)/8 + (3*C)/4))/d - (tan(c/2 + (d*x)/2)^9*(2*A*a^2 + (7*B*a^2)/4 + (
3*C*a^2)/2) - tan(c/2 + (d*x)/2)^7*((28*A*a^2)/3 + (49*B*a^2)/6 + 7*C*a^2) - tan(c/2 + (d*x)/2)^3*((52*A*a^2)/
3 + (79*B*a^2)/6 + 9*C*a^2) + tan(c/2 + (d*x)/2)^5*((56*A*a^2)/3 + (40*B*a^2)/3 + (72*C*a^2)/5) + tan(c/2 + (d
*x)/2)*(6*A*a^2 + (25*B*a^2)/4 + (13*C*a^2)/2))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(
c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1)) - (a^2*log(tan(c/2 + (d*x)/2) - 1)*(8*
A + 7*B + 6*C))/(8*d)