\(\int \sec ^2(c+d x) (a+a \sec (c+d x))^2 (A+C \sec ^2(c+d x)) \, dx\) [93]

Optimal result

Integrand size = 33, antiderivative size = 172 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (4 A+3 C) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^2 (4 A+3 C) \tan (c+d x)}{3 d}+\frac {a^2 (4 A+3 C) \sec (c+d x) \tan (c+d x)}{12 d}+\frac {(10 A+3 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{30 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{10 a d} \]

[Out]

1/4*a^2*(4*A+3*C)*arctanh(sin(d*x+c))/d+1/3*a^2*(4*A+3*C)*tan(d*x+c)/d+1/12*a^2*(4*A+3*C)*sec(d*x+c)*tan(d*x+c
)/d+1/30*(10*A+3*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+1/10*C*
(a+a*sec(d*x+c))^3*tan(d*x+c)/a/d

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {4174, 4095, 4086, 3873, 3852, 8, 4131, 3855} \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 (4 A+3 C) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^2 (4 A+3 C) \tan (c+d x)}{3 d}+\frac {a^2 (4 A+3 C) \tan (c+d x) \sec (c+d x)}{12 d}+\frac {(10 A+3 C) \tan (c+d x) (a \sec (c+d x)+a)^2}{30 d}+\frac {C \tan (c+d x) \sec ^2(c+d x) (a \sec (c+d x)+a)^2}{5 d}+\frac {C \tan (c+d x) (a \sec (c+d x)+a)^3}{10 a d} \]

[In]

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

[Out]

(a^2*(4*A + 3*C)*ArcTanh[Sin[c + d*x]])/(4*d) + (a^2*(4*A + 3*C)*Tan[c + d*x])/(3*d) + (a^2*(4*A + 3*C)*Sec[c
+ d*x]*Tan[c + d*x])/(12*d) + ((10*A + 3*C)*(a + a*Sec[c + d*x])^2*Tan[c + d*x])/(30*d) + (C*Sec[c + d*x]^2*(a
+ a*Sec[c + d*x])^2*Tan[c + d*x])/(5*d) + (C*(a + a*Sec[c + d*x])^3*Tan[c + d*x])/(10*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 4174

Int[((A_.) + 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*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 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)+2 a 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 {C (a+a \sec (c+d x))^3 \tan (c+d x)}{10 a d}+\frac {\int \sec (c+d x) (a+a \sec (c+d x))^2 \left (6 a^2 C+2 a^2 (10 A+3 C) \sec (c+d x)\right ) \, dx}{20 a^2} \\ & = \frac {(10 A+3 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{30 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{10 a d}+\frac {1}{6} (4 A+3 C) \int \sec (c+d x) (a+a \sec (c+d x))^2 \, dx \\ & = \frac {(10 A+3 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{30 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{10 a d}+\frac {1}{6} (4 A+3 C) \int \sec (c+d x) \left (a^2+a^2 \sec ^2(c+d x)\right ) \, dx+\frac {1}{3} \left (a^2 (4 A+3 C)\right ) \int \sec ^2(c+d x) \, dx \\ & = \frac {a^2 (4 A+3 C) \sec (c+d x) \tan (c+d x)}{12 d}+\frac {(10 A+3 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{30 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{10 a d}+\frac {1}{4} \left (a^2 (4 A+3 C)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^2 (4 A+3 C)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d} \\ & = \frac {a^2 (4 A+3 C) \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^2 (4 A+3 C) \tan (c+d x)}{3 d}+\frac {a^2 (4 A+3 C) \sec (c+d x) \tan (c+d x)}{12 d}+\frac {(10 A+3 C) (a+a \sec (c+d x))^2 \tan (c+d x)}{30 d}+\frac {C \sec ^2(c+d x) (a+a \sec (c+d x))^2 \tan (c+d x)}{5 d}+\frac {C (a+a \sec (c+d x))^3 \tan (c+d x)}{10 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.45 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.55 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {a^2 \left (15 (4 A+3 C) \text {arctanh}(\sin (c+d x))+\left (100 A+72 C+15 (4 A+3 C) \sec (c+d x)+4 (5 A+9 C) \sec ^2(c+d x)+30 C \sec ^3(c+d x)+12 C \sec ^4(c+d x)\right ) \tan (c+d x)\right )}{60 d} \]

[In]

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

[Out]

(a^2*(15*(4*A + 3*C)*ArcTanh[Sin[c + d*x]] + (100*A + 72*C + 15*(4*A + 3*C)*Sec[c + d*x] + 4*(5*A + 9*C)*Sec[c
+ d*x]^2 + 30*C*Sec[c + d*x]^3 + 12*C*Sec[c + d*x]^4)*Tan[c + d*x]))/(60*d)

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.06

method	result	size
parts	\(-\frac {\left (a^{2} A +C \,a^{2}\right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}-\frac {C \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} A \tan \left (d x +c \right )}{d}+\frac {2 C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {a^{2} A \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(182\)
derivativedivides	\(\frac {a^{2} A \tan \left (d x +c \right )-C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a^{2} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{2} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(186\)
default	\(\frac {a^{2} A \tan \left (d x +c \right )-C \,a^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+2 a^{2} A \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 C \,a^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-a^{2} A \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-C \,a^{2} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(186\)
norman	\(\frac {\frac {7 a^{2} \left (4 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 d}-\frac {a^{2} \left (4 A +3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{2 d}-\frac {a^{2} \left (12 A +13 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d}-\frac {8 a^{2} \left (35 A +27 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}+\frac {a^{2} \left (52 A +27 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{5}}-\frac {a^{2} \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4 d}+\frac {a^{2} \left (4 A +3 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d}\)	\(201\)
parallelrisch	\(\frac {14 a^{2} \left (-\frac {15 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{7}+\frac {15 \left (\frac {\cos \left (5 d x +5 c \right )}{10}+\frac {\cos \left (3 d x +3 c \right )}{2}+\cos \left (d x +c \right )\right ) \left (A +\frac {3 C}{4}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{7}+\left (\frac {6 A}{7}+\frac {3 C}{2}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {19 A}{14}+\frac {9 C}{7}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {3 A}{7}+\frac {9 C}{28}\right ) \sin \left (4 d x +4 c \right )+\left (\frac {5 A}{14}+\frac {9 C}{35}\right ) \sin \left (5 d x +5 c \right )+\sin \left (d x +c \right ) \left (A +\frac {12 C}{7}\right )\right )}{3 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(217\)
risch	\(-\frac {i a^{2} \left (60 A \,{\mathrm e}^{9 i \left (d x +c \right )}+45 C \,{\mathrm e}^{9 i \left (d x +c \right )}-60 A \,{\mathrm e}^{8 i \left (d x +c \right )}+120 A \,{\mathrm e}^{7 i \left (d x +c \right )}+210 C \,{\mathrm e}^{7 i \left (d x +c \right )}-360 A \,{\mathrm e}^{6 i \left (d x +c \right )}-120 C \,{\mathrm e}^{6 i \left (d x +c \right )}-640 A \,{\mathrm e}^{4 i \left (d x +c \right )}-600 C \,{\mathrm e}^{4 i \left (d x +c \right )}-120 A \,{\mathrm e}^{3 i \left (d x +c \right )}-210 C \,{\mathrm e}^{3 i \left (d x +c \right )}-440 A \,{\mathrm e}^{2 i \left (d x +c \right )}-360 C \,{\mathrm e}^{2 i \left (d x +c \right )}-60 A \,{\mathrm e}^{i \left (d x +c \right )}-45 C \,{\mathrm e}^{i \left (d x +c \right )}-100 A -72 C \right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A}{d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{4 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A}{d}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{4 d}\)	\(298\)

[In]

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

[Out]

-(A*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)+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^2*A/d*tan(d*x+c)*sec(d*x+c)+a^2*A/d*ln(sec(d*x+c)+tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.94 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, {\left (25 \, A + 18 \, C\right )} a^{2} \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} + 4 \, {\left (5 \, A + 9 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 30 \, C a^{2} \cos \left (d x + c\right ) + 12 \, C a^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \]

[In]

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

[Out]

1/120*(15*(4*A + 3*C)*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*A + 3*C)*a^2*cos(d*x + c)^5*log(-sin(d*
x + c) + 1) + 2*(4*(25*A + 18*C)*a^2*cos(d*x + c)^4 + 15*(4*A + 3*C)*a^2*cos(d*x + c)^3 + 4*(5*A + 9*C)*a^2*co
s(d*x + c)^2 + 30*C*a^2*cos(d*x + c) + 12*C*a^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

Sympy [F]

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

[In]

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

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.27 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} - 15 \, C a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, A a^{2} \tan \left (d x + c\right )}{120 \, d} \]

[In]

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

[Out]

1/120*(40*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 8*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))
*C*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^2 - 15*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)) - 60*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)) + 120*A*a^2*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 246, normalized size of antiderivative = 1.43 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {15 \, {\left (4 \, A a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (4 \, A a^{2} + 3 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 45 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 280 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 210 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 560 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 432 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 520 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 270 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 180 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 195 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{60 \, d} \]

[In]

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

[Out]

1/60*(15*(4*A*a^2 + 3*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*A*a^2 + 3*C*a^2)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(60*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 45*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 280*A*a^2*tan(1/2*d*x +
1/2*c)^7 - 210*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 560*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 432*C*a^2*tan(1/2*d*x + 1/2*
c)^5 - 520*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 270*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 180*A*a^2*tan(1/2*d*x + 1/2*c) +
195*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 = 17.62 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.29 \[ \int \sec ^2(c+d x) (a+a \sec (c+d x))^2 \left (A+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (A+\frac {3\,C}{4}\right )}{d}-\frac {\left (2\,A\,a^2+\frac {3\,C\,a^2}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (-\frac {28\,A\,a^2}{3}-7\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {56\,A\,a^2}{3}+\frac {72\,C\,a^2}{5}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {52\,A\,a^2}{3}-9\,C\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,A\,a^2+\frac {13\,C\,a^2}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

(2*a^2*atanh(tan(c/2 + (d*x)/2))*(A + (3*C)/4))/d - (tan(c/2 + (d*x)/2)*(6*A*a^2 + (13*C*a^2)/2) + tan(c/2 + (
d*x)/2)^9*(2*A*a^2 + (3*C*a^2)/2) - tan(c/2 + (d*x)/2)^7*((28*A*a^2)/3 + 7*C*a^2) - tan(c/2 + (d*x)/2)^3*((52*
A*a^2)/3 + 9*C*a^2) + tan(c/2 + (d*x)/2)^5*((56*A*a^2)/3 + (72*C*a^2)/5))/(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))