\(\int \sec (c+d x) (a+b \sec (c+d x))^2 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [776]

Optimal result

Integrand size = 38, antiderivative size = 179 \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (8 a b B+4 a^2 C+3 b^2 C\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (4 a^2 b B+4 b^3 B-a^3 C+8 a b^2 C\right ) \tan (c+d x)}{6 b d}+\frac {\left (8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d} \]

[Out]

1/8*(8*B*a*b+4*C*a^2+3*C*b^2)*arctanh(sin(d*x+c))/d+1/6*(4*B*a^2*b+4*B*b^3-C*a^3+8*C*a*b^2)*tan(d*x+c)/b/d+1/2
4*(8*B*a*b-2*C*a^2+9*C*b^2)*sec(d*x+c)*tan(d*x+c)/d+1/12*(4*B*b-C*a)*(a+b*sec(d*x+c))^2*tan(d*x+c)/b/d+1/4*C*(
a+b*sec(d*x+c))^3*tan(d*x+c)/b/d

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {4157, 4095, 4087, 4082, 3872, 3855, 3852, 8} \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\left (4 a^2 C+8 a b B+3 b^2 C\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (-2 a^2 C+8 a b B+9 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {\left (a^3 (-C)+4 a^2 b B+8 a b^2 C+4 b^3 B\right ) \tan (c+d x)}{6 b d}+\frac {(4 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^2}{12 b d}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d} \]

[In]

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

[Out]

((8*a*b*B + 4*a^2*C + 3*b^2*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((4*a^2*b*B + 4*b^3*B - a^3*C + 8*a*b^2*C)*Tan[c
+ d*x])/(6*b*d) + ((8*a*b*B - 2*a^2*C + 9*b^2*C)*Sec[c + d*x]*Tan[c + d*x])/(24*d) + ((4*b*B - a*C)*(a + b*Se
c[c + d*x])^2*Tan[c + d*x])/(12*b*d) + (C*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*b*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 3872

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

Rule 4082

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

Rule 4087

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[1/(m + 1), Int[Csc[e + f
*x]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /;
FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

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 4157

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

Rubi steps \begin{align*} \text {integral}& = \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (B+C \sec (c+d x)) \, dx \\ & = \frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 (3 b C+(4 b B-a C) \sec (c+d x)) \, dx}{4 b} \\ & = \frac {(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (b (8 b B+7 a C)+\left (8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x)\right ) \, dx}{12 b} \\ & = \frac {\left (8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\int \sec (c+d x) \left (3 b \left (8 a b B+4 a^2 C+3 b^2 C\right )+4 \left (4 a^2 b B+4 b^3 B-a^3 C+8 a b^2 C\right ) \sec (c+d x)\right ) \, dx}{24 b} \\ & = \frac {\left (8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {1}{8} \left (8 a b B+4 a^2 C+3 b^2 C\right ) \int \sec (c+d x) \, dx+\frac {\left (4 a^2 b B+4 b^3 B-a^3 C+8 a b^2 C\right ) \int \sec ^2(c+d x) \, dx}{6 b} \\ & = \frac {\left (8 a b B+4 a^2 C+3 b^2 C\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}-\frac {\left (4 a^2 b B+4 b^3 B-a^3 C+8 a b^2 C\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 b d} \\ & = \frac {\left (8 a b B+4 a^2 C+3 b^2 C\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (4 a^2 b B+4 b^3 B-a^3 C+8 a b^2 C\right ) \tan (c+d x)}{6 b d}+\frac {\left (8 a b B-2 a^2 C+9 b^2 C\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 b B-a C) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {C (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.67 \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \left (8 a b B+4 a^2 C+3 b^2 C\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (24 \left (a^2 B+b^2 B+2 a b C\right )+3 \left (8 a b B+4 a^2 C+3 b^2 C\right ) \sec (c+d x)+6 b^2 C \sec ^3(c+d x)+8 b (b B+2 a C) \tan ^2(c+d x)\right )}{24 d} \]

[In]

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

[Out]

(3*(8*a*b*B + 4*a^2*C + 3*b^2*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(24*(a^2*B + b^2*B + 2*a*b*C) + 3*(8*a*b
*B + 4*a^2*C + 3*b^2*C)*Sec[c + d*x] + 6*b^2*C*Sec[c + d*x]^3 + 8*b*(b*B + 2*a*C)*Tan[c + d*x]^2))/(24*d)

Maple [A] (verified)

Time = 0.98 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.83

method	result	size
parts	\(-\frac {\left (B \,b^{2}+2 C a b \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (2 B a b +C \,a^{2}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {B \,a^{2} \tan \left (d x +c \right )}{d}+\frac {C \,b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(148\)
derivativedivides	\(\frac {B \,a^{2} \tan \left (d x +c \right )+C \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 B a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 C a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-B \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C \,b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(185\)
default	\(\frac {B \,a^{2} \tan \left (d x +c \right )+C \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 B a b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-2 C a b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-B \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+C \,b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\)	\(185\)
parallelrisch	\(\frac {-96 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (B a b +\frac {1}{2} C \,a^{2}+\frac {3}{8} C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+96 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (B a b +\frac {1}{2} C \,a^{2}+\frac {3}{8} C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (48 B \,a^{2}+64 B \,b^{2}+128 C a b \right ) \sin \left (2 d x +2 c \right )+\left (48 B a b +24 C \,a^{2}+18 C \,b^{2}\right ) \sin \left (3 d x +3 c \right )+\left (24 B \,a^{2}+16 B \,b^{2}+32 C a b \right ) \sin \left (4 d x +4 c \right )+48 \sin \left (d x +c \right ) \left (B a b +\frac {1}{2} C \,a^{2}+\frac {11}{8} C \,b^{2}\right )}{24 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(246\)
norman	\(\frac {-\frac {\left (8 B \,a^{2}-8 B a b +8 B \,b^{2}-4 C \,a^{2}+16 C a b -5 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {\left (8 B \,a^{2}+8 B a b +8 B \,b^{2}+4 C \,a^{2}+16 C a b +5 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (72 B \,a^{2}-24 B a b +40 B \,b^{2}-12 C \,a^{2}+80 C a b +9 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}-\frac {\left (72 B \,a^{2}+24 B a b +40 B \,b^{2}+12 C \,a^{2}+80 C a b -9 C \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )^{4}}-\frac {\left (8 B a b +4 C \,a^{2}+3 C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (8 B a b +4 C \,a^{2}+3 C \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(291\)
risch	\(-\frac {i \left (24 B a b \,{\mathrm e}^{7 i \left (d x +c \right )}+12 C \,a^{2} {\mathrm e}^{7 i \left (d x +c \right )}+9 C \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-24 B \,a^{2} {\mathrm e}^{6 i \left (d x +c \right )}+24 B a b \,{\mathrm e}^{5 i \left (d x +c \right )}+12 C \,a^{2} {\mathrm e}^{5 i \left (d x +c \right )}+33 C \,b^{2} {\mathrm e}^{5 i \left (d x +c \right )}-72 B \,a^{2} {\mathrm e}^{4 i \left (d x +c \right )}-48 B \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-96 C a b \,{\mathrm e}^{4 i \left (d x +c \right )}-24 B a b \,{\mathrm e}^{3 i \left (d x +c \right )}-12 C \,a^{2} {\mathrm e}^{3 i \left (d x +c \right )}-33 C \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-72 B \,a^{2} {\mathrm e}^{2 i \left (d x +c \right )}-64 B \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-128 C a b \,{\mathrm e}^{2 i \left (d x +c \right )}-24 B a b \,{\mathrm e}^{i \left (d x +c \right )}-12 C \,a^{2} {\mathrm e}^{i \left (d x +c \right )}-9 C \,b^{2} {\mathrm e}^{i \left (d x +c \right )}-24 B \,a^{2}-16 B \,b^{2}-32 C a b \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a b}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C \,b^{2}}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a b}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C \,b^{2}}{8 d}\)	\(447\)

[In]

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

[Out]

-(B*b^2+2*C*a*b)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+(2*B*a*b+C*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/d*tan(d*x+c)+C*b^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)))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.01 \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (3 \, B a^{2} + 4 \, C a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, C b^{2} + 3 \, {\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

1/48*(3*(4*C*a^2 + 8*B*a*b + 3*C*b^2)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(4*C*a^2 + 8*B*a*b + 3*C*b^2)*c
os(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(8*(3*B*a^2 + 4*C*a*b + 2*B*b^2)*cos(d*x + c)^3 + 6*C*b^2 + 3*(4*C*a^
2 + 8*B*a*b + 3*C*b^2)*cos(d*x + c)^2 + 8*(2*C*a*b + B*b^2)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F]

\[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int \left (B + C \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{2}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

Integral((B + C*sec(c + d*x))*(a + b*sec(c + d*x))**2*sec(c + d*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.27 \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b^{2} - 3 \, C b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, B a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B a^{2} \tan \left (d x + c\right )}{48 \, d} \]

[In]

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

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a*b + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*b^2 - 3*C*b^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)) - 12*C*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
)) - 24*B*a*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 48*B*a^2
*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 478 vs. \(2 (167) = 334\).

Time = 0.33 (sec) , antiderivative size = 478, normalized size of antiderivative = 2.67 \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {3 \, {\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 80 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C 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 )}^{4}}}{24 \, d} \]

[In]

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

[Out]

1/24*(3*(4*C*a^2 + 8*B*a*b + 3*C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*C*a^2 + 8*B*a*b + 3*C*b^2)*log
(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(24*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 12*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 24*B*
a*b*tan(1/2*d*x + 1/2*c)^7 + 48*C*a*b*tan(1/2*d*x + 1/2*c)^7 + 24*B*b^2*tan(1/2*d*x + 1/2*c)^7 - 15*C*b^2*tan(
1/2*d*x + 1/2*c)^7 - 72*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 24*B*a*b*tan(1/2*d*x
+ 1/2*c)^5 - 80*C*a*b*tan(1/2*d*x + 1/2*c)^5 - 40*B*b^2*tan(1/2*d*x + 1/2*c)^5 - 9*C*b^2*tan(1/2*d*x + 1/2*c)^
5 + 72*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 80*C
*a*b*tan(1/2*d*x + 1/2*c)^3 + 40*B*b^2*tan(1/2*d*x + 1/2*c)^3 - 9*C*b^2*tan(1/2*d*x + 1/2*c)^3 - 24*B*a^2*tan(
1/2*d*x + 1/2*c) - 12*C*a^2*tan(1/2*d*x + 1/2*c) - 24*B*a*b*tan(1/2*d*x + 1/2*c) - 48*C*a*b*tan(1/2*d*x + 1/2*
c) - 24*B*b^2*tan(1/2*d*x + 1/2*c) - 15*C*b^2*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 = 20.88 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.77 \[ \int \sec (c+d x) (a+b \sec (c+d x))^2 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {C\,a^2}{2}+B\,a\,b+\frac {3\,C\,b^2}{8}\right )}{2\,C\,a^2+4\,B\,a\,b+\frac {3\,C\,b^2}{2}}\right )\,\left (C\,a^2+2\,B\,a\,b+\frac {3\,C\,b^2}{4}\right )}{d}-\frac {\left (2\,B\,a^2+2\,B\,b^2-C\,a^2-\frac {5\,C\,b^2}{4}-2\,B\,a\,b+4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (C\,a^2-\frac {10\,B\,b^2}{3}-6\,B\,a^2-\frac {3\,C\,b^2}{4}+2\,B\,a\,b-\frac {20\,C\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (6\,B\,a^2+\frac {10\,B\,b^2}{3}+C\,a^2-\frac {3\,C\,b^2}{4}+2\,B\,a\,b+\frac {20\,C\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,B\,a^2-2\,B\,b^2-C\,a^2-\frac {5\,C\,b^2}{4}-2\,B\,a\,b-4\,C\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

[In]

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

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((C*a^2)/2 + (3*C*b^2)/8 + B*a*b))/(2*C*a^2 + (3*C*b^2)/2 + 4*B*a*b))*(C*a^2 + (3
*C*b^2)/4 + 2*B*a*b))/d - (tan(c/2 + (d*x)/2)^7*(2*B*a^2 + 2*B*b^2 - C*a^2 - (5*C*b^2)/4 - 2*B*a*b + 4*C*a*b)
+ tan(c/2 + (d*x)/2)^3*(6*B*a^2 + (10*B*b^2)/3 + C*a^2 - (3*C*b^2)/4 + 2*B*a*b + (20*C*a*b)/3) - tan(c/2 + (d*
x)/2)^5*(6*B*a^2 + (10*B*b^2)/3 - C*a^2 + (3*C*b^2)/4 - 2*B*a*b + (20*C*a*b)/3) - tan(c/2 + (d*x)/2)*(2*B*a^2
+ 2*B*b^2 + C*a^2 + (5*C*b^2)/4 + 2*B*a*b + 4*C*a*b))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*
tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))