\(\int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\) [294]

Optimal result

Integrand size = 31, antiderivative size = 252 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {\left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (15 a^3 A b+60 a A b^3-3 a^4 B+52 a^2 b^2 B+16 b^4 B\right ) \tan (c+d x)}{30 b d}+\frac {\left (30 a^2 A b+45 A b^3-6 a^3 B+71 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {\left (15 a A b-3 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac {(5 A b-a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d} \]

[Out]

1/8*(12*A*a^2*b+3*A*b^3+4*B*a^3+9*B*a*b^2)*arctanh(sin(d*x+c))/d+1/30*(15*A*a^3*b+60*A*a*b^3-3*B*a^4+52*B*a^2*
b^2+16*B*b^4)*tan(d*x+c)/b/d+1/120*(30*A*a^2*b+45*A*b^3-6*B*a^3+71*B*a*b^2)*sec(d*x+c)*tan(d*x+c)/d+1/60*(15*A
*a*b-3*B*a^2+16*B*b^2)*(a+b*sec(d*x+c))^2*tan(d*x+c)/b/d+1/20*(5*A*b-B*a)*(a+b*sec(d*x+c))^3*tan(d*x+c)/b/d+1/
5*B*(a+b*sec(d*x+c))^4*tan(d*x+c)/b/d

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4095, 4087, 4082, 3872, 3855, 3852, 8} \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {\left (-3 a^2 B+15 a A b+16 b^2 B\right ) \tan (c+d x) (a+b \sec (c+d x))^2}{60 b d}+\frac {\left (4 a^3 B+12 a^2 A b+9 a b^2 B+3 A b^3\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (-6 a^3 B+30 a^2 A b+71 a b^2 B+45 A b^3\right ) \tan (c+d x) \sec (c+d x)}{120 d}+\frac {\left (-3 a^4 B+15 a^3 A b+52 a^2 b^2 B+60 a A b^3+16 b^4 B\right ) \tan (c+d x)}{30 b d}+\frac {(5 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^3}{20 b d}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^4}{5 b d} \]

[In]

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

[Out]

((12*a^2*A*b + 3*A*b^3 + 4*a^3*B + 9*a*b^2*B)*ArcTanh[Sin[c + d*x]])/(8*d) + ((15*a^3*A*b + 60*a*A*b^3 - 3*a^4
*B + 52*a^2*b^2*B + 16*b^4*B)*Tan[c + d*x])/(30*b*d) + ((30*a^2*A*b + 45*A*b^3 - 6*a^3*B + 71*a*b^2*B)*Sec[c +
d*x]*Tan[c + d*x])/(120*d) + ((15*a*A*b - 3*a^2*B + 16*b^2*B)*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(60*b*d) +
((5*A*b - a*B)*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(20*b*d) + (B*(a + b*Sec[c + d*x])^4*Tan[c + d*x])/(5*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]

Rubi steps \begin{align*} \text {integral}& = \frac {B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^3 (4 b B+(5 A b-a B) \sec (c+d x)) \, dx}{5 b} \\ & = \frac {(5 A b-a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 \left (b (15 A b+13 a B)+\left (15 a A b-3 a^2 B+16 b^2 B\right ) \sec (c+d x)\right ) \, dx}{20 b} \\ & = \frac {\left (15 a A b-3 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac {(5 A b-a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (b \left (75 a A b+33 a^2 B+32 b^2 B\right )+\left (30 a^2 A b+45 A b^3-6 a^3 B+71 a b^2 B\right ) \sec (c+d x)\right ) \, dx}{60 b} \\ & = \frac {\left (30 a^2 A b+45 A b^3-6 a^3 B+71 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {\left (15 a A b-3 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac {(5 A b-a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {\int \sec (c+d x) \left (15 b \left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right )+4 \left (15 a^3 A b+60 a A b^3-3 a^4 B+52 a^2 b^2 B+16 b^4 B\right ) \sec (c+d x)\right ) \, dx}{120 b} \\ & = \frac {\left (30 a^2 A b+45 A b^3-6 a^3 B+71 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {\left (15 a A b-3 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac {(5 A b-a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}+\frac {1}{8} \left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right ) \int \sec (c+d x) \, dx+\frac {\left (15 a^3 A b+60 a A b^3-3 a^4 B+52 a^2 b^2 B+16 b^4 B\right ) \int \sec ^2(c+d x) \, dx}{30 b} \\ & = \frac {\left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (30 a^2 A b+45 A b^3-6 a^3 B+71 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {\left (15 a A b-3 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac {(5 A b-a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d}-\frac {\left (15 a^3 A b+60 a A b^3-3 a^4 B+52 a^2 b^2 B+16 b^4 B\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{30 b d} \\ & = \frac {\left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right ) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {\left (15 a^3 A b+60 a A b^3-3 a^4 B+52 a^2 b^2 B+16 b^4 B\right ) \tan (c+d x)}{30 b d}+\frac {\left (30 a^2 A b+45 A b^3-6 a^3 B+71 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{120 d}+\frac {\left (15 a A b-3 a^2 B+16 b^2 B\right ) (a+b \sec (c+d x))^2 \tan (c+d x)}{60 b d}+\frac {(5 A b-a B) (a+b \sec (c+d x))^3 \tan (c+d x)}{20 b d}+\frac {B (a+b \sec (c+d x))^4 \tan (c+d x)}{5 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.74 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.72 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {15 \left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right ) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (15 \left (12 a^2 A b+3 A b^3+4 a^3 B+9 a b^2 B\right ) \sec (c+d x)+30 b^2 (A b+3 a B) \sec ^3(c+d x)+8 \left (15 \left (a^3 A+3 a A b^2+3 a^2 b B+b^3 B\right )+5 b \left (3 a A b+3 a^2 B+2 b^2 B\right ) \tan ^2(c+d x)+3 b^3 B \tan ^4(c+d x)\right )\right )}{120 d} \]

[In]

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

[Out]

(15*(12*a^2*A*b + 3*A*b^3 + 4*a^3*B + 9*a*b^2*B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(12*a^2*A*b + 3*A*b^
3 + 4*a^3*B + 9*a*b^2*B)*Sec[c + d*x] + 30*b^2*(A*b + 3*a*B)*Sec[c + d*x]^3 + 8*(15*(a^3*A + 3*a*A*b^2 + 3*a^2
*b*B + b^3*B) + 5*b*(3*a*A*b + 3*a^2*B + 2*b^2*B)*Tan[c + d*x]^2 + 3*b^3*B*Tan[c + d*x]^4)))/(120*d)

Maple [A] (verified)

Time = 6.41 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.79

method	result	size
parts	\(\frac {\left (A \,b^{3}+3 B a \,b^{2}\right ) \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (3 A a \,b^{2}+3 B \,a^{2} b \right ) \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 A \,a^{2} b +B \,a^{3}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {B \,b^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}+\frac {a^{3} A \tan \left (d x +c \right )}{d}\)	\(200\)
derivativedivides	\(\frac {a^{3} A \tan \left (d x +c \right )+B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \,a^{2} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 B \,a^{2} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-3 A a \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 B a \,b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A \,b^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,b^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(275\)
default	\(\frac {a^{3} A \tan \left (d x +c \right )+B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 A \,a^{2} b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )-3 B \,a^{2} b \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )-3 A a \,b^{2} \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+3 B a \,b^{2} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+A \,b^{3} \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,b^{3} \left (-\frac {8}{15}-\frac {\sec \left (d x +c \right )^{4}}{5}-\frac {4 \sec \left (d x +c \right )^{2}}{15}\right ) \tan \left (d x +c \right )}{d}\)	\(275\)
parallelrisch	\(\frac {-180 \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (A \,a^{2} b +\frac {1}{4} A \,b^{3}+\frac {1}{3} B \,a^{3}+\frac {3}{4} B a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+180 \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right ) \left (A \,a^{2} b +\frac {1}{4} A \,b^{3}+\frac {1}{3} B \,a^{3}+\frac {3}{4} B a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (720 A \,a^{2} b +420 A \,b^{3}+240 B \,a^{3}+1260 B a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+\left (360 a^{3} A +1200 A a \,b^{2}+1200 B \,a^{2} b +320 B \,b^{3}\right ) \sin \left (3 d x +3 c \right )+\left (360 A \,a^{2} b +90 A \,b^{3}+120 B \,a^{3}+270 B a \,b^{2}\right ) \sin \left (4 d x +4 c \right )+\left (120 a^{3} A +240 A a \,b^{2}+240 B \,a^{2} b +64 B \,b^{3}\right ) \sin \left (5 d x +5 c \right )+240 \left (a^{3} A +4 A a \,b^{2}+4 B \,a^{2} b +\frac {8}{3} B \,b^{3}\right ) \sin \left (d x +c \right )}{120 d \left (\cos \left (5 d x +5 c \right )+5 \cos \left (3 d x +3 c \right )+10 \cos \left (d x +c \right )\right )}\)	\(358\)
norman	\(\frac {-\frac {4 \left (45 a^{3} A +75 A a \,b^{2}+75 B \,a^{2} b +29 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{15 d}-\frac {\left (8 a^{3} A -12 A \,a^{2} b +24 A a \,b^{2}-5 A \,b^{3}-4 B \,a^{3}+24 B \,a^{2} b -15 B a \,b^{2}+8 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{4 d}-\frac {\left (8 a^{3} A +12 A \,a^{2} b +24 A a \,b^{2}+5 A \,b^{3}+4 B \,a^{3}+24 B \,a^{2} b +15 B a \,b^{2}+8 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (48 a^{3} A -36 A \,a^{2} b +96 A a \,b^{2}-3 A \,b^{3}-12 B \,a^{3}+96 B \,a^{2} b -9 B a \,b^{2}+16 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 d}+\frac {\left (48 a^{3} A +36 A \,a^{2} b +96 A a \,b^{2}+3 A \,b^{3}+12 B \,a^{3}+96 B \,a^{2} b +9 B a \,b^{2}+16 B \,b^{3}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{5}}-\frac {\left (12 A \,a^{2} b +3 A \,b^{3}+4 B \,a^{3}+9 B a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (12 A \,a^{2} b +3 A \,b^{3}+4 B \,a^{3}+9 B a \,b^{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(424\)
risch	\(-\frac {i \left (-240 B \,a^{2} b -240 A a \,b^{2}-64 B \,b^{3}-120 a^{3} A -480 A \,a^{3} {\mathrm e}^{6 i \left (d x +c \right )}-45 A \,b^{3} {\mathrm e}^{i \left (d x +c \right )}-60 B \,a^{3} {\mathrm e}^{i \left (d x +c \right )}+120 B \,a^{3} {\mathrm e}^{7 i \left (d x +c \right )}-720 A \,a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-480 a^{3} A \,{\mathrm e}^{2 i \left (d x +c \right )}+45 A \,b^{3} {\mathrm e}^{9 i \left (d x +c \right )}-640 B \,b^{3} {\mathrm e}^{4 i \left (d x +c \right )}+210 A \,b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+60 B \,a^{3} {\mathrm e}^{9 i \left (d x +c \right )}-120 A \,a^{3} {\mathrm e}^{8 i \left (d x +c \right )}-120 B \,a^{3} {\mathrm e}^{3 i \left (d x +c \right )}-210 A \,b^{3} {\mathrm e}^{3 i \left (d x +c \right )}-320 B \,b^{3} {\mathrm e}^{2 i \left (d x +c \right )}+180 A \,a^{2} b \,{\mathrm e}^{9 i \left (d x +c \right )}-720 B \,a^{2} b \,{\mathrm e}^{6 i \left (d x +c \right )}-1680 A a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}-630 B a \,b^{2} {\mathrm e}^{3 i \left (d x +c \right )}-1200 A a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-1200 B \,a^{2} b \,{\mathrm e}^{2 i \left (d x +c \right )}-180 A \,a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}-135 B a \,b^{2} {\mathrm e}^{i \left (d x +c \right )}+135 B a \,b^{2} {\mathrm e}^{9 i \left (d x +c \right )}+360 A \,a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}+630 B a \,b^{2} {\mathrm e}^{7 i \left (d x +c \right )}-720 A a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-1680 B \,a^{2} b \,{\mathrm e}^{4 i \left (d x +c \right )}-360 A \,a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}\right )}{60 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,a^{2} b}{2 d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) A \,b^{3}}{8 d}-\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}-\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B a \,b^{2}}{8 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,a^{2} b}{2 d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) A \,b^{3}}{8 d}+\frac {a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}+\frac {9 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B a \,b^{2}}{8 d}\)	\(662\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 249, normalized size of antiderivative = 0.99 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {15 \, {\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (15 \, A a^{3} + 30 \, B a^{2} b + 30 \, A a b^{2} + 8 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} + 24 \, B b^{3} + 15 \, {\left (4 \, B a^{3} + 12 \, A a^{2} b + 9 \, B a b^{2} + 3 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (15 \, B a^{2} b + 15 \, A a b^{2} + 4 \, B b^{3}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

[In]

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

[Out]

1/240*(15*(4*B*a^3 + 12*A*a^2*b + 9*B*a*b^2 + 3*A*b^3)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*B*a^3 + 12
*A*a^2*b + 9*B*a*b^2 + 3*A*b^3)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(8*(15*A*a^3 + 30*B*a^2*b + 30*A*a*b
^2 + 8*B*b^3)*cos(d*x + c)^4 + 24*B*b^3 + 15*(4*B*a^3 + 12*A*a^2*b + 9*B*a*b^2 + 3*A*b^3)*cos(d*x + c)^3 + 8*(
15*B*a^2*b + 15*A*a*b^2 + 4*B*b^3)*cos(d*x + c)^2 + 30*(3*B*a*b^2 + A*b^3)*cos(d*x + c))*sin(d*x + c))/(d*cos(
d*x + c)^5)

Sympy [F]

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

[In]

integrate(sec(d*x+c)**2*(a+b*sec(d*x+c))**3*(A+B*sec(d*x+c)),x)

[Out]

Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))**3*sec(c + d*x)**2, x)

Maxima [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.35 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{2} + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B b^{3} - 45 \, B a b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, A b^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, A a^{2} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, A a^{3} \tan \left (d x + c\right )}{240 \, d} \]

[In]

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

[Out]

1/240*(240*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2*b + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b^2 + 16*(3*t
an(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*b^3 - 45*B*a*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)) - 15*A*b^3*(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(si
n(d*x + c) - 1)) - 60*B*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) -
1)) - 180*A*a^2*b*(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^3*tan(d*x + c))/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 722 vs. \(2 (239) = 478\).

Time = 0.37 (sec) , antiderivative size = 722, normalized size of antiderivative = 2.87 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/120*(15*(4*B*a^3 + 12*A*a^2*b + 9*B*a*b^2 + 3*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*B*a^3 + 12*A
*a^2*b + 9*B*a*b^2 + 3*A*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a^3*tan(1/2*d*x + 1/2*c)^9 - 60*B*
a^3*tan(1/2*d*x + 1/2*c)^9 - 180*A*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*B*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*A*a
*b^2*tan(1/2*d*x + 1/2*c)^9 - 225*B*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 75*A*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*B*b^3
*tan(1/2*d*x + 1/2*c)^9 - 480*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 120*B*a^3*tan(1/2*d*x + 1/2*c)^7 + 360*A*a^2*b*ta
n(1/2*d*x + 1/2*c)^7 - 960*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 - 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 + 90*B*a*b^2*ta
n(1/2*d*x + 1/2*c)^7 + 30*A*b^3*tan(1/2*d*x + 1/2*c)^7 - 160*B*b^3*tan(1/2*d*x + 1/2*c)^7 + 720*A*a^3*tan(1/2*
d*x + 1/2*c)^5 + 1200*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 1200*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 464*B*b^3*tan(1/2
*d*x + 1/2*c)^5 - 480*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 120*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 360*A*a^2*b*tan(1/2*d*
x + 1/2*c)^3 - 960*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 90*B*a*b^2*tan(1/2*d*
x + 1/2*c)^3 - 30*A*b^3*tan(1/2*d*x + 1/2*c)^3 - 160*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*A*a^3*tan(1/2*d*x + 1/
2*c) + 60*B*a^3*tan(1/2*d*x + 1/2*c) + 180*A*a^2*b*tan(1/2*d*x + 1/2*c) + 360*B*a^2*b*tan(1/2*d*x + 1/2*c) + 3
60*A*a*b^2*tan(1/2*d*x + 1/2*c) + 225*B*a*b^2*tan(1/2*d*x + 1/2*c) + 75*A*b^3*tan(1/2*d*x + 1/2*c) + 120*B*b^3
*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 = 18.01 (sec) , antiderivative size = 470, normalized size of antiderivative = 1.87 \[ \int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx=\frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B\,a^3}{2}+\frac {3\,A\,a^2\,b}{2}+\frac {9\,B\,a\,b^2}{8}+\frac {3\,A\,b^3}{8}\right )}{2\,B\,a^3+6\,A\,a^2\,b+\frac {9\,B\,a\,b^2}{2}+\frac {3\,A\,b^3}{2}}\right )\,\left (B\,a^3+3\,A\,a^2\,b+\frac {9\,B\,a\,b^2}{4}+\frac {3\,A\,b^3}{4}\right )}{d}-\frac {\left (2\,A\,a^3-\frac {5\,A\,b^3}{4}-B\,a^3+2\,B\,b^3+6\,A\,a\,b^2-3\,A\,a^2\,b-\frac {15\,B\,a\,b^2}{4}+6\,B\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (\frac {A\,b^3}{2}-8\,A\,a^3+2\,B\,a^3-\frac {8\,B\,b^3}{3}-16\,A\,a\,b^2+6\,A\,a^2\,b+\frac {3\,B\,a\,b^2}{2}-16\,B\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (12\,A\,a^3+20\,B\,a^2\,b+20\,A\,a\,b^2+\frac {116\,B\,b^3}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-8\,A\,a^3-\frac {A\,b^3}{2}-2\,B\,a^3-\frac {8\,B\,b^3}{3}-16\,A\,a\,b^2-6\,A\,a^2\,b-\frac {3\,B\,a\,b^2}{2}-16\,B\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^3+\frac {5\,A\,b^3}{4}+B\,a^3+2\,B\,b^3+6\,A\,a\,b^2+3\,A\,a^2\,b+\frac {15\,B\,a\,b^2}{4}+6\,B\,a^2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

[In]

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

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((3*A*b^3)/8 + (B*a^3)/2 + (3*A*a^2*b)/2 + (9*B*a*b^2)/8))/((3*A*b^3)/2 + 2*B*a^3
+ 6*A*a^2*b + (9*B*a*b^2)/2))*((3*A*b^3)/4 + B*a^3 + 3*A*a^2*b + (9*B*a*b^2)/4))/d - (tan(c/2 + (d*x)/2)*(2*A
*a^3 + (5*A*b^3)/4 + B*a^3 + 2*B*b^3 + 6*A*a*b^2 + 3*A*a^2*b + (15*B*a*b^2)/4 + 6*B*a^2*b) + tan(c/2 + (d*x)/2
)^5*(12*A*a^3 + (116*B*b^3)/15 + 20*A*a*b^2 + 20*B*a^2*b) + tan(c/2 + (d*x)/2)^9*(2*A*a^3 - (5*A*b^3)/4 - B*a^
3 + 2*B*b^3 + 6*A*a*b^2 - 3*A*a^2*b - (15*B*a*b^2)/4 + 6*B*a^2*b) - tan(c/2 + (d*x)/2)^3*(8*A*a^3 + (A*b^3)/2
+ 2*B*a^3 + (8*B*b^3)/3 + 16*A*a*b^2 + 6*A*a^2*b + (3*B*a*b^2)/2 + 16*B*a^2*b) - tan(c/2 + (d*x)/2)^7*(8*A*a^3
- (A*b^3)/2 - 2*B*a^3 + (8*B*b^3)/3 + 16*A*a*b^2 - 6*A*a^2*b - (3*B*a*b^2)/2 + 16*B*a^2*b))/(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))