3.784 \(\int \sec ^2(c+d x) (a+b \sec (c+d x))^3 (B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=278 \[ \frac {b \left (14 a^2 C+18 a b B+5 b^2 C\right ) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {\left (5 a^3 C+15 a^2 b B+12 a b^2 C+4 b^3 B\right ) \tan ^3(c+d x)}{15 d}+\frac {\left (5 a^3 C+15 a^2 b B+12 a b^2 C+4 b^3 B\right ) \tan (c+d x)}{5 d}+\frac {\left (8 a^3 B+18 a^2 b C+18 a b^2 B+5 b^3 C\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (8 a^3 B+18 a^2 b C+18 a b^2 B+5 b^3 C\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {b^2 (4 a C+3 b B) \tan (c+d x) \sec ^4(c+d x)}{15 d}+\frac {b C \tan (c+d x) \sec ^3(c+d x) (a+b \sec (c+d x))^2}{6 d} \]

[Out]

1/16*(8*B*a^3+18*B*a*b^2+18*C*a^2*b+5*C*b^3)*arctanh(sin(d*x+c))/d+1/5*(15*B*a^2*b+4*B*b^3+5*C*a^3+12*C*a*b^2)
*tan(d*x+c)/d+1/16*(8*B*a^3+18*B*a*b^2+18*C*a^2*b+5*C*b^3)*sec(d*x+c)*tan(d*x+c)/d+1/24*b*(18*B*a*b+14*C*a^2+5
*C*b^2)*sec(d*x+c)^3*tan(d*x+c)/d+1/15*b^2*(3*B*b+4*C*a)*sec(d*x+c)^4*tan(d*x+c)/d+1/6*b*C*sec(d*x+c)^3*(a+b*s
ec(d*x+c))^2*tan(d*x+c)/d+1/15*(15*B*a^2*b+4*B*b^3+5*C*a^3+12*C*a*b^2)*tan(d*x+c)^3/d

________________________________________________________________________________________

Rubi [A]  time = 0.61, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4072, 4026, 4076, 4047, 3767, 4046, 3768, 3770} \[ \frac {\left (15 a^2 b B+5 a^3 C+12 a b^2 C+4 b^3 B\right ) \tan ^3(c+d x)}{15 d}+\frac {\left (15 a^2 b B+5 a^3 C+12 a b^2 C+4 b^3 B\right ) \tan (c+d x)}{5 d}+\frac {\left (18 a^2 b C+8 a^3 B+18 a b^2 B+5 b^3 C\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {b \left (14 a^2 C+18 a b B+5 b^2 C\right ) \tan (c+d x) \sec ^3(c+d x)}{24 d}+\frac {\left (18 a^2 b C+8 a^3 B+18 a b^2 B+5 b^3 C\right ) \tan (c+d x) \sec (c+d x)}{16 d}+\frac {b^2 (4 a C+3 b B) \tan (c+d x) \sec ^4(c+d x)}{15 d}+\frac {b C \tan (c+d x) \sec ^3(c+d x) (a+b \sec (c+d x))^2}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^3*B + 18*a*b^2*B + 18*a^2*b*C + 5*b^3*C)*ArcTanh[Sin[c + d*x]])/(16*d) + ((15*a^2*b*B + 4*b^3*B + 5*a^3*
C + 12*a*b^2*C)*Tan[c + d*x])/(5*d) + ((8*a^3*B + 18*a*b^2*B + 18*a^2*b*C + 5*b^3*C)*Sec[c + d*x]*Tan[c + d*x]
)/(16*d) + (b*(18*a*b*B + 14*a^2*C + 5*b^2*C)*Sec[c + d*x]^3*Tan[c + d*x])/(24*d) + (b^2*(3*b*B + 4*a*C)*Sec[c
+ d*x]^4*Tan[c + d*x])/(15*d) + (b*C*Sec[c + d*x]^3*(a + b*Sec[c + d*x])^2*Tan[c + d*x])/(6*d) + ((15*a^2*b*B
+ 4*b^3*B + 5*a^3*C + 12*a*b^2*C)*Tan[c + d*x]^3)/(15*d)

Rule 3767

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 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 4026

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

Rule 4046

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 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, x]

Rule 4072

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]

Rule 4076

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

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^3 \left (B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\int \sec ^3(c+d x) (a+b \sec (c+d x))^3 (B+C \sec (c+d x)) \, dx\\ &=\frac {b C \sec ^3(c+d x) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {1}{6} \int \sec ^3(c+d x) (a+b \sec (c+d x)) \left (3 a (2 a B+b C)+\left (5 b^2 C+6 a (2 b B+a C)\right ) \sec (c+d x)+2 b (3 b B+4 a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b^2 (3 b B+4 a C) \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {b C \sec ^3(c+d x) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {1}{30} \int \sec ^3(c+d x) \left (15 a^2 (2 a B+b C)+6 \left (15 a^2 b B+4 b^3 B+5 a^3 C+12 a b^2 C\right ) \sec (c+d x)+5 b \left (18 a b B+14 a^2 C+5 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b^2 (3 b B+4 a C) \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {b C \sec ^3(c+d x) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {1}{30} \int \sec ^3(c+d x) \left (15 a^2 (2 a B+b C)+5 b \left (18 a b B+14 a^2 C+5 b^2 C\right ) \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} \left (15 a^2 b B+4 b^3 B+5 a^3 C+12 a b^2 C\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {b \left (18 a b B+14 a^2 C+5 b^2 C\right ) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {b^2 (3 b B+4 a C) \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {b C \sec ^3(c+d x) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {1}{8} \left (8 a^3 B+18 a b^2 B+18 a^2 b C+5 b^3 C\right ) \int \sec ^3(c+d x) \, dx-\frac {\left (15 a^2 b B+4 b^3 B+5 a^3 C+12 a b^2 C\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {\left (15 a^2 b B+4 b^3 B+5 a^3 C+12 a b^2 C\right ) \tan (c+d x)}{5 d}+\frac {\left (8 a^3 B+18 a b^2 B+18 a^2 b C+5 b^3 C\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {b \left (18 a b B+14 a^2 C+5 b^2 C\right ) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {b^2 (3 b B+4 a C) \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {b C \sec ^3(c+d x) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {\left (15 a^2 b B+4 b^3 B+5 a^3 C+12 a b^2 C\right ) \tan ^3(c+d x)}{15 d}+\frac {1}{16} \left (8 a^3 B+18 a b^2 B+18 a^2 b C+5 b^3 C\right ) \int \sec (c+d x) \, dx\\ &=\frac {\left (8 a^3 B+18 a b^2 B+18 a^2 b C+5 b^3 C\right ) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {\left (15 a^2 b B+4 b^3 B+5 a^3 C+12 a b^2 C\right ) \tan (c+d x)}{5 d}+\frac {\left (8 a^3 B+18 a b^2 B+18 a^2 b C+5 b^3 C\right ) \sec (c+d x) \tan (c+d x)}{16 d}+\frac {b \left (18 a b B+14 a^2 C+5 b^2 C\right ) \sec ^3(c+d x) \tan (c+d x)}{24 d}+\frac {b^2 (3 b B+4 a C) \sec ^4(c+d x) \tan (c+d x)}{15 d}+\frac {b C \sec ^3(c+d x) (a+b \sec (c+d x))^2 \tan (c+d x)}{6 d}+\frac {\left (15 a^2 b B+4 b^3 B+5 a^3 C+12 a b^2 C\right ) \tan ^3(c+d x)}{15 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 5.38, size = 214, normalized size = 0.77 \[ \frac {15 \left (8 a^3 B+18 a^2 b C+18 a b^2 B+5 b^3 C\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (10 b \left (18 a^2 C+18 a b B+5 b^2 C\right ) \sec ^3(c+d x)+80 \left (a^3 C+3 a^2 b B+6 a b^2 C+2 b^3 B\right ) \tan ^2(c+d x)+15 \left (8 a^3 B+18 a^2 b C+18 a b^2 B+5 b^3 C\right ) \sec (c+d x)+240 \left (a^3 C+3 a^2 b B+3 a b^2 C+b^3 B\right )+48 b^2 (3 a C+b B) \tan ^4(c+d x)+40 b^3 C \sec ^5(c+d x)\right )}{240 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(8*a^3*B + 18*a*b^2*B + 18*a^2*b*C + 5*b^3*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(240*(3*a^2*b*B + b^3*B
+ a^3*C + 3*a*b^2*C) + 15*(8*a^3*B + 18*a*b^2*B + 18*a^2*b*C + 5*b^3*C)*Sec[c + d*x] + 10*b*(18*a*b*B + 18*a^
2*C + 5*b^2*C)*Sec[c + d*x]^3 + 40*b^3*C*Sec[c + d*x]^5 + 80*(3*a^2*b*B + 2*b^3*B + a^3*C + 6*a*b^2*C)*Tan[c +
d*x]^2 + 48*b^2*(b*B + 3*a*C)*Tan[c + d*x]^4))/(240*d)

________________________________________________________________________________________

fricas [A]  time = 0.46, size = 286, normalized size = 1.03 \[ \frac {15 \, {\left (8 \, B a^{3} + 18 \, C a^{2} b + 18 \, B a b^{2} + 5 \, C b^{3}\right )} \cos \left (d x + c\right )^{6} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (8 \, B a^{3} + 18 \, C a^{2} b + 18 \, B a b^{2} + 5 \, C b^{3}\right )} \cos \left (d x + c\right )^{6} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (32 \, {\left (5 \, C a^{3} + 15 \, B a^{2} b + 12 \, C a b^{2} + 4 \, B b^{3}\right )} \cos \left (d x + c\right )^{5} + 15 \, {\left (8 \, B a^{3} + 18 \, C a^{2} b + 18 \, B a b^{2} + 5 \, C b^{3}\right )} \cos \left (d x + c\right )^{4} + 40 \, C b^{3} + 16 \, {\left (5 \, C a^{3} + 15 \, B a^{2} b + 12 \, C a b^{2} + 4 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} + 10 \, {\left (18 \, C a^{2} b + 18 \, B a b^{2} + 5 \, C b^{3}\right )} \cos \left (d x + c\right )^{2} + 48 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{480 \, d \cos \left (d x + c\right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(15*(8*B*a^3 + 18*C*a^2*b + 18*B*a*b^2 + 5*C*b^3)*cos(d*x + c)^6*log(sin(d*x + c) + 1) - 15*(8*B*a^3 + 1
8*C*a^2*b + 18*B*a*b^2 + 5*C*b^3)*cos(d*x + c)^6*log(-sin(d*x + c) + 1) + 2*(32*(5*C*a^3 + 15*B*a^2*b + 12*C*a
*b^2 + 4*B*b^3)*cos(d*x + c)^5 + 15*(8*B*a^3 + 18*C*a^2*b + 18*B*a*b^2 + 5*C*b^3)*cos(d*x + c)^4 + 40*C*b^3 +
16*(5*C*a^3 + 15*B*a^2*b + 12*C*a*b^2 + 4*B*b^3)*cos(d*x + c)^3 + 10*(18*C*a^2*b + 18*B*a*b^2 + 5*C*b^3)*cos(d
*x + c)^2 + 48*(3*C*a*b^2 + B*b^3)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^6)

________________________________________________________________________________________

giac [B]  time = 0.38, size = 932, normalized size = 3.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(8*B*a^3 + 18*C*a^2*b + 18*B*a*b^2 + 5*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(8*B*a^3 + 18*
C*a^2*b + 18*B*a*b^2 + 5*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(120*B*a^3*tan(1/2*d*x + 1/2*c)^11 - 24
0*C*a^3*tan(1/2*d*x + 1/2*c)^11 - 720*B*a^2*b*tan(1/2*d*x + 1/2*c)^11 + 450*C*a^2*b*tan(1/2*d*x + 1/2*c)^11 +
450*B*a*b^2*tan(1/2*d*x + 1/2*c)^11 - 720*C*a*b^2*tan(1/2*d*x + 1/2*c)^11 - 240*B*b^3*tan(1/2*d*x + 1/2*c)^11
+ 165*C*b^3*tan(1/2*d*x + 1/2*c)^11 - 360*B*a^3*tan(1/2*d*x + 1/2*c)^9 + 880*C*a^3*tan(1/2*d*x + 1/2*c)^9 + 26
40*B*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 630*C*a^2*b*tan(1/2*d*x + 1/2*c)^9 - 630*B*a*b^2*tan(1/2*d*x + 1/2*c)^9 +
1680*C*a*b^2*tan(1/2*d*x + 1/2*c)^9 + 560*B*b^3*tan(1/2*d*x + 1/2*c)^9 + 25*C*b^3*tan(1/2*d*x + 1/2*c)^9 + 240
*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 1440*C*a^3*tan(1/2*d*x + 1/2*c)^7 - 4320*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 180*
C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 180*B*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 3744*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 12
48*B*b^3*tan(1/2*d*x + 1/2*c)^7 + 450*C*b^3*tan(1/2*d*x + 1/2*c)^7 + 240*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 1440*C
*a^3*tan(1/2*d*x + 1/2*c)^5 + 4320*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 180*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 180*B
*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 3744*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 1248*B*b^3*tan(1/2*d*x + 1/2*c)^5 + 450*
C*b^3*tan(1/2*d*x + 1/2*c)^5 - 360*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 880*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 2640*B*a^
2*b*tan(1/2*d*x + 1/2*c)^3 - 630*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 630*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 1680*C*
a*b^2*tan(1/2*d*x + 1/2*c)^3 - 560*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 25*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*B*a^3*
tan(1/2*d*x + 1/2*c) + 240*C*a^3*tan(1/2*d*x + 1/2*c) + 720*B*a^2*b*tan(1/2*d*x + 1/2*c) + 450*C*a^2*b*tan(1/2
*d*x + 1/2*c) + 450*B*a*b^2*tan(1/2*d*x + 1/2*c) + 720*C*a*b^2*tan(1/2*d*x + 1/2*c) + 240*B*b^3*tan(1/2*d*x +
1/2*c) + 165*C*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^6)/d

________________________________________________________________________________________

maple [A]  time = 1.70, size = 478, normalized size = 1.72 \[ \frac {a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 a^{3} C \tan \left (d x +c \right )}{3 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {2 a^{2} b B \tan \left (d x +c \right )}{d}+\frac {a^{2} b B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 C \,a^{2} b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {9 C \,a^{2} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {9 C \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 B a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {9 B a \,b^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {9 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {8 C a \,b^{2} \tan \left (d x +c \right )}{5 d}+\frac {3 C a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 C a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{5 d}+\frac {8 b^{3} B \tan \left (d x +c \right )}{15 d}+\frac {b^{3} B \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 b^{3} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {b^{3} C \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {5 b^{3} C \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {5 b^{3} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {5 b^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(a+b*sec(d*x+c))^3*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/2/d*a^3*B*sec(d*x+c)*tan(d*x+c)+1/2/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+2/3*a^3*C*tan(d*x+c)/d+1/3/d*C*a^3*tan
(d*x+c)*sec(d*x+c)^2+2/d*a^2*b*B*tan(d*x+c)+1/d*a^2*b*B*tan(d*x+c)*sec(d*x+c)^2+3/4/d*C*a^2*b*tan(d*x+c)*sec(d
*x+c)^3+9/8/d*C*a^2*b*sec(d*x+c)*tan(d*x+c)+9/8/d*C*a^2*b*ln(sec(d*x+c)+tan(d*x+c))+3/4/d*B*a*b^2*tan(d*x+c)*s
ec(d*x+c)^3+9/8/d*B*a*b^2*sec(d*x+c)*tan(d*x+c)+9/8/d*B*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+8/5/d*C*a*b^2*tan(d*x+
c)+3/5/d*C*a*b^2*tan(d*x+c)*sec(d*x+c)^4+4/5/d*C*a*b^2*tan(d*x+c)*sec(d*x+c)^2+8/15/d*b^3*B*tan(d*x+c)+1/5/d*b
^3*B*tan(d*x+c)*sec(d*x+c)^4+4/15/d*b^3*B*tan(d*x+c)*sec(d*x+c)^2+1/6/d*b^3*C*tan(d*x+c)*sec(d*x+c)^5+5/24/d*b
^3*C*tan(d*x+c)*sec(d*x+c)^3+5/16/d*b^3*C*sec(d*x+c)*tan(d*x+c)+5/16/d*b^3*C*ln(sec(d*x+c)+tan(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.34, size = 409, normalized size = 1.47 \[ \frac {160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} + 480 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b + 96 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} C a b^{2} + 32 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B b^{3} - 5 \, C b^{3} {\left (\frac {2 \, {\left (15 \, \sin \left (d x + c\right )^{5} - 40 \, \sin \left (d x + c\right )^{3} + 33 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{6} - 3 \, \sin \left (d x + c\right )^{4} + 3 \, \sin \left (d x + c\right )^{2} - 1} - 15 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 15 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 90 \, C a^{2} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 90 \, B a b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 120 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/480*(160*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 + 480*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2*b + 96*(3*tan
(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*C*a*b^2 + 32*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan
(d*x + c))*B*b^3 - 5*C*b^3*(2*(15*sin(d*x + c)^5 - 40*sin(d*x + c)^3 + 33*sin(d*x + c))/(sin(d*x + c)^6 - 3*si
n(d*x + c)^4 + 3*sin(d*x + c)^2 - 1) - 15*log(sin(d*x + c) + 1) + 15*log(sin(d*x + c) - 1)) - 90*C*a^2*b*(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)) - 90*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)) - 120*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)))/d

________________________________________________________________________________________

mupad [B]  time = 7.54, size = 571, normalized size = 2.05 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B\,a^3}{2}+\frac {9\,C\,a^2\,b}{8}+\frac {9\,B\,a\,b^2}{8}+\frac {5\,C\,b^3}{16}\right )}{2\,B\,a^3+\frac {9\,C\,a^2\,b}{2}+\frac {9\,B\,a\,b^2}{2}+\frac {5\,C\,b^3}{4}}\right )\,\left (B\,a^3+\frac {9\,C\,a^2\,b}{4}+\frac {9\,B\,a\,b^2}{4}+\frac {5\,C\,b^3}{8}\right )}{d}+\frac {\left (B\,a^3-2\,B\,b^3-2\,C\,a^3+\frac {11\,C\,b^3}{8}+\frac {15\,B\,a\,b^2}{4}-6\,B\,a^2\,b-6\,C\,a\,b^2+\frac {15\,C\,a^2\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{11}+\left (\frac {14\,B\,b^3}{3}-3\,B\,a^3+\frac {22\,C\,a^3}{3}+\frac {5\,C\,b^3}{24}-\frac {21\,B\,a\,b^2}{4}+22\,B\,a^2\,b+14\,C\,a\,b^2-\frac {21\,C\,a^2\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,B\,a^3-\frac {52\,B\,b^3}{5}-12\,C\,a^3+\frac {15\,C\,b^3}{4}+\frac {3\,B\,a\,b^2}{2}-36\,B\,a^2\,b-\frac {156\,C\,a\,b^2}{5}+\frac {3\,C\,a^2\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (2\,B\,a^3+\frac {52\,B\,b^3}{5}+12\,C\,a^3+\frac {15\,C\,b^3}{4}+\frac {3\,B\,a\,b^2}{2}+36\,B\,a^2\,b+\frac {156\,C\,a\,b^2}{5}+\frac {3\,C\,a^2\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {5\,C\,b^3}{24}-\frac {14\,B\,b^3}{3}-\frac {22\,C\,a^3}{3}-3\,B\,a^3-\frac {21\,B\,a\,b^2}{4}-22\,B\,a^2\,b-14\,C\,a\,b^2-\frac {21\,C\,a^2\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (B\,a^3+2\,B\,b^3+2\,C\,a^3+\frac {11\,C\,b^3}{8}+\frac {15\,B\,a\,b^2}{4}+6\,B\,a^2\,b+6\,C\,a\,b^2+\frac {15\,C\,a^2\,b}{4}\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 )}^{12}-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+15\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((B*a^3)/2 + (5*C*b^3)/16 + (9*B*a*b^2)/8 + (9*C*a^2*b)/8))/(2*B*a^3 + (5*C*b^3)/
4 + (9*B*a*b^2)/2 + (9*C*a^2*b)/2))*(B*a^3 + (5*C*b^3)/8 + (9*B*a*b^2)/4 + (9*C*a^2*b)/4))/d + (tan(c/2 + (d*x
)/2)*(B*a^3 + 2*B*b^3 + 2*C*a^3 + (11*C*b^3)/8 + (15*B*a*b^2)/4 + 6*B*a^2*b + 6*C*a*b^2 + (15*C*a^2*b)/4) + ta
n(c/2 + (d*x)/2)^11*(B*a^3 - 2*B*b^3 - 2*C*a^3 + (11*C*b^3)/8 + (15*B*a*b^2)/4 - 6*B*a^2*b - 6*C*a*b^2 + (15*C
*a^2*b)/4) - tan(c/2 + (d*x)/2)^3*(3*B*a^3 + (14*B*b^3)/3 + (22*C*a^3)/3 - (5*C*b^3)/24 + (21*B*a*b^2)/4 + 22*
B*a^2*b + 14*C*a*b^2 + (21*C*a^2*b)/4) + tan(c/2 + (d*x)/2)^9*((14*B*b^3)/3 - 3*B*a^3 + (22*C*a^3)/3 + (5*C*b^
3)/24 - (21*B*a*b^2)/4 + 22*B*a^2*b + 14*C*a*b^2 - (21*C*a^2*b)/4) + tan(c/2 + (d*x)/2)^5*(2*B*a^3 + (52*B*b^3
)/5 + 12*C*a^3 + (15*C*b^3)/4 + (3*B*a*b^2)/2 + 36*B*a^2*b + (156*C*a*b^2)/5 + (3*C*a^2*b)/2) + tan(c/2 + (d*x
)/2)^7*(2*B*a^3 - (52*B*b^3)/5 - 12*C*a^3 + (15*C*b^3)/4 + (3*B*a*b^2)/2 - 36*B*a^2*b - (156*C*a*b^2)/5 + (3*C
*a^2*b)/2))/(d*(15*tan(c/2 + (d*x)/2)^4 - 6*tan(c/2 + (d*x)/2)^2 - 20*tan(c/2 + (d*x)/2)^6 + 15*tan(c/2 + (d*x
)/2)^8 - 6*tan(c/2 + (d*x)/2)^10 + tan(c/2 + (d*x)/2)^12 + 1))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (B + C \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{3} \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________