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

Optimal. Leaf size=198 \[ \frac {\left (4 a^2 A+6 a b B+3 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (4 a^2 A+6 a b B+3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {\left (5 a (a B+2 A b)+4 b^2 B\right ) \tan ^3(c+d x)}{15 d}+\frac {\left (5 a (a B+2 A b)+4 b^2 B\right ) \tan (c+d x)}{5 d}+\frac {b (6 a B+5 A b) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {b B \tan (c+d x) \sec ^3(c+d x) (a+b \sec (c+d x))}{5 d} \]

[Out]

1/8*(4*A*a^2+3*A*b^2+6*B*a*b)*arctanh(sin(d*x+c))/d+1/5*(4*b^2*B+5*a*(2*A*b+B*a))*tan(d*x+c)/d+1/8*(4*A*a^2+3*
A*b^2+6*B*a*b)*sec(d*x+c)*tan(d*x+c)/d+1/20*b*(5*A*b+6*B*a)*sec(d*x+c)^3*tan(d*x+c)/d+1/5*b*B*sec(d*x+c)^3*(a+
b*sec(d*x+c))*tan(d*x+c)/d+1/15*(4*b^2*B+5*a*(2*A*b+B*a))*tan(d*x+c)^3/d

________________________________________________________________________________________

Rubi [A]  time = 0.29, antiderivative size = 198, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {4026, 4047, 3767, 4046, 3768, 3770} \[ \frac {\left (4 a^2 A+6 a b B+3 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (4 a^2 A+6 a b B+3 A b^2\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {\left (5 a (a B+2 A b)+4 b^2 B\right ) \tan ^3(c+d x)}{15 d}+\frac {\left (5 a (a B+2 A b)+4 b^2 B\right ) \tan (c+d x)}{5 d}+\frac {b (6 a B+5 A b) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {b B \tan (c+d x) \sec ^3(c+d x) (a+b \sec (c+d x))}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a^2*A + 3*A*b^2 + 6*a*b*B)*ArcTanh[Sin[c + d*x]])/(8*d) + ((4*b^2*B + 5*a*(2*A*b + a*B))*Tan[c + d*x])/(5*
d) + ((4*a^2*A + 3*A*b^2 + 6*a*b*B)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (b*(5*A*b + 6*a*B)*Sec[c + d*x]^3*Tan[c
+ d*x])/(20*d) + (b*B*Sec[c + d*x]^3*(a + b*Sec[c + d*x])*Tan[c + d*x])/(5*d) + ((4*b^2*B + 5*a*(2*A*b + a*B)
)*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]

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac {b B \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) \left (a (5 a A+3 b B)+\left (4 b^2 B+5 a (2 A b+a B)\right ) \sec (c+d x)+b (5 A b+6 a B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b B \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac {1}{5} \int \sec ^3(c+d x) \left (a (5 a A+3 b B)+b (5 A b+6 a B) \sec ^2(c+d x)\right ) \, dx+\frac {1}{5} \left (4 b^2 B+5 a (2 A b+a B)\right ) \int \sec ^4(c+d x) \, dx\\ &=\frac {b (5 A b+6 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {b B \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac {1}{4} \left (4 a^2 A+3 A b^2+6 a b B\right ) \int \sec ^3(c+d x) \, dx-\frac {\left (4 b^2 B+5 a (2 A b+a B)\right ) \operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{5 d}\\ &=\frac {\left (4 b^2 B+5 a (2 A b+a B)\right ) \tan (c+d x)}{5 d}+\frac {\left (4 a^2 A+3 A b^2+6 a b B\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b (5 A b+6 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {b B \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac {\left (4 b^2 B+5 a (2 A b+a B)\right ) \tan ^3(c+d x)}{15 d}+\frac {1}{8} \left (4 a^2 A+3 A b^2+6 a b B\right ) \int \sec (c+d x) \, dx\\ &=\frac {\left (4 a^2 A+3 A b^2+6 a b B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (4 b^2 B+5 a (2 A b+a B)\right ) \tan (c+d x)}{5 d}+\frac {\left (4 a^2 A+3 A b^2+6 a b B\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b (5 A b+6 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {b B \sec ^3(c+d x) (a+b \sec (c+d x)) \tan (c+d x)}{5 d}+\frac {\left (4 b^2 B+5 a (2 A b+a B)\right ) \tan ^3(c+d x)}{15 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.56, size = 150, normalized size = 0.76 \[ \frac {15 \left (4 a^2 A+6 a b B+3 A b^2\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (5 \left (a^2 B+2 a A b+2 b^2 B\right ) \tan ^2(c+d x)+15 \left (a^2 B+2 a A b+b^2 B\right )+3 b^2 B \tan ^4(c+d x)\right )+15 \left (4 a^2 A+6 a b B+3 A b^2\right ) \sec (c+d x)+30 b (2 a B+A b) \sec ^3(c+d x)\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(4*a^2*A + 3*A*b^2 + 6*a*b*B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(4*a^2*A + 3*A*b^2 + 6*a*b*B)*Sec[c
+ d*x] + 30*b*(A*b + 2*a*B)*Sec[c + d*x]^3 + 8*(15*(2*a*A*b + a^2*B + b^2*B) + 5*(2*a*A*b + a^2*B + 2*b^2*B)*
Tan[c + d*x]^2 + 3*b^2*B*Tan[c + d*x]^4)))/(120*d)

________________________________________________________________________________________

fricas [A]  time = 0.54, size = 208, normalized size = 1.05 \[ \frac {15 \, {\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (5 \, B a^{2} + 10 \, A a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{4} + 15 \, {\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{2}\right )} \cos \left (d x + c\right )^{3} + 24 \, B b^{2} + 8 \, {\left (5 \, B a^{2} + 10 \, A a b + 4 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(4*A*a^2 + 6*B*a*b + 3*A*b^2)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(4*A*a^2 + 6*B*a*b + 3*A*b^2
)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(16*(5*B*a^2 + 10*A*a*b + 4*B*b^2)*cos(d*x + c)^4 + 15*(4*A*a^2 +
6*B*a*b + 3*A*b^2)*cos(d*x + c)^3 + 24*B*b^2 + 8*(5*B*a^2 + 10*A*a*b + 4*B*b^2)*cos(d*x + c)^2 + 30*(2*B*a*b +
A*b^2)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)

________________________________________________________________________________________

giac [B]  time = 0.33, size = 528, normalized size = 2.67 \[ \frac {15 \, {\left (4 \, A a^{2} + 6 \, B a b + 3 \, A b^{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} + 6 \, B a b + 3 \, A b^{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} - 120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 240 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 150 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 75 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 120 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 320 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 640 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 60 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 160 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 400 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 800 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 464 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 120 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 320 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 640 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 160 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 240 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 150 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 75 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, B 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 )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(15*(4*A*a^2 + 6*B*a*b + 3*A*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(4*A*a^2 + 6*B*a*b + 3*A*b^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 - 120*B*a^2*tan(1/2*d*x + 1/2*c)^9 - 2
40*A*a*b*tan(1/2*d*x + 1/2*c)^9 + 150*B*a*b*tan(1/2*d*x + 1/2*c)^9 + 75*A*b^2*tan(1/2*d*x + 1/2*c)^9 - 120*B*b
^2*tan(1/2*d*x + 1/2*c)^9 - 120*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 320*B*a^2*tan(1/2*d*x + 1/2*c)^7 + 640*A*a*b*ta
n(1/2*d*x + 1/2*c)^7 - 60*B*a*b*tan(1/2*d*x + 1/2*c)^7 - 30*A*b^2*tan(1/2*d*x + 1/2*c)^7 + 160*B*b^2*tan(1/2*d
*x + 1/2*c)^7 - 400*B*a^2*tan(1/2*d*x + 1/2*c)^5 - 800*A*a*b*tan(1/2*d*x + 1/2*c)^5 - 464*B*b^2*tan(1/2*d*x +
1/2*c)^5 + 120*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 320*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 640*A*a*b*tan(1/2*d*x + 1/2*c
)^3 + 60*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 30*A*b^2*tan(1/2*d*x + 1/2*c)^3 + 160*B*b^2*tan(1/2*d*x + 1/2*c)^3 - 6
0*A*a^2*tan(1/2*d*x + 1/2*c) - 120*B*a^2*tan(1/2*d*x + 1/2*c) - 240*A*a*b*tan(1/2*d*x + 1/2*c) - 150*B*a*b*tan
(1/2*d*x + 1/2*c) - 75*A*b^2*tan(1/2*d*x + 1/2*c) - 120*B*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 -
1)^5)/d

________________________________________________________________________________________

maple [A]  time = 1.52, size = 312, normalized size = 1.58 \[ \frac {a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 a^{2} B \tan \left (d x +c \right )}{3 d}+\frac {a^{2} B \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {4 a A b \tan \left (d x +c \right )}{3 d}+\frac {2 a A b \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {B a b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{2 d}+\frac {3 B a b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {A \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 A \,b^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {8 b^{2} B \tan \left (d x +c \right )}{15 d}+\frac {b^{2} B \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 b^{2} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.67, size = 276, normalized size = 1.39 \[ \frac {80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} + 160 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b + 16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} B b^{2} - 30 \, B a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 15 \, A b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(80*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 + 160*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b + 16*(3*tan(d*
x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*B*b^2 - 30*B*a*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(sin(d*x + c) - 1)) - 15*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)) - 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)))/d

________________________________________________________________________________________

mupad [B]  time = 5.71, size = 359, normalized size = 1.81 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A\,a^2}{2}+\frac {3\,B\,a\,b}{4}+\frac {3\,A\,b^2}{8}\right )}{2\,A\,a^2+3\,B\,a\,b+\frac {3\,A\,b^2}{2}}\right )\,\left (A\,a^2+\frac {3\,B\,a\,b}{2}+\frac {3\,A\,b^2}{4}\right )}{d}-\frac {\left (2\,B\,a^2-\frac {5\,A\,b^2}{4}-A\,a^2+2\,B\,b^2+4\,A\,a\,b-\frac {5\,B\,a\,b}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,A\,a^2+\frac {A\,b^2}{2}-\frac {16\,B\,a^2}{3}-\frac {8\,B\,b^2}{3}-\frac {32\,A\,a\,b}{3}+B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {20\,B\,a^2}{3}+\frac {40\,A\,a\,b}{3}+\frac {116\,B\,b^2}{15}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-2\,A\,a^2-\frac {A\,b^2}{2}-\frac {16\,B\,a^2}{3}-\frac {8\,B\,b^2}{3}-\frac {32\,A\,a\,b}{3}-B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (A\,a^2+\frac {5\,A\,b^2}{4}+2\,B\,a^2+2\,B\,b^2+4\,A\,a\,b+\frac {5\,B\,a\,b}{2}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((A*a^2)/2 + (3*A*b^2)/8 + (3*B*a*b)/4))/(2*A*a^2 + (3*A*b^2)/2 + 3*B*a*b))*(A*a^
2 + (3*A*b^2)/4 + (3*B*a*b)/2))/d - (tan(c/2 + (d*x)/2)^5*((20*B*a^2)/3 + (116*B*b^2)/15 + (40*A*a*b)/3) - tan
(c/2 + (d*x)/2)^9*(A*a^2 + (5*A*b^2)/4 - 2*B*a^2 - 2*B*b^2 - 4*A*a*b + (5*B*a*b)/2) - tan(c/2 + (d*x)/2)^3*(2*
A*a^2 + (A*b^2)/2 + (16*B*a^2)/3 + (8*B*b^2)/3 + (32*A*a*b)/3 + B*a*b) + tan(c/2 + (d*x)/2)^7*(2*A*a^2 + (A*b^
2)/2 - (16*B*a^2)/3 - (8*B*b^2)/3 - (32*A*a*b)/3 + B*a*b) + tan(c/2 + (d*x)/2)*(A*a^2 + (5*A*b^2)/4 + 2*B*a^2
+ 2*B*b^2 + 4*A*a*b + (5*B*a*b)/2))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/
2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________