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

[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)

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Rubi [A]  time = 0.608875, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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 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 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]

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 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 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 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]

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 = 2.61277, size = 214, normalized size = 0.77 \[ \frac{15 \left (18 a^2 b C+8 a^3 B+18 a b^2 B+5 b^3 C\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (80 \left (3 a^2 b B+a^3 C+6 a b^2 C+2 b^3 B\right ) \tan ^2(c+d x)+10 b \left (18 a^2 C+18 a b B+5 b^2 C\right ) \sec ^3(c+d x)+15 \left (18 a^2 b C+8 a^3 B+18 a b^2 B+5 b^3 C\right ) \sec (c+d x)+240 \left (3 a^2 b B+a^3 C+3 a b^2 C+b^3 B\right )+48 b^2 (3 a C+b B) \tan ^4(c+d x)+40 b^3 C \sec ^5(c+d x)\right )}{240 d} \]

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)

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Maple [A]  time = 0.052, size = 478, normalized size = 1.7 \begin{align*}{\frac{B{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{B{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{2\,{a}^{3}C\tan \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{3}C\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3\,d}}+2\,{\frac{B{a}^{2}b\tan \left ( dx+c \right ) }{d}}+{\frac{B{a}^{2}b\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{3\,{a}^{2}bC\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{9\,{a}^{2}bC\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{9\,{a}^{2}bC\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{3\,Ba{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{9\,Ba{b}^{2}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{9\,Ba{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{8\,Ca{b}^{2}\tan \left ( dx+c \right ) }{5\,d}}+{\frac{3\,Ca{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,Ca{b}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{5\,d}}+{\frac{8\,B{b}^{3}\tan \left ( dx+c \right ) }{15\,d}}+{\frac{B{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}}{5\,d}}+{\frac{4\,B{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{15\,d}}+{\frac{C{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{5\,C{b}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{5\,C{b}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}}+{\frac{5\,C{b}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{16\,d}} \end{align*}

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

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Maxima [A]  time = 0.99116, size = 552, normalized size = 1.99 \begin{align*} \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} \end{align*}

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

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Fricas [A]  time = 0.579748, size = 706, normalized size = 2.54 \begin{align*} \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}} \end{align*}

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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

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)

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Giac [B]  time = 1.29572, size = 1258, normalized size = 4.53 \begin{align*} \text{result too large to display} \end{align*}

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