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

Optimal. Leaf size=137 \[ \frac {\tan (c+d x) (3 a A+2 a C+2 b B)}{3 d}+\frac {(4 a B+4 A b+3 b C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec (c+d x) (4 a B+4 A b+3 b C)}{8 d}+\frac {(a C+b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {b C \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 39, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.180, Rules used = {4076, 4047, 3768, 3770, 4046, 3767, 8} \[ \frac {\tan (c+d x) (3 a A+2 a C+2 b B)}{3 d}+\frac {(4 a B+4 A b+3 b C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\tan (c+d x) \sec (c+d x) (4 a B+4 A b+3 b C)}{8 d}+\frac {(a C+b B) \tan (c+d x) \sec ^2(c+d x)}{3 d}+\frac {b C \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 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 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)) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {b C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int \sec ^2(c+d x) \left (4 a A+(4 A b+4 a B+3 b C) \sec (c+d x)+4 (b B+a C) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {b C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int \sec ^2(c+d x) \left (4 a A+4 (b B+a C) \sec ^2(c+d x)\right ) \, dx+\frac {1}{4} (4 A b+4 a B+3 b C) \int \sec ^3(c+d x) \, dx\\ &=\frac {(4 A b+4 a B+3 b C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(b B+a C) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {b C \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{3} (3 a A+2 b B+2 a C) \int \sec ^2(c+d x) \, dx+\frac {1}{8} (4 A b+4 a B+3 b C) \int \sec (c+d x) \, dx\\ &=\frac {(4 A b+4 a B+3 b C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(4 A b+4 a B+3 b C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(b B+a C) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {b C \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {(3 a A+2 b B+2 a C) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {(4 A b+4 a B+3 b C) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {(3 a A+2 b B+2 a C) \tan (c+d x)}{3 d}+\frac {(4 A b+4 a B+3 b C) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {(b B+a C) \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {b C \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.68, size = 100, normalized size = 0.73 \[ \frac {3 (4 a B+4 A b+3 b C) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 \left (3 a (A+C)+(a C+b B) \tan ^2(c+d x)+3 b B\right )+3 \sec (c+d x) (4 a B+4 A b+3 b C)+6 b C \sec ^3(c+d x)\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.88, size = 158, normalized size = 1.15 \[ \frac {3 \, {\left (4 \, B a + {\left (4 \, A + 3 \, C\right )} b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, B a + {\left (4 \, A + 3 \, C\right )} b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left ({\left (3 \, A + 2 \, C\right )} a + 2 \, B b\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (4 \, B a + {\left (4 \, A + 3 \, C\right )} b\right )} \cos \left (d x + c\right )^{2} + 6 \, C b + 8 \, {\left (C a + B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.26, size = 428, normalized size = 3.12 \[ \frac {3 \, {\left (4 \, B a + 4 \, A b + 3 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, B a + 4 \, A b + 3 \, C b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (24 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C b \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} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(4*B*a + 4*A*b + 3*C*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*B*a + 4*A*b + 3*C*b)*log(abs(tan(1/2
*d*x + 1/2*c) - 1)) - 2*(24*A*a*tan(1/2*d*x + 1/2*c)^7 - 12*B*a*tan(1/2*d*x + 1/2*c)^7 + 24*C*a*tan(1/2*d*x +
1/2*c)^7 - 12*A*b*tan(1/2*d*x + 1/2*c)^7 + 24*B*b*tan(1/2*d*x + 1/2*c)^7 - 15*C*b*tan(1/2*d*x + 1/2*c)^7 - 72*
A*a*tan(1/2*d*x + 1/2*c)^5 + 12*B*a*tan(1/2*d*x + 1/2*c)^5 - 40*C*a*tan(1/2*d*x + 1/2*c)^5 + 12*A*b*tan(1/2*d*
x + 1/2*c)^5 - 40*B*b*tan(1/2*d*x + 1/2*c)^5 - 9*C*b*tan(1/2*d*x + 1/2*c)^5 + 72*A*a*tan(1/2*d*x + 1/2*c)^3 +
12*B*a*tan(1/2*d*x + 1/2*c)^3 + 40*C*a*tan(1/2*d*x + 1/2*c)^3 + 12*A*b*tan(1/2*d*x + 1/2*c)^3 + 40*B*b*tan(1/2
*d*x + 1/2*c)^3 - 9*C*b*tan(1/2*d*x + 1/2*c)^3 - 24*A*a*tan(1/2*d*x + 1/2*c) - 12*B*a*tan(1/2*d*x + 1/2*c) - 2
4*C*a*tan(1/2*d*x + 1/2*c) - 12*A*b*tan(1/2*d*x + 1/2*c) - 24*B*b*tan(1/2*d*x + 1/2*c) - 15*C*b*tan(1/2*d*x +
1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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maple [A]  time = 1.36, size = 223, normalized size = 1.63 \[ \frac {a A \tan \left (d x +c \right )}{d}+\frac {a B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 a C \tan \left (d x +c \right )}{3 d}+\frac {a C \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {A b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {A b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 b B \tan \left (d x +c \right )}{3 d}+\frac {b B \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{3 d}+\frac {b C \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{4 d}+\frac {3 b C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 C b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.37, size = 218, normalized size = 1.59 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b - 3 \, C b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 7.72, size = 260, normalized size = 1.90 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {A\,b}{2}+\frac {B\,a}{2}+\frac {3\,C\,b}{8}\right )}{2\,A\,b+2\,B\,a+\frac {3\,C\,b}{2}}\right )\,\left (A\,b+B\,a+\frac {3\,C\,b}{4}\right )}{d}-\frac {\left (2\,A\,a-A\,b-B\,a+2\,B\,b+2\,C\,a-\frac {5\,C\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (A\,b-6\,A\,a+B\,a-\frac {10\,B\,b}{3}-\frac {10\,C\,a}{3}-\frac {3\,C\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (6\,A\,a+A\,b+B\,a+\frac {10\,B\,b}{3}+\frac {10\,C\,a}{3}-\frac {3\,C\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,A\,a-A\,b-B\,a-2\,B\,b-2\,C\,a-\frac {5\,C\,b}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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