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

Optimal. Leaf size=179 \[ \frac {\left (4 a^2 B+8 a A b+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (-2 a^2 B+8 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {\left (a^3 (-B)+4 a^2 A b+8 a b^2 B+4 A b^3\right ) \tan (c+d x)}{6 b d}+\frac {(4 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^2}{12 b d}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d} \]

[Out]

1/8*(8*A*a*b+4*B*a^2+3*B*b^2)*arctanh(sin(d*x+c))/d+1/6*(4*A*a^2*b+4*A*b^3-B*a^3+8*B*a*b^2)*tan(d*x+c)/b/d+1/2
4*(8*A*a*b-2*B*a^2+9*B*b^2)*sec(d*x+c)*tan(d*x+c)/d+1/12*(4*A*b-B*a)*(a+b*sec(d*x+c))^2*tan(d*x+c)/b/d+1/4*B*(
a+b*sec(d*x+c))^3*tan(d*x+c)/b/d

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Rubi [A]  time = 0.32, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4010, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac {\left (4 a^2 A b+a^3 (-B)+8 a b^2 B+4 A b^3\right ) \tan (c+d x)}{6 b d}+\frac {\left (4 a^2 B+8 a A b+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (-2 a^2 B+8 a A b+9 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {(4 A b-a B) \tan (c+d x) (a+b \sec (c+d x))^2}{12 b d}+\frac {B \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a*A*b + 4*a^2*B + 3*b^2*B)*ArcTanh[Sin[c + d*x]])/(8*d) + ((4*a^2*A*b + 4*A*b^3 - a^3*B + 8*a*b^2*B)*Tan[c
+ d*x])/(6*b*d) + ((8*a*A*b - 2*a^2*B + 9*b^2*B)*Sec[c + d*x]*Tan[c + d*x])/(24*d) + ((4*A*b - a*B)*(a + b*Se
c[c + d*x])^2*Tan[c + d*x])/(12*b*d) + (B*(a + b*Sec[c + d*x])^3*Tan[c + d*x])/(4*b*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 3770

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

Rule 3787

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 3997

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*C
sc[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 4002

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] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4010

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)), I
nt[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] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+b \sec (c+d x))^2 (A+B \sec (c+d x)) \, dx &=\frac {B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x))^2 (3 b B+(4 A b-a B) \sec (c+d x)) \, dx}{4 b}\\ &=\frac {(4 A b-a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\int \sec (c+d x) (a+b \sec (c+d x)) \left (b (8 A b+7 a B)+\left (8 a A b-2 a^2 B+9 b^2 B\right ) \sec (c+d x)\right ) \, dx}{12 b}\\ &=\frac {\left (8 a A b-2 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 A b-a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {\int \sec (c+d x) \left (3 b \left (8 a A b+4 a^2 B+3 b^2 B\right )+4 \left (4 a^2 A b+4 A b^3-a^3 B+8 a b^2 B\right ) \sec (c+d x)\right ) \, dx}{24 b}\\ &=\frac {\left (8 a A b-2 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 A b-a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}+\frac {1}{8} \left (8 a A b+4 a^2 B+3 b^2 B\right ) \int \sec (c+d x) \, dx+\frac {\left (4 a^2 A b+4 A b^3-a^3 B+8 a b^2 B\right ) \int \sec ^2(c+d x) \, dx}{6 b}\\ &=\frac {\left (8 a A b+4 a^2 B+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (8 a A b-2 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 A b-a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}-\frac {\left (4 a^2 A b+4 A b^3-a^3 B+8 a b^2 B\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{6 b d}\\ &=\frac {\left (8 a A b+4 a^2 B+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (4 a^2 A b+4 A b^3-a^3 B+8 a b^2 B\right ) \tan (c+d x)}{6 b d}+\frac {\left (8 a A b-2 a^2 B+9 b^2 B\right ) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(4 A b-a B) (a+b \sec (c+d x))^2 \tan (c+d x)}{12 b d}+\frac {B (a+b \sec (c+d x))^3 \tan (c+d x)}{4 b d}\\ \end {align*}

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Mathematica [A]  time = 0.75, size = 120, normalized size = 0.67 \[ \frac {3 \left (4 a^2 B+8 a A b+3 b^2 B\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 \left (4 a^2 B+8 a A b+3 b^2 B\right ) \sec (c+d x)+24 \left (a^2 A+2 a b B+A b^2\right )+8 b (2 a B+A b) \tan ^2(c+d x)+6 b^2 B \sec ^3(c+d x)\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 180, normalized size = 1.01 \[ \frac {3 \, {\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (3 \, A a^{2} + 4 \, B a b + 2 \, A b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, B b^{2} + 3 \, {\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.29, size = 478, normalized size = 2.67 \[ \frac {3 \, {\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, B a^{2} + 8 \, A a b + 3 \, B b^{2}\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^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 48 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 80 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, 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 )}^{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))^2*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/24*(3*(4*B*a^2 + 8*A*a*b + 3*B*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*B*a^2 + 8*A*a*b + 3*B*b^2)*log
(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(24*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 12*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 24*A*
a*b*tan(1/2*d*x + 1/2*c)^7 + 48*B*a*b*tan(1/2*d*x + 1/2*c)^7 + 24*A*b^2*tan(1/2*d*x + 1/2*c)^7 - 15*B*b^2*tan(
1/2*d*x + 1/2*c)^7 - 72*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 12*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 24*A*a*b*tan(1/2*d*x
+ 1/2*c)^5 - 80*B*a*b*tan(1/2*d*x + 1/2*c)^5 - 40*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 9*B*b^2*tan(1/2*d*x + 1/2*c)^
5 + 72*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*A*a*b*tan(1/2*d*x + 1/2*c)^3 + 80*B
*a*b*tan(1/2*d*x + 1/2*c)^3 + 40*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 9*B*b^2*tan(1/2*d*x + 1/2*c)^3 - 24*A*a^2*tan(
1/2*d*x + 1/2*c) - 12*B*a^2*tan(1/2*d*x + 1/2*c) - 24*A*a*b*tan(1/2*d*x + 1/2*c) - 48*B*a*b*tan(1/2*d*x + 1/2*
c) - 24*A*b^2*tan(1/2*d*x + 1/2*c) - 15*B*b^2*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 = 1.43, size = 241, normalized size = 1.35 \[ \frac {a^{2} A \tan \left (d x +c \right )}{d}+\frac {a^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a A b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 B a b \tan \left (d x +c \right )}{3 d}+\frac {2 B a b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {2 A \,b^{2} \tan \left (d x +c \right )}{3 d}+\frac {A \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {b^{2} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 b^{2} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 b^{2} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.96, size = 228, normalized size = 1.27 \[ \frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a b + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} - 3 \, B b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, A a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{2} \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a*b + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b^2 - 3*B*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)) - 12*B*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
)) - 24*A*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^2
*tan(d*x + c))/d

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mupad [B]  time = 5.69, size = 317, normalized size = 1.77 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {B\,a^2}{2}+A\,a\,b+\frac {3\,B\,b^2}{8}\right )}{2\,B\,a^2+4\,A\,a\,b+\frac {3\,B\,b^2}{2}}\right )\,\left (B\,a^2+2\,A\,a\,b+\frac {3\,B\,b^2}{4}\right )}{d}-\frac {\left (2\,A\,a^2+2\,A\,b^2-B\,a^2-\frac {5\,B\,b^2}{4}-2\,A\,a\,b+4\,B\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (B\,a^2-\frac {10\,A\,b^2}{3}-6\,A\,a^2-\frac {3\,B\,b^2}{4}+2\,A\,a\,b-\frac {20\,B\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (6\,A\,a^2+\frac {10\,A\,b^2}{3}+B\,a^2-\frac {3\,B\,b^2}{4}+2\,A\,a\,b+\frac {20\,B\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,A\,a^2-2\,A\,b^2-B\,a^2-\frac {5\,B\,b^2}{4}-2\,A\,a\,b-4\,B\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((B*a^2)/2 + (3*B*b^2)/8 + A*a*b))/(2*B*a^2 + (3*B*b^2)/2 + 4*A*a*b))*(B*a^2 + (3
*B*b^2)/4 + 2*A*a*b))/d - (tan(c/2 + (d*x)/2)^7*(2*A*a^2 + 2*A*b^2 - B*a^2 - (5*B*b^2)/4 - 2*A*a*b + 4*B*a*b)
+ tan(c/2 + (d*x)/2)^3*(6*A*a^2 + (10*A*b^2)/3 + B*a^2 - (3*B*b^2)/4 + 2*A*a*b + (20*B*a*b)/3) - tan(c/2 + (d*
x)/2)^5*(6*A*a^2 + (10*A*b^2)/3 - B*a^2 + (3*B*b^2)/4 - 2*A*a*b + (20*B*a*b)/3) - tan(c/2 + (d*x)/2)*(2*A*a^2
+ 2*A*b^2 + B*a^2 + (5*B*b^2)/4 + 2*A*a*b + 4*B*a*b))/(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 \[ \int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right )^{2} \sec ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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