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

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

[Out]

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

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Rubi [A]  time = 0.35, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {4072, 4010, 4002, 3997, 3787, 3770, 3767, 8} \[ \frac {\left (4 a^2 b B+a^3 (-C)+8 a b^2 C+4 b^3 B\right ) \tan (c+d x)}{6 b d}+\frac {\left (4 a^2 C+8 a b B+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (-2 a^2 C+8 a b B+9 b^2 C\right ) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {(4 b B-a C) \tan (c+d x) (a+b \sec (c+d x))^2}{12 b d}+\frac {C \tan (c+d x) (a+b \sec (c+d x))^3}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.80, size = 120, normalized size = 0.67 \[ \frac {3 \left (4 a^2 C+8 a b B+3 b^2 C\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (3 \left (4 a^2 C+8 a b B+3 b^2 C\right ) \sec (c+d x)+24 \left (a^2 B+2 a b C+b^2 B\right )+8 b (2 a C+b B) \tan ^2(c+d x)+6 b^2 C \sec ^3(c+d x)\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.92, size = 180, normalized size = 1.01 \[ \frac {3 \, {\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (3 \, B a^{2} + 4 \, C a b + 2 \, B b^{2}\right )} \cos \left (d x + c\right )^{3} + 6 \, C b^{2} + 3 \, {\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.33, size = 478, normalized size = 2.67 \[ \frac {3 \, {\left (4 \, C a^{2} + 8 \, B a b + 3 \, C b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (4 \, C a^{2} + 8 \, B a b + 3 \, C 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 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 80 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 40 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 48 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C 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)*(a+b*sec(d*x+c))^2*(B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(3*(4*C*a^2 + 8*B*a*b + 3*C*b^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(4*C*a^2 + 8*B*a*b + 3*C*b^2)*log
(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(24*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 12*C*a^2*tan(1/2*d*x + 1/2*c)^7 - 24*B*
a*b*tan(1/2*d*x + 1/2*c)^7 + 48*C*a*b*tan(1/2*d*x + 1/2*c)^7 + 24*B*b^2*tan(1/2*d*x + 1/2*c)^7 - 15*C*b^2*tan(
1/2*d*x + 1/2*c)^7 - 72*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 24*B*a*b*tan(1/2*d*x
+ 1/2*c)^5 - 80*C*a*b*tan(1/2*d*x + 1/2*c)^5 - 40*B*b^2*tan(1/2*d*x + 1/2*c)^5 - 9*C*b^2*tan(1/2*d*x + 1/2*c)^
5 + 72*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 12*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 80*C
*a*b*tan(1/2*d*x + 1/2*c)^3 + 40*B*b^2*tan(1/2*d*x + 1/2*c)^3 - 9*C*b^2*tan(1/2*d*x + 1/2*c)^3 - 24*B*a^2*tan(
1/2*d*x + 1/2*c) - 12*C*a^2*tan(1/2*d*x + 1/2*c) - 24*B*a*b*tan(1/2*d*x + 1/2*c) - 48*C*a*b*tan(1/2*d*x + 1/2*
c) - 24*B*b^2*tan(1/2*d*x + 1/2*c) - 15*C*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 = 1.34, size = 241, normalized size = 1.35 \[ \frac {a^{2} B \tan \left (d x +c \right )}{d}+\frac {a^{2} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {B a b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {4 C a b \tan \left (d x +c \right )}{3 d}+\frac {2 C a b \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {2 b^{2} B \tan \left (d x +c \right )}{3 d}+\frac {b^{2} B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {b^{2} C \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 b^{2} C \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 b^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 228, normalized size = 1.27 \[ \frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a b + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B b^{2} - 3 \, C b^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, C a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, B a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B a^{2} \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 7.69, size = 317, normalized size = 1.77 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {C\,a^2}{2}+B\,a\,b+\frac {3\,C\,b^2}{8}\right )}{2\,C\,a^2+4\,B\,a\,b+\frac {3\,C\,b^2}{2}}\right )\,\left (C\,a^2+2\,B\,a\,b+\frac {3\,C\,b^2}{4}\right )}{d}-\frac {\left (2\,B\,a^2+2\,B\,b^2-C\,a^2-\frac {5\,C\,b^2}{4}-2\,B\,a\,b+4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (C\,a^2-\frac {10\,B\,b^2}{3}-6\,B\,a^2-\frac {3\,C\,b^2}{4}+2\,B\,a\,b-\frac {20\,C\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (6\,B\,a^2+\frac {10\,B\,b^2}{3}+C\,a^2-\frac {3\,C\,b^2}{4}+2\,B\,a\,b+\frac {20\,C\,a\,b}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-2\,B\,a^2-2\,B\,b^2-C\,a^2-\frac {5\,C\,b^2}{4}-2\,B\,a\,b-4\,C\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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