3.805 \(\int \frac {\cos (c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^2} \, dx\)

Optimal. Leaf size=124 \[ \frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {B x}{a^2}-\frac {2 \left (a^3 (-C)+2 a^2 b B-b^3 B\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}} \]

[Out]

B*x/a^2-2*(2*B*a^2*b-B*b^3-C*a^3)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^2/(a-b)^(3/2)/(a+b)^(3
/2)/d+b*(B*b-C*a)*tan(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))

________________________________________________________________________________________

Rubi [A]  time = 0.27, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4072, 3923, 3919, 3831, 2659, 208} \[ -\frac {2 \left (2 a^2 b B+a^3 (-C)-b^3 B\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {b (b B-a C) \tan (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {B x}{a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*x)/a^2 - (2*(2*a^2*b*B - b^3*B - a^3*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(
3/2)*(a + b)^(3/2)*d) + (b*(b*B - a*C)*Tan[c + d*x])/(a*(a^2 - b^2)*d*(a + b*Sec[c + d*x]))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 2659

Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x
]}, Dist[(2*e)/d, Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}
, x] && NeQ[a^2 - b^2, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3923

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[(b*(
b*c - a*d)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 -
b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[c*(a^2 - b^2)*(m + 1) - (a*(b*c - a*d)*(m + 1))*Csc[e + f*x] + b
*(b*c - a*d)*(m + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m,
-1] && NeQ[a^2 - b^2, 0] && IntegerQ[2*m]

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 \frac {\cos (c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^2} \, dx &=\int \frac {B+C \sec (c+d x)}{(a+b \sec (c+d x))^2} \, dx\\ &=\frac {b (b B-a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {-\left (a^2-b^2\right ) B+a (b B-a C) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {B x}{a^2}+\frac {b (b B-a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (2 a^2 b B-b^3 B-a^3 C\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=\frac {B x}{a^2}+\frac {b (b B-a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (2 a^2 b B-b^3 B-a^3 C\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{a^2 b \left (a^2-b^2\right )}\\ &=\frac {B x}{a^2}+\frac {b (b B-a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (2 \left (2 a^2 b B-b^3 B-a^3 C\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1+\frac {a}{b}+\left (1-\frac {a}{b}\right ) x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^2 b \left (a^2-b^2\right ) d}\\ &=\frac {B x}{a^2}-\frac {2 \left (2 a^2 b B-b^3 B-a^3 C\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {b (b B-a C) \tan (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 8.28, size = 119, normalized size = 0.96 \[ \frac {-\frac {2 \left (a^3 C-2 a^2 b B+b^3 B\right ) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+\frac {a b (b B-a C) \sin (c+d x)}{(a-b) (a+b) (a \cos (c+d x)+b)}+B (c+d x)}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(c + d*x) - (2*(-2*a^2*b*B + b^3*B + a^3*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2 - b^
2)^(3/2) + (a*b*(b*B - a*C)*Sin[c + d*x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])))/(a^2*d)

________________________________________________________________________________________

fricas [B]  time = 0.83, size = 561, normalized size = 4.52 \[ \left [\frac {2 \, {\left (B a^{5} - 2 \, B a^{3} b^{2} + B a b^{4}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (B a^{4} b - 2 \, B a^{2} b^{3} + B b^{5}\right )} d x - {\left (C a^{3} b - 2 \, B a^{2} b^{2} + B b^{4} + {\left (C a^{4} - 2 \, B a^{3} b + B a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 2 \, {\left (C a^{4} b - B a^{3} b^{2} - C a^{2} b^{3} + B a b^{4}\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d\right )}}, \frac {{\left (B a^{5} - 2 \, B a^{3} b^{2} + B a b^{4}\right )} d x \cos \left (d x + c\right ) + {\left (B a^{4} b - 2 \, B a^{2} b^{3} + B b^{5}\right )} d x + {\left (C a^{3} b - 2 \, B a^{2} b^{2} + B b^{4} + {\left (C a^{4} - 2 \, B a^{3} b + B a b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (C a^{4} b - B a^{3} b^{2} - C a^{2} b^{3} + B a b^{4}\right )} \sin \left (d x + c\right )}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(2*(B*a^5 - 2*B*a^3*b^2 + B*a*b^4)*d*x*cos(d*x + c) + 2*(B*a^4*b - 2*B*a^2*b^3 + B*b^5)*d*x - (C*a^3*b -
2*B*a^2*b^2 + B*b^4 + (C*a^4 - 2*B*a^3*b + B*a*b^3)*cos(d*x + c))*sqrt(a^2 - b^2)*log((2*a*b*cos(d*x + c) - (a
^2 - 2*b^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x +
c)^2 + 2*a*b*cos(d*x + c) + b^2)) - 2*(C*a^4*b - B*a^3*b^2 - C*a^2*b^3 + B*a*b^4)*sin(d*x + c))/((a^7 - 2*a^5
*b^2 + a^3*b^4)*d*cos(d*x + c) + (a^6*b - 2*a^4*b^3 + a^2*b^5)*d), ((B*a^5 - 2*B*a^3*b^2 + B*a*b^4)*d*x*cos(d*
x + c) + (B*a^4*b - 2*B*a^2*b^3 + B*b^5)*d*x + (C*a^3*b - 2*B*a^2*b^2 + B*b^4 + (C*a^4 - 2*B*a^3*b + B*a*b^3)*
cos(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (C*
a^4*b - B*a^3*b^2 - C*a^2*b^3 + B*a*b^4)*sin(d*x + c))/((a^7 - 2*a^5*b^2 + a^3*b^4)*d*cos(d*x + c) + (a^6*b -
2*a^4*b^3 + a^2*b^5)*d)]

________________________________________________________________________________________

giac [A]  time = 0.28, size = 201, normalized size = 1.62 \[ \frac {\frac {2 \, {\left (C a^{3} - 2 \, B a^{2} b + B b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} + \frac {{\left (d x + c\right )} B}{a^{2}} + \frac {2 \, {\left (C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (a^{3} - a b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(C*a^3 - 2*B*a^2*b + B*b^3)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2*d*x + 1/
2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))/((a^4 - a^2*b^2)*sqrt(-a^2 + b^2)) + (d*x + c)*B/a^2 + 2*(C*
a*b*tan(1/2*d*x + 1/2*c) - B*b^2*tan(1/2*d*x + 1/2*c))/((a^3 - a*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*
x + 1/2*c)^2 - a - b)))/d

________________________________________________________________________________________

maple [B]  time = 1.12, size = 328, normalized size = 2.65 \[ -\frac {2 b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B}{d a \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}+\frac {2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C}{d \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}-\frac {4 b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) B}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) b^{3} B}{d \,a^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C a}{d \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/a*b^2/(a^2-b^2)*tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b-a-b)*B+2/d*b/(a^2-b^2)*
tan(1/2*d*x+1/2*c)/(a*tan(1/2*d*x+1/2*c)^2-tan(1/2*d*x+1/2*c)^2*b-a-b)*C-4/d*b/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)
*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*B+2/d/a^2/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctanh(tan(1
/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*b^3*B+2/d/(a-b)/(a+b)/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*
(a-b)/((a-b)*(a+b))^(1/2))*C*a+2/d/a^2*arctan(tan(1/2*d*x+1/2*c))*B

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?`
for more details)Is 4*a^2-4*b^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 12.09, size = 3763, normalized size = 30.35 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*B*atan(((B*((B*((32*(B*a^4*b^5 - C*a^9 - B*a^9 - 3*B*a^6*b^3 + B*a^7*b^2 - C*a^6*b^3 + C*a^7*b^2 + 2*B*a^8*
b + C*a^8*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (B*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a
^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))*1i)/a^2 + (32*tan(c/2 + (d*x)/2)
*(B^2*a^6 + 2*B^2*b^6 + C^2*a^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b - 5*B^2*a^2*b^4 + 4*B^2*a^3*b^3 + 3*B^2*a^4*b^2 -
4*B*C*a^5*b + 2*B*C*a^3*b^3))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))/a^2 - (B*((B*((32*(B*a^4*b^5 - C*a^9 - B*a^9
- 3*B*a^6*b^3 + B*a^7*b^2 - C*a^6*b^3 + C*a^7*b^2 + 2*B*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) +
(B*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b
+ a^5 - a^2*b^3 - a^3*b^2)))*1i)/a^2 - (32*tan(c/2 + (d*x)/2)*(B^2*a^6 + 2*B^2*b^6 + C^2*a^6 - 2*B^2*a*b^5 - 2
*B^2*a^5*b - 5*B^2*a^2*b^4 + 4*B^2*a^3*b^3 + 3*B^2*a^4*b^2 - 4*B*C*a^5*b + 2*B*C*a^3*b^3))/(a^4*b + a^5 - a^2*
b^3 - a^3*b^2)))/a^2)/((64*(B^3*b^5 + B*C^2*a^5 - B^2*C*a^5 - B^3*a*b^4 + 2*B^3*a^4*b - 3*B^3*a^2*b^3 + 2*B^3*
a^3*b^2 - 3*B^2*C*a^4*b + B^2*C*a^2*b^3 + B^2*C*a^3*b^2))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (B*((B*((32*(B*a
^4*b^5 - C*a^9 - B*a^9 - 3*B*a^6*b^3 + B*a^7*b^2 - C*a^6*b^3 + C*a^7*b^2 + 2*B*a^8*b + C*a^8*b))/(a^5*b + a^6
- a^3*b^3 - a^4*b^2) - (B*tan(c/2 + (d*x)/2)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*
b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^2)))*1i)/a^2 + (32*tan(c/2 + (d*x)/2)*(B^2*a^6 + 2*B^2*b^6 + C^2
*a^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b - 5*B^2*a^2*b^4 + 4*B^2*a^3*b^3 + 3*B^2*a^4*b^2 - 4*B*C*a^5*b + 2*B*C*a^3*b^3
))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2))*1i)/a^2 + (B*((B*((32*(B*a^4*b^5 - C*a^9 - B*a^9 - 3*B*a^6*b^3 + B*a^7*b
^2 - C*a^6*b^3 + C*a^7*b^2 + 2*B*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (B*tan(c/2 + (d*x)/2)*(
2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2)*32i)/(a^2*(a^4*b + a^5 - a^2*b^3 - a^3*b^
2)))*1i)/a^2 - (32*tan(c/2 + (d*x)/2)*(B^2*a^6 + 2*B^2*b^6 + C^2*a^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b - 5*B^2*a^2*b
^4 + 4*B^2*a^3*b^3 + 3*B^2*a^4*b^2 - 4*B*C*a^5*b + 2*B*C*a^3*b^3))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2))*1i)/a^2)
))/(a^2*d) + (atan(((((32*tan(c/2 + (d*x)/2)*(B^2*a^6 + 2*B^2*b^6 + C^2*a^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b - 5*B^
2*a^2*b^4 + 4*B^2*a^3*b^3 + 3*B^2*a^4*b^2 - 4*B*C*a^5*b + 2*B*C*a^3*b^3))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) +
(((32*(B*a^4*b^5 - C*a^9 - B*a^9 - 3*B*a^6*b^3 + B*a^7*b^2 - C*a^6*b^3 + C*a^7*b^2 + 2*B*a^8*b + C*a^8*b))/(a^
5*b + a^6 - a^3*b^3 - a^4*b^2) - (32*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*B*a^2*b
)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/((a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a
^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*B*a^2*b))/(a^8 - a^2*b^
6 + 3*a^4*b^4 - 3*a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*B*a^2*b)*1i)/(a^8 - a^2*b^6 + 3*a^4
*b^4 - 3*a^6*b^2) + (((32*tan(c/2 + (d*x)/2)*(B^2*a^6 + 2*B^2*b^6 + C^2*a^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b - 5*B^
2*a^2*b^4 + 4*B^2*a^3*b^3 + 3*B^2*a^4*b^2 - 4*B*C*a^5*b + 2*B*C*a^3*b^3))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) -
(((32*(B*a^4*b^5 - C*a^9 - B*a^9 - 3*B*a^6*b^3 + B*a^7*b^2 - C*a^6*b^3 + C*a^7*b^2 + 2*B*a^8*b + C*a^8*b))/(a^
5*b + a^6 - a^3*b^3 - a^4*b^2) + (32*tan(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*B*a^2*b
)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/((a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a
^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*B*a^2*b))/(a^8 - a^2*b^
6 + 3*a^4*b^4 - 3*a^6*b^2))*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*B*a^2*b)*1i)/(a^8 - a^2*b^6 + 3*a^4
*b^4 - 3*a^6*b^2))/((64*(B^3*b^5 + B*C^2*a^5 - B^2*C*a^5 - B^3*a*b^4 + 2*B^3*a^4*b - 3*B^3*a^2*b^3 + 2*B^3*a^3
*b^2 - 3*B^2*C*a^4*b + B^2*C*a^2*b^3 + B^2*C*a^3*b^2))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (((32*tan(c/2 + (d*
x)/2)*(B^2*a^6 + 2*B^2*b^6 + C^2*a^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b - 5*B^2*a^2*b^4 + 4*B^2*a^3*b^3 + 3*B^2*a^4*b
^2 - 4*B*C*a^5*b + 2*B*C*a^3*b^3))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) + (((32*(B*a^4*b^5 - C*a^9 - B*a^9 - 3*B*
a^6*b^3 + B*a^7*b^2 - C*a^6*b^3 + C*a^7*b^2 + 2*B*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) - (32*ta
n(c/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*B*a^2*b)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*
a^6*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/((a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))
*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*B*a^2*b))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*((a + b)^3*
(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*B*a^2*b))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2) - (((32*tan(c/2 + (d*x)/
2)*(B^2*a^6 + 2*B^2*b^6 + C^2*a^6 - 2*B^2*a*b^5 - 2*B^2*a^5*b - 5*B^2*a^2*b^4 + 4*B^2*a^3*b^3 + 3*B^2*a^4*b^2
- 4*B*C*a^5*b + 2*B*C*a^3*b^3))/(a^4*b + a^5 - a^2*b^3 - a^3*b^2) - (((32*(B*a^4*b^5 - C*a^9 - B*a^9 - 3*B*a^6
*b^3 + B*a^7*b^2 - C*a^6*b^3 + C*a^7*b^2 + 2*B*a^8*b + C*a^8*b))/(a^5*b + a^6 - a^3*b^3 - a^4*b^2) + (32*tan(c
/2 + (d*x)/2)*((a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*B*a^2*b)*(2*a^9*b - 2*a^4*b^6 + 2*a^5*b^5 + 4*a^6
*b^4 - 4*a^7*b^3 - 2*a^8*b^2))/((a^4*b + a^5 - a^2*b^3 - a^3*b^2)*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*((
a + b)^3*(a - b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*B*a^2*b))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2))*((a + b)^3*(a
- b)^3)^(1/2)*(B*b^3 + C*a^3 - 2*B*a^2*b))/(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)))*((a + b)^3*(a - b)^3)^(1/
2)*(B*b^3 + C*a^3 - 2*B*a^2*b)*2i)/(d*(a^8 - a^2*b^6 + 3*a^4*b^4 - 3*a^6*b^2)) - (2*tan(c/2 + (d*x)/2)*(B*b^2
- C*a*b))/(d*(a + b)*(a*b - a^2)*(a + b - tan(c/2 + (d*x)/2)^2*(a - b)))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________