3.681 \(\int \frac {\cos ^2(c+d x) (A+C \sec ^2(c+d x))}{a+b \sec (c+d x)} \, dx\)
Optimal. Leaf size=128 \[ -\frac {A b \sin (c+d x)}{a^2 d}-\frac {2 b \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d \sqrt {a-b} \sqrt {a+b}}+\frac {x \left (a^2 (A+2 C)+2 A b^2\right )}{2 a^3}+\frac {A \sin (c+d x) \cos (c+d x)}{2 a d} \]
[Out]
1/2*(2*A*b^2+a^2*(A+2*C))*x/a^3-A*b*sin(d*x+c)/a^2/d+1/2*A*cos(d*x+c)*sin(d*x+c)/a/d-2*b*(A*b^2+C*a^2)*arctanh
((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^3/d/(a-b)^(1/2)/(a+b)^(1/2)
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Rubi [A] time = 0.39, antiderivative size = 126, normalized size of antiderivative = 0.98,
number of steps used = 6, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used =
{4105, 4104, 3919, 3831, 2659, 208} \[ -\frac {2 b \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d \sqrt {a-b} \sqrt {a+b}}+\frac {x \left (\frac {2 A b^2}{a^2}+A+2 C\right )}{2 a}-\frac {A b \sin (c+d x)}{a^2 d}+\frac {A \sin (c+d x) \cos (c+d x)}{2 a d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^2*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]
[Out]
((A + (2*A*b^2)/a^2 + 2*C)*x)/(2*a) - (2*b*(A*b^2 + a^2*C)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]]
)/(a^3*Sqrt[a - b]*Sqrt[a + b]*d) - (A*b*Sin[c + d*x])/(a^2*d) + (A*Cos[c + d*x]*Sin[c + d*x])/(2*a*d)
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 4104
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_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
Rule 4105
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n)/(a*f*n),
x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[-(A*b*(m + n + 1)) + a*(A + A*n
+ C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, C, m}, x] && NeQ[
a^2 - b^2, 0] && LeQ[n, -1]
Rubi steps
\begin {align*} \int \frac {\cos ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx &=\frac {A \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {\int \frac {\cos (c+d x) \left (2 A b-a (A+2 C) \sec (c+d x)-A b \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 a}\\ &=-\frac {A b \sin (c+d x)}{a^2 d}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d}+\frac {\int \frac {2 A b^2+a^2 (A+2 C)+a A b \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^2}\\ &=\frac {\left (2 A b^2+a^2 (A+2 C)\right ) x}{2 a^3}-\frac {A b \sin (c+d x)}{a^2 d}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {\left (b \left (A b^2+a^2 C\right )\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^3}\\ &=\frac {\left (2 A b^2+a^2 (A+2 C)\right ) x}{2 a^3}-\frac {A b \sin (c+d x)}{a^2 d}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {\left (A b^2+a^2 C\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{a^3}\\ &=\frac {\left (2 A b^2+a^2 (A+2 C)\right ) x}{2 a^3}-\frac {A b \sin (c+d x)}{a^2 d}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {\left (2 \left (A b^2+a^2 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^3 d}\\ &=\frac {\left (2 A b^2+a^2 (A+2 C)\right ) x}{2 a^3}-\frac {2 b \left (A b^2+a^2 C\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 \sqrt {a-b} \sqrt {a+b} d}-\frac {A b \sin (c+d x)}{a^2 d}+\frac {A \cos (c+d x) \sin (c+d x)}{2 a d}\\ \end {align*}
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Mathematica [A] time = 0.38, size = 115, normalized size = 0.90 \[ \frac {2 (c+d x) \left (a^2 (A+2 C)+2 A b^2\right )+\frac {8 b \left (a^2 C+A b^2\right ) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+a^2 A \sin (2 (c+d x))-4 a A b \sin (c+d x)}{4 a^3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^2*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x]),x]
[Out]
(2*(2*A*b^2 + a^2*(A + 2*C))*(c + d*x) + (8*b*(A*b^2 + a^2*C)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b
^2]])/Sqrt[a^2 - b^2] - 4*a*A*b*Sin[c + d*x] + a^2*A*Sin[2*(c + d*x)])/(4*a^3*d)
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fricas [A] time = 0.51, size = 386, normalized size = 3.02 \[ \left [\frac {{\left ({\left (A + 2 \, C\right )} a^{4} + {\left (A - 2 \, C\right )} a^{2} b^{2} - 2 \, A b^{4}\right )} d x + {\left (C a^{2} b + A b^{3}\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - {\left (2 \, A a^{3} b - 2 \, A a b^{3} - {\left (A a^{4} - A a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{5} - a^{3} b^{2}\right )} d}, \frac {{\left ({\left (A + 2 \, C\right )} a^{4} + {\left (A - 2 \, C\right )} a^{2} b^{2} - 2 \, A b^{4}\right )} d x - 2 \, {\left (C a^{2} b + A b^{3}\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (2 \, A a^{3} b - 2 \, A a b^{3} - {\left (A a^{4} - A a^{2} b^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{5} - a^{3} b^{2}\right )} d}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="fricas")
[Out]
[1/2*(((A + 2*C)*a^4 + (A - 2*C)*a^2*b^2 - 2*A*b^4)*d*x + (C*a^2*b + A*b^3)*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*A*a^3*b - 2*A*a*b^3 - (A*a^4 - A*a^2*b^2)*cos(d*x + c))*sin(
d*x + c))/((a^5 - a^3*b^2)*d), 1/2*(((A + 2*C)*a^4 + (A - 2*C)*a^2*b^2 - 2*A*b^4)*d*x - 2*(C*a^2*b + A*b^3)*sq
rt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (2*A*a^3*b - 2*A*a*
b^3 - (A*a^4 - A*a^2*b^2)*cos(d*x + c))*sin(d*x + c))/((a^5 - a^3*b^2)*d)]
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giac [A] time = 0.25, size = 199, normalized size = 1.55 \[ \frac {\frac {{\left (A a^{2} + 2 \, C a^{2} + 2 \, A b^{2}\right )} {\left (d x + c\right )}}{a^{3}} - \frac {4 \, {\left (C a^{2} b + A b^{3}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{\sqrt {-a^{2} + b^{2}} a^{3}} - \frac {2 \, {\left (A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{2}}}{2 \, d} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x, algorithm="giac")
[Out]
1/2*((A*a^2 + 2*C*a^2 + 2*A*b^2)*(d*x + c)/a^3 - 4*(C*a^2*b + A*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)))/(sqrt(-a^2 + b^2)*a^3)
- 2*(A*a*tan(1/2*d*x + 1/2*c)^3 + 2*A*b*tan(1/2*d*x + 1/2*c)^3 - A*a*tan(1/2*d*x + 1/2*c) + 2*A*b*tan(1/2*d*x
+ 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^2))/d
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maple [B] time = 1.58, size = 296, normalized size = 2.31 \[ -\frac {2 b^{3} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) A}{d \,a^{3} \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {2 b \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right ) C}{d a \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A b}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A}{d a \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) A b}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A \,b^{2}}{d \,a^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d a} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^2*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),x)
[Out]
-2/d*b^3/a^3/((a-b)*(a+b))^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*A-2/d*b/a/((a-b)*(a+b))
^(1/2)*arctanh(tan(1/2*d*x+1/2*c)*(a-b)/((a-b)*(a+b))^(1/2))*C-1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/
2*c)^3*A-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*A*b+1/d/a/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*
d*x+1/2*c)*A-2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)*A*b+1/d*A/a*arctan(tan(1/2*d*x+1/2*c))+2/d/
a^3*arctan(tan(1/2*d*x+1/2*c))*A*b^2+2/d/a*arctan(tan(1/2*d*x+1/2*c))*C
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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)^2*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c)),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?
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mupad [B] time = 6.22, size = 2478, normalized size = 19.36 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((cos(c + d*x)^2*(A + C/cos(c + d*x)^2))/(a + b/cos(c + d*x)),x)
[Out]
(A*a*sin(2*c + 2*d*x))/(4*d*(a^2 - b^2)) + (A*a*atan((A^2*a^6*sin(c/2 + (d*x)/2) + 4*C^2*a^6*sin(c/2 + (d*x)/2
) + 3*A^2*a^4*b^2*sin(c/2 + (d*x)/2) + 4*A*C*a^6*sin(c/2 + (d*x)/2) + 4*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(a*cos
(c/2 + (d*x)/2)*(A^2*a^5 + 4*C^2*a^5 + 3*A^2*a^3*b^2 + 4*A*C*a^5 + 4*A*C*a^3*b^2))))/(d*(a^2 - b^2)) + (2*C*a*
atan((A^2*a^6*sin(c/2 + (d*x)/2) + 4*C^2*a^6*sin(c/2 + (d*x)/2) + 3*A^2*a^4*b^2*sin(c/2 + (d*x)/2) + 4*A*C*a^6
*sin(c/2 + (d*x)/2) + 4*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(a*cos(c/2 + (d*x)/2)*(A^2*a^5 + 4*C^2*a^5 + 3*A^2*a^3
*b^2 + 4*A*C*a^5 + 4*A*C*a^3*b^2))))/(d*(a^2 - b^2)) - (A*b*sin(c + d*x))/(d*(a^2 - b^2)) - (A*b^2*sin(2*c + 2
*d*x))/(4*a*d*(a^2 - b^2)) + (A*b^2*atan((A^2*a^6*sin(c/2 + (d*x)/2) + 4*C^2*a^6*sin(c/2 + (d*x)/2) + 3*A^2*a^
4*b^2*sin(c/2 + (d*x)/2) + 4*A*C*a^6*sin(c/2 + (d*x)/2) + 4*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(a*cos(c/2 + (d*x)
/2)*(A^2*a^5 + 4*C^2*a^5 + 3*A^2*a^3*b^2 + 4*A*C*a^5 + 4*A*C*a^3*b^2))))/(a*d*(a^2 - b^2)) - (2*A*b^4*atan((A^
2*a^6*sin(c/2 + (d*x)/2) + 4*C^2*a^6*sin(c/2 + (d*x)/2) + 3*A^2*a^4*b^2*sin(c/2 + (d*x)/2) + 4*A*C*a^6*sin(c/2
+ (d*x)/2) + 4*A*C*a^4*b^2*sin(c/2 + (d*x)/2))/(a*cos(c/2 + (d*x)/2)*(A^2*a^5 + 4*C^2*a^5 + 3*A^2*a^3*b^2 + 4
*A*C*a^5 + 4*A*C*a^3*b^2))))/(a^3*d*(a^2 - b^2)) - (2*C*b^2*atan((A^2*a^6*sin(c/2 + (d*x)/2) + 4*C^2*a^6*sin(c
/2 + (d*x)/2) + 3*A^2*a^4*b^2*sin(c/2 + (d*x)/2) + 4*A*C*a^6*sin(c/2 + (d*x)/2) + 4*A*C*a^4*b^2*sin(c/2 + (d*x
)/2))/(a*cos(c/2 + (d*x)/2)*(A^2*a^5 + 4*C^2*a^5 + 3*A^2*a^3*b^2 + 4*A*C*a^5 + 4*A*C*a^3*b^2))))/(a*d*(a^2 - b
^2)) + (A*b^3*sin(c + d*x))/(a^2*d*(a^2 - b^2)) + (C*b*atan(((8*A^2*b^7*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(3/2) -
A^2*a^9*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 8*A^2*b^9*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 4*C^2*a^9*sin
(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 4*A*C*a^9*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + A^2*a^8*b*sin(c/2 + (d*x)
/2)*(a^2 - b^2)^(1/2) + 4*C^2*a^8*b*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 8*A^2*a^2*b^7*sin(c/2 + (d*x)/2)*(a
^2 - b^2)^(1/2) - 3*A^2*a^4*b^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 3*A^2*a^5*b^4*sin(c/2 + (d*x)/2)*(a^2 -
b^2)^(1/2) + 2*A^2*a^6*b^3*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 2*A^2*a^7*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2
)^(1/2) + 8*C^2*a^4*b^3*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(3/2) + 8*C^2*a^4*b^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1
/2) - 12*C^2*a^6*b^3*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 4*C^2*a^7*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)
+ 4*A*C*a^8*b*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 16*A*C*a^2*b^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(3/2) + 16
*A*C*a^2*b^7*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 20*A*C*a^4*b^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 4*A*
C*a^5*b^4*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))*1i)/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)*(A^2*a^7 + 4*C^2*a^7 - 3
*A^2*a^3*b^4 + 2*A^2*a^5*b^2 - 4*C^2*a^5*b^2 + 4*A*C*a^7 - 4*A*C*a^3*b^4)))*((a + b)*(a - b))^(1/2)*2i)/(a*d*(
a^2 - b^2)) + (A*b^3*atan(((8*A^2*b^7*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(3/2) - A^2*a^9*sin(c/2 + (d*x)/2)*(a^2 -
b^2)^(1/2) + 8*A^2*b^9*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 4*C^2*a^9*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)
- 4*A*C*a^9*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + A^2*a^8*b*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 4*C^2*a^8*
b*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 8*A^2*a^2*b^7*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 3*A^2*a^4*b^5*si
n(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 3*A^2*a^5*b^4*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 2*A^2*a^6*b^3*sin(c/
2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 2*A^2*a^7*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 8*C^2*a^4*b^3*sin(c/2 +
(d*x)/2)*(a^2 - b^2)^(3/2) + 8*C^2*a^4*b^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) - 12*C^2*a^6*b^3*sin(c/2 + (d*
x)/2)*(a^2 - b^2)^(1/2) + 4*C^2*a^7*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 4*A*C*a^8*b*sin(c/2 + (d*x)/2)*
(a^2 - b^2)^(1/2) + 16*A*C*a^2*b^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(3/2) + 16*A*C*a^2*b^7*sin(c/2 + (d*x)/2)*(a
^2 - b^2)^(1/2) - 20*A*C*a^4*b^5*sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2) + 4*A*C*a^5*b^4*sin(c/2 + (d*x)/2)*(a^2
- b^2)^(1/2))*1i)/(cos(c/2 + (d*x)/2)*(a*b^2 - a^3)*(A^2*a^7 + 4*C^2*a^7 - 3*A^2*a^3*b^4 + 2*A^2*a^5*b^2 - 4*C
^2*a^5*b^2 + 4*A*C*a^7 - 4*A*C*a^3*b^4)))*((a + b)*(a - b))^(1/2)*2i)/(a^3*d*(a^2 - b^2))
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{2}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**2*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c)),x)
[Out]
Integral((A + C*sec(c + d*x)**2)*cos(c + d*x)**2/(a + b*sec(c + d*x)), x)
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