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

Optimal. Leaf size=208 \[ \frac {\left (a^2-3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x) \cos (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {x \left (a^2+6 b^2\right )}{2 a^4}-\frac {2 b^3 \left (4 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}-\frac {b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 d \left (a^2-b^2\right )} \]

[Out]

1/2*(a^2+6*b^2)*x/a^4-2*b^3*(4*a^2-3*b^2)*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^(1/2))/a^4/(a-b)^(3/2)/
(a+b)^(3/2)/d-b*(2*a^2-3*b^2)*sin(d*x+c)/a^3/(a^2-b^2)/d+1/2*(a^2-3*b^2)*cos(d*x+c)*sin(d*x+c)/a^2/(a^2-b^2)/d
+b^2*cos(d*x+c)*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))

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Rubi [A]  time = 0.59, antiderivative size = 208, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3847, 4104, 3919, 3831, 2659, 208} \[ -\frac {b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 d \left (a^2-b^2\right )}+\frac {\left (a^2-3 b^2\right ) \sin (c+d x) \cos (c+d x)}{2 a^2 d \left (a^2-b^2\right )}-\frac {2 b^3 \left (4 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {b^2 \sin (c+d x) \cos (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {x \left (a^2+6 b^2\right )}{2 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a^2 + 6*b^2)*x)/(2*a^4) - (2*b^3*(4*a^2 - 3*b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^4*(
a - b)^(3/2)*(a + b)^(3/2)*d) - (b*(2*a^2 - 3*b^2)*Sin[c + d*x])/(a^3*(a^2 - b^2)*d) + ((a^2 - 3*b^2)*Cos[c +
d*x]*Sin[c + d*x])/(2*a^2*(a^2 - b^2)*d) + (b^2*Cos[c + d*x]*Sin[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 3847

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

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]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\frac {b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {\cos ^2(c+d x) \left (-a^2+3 b^2+a b \sec (c+d x)-2 b^2 \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\int \frac {\cos (c+d x) \left (-2 b \left (2 a^2-3 b^2\right )+a \left (a^2+b^2\right ) \sec (c+d x)+b \left (a^2-3 b^2\right ) \sec ^2(c+d x)\right )}{a+b \sec (c+d x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=-\frac {b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\int \frac {-a^4-5 a^2 b^2+6 b^4-a b \left (a^2-3 b^2\right ) \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {\left (a^2+6 b^2\right ) x}{2 a^4}-\frac {b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (b^3 \left (4 a^2-3 b^2\right )\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^4 \left (a^2-b^2\right )}\\ &=\frac {\left (a^2+6 b^2\right ) x}{2 a^4}-\frac {b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (b^2 \left (4 a^2-3 b^2\right )\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{a^4 \left (a^2-b^2\right )}\\ &=\frac {\left (a^2+6 b^2\right ) x}{2 a^4}-\frac {b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}-\frac {\left (2 b^2 \left (4 a^2-3 b^2\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^4 \left (a^2-b^2\right ) d}\\ &=\frac {\left (a^2+6 b^2\right ) x}{2 a^4}-\frac {2 b^3 \left (4 a^2-3 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^4 (a-b)^{3/2} (a+b)^{3/2} d}-\frac {b \left (2 a^2-3 b^2\right ) \sin (c+d x)}{a^3 \left (a^2-b^2\right ) d}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x) \sin (c+d x)}{2 a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \cos (c+d x) \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.77, size = 144, normalized size = 0.69 \[ \frac {2 \left (a^2+6 b^2\right ) (c+d x)-\frac {8 b^3 \left (3 b^2-4 a^2\right ) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}+a^2 \sin (2 (c+d x))+\frac {4 a b^4 \sin (c+d x)}{(a-b) (a+b) (a \cos (c+d x)+b)}-8 a b \sin (c+d x)}{4 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a^2 + 6*b^2)*(c + d*x) - (8*b^3*(-4*a^2 + 3*b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^
2 - b^2)^(3/2) - 8*a*b*Sin[c + d*x] + (4*a*b^4*Sin[c + d*x])/((a - b)*(a + b)*(b + a*Cos[c + d*x])) + a^2*Sin[
2*(c + d*x)])/(4*a^4*d)

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fricas [A]  time = 0.56, size = 660, normalized size = 3.17 \[ \left [\frac {{\left (a^{7} + 4 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + 6 \, a b^{6}\right )} d x \cos \left (d x + c\right ) + {\left (a^{6} b + 4 \, a^{4} b^{3} - 11 \, a^{2} b^{5} + 6 \, b^{7}\right )} d x + {\left (4 \, a^{2} b^{4} - 3 \, b^{6} + {\left (4 \, a^{3} b^{3} - 3 \, a b^{5}\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 ) - {\left (4 \, a^{5} b^{2} - 10 \, a^{3} b^{4} + 6 \, a b^{6} - {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{9} - 2 \, a^{7} b^{2} + a^{5} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b - 2 \, a^{6} b^{3} + a^{4} b^{5}\right )} d\right )}}, \frac {{\left (a^{7} + 4 \, a^{5} b^{2} - 11 \, a^{3} b^{4} + 6 \, a b^{6}\right )} d x \cos \left (d x + c\right ) + {\left (a^{6} b + 4 \, a^{4} b^{3} - 11 \, a^{2} b^{5} + 6 \, b^{7}\right )} d x - 2 \, {\left (4 \, a^{2} b^{4} - 3 \, b^{6} + {\left (4 \, a^{3} b^{3} - 3 \, a b^{5}\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 (4 \, a^{5} b^{2} - 10 \, a^{3} b^{4} + 6 \, a b^{6} - {\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{9} - 2 \, a^{7} b^{2} + a^{5} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{8} b - 2 \, a^{6} b^{3} + a^{4} b^{5}\right )} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((a^7 + 4*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6)*d*x*cos(d*x + c) + (a^6*b + 4*a^4*b^3 - 11*a^2*b^5 + 6*b^7)*d*x
+ (4*a^2*b^4 - 3*b^6 + (4*a^3*b^3 - 3*a*b^5)*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)) - (4*a^5*b^2 - 10*a^3*b^4 + 6*a*b^6 - (a^7 - 2*a^5*b^2 + a^3*b^4)*cos(d*x + c)^2
+ 3*(a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))*sin(d*x + c))/((a^9 - 2*a^7*b^2 + a^5*b^4)*d*cos(d*x + c) + (a
^8*b - 2*a^6*b^3 + a^4*b^5)*d), 1/2*((a^7 + 4*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6)*d*x*cos(d*x + c) + (a^6*b + 4*a^
4*b^3 - 11*a^2*b^5 + 6*b^7)*d*x - 2*(4*a^2*b^4 - 3*b^6 + (4*a^3*b^3 - 3*a*b^5)*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))) - (4*a^5*b^2 - 10*a^3*b^4 + 6*a*b^6
- (a^7 - 2*a^5*b^2 + a^3*b^4)*cos(d*x + c)^2 + 3*(a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c))*sin(d*x + c))/((a
^9 - 2*a^7*b^2 + a^5*b^4)*d*cos(d*x + c) + (a^8*b - 2*a^6*b^3 + a^4*b^5)*d)]

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giac [A]  time = 0.21, size = 264, normalized size = 1.27 \[ -\frac {\frac {4 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a^{5} - a^{3} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )}} + \frac {4 \, {\left (4 \, a^{2} b^{3} - 3 \, b^{5}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt {-a^{2} + b^{2}}} - \frac {{\left (a^{2} + 6 \, b^{2}\right )} {\left (d x + c\right )}}{a^{4}} + \frac {2 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*b^4*tan(1/2*d*x + 1/2*c)/((a^5 - a^3*b^2)*(a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c)^2 - a - b
)) + 4*(4*a^2*b^3 - 3*b^5)*(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^6 - a^4*b^2)*sqrt(-a^2 + b^2)) - (a^2 + 6*b^2)*(d*x + c)/a^4
+ 2*(a*tan(1/2*d*x + 1/2*c)^3 + 4*b*tan(1/2*d*x + 1/2*c)^3 - a*tan(1/2*d*x + 1/2*c) + 4*b*tan(1/2*d*x + 1/2*c
))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^3))/d

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maple [A]  time = 0.73, size = 362, normalized size = 1.74 \[ -\frac {2 b^{4} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}-\frac {8 b^{3} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,a^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {6 b^{5} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,a^{4} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {\arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}+\frac {6 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d*b^4/a^3/(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)-8/d*b^3/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))+6/d*b^5/a^4/(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))-1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1
/2*d*x+1/2*c)^3-4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*b+1/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^2*t
an(1/2*d*x+1/2*c)-4/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)*b+1/d/a^2*arctan(tan(1/2*d*x+1/2*c))+6
/d/a^4*arctan(tan(1/2*d*x+1/2*c))*b^2

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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+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?

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mupad [B]  time = 8.32, size = 3738, normalized size = 17.97 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan(((((8*tan(c/2 + (d*x)/2)*(a^10 - 2*a^9*b - 72*a*b^9 + 72*b^10 - 120*a^2*b^8 + 120*a^3*b^7 + 17*a^4*b^6 -
26*a^5*b^5 + 23*a^6*b^4 - 20*a^7*b^3 + 11*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + ((a^2*1i + b^2*6i)*((
8*(2*a^15 - 12*a^8*b^7 + 6*a^9*b^6 + 28*a^10*b^5 - 14*a^11*b^4 - 16*a^12*b^3 + 6*a^13*b^2))/(a^11*b + a^12 - a
^9*b^3 - a^10*b^2) - (4*tan(c/2 + (d*x)/2)*(a^2*1i + b^2*6i)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 -
16*a^11*b^3 - 8*a^12*b^2))/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b^2))))/(2*a^4))*(a^2*1i + b^2*6i)*1i)/(2*a^4) +
(((8*tan(c/2 + (d*x)/2)*(a^10 - 2*a^9*b - 72*a*b^9 + 72*b^10 - 120*a^2*b^8 + 120*a^3*b^7 + 17*a^4*b^6 - 26*a^
5*b^5 + 23*a^6*b^4 - 20*a^7*b^3 + 11*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - ((a^2*1i + b^2*6i)*((8*(2*a
^15 - 12*a^8*b^7 + 6*a^9*b^6 + 28*a^10*b^5 - 14*a^11*b^4 - 16*a^12*b^3 + 6*a^13*b^2))/(a^11*b + a^12 - a^9*b^3
- a^10*b^2) + (4*tan(c/2 + (d*x)/2)*(a^2*1i + b^2*6i)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^
11*b^3 - 8*a^12*b^2))/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b^2))))/(2*a^4))*(a^2*1i + b^2*6i)*1i)/(2*a^4))/((16*(
108*b^11 - 54*a*b^10 - 216*a^2*b^9 + 81*a^3*b^8 + 63*a^4*b^7 - 9*a^5*b^6 + 41*a^6*b^5 - 4*a^7*b^4 + 4*a^8*b^3)
)/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (((8*tan(c/2 + (d*x)/2)*(a^10 - 2*a^9*b - 72*a*b^9 + 72*b^10 - 120*a^
2*b^8 + 120*a^3*b^7 + 17*a^4*b^6 - 26*a^5*b^5 + 23*a^6*b^4 - 20*a^7*b^3 + 11*a^8*b^2))/(a^8*b + a^9 - a^6*b^3
- a^7*b^2) + ((a^2*1i + b^2*6i)*((8*(2*a^15 - 12*a^8*b^7 + 6*a^9*b^6 + 28*a^10*b^5 - 14*a^11*b^4 - 16*a^12*b^3
+ 6*a^13*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (4*tan(c/2 + (d*x)/2)*(a^2*1i + b^2*6i)*(8*a^13*b - 8*a
^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b^2))))/(2*a^4
))*(a^2*1i + b^2*6i))/(2*a^4) + (((8*tan(c/2 + (d*x)/2)*(a^10 - 2*a^9*b - 72*a*b^9 + 72*b^10 - 120*a^2*b^8 + 1
20*a^3*b^7 + 17*a^4*b^6 - 26*a^5*b^5 + 23*a^6*b^4 - 20*a^7*b^3 + 11*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2
) - ((a^2*1i + b^2*6i)*((8*(2*a^15 - 12*a^8*b^7 + 6*a^9*b^6 + 28*a^10*b^5 - 14*a^11*b^4 - 16*a^12*b^3 + 6*a^13
*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) + (4*tan(c/2 + (d*x)/2)*(a^2*1i + b^2*6i)*(8*a^13*b - 8*a^8*b^6 +
8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/(a^4*(a^8*b + a^9 - a^6*b^3 - a^7*b^2))))/(2*a^4))*(a^2*1
i + b^2*6i))/(2*a^4)))*(a^2*1i + b^2*6i)*1i)/(a^4*d) - ((tan(c/2 + (d*x)/2)*(3*a*b^3 - 3*a^3*b + a^4 + 6*b^4 -
5*a^2*b^2))/((a^3*b - a^4)*(a + b)) + (tan(c/2 + (d*x)/2)^5*(3*a^3*b - 3*a*b^3 + a^4 + 6*b^4 - 5*a^2*b^2))/((
a^3*b - a^4)*(a + b)) - (2*tan(c/2 + (d*x)/2)^3*(a^4 - 6*b^4 + 3*a^2*b^2))/(a*(a^2*b - a^3)*(a + b)))/(d*(a +
b + tan(c/2 + (d*x)/2)^2*(a + 3*b) - tan(c/2 + (d*x)/2)^4*(a - 3*b) - tan(c/2 + (d*x)/2)^6*(a - b))) + (b^3*at
an(((b^3*(4*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2)*((8*tan(c/2 + (d*x)/2)*(a^10 - 2*a^9*b - 72*a*b^9 + 72*b^
10 - 120*a^2*b^8 + 120*a^3*b^7 + 17*a^4*b^6 - 26*a^5*b^5 + 23*a^6*b^4 - 20*a^7*b^3 + 11*a^8*b^2))/(a^8*b + a^9
- a^6*b^3 - a^7*b^2) + (b^3*((8*(2*a^15 - 12*a^8*b^7 + 6*a^9*b^6 + 28*a^10*b^5 - 14*a^11*b^4 - 16*a^12*b^3 +
6*a^13*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (8*b^3*tan(c/2 + (d*x)/2)*(4*a^2 - 3*b^2)*((a + b)^3*(a -
b)^3)^(1/2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/((a^8*b + a^9 - a^6*b
^3 - a^7*b^2)*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*(4*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2))/(a^10 -
a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))*1i)/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2) + (b^3*(4*a^2 - 3*b^2)*((a + b)
^3*(a - b)^3)^(1/2)*((8*tan(c/2 + (d*x)/2)*(a^10 - 2*a^9*b - 72*a*b^9 + 72*b^10 - 120*a^2*b^8 + 120*a^3*b^7 +
17*a^4*b^6 - 26*a^5*b^5 + 23*a^6*b^4 - 20*a^7*b^3 + 11*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (b^3*((8*
(2*a^15 - 12*a^8*b^7 + 6*a^9*b^6 + 28*a^10*b^5 - 14*a^11*b^4 - 16*a^12*b^3 + 6*a^13*b^2))/(a^11*b + a^12 - a^9
*b^3 - a^10*b^2) + (8*b^3*tan(c/2 + (d*x)/2)*(4*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(8*a^13*b - 8*a^8*b^6
+ 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/((a^8*b + a^9 - a^6*b^3 - a^7*b^2)*(a^10 - a^4*b^6 + 3
*a^6*b^4 - 3*a^8*b^2)))*(4*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))
*1i)/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))/((16*(108*b^11 - 54*a*b^10 - 216*a^2*b^9 + 81*a^3*b^8 + 63*a^4*
b^7 - 9*a^5*b^6 + 41*a^6*b^5 - 4*a^7*b^4 + 4*a^8*b^3))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) - (b^3*(4*a^2 - 3*
b^2)*((a + b)^3*(a - b)^3)^(1/2)*((8*tan(c/2 + (d*x)/2)*(a^10 - 2*a^9*b - 72*a*b^9 + 72*b^10 - 120*a^2*b^8 + 1
20*a^3*b^7 + 17*a^4*b^6 - 26*a^5*b^5 + 23*a^6*b^4 - 20*a^7*b^3 + 11*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2
) + (b^3*((8*(2*a^15 - 12*a^8*b^7 + 6*a^9*b^6 + 28*a^10*b^5 - 14*a^11*b^4 - 16*a^12*b^3 + 6*a^13*b^2))/(a^11*b
+ a^12 - a^9*b^3 - a^10*b^2) - (8*b^3*tan(c/2 + (d*x)/2)*(4*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(8*a^13*
b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*b^4 - 16*a^11*b^3 - 8*a^12*b^2))/((a^8*b + a^9 - a^6*b^3 - a^7*b^2)*(a^10
- a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*(4*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2))/(a^10 - a^4*b^6 + 3*a^6*b^4
- 3*a^8*b^2)))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2) + (b^3*(4*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2)*((8
*tan(c/2 + (d*x)/2)*(a^10 - 2*a^9*b - 72*a*b^9 + 72*b^10 - 120*a^2*b^8 + 120*a^3*b^7 + 17*a^4*b^6 - 26*a^5*b^5
+ 23*a^6*b^4 - 20*a^7*b^3 + 11*a^8*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (b^3*((8*(2*a^15 - 12*a^8*b^7 +
6*a^9*b^6 + 28*a^10*b^5 - 14*a^11*b^4 - 16*a^12*b^3 + 6*a^13*b^2))/(a^11*b + a^12 - a^9*b^3 - a^10*b^2) + (8*b
^3*tan(c/2 + (d*x)/2)*(4*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(8*a^13*b - 8*a^8*b^6 + 8*a^9*b^5 + 16*a^10*
b^4 - 16*a^11*b^3 - 8*a^12*b^2))/((a^8*b + a^9 - a^6*b^3 - a^7*b^2)*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))
*(4*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))/(a^10 - a^4*b^6 + 3*a
^6*b^4 - 3*a^8*b^2)))*(4*a^2 - 3*b^2)*((a + b)^3*(a - b)^3)^(1/2)*2i)/(d*(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b
^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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