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

Optimal. Leaf size=146 \[ -\frac {2 b x}{a^3}+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}+\frac {2 b^2 \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.33, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3847, 4104, 3919, 3831, 2659, 208} \[ \frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{a^2 d \left (a^2-b^2\right )}+\frac {2 b^2 \left (3 a^2-2 b^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^3 d (a-b)^{3/2} (a+b)^{3/2}}+\frac {b^2 \sin (c+d x)}{a d \left (a^2-b^2\right ) (a+b \sec (c+d x))}-\frac {2 b x}{a^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*x)/a^3 + (2*b^2*(3*a^2 - 2*b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^3*(a - b)^(3/2)*
(a + b)^(3/2)*d) + ((a^2 - 2*b^2)*Sin[c + d*x])/(a^2*(a^2 - b^2)*d) + (b^2*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 (c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\frac {b^2 \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 (-a^2+2 b^2+a b \sec (c+d x)-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^2-2 b^2\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\int \frac {-2 b \left (a^2-b^2\right )+a b^2 \sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=-\frac {2 b x}{a^3}+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (b^2 \left (3 a^2-2 b^2\right )\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=-\frac {2 b x}{a^3}+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (b \left (3 a^2-2 b^2\right )\right ) \int \frac {1}{1+\frac {a \cos (c+d x)}{b}} \, dx}{a^3 \left (a^2-b^2\right )}\\ &=-\frac {2 b x}{a^3}+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \sin (c+d x)}{a \left (a^2-b^2\right ) d (a+b \sec (c+d x))}+\frac {\left (2 b \left (3 a^2-2 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^3 \left (a^2-b^2\right ) d}\\ &=-\frac {2 b x}{a^3}+\frac {2 b^2 \left (3 a^2-2 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 (a-b)^{3/2} (a+b)^{3/2} d}+\frac {\left (a^2-2 b^2\right ) \sin (c+d x)}{a^2 \left (a^2-b^2\right ) d}+\frac {b^2 \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 = 0.76, size = 172, normalized size = 1.18 \[ \frac {\frac {2 a b \left (a^2-2 b^2\right ) \sin (c+d x)+\left (a^2-b^2\right ) \left (a^2 \sin (2 (c+d x))-4 b^2 (c+d x)\right )-4 a b \left (a^2-b^2\right ) (c+d x) \cos (c+d x)}{a \cos (c+d x)+b}+\frac {4 b^2 \left (2 b^2-3 a^2\right ) \tanh ^{-1}\left (\frac {(b-a) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}}{2 a^3 d (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.57, size = 565, normalized size = 3.87 \[ \left [-\frac {4 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d x \cos \left (d x + c\right ) + 4 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d x - {\left (3 \, a^{2} b^{3} - 2 \, b^{5} + {\left (3 \, a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {a^{2} - b^{2}} \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) - 2 \, {\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5} + {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{2 \, {\left ({\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} d\right )}}, -\frac {2 \, {\left (a^{5} b - 2 \, a^{3} b^{3} + a b^{5}\right )} d x \cos \left (d x + c\right ) + 2 \, {\left (a^{4} b^{2} - 2 \, a^{2} b^{4} + b^{6}\right )} d x - {\left (3 \, a^{2} b^{3} - 2 \, b^{5} + {\left (3 \, a^{3} b^{2} - 2 \, a b^{4}\right )} \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}} \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (a^{5} b - 3 \, a^{3} b^{3} + 2 \, a b^{5} + {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{{\left (a^{8} - 2 \, a^{6} b^{2} + a^{4} b^{4}\right )} d \cos \left (d x + c\right ) + {\left (a^{7} b - 2 \, a^{5} b^{3} + a^{3} b^{5}\right )} d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.39, size = 837, normalized size = 5.73 \[ -\frac {\frac {{\left (2 \, a^{7} b - 5 \, a^{6} b^{2} - 4 \, a^{5} b^{3} + 9 \, a^{4} b^{4} + 2 \, a^{3} b^{5} - 4 \, a^{2} b^{6} - 2 \, a^{2} b {\left | -a^{5} + a^{3} b^{2} \right |} - a b^{2} {\left | -a^{5} + a^{3} b^{2} \right |} + 2 \, b^{3} {\left | -a^{5} + a^{3} b^{2} \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {a^{4} b - a^{2} b^{3} + \sqrt {{\left (a^{5} + a^{4} b - a^{3} b^{2} - a^{2} b^{3}\right )} {\left (a^{5} - a^{4} b - a^{3} b^{2} + a^{2} b^{3}\right )} + {\left (a^{4} b - a^{2} b^{3}\right )}^{2}}}{a^{5} - a^{4} b - a^{3} b^{2} + a^{2} b^{3}}}}\right )\right )}}{a^{4} b {\left | -a^{5} + a^{3} b^{2} \right |} - a^{2} b^{3} {\left | -a^{5} + a^{3} b^{2} \right |} + {\left (a^{5} - a^{3} b^{2}\right )}^{2}} + \frac {{\left ({\left (2 \, a^{2} b + a b^{2} - 2 \, b^{3}\right )} \sqrt {-a^{2} + b^{2}} {\left | -a^{5} + a^{3} b^{2} \right |} {\left | -a + b \right |} + {\left (2 \, a^{7} b - 5 \, a^{6} b^{2} - 4 \, a^{5} b^{3} + 9 \, a^{4} b^{4} + 2 \, a^{3} b^{5} - 4 \, a^{2} b^{6}\right )} \sqrt {-a^{2} + b^{2}} {\left | -a + b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {a^{4} b - a^{2} b^{3} - \sqrt {{\left (a^{5} + a^{4} b - a^{3} b^{2} - a^{2} b^{3}\right )} {\left (a^{5} - a^{4} b - a^{3} b^{2} + a^{2} b^{3}\right )} + {\left (a^{4} b - a^{2} b^{3}\right )}^{2}}}{a^{5} - a^{4} b - a^{3} b^{2} + a^{2} b^{3}}}}\right )\right )}}{{\left (a^{5} - a^{3} b^{2}\right )}^{2} {\left (a^{2} - 2 \, a b + b^{2}\right )} - {\left (a^{6} b - 2 \, a^{5} b^{2} + 2 \, a^{3} b^{4} - a^{2} b^{5}\right )} {\left | -a^{5} + a^{3} b^{2} \right |}} - \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (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 )^{4} - 2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a - b\right )} {\left (a^{4} - a^{2} b^{2}\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((2*a^7*b - 5*a^6*b^2 - 4*a^5*b^3 + 9*a^4*b^4 + 2*a^3*b^5 - 4*a^2*b^6 - 2*a^2*b*abs(-a^5 + a^3*b^2) - a*b^2*a
bs(-a^5 + a^3*b^2) + 2*b^3*abs(-a^5 + a^3*b^2))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c
)/sqrt(-(a^4*b - a^2*b^3 + sqrt((a^5 + a^4*b - a^3*b^2 - a^2*b^3)*(a^5 - a^4*b - a^3*b^2 + a^2*b^3) + (a^4*b -
a^2*b^3)^2))/(a^5 - a^4*b - a^3*b^2 + a^2*b^3))))/(a^4*b*abs(-a^5 + a^3*b^2) - a^2*b^3*abs(-a^5 + a^3*b^2) +
(a^5 - a^3*b^2)^2) + ((2*a^2*b + a*b^2 - 2*b^3)*sqrt(-a^2 + b^2)*abs(-a^5 + a^3*b^2)*abs(-a + b) + (2*a^7*b -
5*a^6*b^2 - 4*a^5*b^3 + 9*a^4*b^4 + 2*a^3*b^5 - 4*a^2*b^6)*sqrt(-a^2 + b^2)*abs(-a + b))*(pi*floor(1/2*(d*x +
c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(a^4*b - a^2*b^3 - sqrt((a^5 + a^4*b - a^3*b^2 - a^2*b^3)*(a^
5 - a^4*b - a^3*b^2 + a^2*b^3) + (a^4*b - a^2*b^3)^2))/(a^5 - a^4*b - a^3*b^2 + a^2*b^3))))/((a^5 - a^3*b^2)^2
*(a^2 - 2*a*b + b^2) - (a^6*b - 2*a^5*b^2 + 2*a^3*b^4 - a^2*b^5)*abs(-a^5 + a^3*b^2)) - 2*(a^3*tan(1/2*d*x + 1
/2*c)^3 - a^2*b*tan(1/2*d*x + 1/2*c)^3 - a*b^2*tan(1/2*d*x + 1/2*c)^3 + 2*b^3*tan(1/2*d*x + 1/2*c)^3 - a^3*tan
(1/2*d*x + 1/2*c) - a^2*b*tan(1/2*d*x + 1/2*c) + a*b^2*tan(1/2*d*x + 1/2*c) + 2*b^3*tan(1/2*d*x + 1/2*c))/((a*
tan(1/2*d*x + 1/2*c)^4 - b*tan(1/2*d*x + 1/2*c)^4 - 2*b*tan(1/2*d*x + 1/2*c)^2 - a - b)*(a^4 - a^2*b^2)))/d

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maple [A]  time = 0.56, size = 242, normalized size = 1.66 \[ \frac {2 b^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (a^{2}-b^{2}\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b -a -b \right )}+\frac {6 b^{2} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d a \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}-\frac {4 b^{4} \arctanh \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (a -b \right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{d \,a^{3} \left (a -b \right ) \left (a +b \right ) \sqrt {\left (a -b \right ) \left (a +b \right )}}+\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}-\frac {4 b \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/d*b^3/a^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)+6/d*b^2/a/(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))-4/d*b^4/a^3/(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))+2/d/a^2*tan(1/2*d*x+1/2*c)/(1+tan(1/2*d*x+1
/2*c)^2)-4/d/a^3*b*arctan(tan(1/2*d*x+1/2*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)/(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 = 7.01, size = 3169, normalized size = 21.71 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*tan(c/2 + (d*x)/2)^3*(a*b^2 + a^2*b - a^3 - 2*b^3))/(a^2*(a + b)*(a - b)) - (2*tan(c/2 + (d*x)/2)*(a*b^2 -
a^2*b - a^3 + 2*b^3))/(a^2*(a + b)*(a - b)))/(d*(a + b - tan(c/2 + (d*x)/2)^4*(a - b) + 2*b*tan(c/2 + (d*x)/2
)^2)) - (4*b*atan(((2*b*((32*tan(c/2 + (d*x)/2)*(8*b^8 - 8*a*b^7 - 16*a^2*b^6 + 16*a^3*b^5 + 5*a^4*b^4 - 8*a^5
*b^3 + 4*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) + (b*((32*(2*a^11*b - 2*a^6*b^6 + a^7*b^5 + 5*a^8*b^4 - 3
*a^9*b^3 - 3*a^10*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (b*tan(c/2 + (d*x)/2)*(2*a^11*b - 2*a^6*b^6 + 2*a^
7*b^5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2)*64i)/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2)))*2i)/a^3))/a^3 + (2*b
*((32*tan(c/2 + (d*x)/2)*(8*b^8 - 8*a*b^7 - 16*a^2*b^6 + 16*a^3*b^5 + 5*a^4*b^4 - 8*a^5*b^3 + 4*a^6*b^2))/(a^6
*b + a^7 - a^4*b^3 - a^5*b^2) - (b*((32*(2*a^11*b - 2*a^6*b^6 + a^7*b^5 + 5*a^8*b^4 - 3*a^9*b^3 - 3*a^10*b^2))
/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (b*tan(c/2 + (d*x)/2)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a
^9*b^3 - 2*a^10*b^2)*64i)/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*b^2)))*2i)/a^3))/a^3)/((64*(8*b^8 - 4*a*b^7 - 20*a
^2*b^6 + 6*a^3*b^5 + 12*a^4*b^4))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (b*((32*tan(c/2 + (d*x)/2)*(8*b^8 - 8*a*
b^7 - 16*a^2*b^6 + 16*a^3*b^5 + 5*a^4*b^4 - 8*a^5*b^3 + 4*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) + (b*((3
2*(2*a^11*b - 2*a^6*b^6 + a^7*b^5 + 5*a^8*b^4 - 3*a^9*b^3 - 3*a^10*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (
b*tan(c/2 + (d*x)/2)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2)*64i)/(a^3*(a^6*b
+ a^7 - a^4*b^3 - a^5*b^2)))*2i)/a^3)*2i)/a^3 + (b*((32*tan(c/2 + (d*x)/2)*(8*b^8 - 8*a*b^7 - 16*a^2*b^6 + 16*
a^3*b^5 + 5*a^4*b^4 - 8*a^5*b^3 + 4*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) - (b*((32*(2*a^11*b - 2*a^6*b^
6 + a^7*b^5 + 5*a^8*b^4 - 3*a^9*b^3 - 3*a^10*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (b*tan(c/2 + (d*x)/2)*(
2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b^3 - 2*a^10*b^2)*64i)/(a^3*(a^6*b + a^7 - a^4*b^3 - a^5*
b^2)))*2i)/a^3)*2i)/a^3)))/(a^3*d) - (b^2*atan(((b^2*((32*tan(c/2 + (d*x)/2)*(8*b^8 - 8*a*b^7 - 16*a^2*b^6 + 1
6*a^3*b^5 + 5*a^4*b^4 - 8*a^5*b^3 + 4*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) + (b^2*(3*a^2 - 2*b^2)*((32*
(2*a^11*b - 2*a^6*b^6 + a^7*b^5 + 5*a^8*b^4 - 3*a^9*b^3 - 3*a^10*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (32
*b^2*tan(c/2 + (d*x)/2)*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4*a^8*
b^4 - 4*a^9*b^3 - 2*a^10*b^2))/((a^6*b + a^7 - a^4*b^3 - a^5*b^2)*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)))*((
a + b)^3*(a - b)^3)^(1/2))/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2))*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2
)*1i)/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2) + (b^2*((32*tan(c/2 + (d*x)/2)*(8*b^8 - 8*a*b^7 - 16*a^2*b^6 + 1
6*a^3*b^5 + 5*a^4*b^4 - 8*a^5*b^3 + 4*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) - (b^2*(3*a^2 - 2*b^2)*((32*
(2*a^11*b - 2*a^6*b^6 + a^7*b^5 + 5*a^8*b^4 - 3*a^9*b^3 - 3*a^10*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (32
*b^2*tan(c/2 + (d*x)/2)*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4*a^8*
b^4 - 4*a^9*b^3 - 2*a^10*b^2))/((a^6*b + a^7 - a^4*b^3 - a^5*b^2)*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)))*((
a + b)^3*(a - b)^3)^(1/2))/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2))*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2
)*1i)/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2))/((64*(8*b^8 - 4*a*b^7 - 20*a^2*b^6 + 6*a^3*b^5 + 12*a^4*b^4))/(
a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (b^2*((32*tan(c/2 + (d*x)/2)*(8*b^8 - 8*a*b^7 - 16*a^2*b^6 + 16*a^3*b^5 + 5
*a^4*b^4 - 8*a^5*b^3 + 4*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) + (b^2*(3*a^2 - 2*b^2)*((32*(2*a^11*b - 2
*a^6*b^6 + a^7*b^5 + 5*a^8*b^4 - 3*a^9*b^3 - 3*a^10*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) - (32*b^2*tan(c/2
+ (d*x)/2)*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b
^3 - 2*a^10*b^2))/((a^6*b + a^7 - a^4*b^3 - a^5*b^2)*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)))*((a + b)^3*(a -
b)^3)^(1/2))/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2))*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2))/(a^9 - a^3
*b^6 + 3*a^5*b^4 - 3*a^7*b^2) + (b^2*((32*tan(c/2 + (d*x)/2)*(8*b^8 - 8*a*b^7 - 16*a^2*b^6 + 16*a^3*b^5 + 5*a^
4*b^4 - 8*a^5*b^3 + 4*a^6*b^2))/(a^6*b + a^7 - a^4*b^3 - a^5*b^2) - (b^2*(3*a^2 - 2*b^2)*((32*(2*a^11*b - 2*a^
6*b^6 + a^7*b^5 + 5*a^8*b^4 - 3*a^9*b^3 - 3*a^10*b^2))/(a^8*b + a^9 - a^6*b^3 - a^7*b^2) + (32*b^2*tan(c/2 + (
d*x)/2)*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2)*(2*a^11*b - 2*a^6*b^6 + 2*a^7*b^5 + 4*a^8*b^4 - 4*a^9*b^3
- 2*a^10*b^2))/((a^6*b + a^7 - a^4*b^3 - a^5*b^2)*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2)))*((a + b)^3*(a - b)
^3)^(1/2))/(a^9 - a^3*b^6 + 3*a^5*b^4 - 3*a^7*b^2))*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2))/(a^9 - a^3*b^
6 + 3*a^5*b^4 - 3*a^7*b^2)))*(3*a^2 - 2*b^2)*((a + b)^3*(a - b)^3)^(1/2)*2i)/(d*(a^9 - a^3*b^6 + 3*a^5*b^4 - 3
*a^7*b^2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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