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

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

[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]))

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Rubi [A]  time = 0.32812, antiderivative size = 146, normalized size of antiderivative = 1., 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 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 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 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 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 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 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]

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*}

Mathematica [A]  time = 0.727932, 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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Maple [A]  time = 0.083, size = 242, normalized size = 1.7 \begin{align*} 2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-4\,{\frac{b\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}+2\,{\frac{{b}^{3}\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{2} \left ({a}^{2}-{b}^{2} \right ) \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a- \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b-a-b \right ) }}+6\,{\frac{{b}^{2}}{da \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) }-4\,{\frac{{b}^{4}}{d{a}^{3} \left ( a+b \right ) \left ( a-b \right ) \sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}{\it Artanh} \left ({\frac{ \left ( a-b \right ) \tan \left ( 1/2\,dx+c/2 \right ) }{\sqrt{ \left ( a+b \right ) \left ( a-b \right ) }}} \right ) } \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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

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Fricas [A]  time = 1.93909, size = 1237, normalized size = 8.47 \begin{align*} \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 ] \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{2}}\, dx \end{align*}

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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Giac [B]  time = 1.29059, size = 414, normalized size = 2.84 \begin{align*} \frac{2 \,{\left (\frac{{\left (3 \, a^{2} b^{2} - 2 \, b^{4}\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^{5} - a^{3} b^{2}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{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 )}{{\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 )}} - \frac{{\left (d x + c\right )} b}{a^{3}}\right )}}{d} \end{align*}

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