3.258 \(\int \frac{\cos ^2(c+d x)}{(a \sin (c+d x)+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=227 \[ \frac{b^4 \sin (c+d x)}{a d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac{2 b^5 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{4 b^3 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{x}{a^2}-\frac{\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)} \]

[Out]

-(x/a^2) - (2*b^5*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(5/2)*(a + b)^(5/2)*d) - (
4*b^3*(2*a^2 - b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(5/2)*(a + b)^(5/2)*d) -
Sin[c + d*x]/(2*(a + b)^2*d*(1 - Cos[c + d*x])) + Sin[c + d*x]/(2*(a - b)^2*d*(1 + Cos[c + d*x])) + (b^4*Sin[
c + d*x])/(a*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x]))

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Rubi [A]  time = 0.491171, antiderivative size = 227, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {4397, 2897, 2648, 2664, 12, 2659, 208} \[ \frac{b^4 \sin (c+d x)}{a d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}-\frac{2 b^5 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{4 b^3 \left (2 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a^2 d (a-b)^{5/2} (a+b)^{5/2}}-\frac{x}{a^2}-\frac{\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}+\frac{\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a^2) - (2*b^5*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(5/2)*(a + b)^(5/2)*d) - (
4*b^3*(2*a^2 - b^2)*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt[a + b]])/(a^2*(a - b)^(5/2)*(a + b)^(5/2)*d) -
Sin[c + d*x]/(2*(a + b)^2*d*(1 - Cos[c + d*x])) + Sin[c + d*x]/(2*(a - b)^2*d*(1 + Cos[c + d*x])) + (b^4*Sin[
c + d*x])/(a*(a^2 - b^2)^2*d*(b + a*Cos[c + d*x]))

Rule 4397

Int[u_, x_Symbol] :> Int[TrigSimplify[u], x] /; TrigSimplifyQ[u]

Rule 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2664

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n +
1))/(d*(n + 1)*(a^2 - b^2)), x] + Dist[1/((n + 1)*(a^2 - b^2)), Int[(a + b*Sin[c + d*x])^(n + 1)*Simp[a*(n + 1
) - b*(n + 2)*Sin[c + d*x], x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && Integer
Q[2*n]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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

Mathematica [A]  time = 2.16558, size = 151, normalized size = 0.67 \[ \frac{-\frac{4 b^3 \left (b^2-4 a^2\right ) \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{5/2}}-\frac{2 (c+d x)}{a^2}+\frac{2 b^4 \sin (c+d x)}{a (a-b)^2 (a+b)^2 (a \cos (c+d x)+b)}+\frac{\tan \left (\frac{1}{2} (c+d x)\right )}{(a-b)^2}-\frac{\cot \left (\frac{1}{2} (c+d x)\right )}{(a+b)^2}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*(c + d*x))/a^2 - (4*b^3*(-4*a^2 + b^2)*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 -
b^2)^(5/2)) - Cot[(c + d*x)/2]/(a + b)^2 + (2*b^4*Sin[c + d*x])/(a*(a - b)^2*(a + b)^2*(b + a*Cos[c + d*x])) +
Tan[(c + d*x)/2]/(a - b)^2)/(2*d)

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Maple [A]  time = 0.171, size = 255, normalized size = 1.1 \begin{align*}{\frac{1}{2\,d \left ({a}^{2}-2\,ab+{b}^{2} \right ) }\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{{a}^{2}d}}-2\,{\frac{{b}^{4}\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}a \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 ) }}-8\,{\frac{{b}^{3}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}\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 ) }+2\,{\frac{{b}^{5}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}{a}^{2}\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 ) }-{\frac{1}{2\,d \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/d/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)-2/d/a^2*arctan(tan(1/2*d*x+1/2*c))-2/d*b^4/(a+b)^2/(a-b)^2/a*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)-8/d*b^3/(a+b)^2/(a-b)^2/((a+b)*(a-b))^(1/2)*arc
tanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))+2/d*b^5/(a+b)^2/(a-b)^2/a^2/((a+b)*(a-b))^(1/2)*arctanh((a-
b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2))-1/2/d/(a+b)^2/tan(1/2*d*x+1/2*c)

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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)^2/(a*sin(d*x+c)+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.634744, size = 1507, normalized size = 6.64 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(4*a^5*b^2 - 2*a^3*b^4 - 2*a*b^6 - (4*a^2*b^4 - b^6 + (4*a^3*b^3 - a*b^5)*cos(d*x + c))*sqrt(a^2 - b^2)*l
og((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))*sin(d*x + c) - 2*(a^7 - a*b^6)*cos(d*x + c)^2 +
2*(a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c) - 2*((a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*x*cos(d*x + c) + (a^
6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*x)*sin(d*x + c))/(((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*cos(d*x + c)
+ (a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*d)*sin(d*x + c)), (2*a^5*b^2 - a^3*b^4 - a*b^6 - (4*a^2*b^4 - b^6
+ (4*a^3*b^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)))*sin(d*x + c) - (a^7 - a*b^6)*cos(d*x + c)^2 + (a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(d*x + c) - (
(a^7 - 3*a^5*b^2 + 3*a^3*b^4 - a*b^6)*d*x*cos(d*x + c) + (a^6*b - 3*a^4*b^3 + 3*a^2*b^5 - b^7)*d*x)*sin(d*x +
c))/(((a^9 - 3*a^7*b^2 + 3*a^5*b^4 - a^3*b^6)*d*cos(d*x + c) + (a^8*b - 3*a^6*b^3 + 3*a^4*b^5 - a^2*b^7)*d)*si
n(d*x + c))]

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

[Out]

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

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Giac [A]  time = 1.22763, size = 447, normalized size = 1.97 \begin{align*} \frac{\frac{4 \,{\left (4 \, a^{2} b^{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} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} \sqrt{-a^{2} + b^{2}}} + \frac{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2} - 2 \, a b + b^{2}} - \frac{a^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a^{3} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a^{2} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 4 \, b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{4} + a^{3} b + a^{2} b^{2} - a b^{3}}{{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 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 ) - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}} - \frac{2 \,{\left (d x + c\right )}}{a^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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