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

Optimal. Leaf size=219 \[ -\frac{b^3 \sin (c+d x)}{d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac{2 b^2 \left (3 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d (a-b)^{5/2} (a+b)^{5/2}}+\frac{2 b^4 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}-\frac{\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)} \]

[Out]

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

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Rubi [A]  time = 0.384874, antiderivative size = 219, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 7, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.269, Rules used = {4397, 2897, 2648, 2659, 208, 2664, 12} \[ -\frac{b^3 \sin (c+d x)}{d \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac{2 b^2 \left (3 a^2-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d (a-b)^{5/2} (a+b)^{5/2}}+\frac{2 b^4 \tanh ^{-1}\left (\frac{\sqrt{a-b} \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a+b}}\right )}{a d (a-b)^{5/2} (a+b)^{5/2}}-\frac{\sin (c+d x)}{2 d (a+b)^2 (1-\cos (c+d x))}-\frac{\sin (c+d x)}{2 d (a-b)^2 (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]]

Rubi steps

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

Mathematica [A]  time = 1.2941, size = 131, normalized size = 0.6 \[ \frac{\frac{\csc (c+d x) \left (-2 a \left (a^2-b^2\right ) \cos (c+d x)+\left (2 a^2 b+b^3\right ) \cos (2 (c+d x))-3 b^3\right )}{a \cos (c+d x)+b}-\frac{12 a b^2 \tanh ^{-1}\left (\frac{(b-a) \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}}{2 d (a-b)^2 (a+b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-12*a*b^2*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + ((-3*b^3 - 2*a*(a^2 - b^2)
*Cos[c + d*x] + (2*a^2*b + b^3)*Cos[2*(c + d*x)])*Csc[c + d*x])/(b + a*Cos[c + d*x]))/(2*(a - b)^2*(a + b)^2*d
)

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Maple [A]  time = 0.145, size = 155, normalized size = 0.7 \begin{align*}{\frac{1}{d} \left ( -{\frac{1}{2\,{a}^{2}-4\,ab+2\,{b}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-2\,{\frac{{b}^{2}}{ \left ( a-b \right ) ^{2} \left ( a+b \right ) ^{2}} \left ( -{\frac{\tan \left ( 1/2\,dx+c/2 \right ) b}{ \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}}-3\,{\frac{a}{\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 ) } \right ) }-{\frac{1}{2\, \left ( a+b \right ) ^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(-1/2/(a^2-2*a*b+b^2)*tan(1/2*d*x+1/2*c)-2*b^2/(a-b)^2/(a+b)^2*(-tan(1/2*d*x+1/2*c)*b/(tan(1/2*d*x+1/2*c)^
2*a-tan(1/2*d*x+1/2*c)^2*b-a-b)-3*a/((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/(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)/(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 = 0.595528, size = 1134, normalized size = 5.18 \begin{align*} \left [-\frac{2 \, a^{4} b + 2 \, a^{2} b^{3} - 4 \, b^{5} - 3 \,{\left (a^{2} b^{2} \cos \left (d x + c\right ) + 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 ) \sin \left (d x + c\right ) - 2 \,{\left (2 \, a^{4} b - a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} + 2 \,{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )}{2 \,{\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )} \sin \left (d x + c\right )}, -\frac{a^{4} b + a^{2} b^{3} - 2 \, b^{5} - 3 \,{\left (a^{2} b^{2} \cos \left (d x + c\right ) + 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 ) \sin \left (d x + c\right ) -{\left (2 \, a^{4} b - a^{2} b^{3} - b^{5}\right )} \cos \left (d x + c\right )^{2} +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )}{{\left ({\left (a^{7} - 3 \, a^{5} b^{2} + 3 \, a^{3} b^{4} - a b^{6}\right )} d \cos \left (d x + c\right ) +{\left (a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}\right )} d\right )} \sin \left (d x + c\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Giac [A]  time = 1.21165, size = 381, normalized size = 1.74 \begin{align*} -\frac{\frac{12 \,{\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 )} a b^{2}}{{\left (a^{4} - 2 \, a^{2} b^{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^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a^{2} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 3 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 5 \, b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{3} + a^{2} b + a b^{2} - b^{3}}{{\left (a^{4} - 2 \, a^{2} b^{2} + 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 )}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(12*(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*b^2/((a^4 - 2*a^2*b^2 + b^4)*sqrt(-a^2 + b^2)) + tan(1/2*d*x + 1/2*c)/(a^2 - 2*a*b
+ b^2) + (a^3*tan(1/2*d*x + 1/2*c)^2 - 3*a^2*b*tan(1/2*d*x + 1/2*c)^2 + 3*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 5*b^
3*tan(1/2*d*x + 1/2*c)^2 - a^3 + a^2*b + a*b^2 - b^3)/((a^4 - 2*a^2*b^2 + b^4)*(a*tan(1/2*d*x + 1/2*c)^3 - b*t
an(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))))/d