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

Optimal. Leaf size=84 \[ \frac{2 a \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 d \sqrt{a^2-b^2}}-\frac{\cos (c+d x)}{b d (a+b \sin (c+d x))}-\frac{x}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.110158, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2693, 2735, 2660, 618, 204} \[ \frac{2 a \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 d \sqrt{a^2-b^2}}-\frac{\cos (c+d x)}{b d (a+b \sin (c+d x))}-\frac{x}{b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2693

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Dist[(g^2*(p - 1))/(b*(m + 1)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1)*Sin[e + f*x], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a
^2 - b^2, 0] && LtQ[m, -1] && GtQ[p, 1] && IntegersQ[2*m, 2*p]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [B]  time = 5.30223, size = 494, normalized size = 5.88 \[ \frac{\cos (c+d x) \left (\sqrt{a+b} \left (\sqrt{a-b} \sqrt{1-\sin (c+d x)} \left (b \left (b^2-a^2\right ) \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}} \sqrt{\frac{b (\sin (c+d x)+1)}{b-a}}+2 a \left (a \sqrt{-b^2}+\sqrt{-b} b^{3/2} \sin (c+d x)\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\frac{b (\sin (c+d x)+1)}{b-a}}}{\sqrt{-b} \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}}}\right )\right )+2 \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}} \left (a \left (a \sqrt{-b} \sqrt{-b^2}+b^{5/2}\right )-b^2 \left (a \sqrt{b}+\sqrt{-b} \sqrt{-b^2}\right ) \sin (c+d x)\right ) \sinh ^{-1}\left (\frac{\sqrt{a-b} \sqrt{-\frac{b (\sin (c+d x)+1)}{a-b}}}{\sqrt{2} \sqrt{b}}\right )\right )-2 a b (a-b) \sqrt{1-\sin (c+d x)} (a+b \sin (c+d x)) \tanh ^{-1}\left (\frac{\sqrt{a-b} \sqrt{-\frac{b (\sin (c+d x)+1)}{a-b}}}{\sqrt{a+b} \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}}}\right )\right )}{b^2 d (a-b)^{3/2} (a+b)^{3/2} \sqrt{1-\sin (c+d x)} \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}} \sqrt{-\frac{b (\sin (c+d x)+1)}{a-b}} (a+b \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[c + d*x]*(-2*a*(a - b)*b*ArcTanh[(Sqrt[a - b]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))])/(Sqrt[a + b]*Sqrt[
-((b*(-1 + Sin[c + d*x]))/(a + b))])]*Sqrt[1 - Sin[c + d*x]]*(a + b*Sin[c + d*x]) + Sqrt[a + b]*(2*ArcSinh[(Sq
rt[a - b]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))])/(Sqrt[2]*Sqrt[b])]*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]
*(a*(b^(5/2) + a*Sqrt[-b]*Sqrt[-b^2]) - b^2*(a*Sqrt[b] + Sqrt[-b]*Sqrt[-b^2])*Sin[c + d*x]) + Sqrt[a - b]*Sqrt
[1 - Sin[c + d*x]]*(b*(-a^2 + b^2)*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]*Sqrt[(b*(1 + Sin[c + d*x]))/(-a +
b)] + 2*a*ArcTan[(Sqrt[b]*Sqrt[(b*(1 + Sin[c + d*x]))/(-a + b)])/(Sqrt[-b]*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a +
b))])]*(a*Sqrt[-b^2] + Sqrt[-b]*b^(3/2)*Sin[c + d*x])))))/((a - b)^(3/2)*b^2*(a + b)^(3/2)*d*Sqrt[1 - Sin[c +
d*x]]*Sqrt[-((b*(-1 + Sin[c + d*x]))/(a + b))]*Sqrt[-((b*(1 + Sin[c + d*x]))/(a - b))]*(a + b*Sin[c + d*x]))

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Maple [A]  time = 0.072, size = 153, normalized size = 1.8 \begin{align*} -2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{b}^{2}}}-2\,{\frac{\tan \left ( 1/2\,dx+c/2 \right ) }{d \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) a}}-2\,{\frac{1}{bd \left ( \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}a+2\,\tan \left ( 1/2\,dx+c/2 \right ) b+a \right ) }}+2\,{\frac{a}{d{b}^{2}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/d/b^2*arctan(tan(1/2*d*x+1/2*c))-2/d/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)/a*tan(1/2*d*x+1/2*c)
-2/d/b/(tan(1/2*d*x+1/2*c)^2*a+2*tan(1/2*d*x+1/2*c)*b+a)+2/d/b^2*a/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x
+1/2*c)+2*b)/(a^2-b^2)^(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)^2/(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.76553, size = 852, normalized size = 10.14 \begin{align*} \left [-\frac{2 \,{\left (a^{2} b - b^{3}\right )} d x \sin \left (d x + c\right ) + 2 \,{\left (a^{3} - a b^{2}\right )} d x +{\left (a b \sin \left (d x + c\right ) + a^{2}\right )} \sqrt{-a^{2} + b^{2}} \log \left (\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \,{\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) + 2 \,{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{2 \,{\left ({\left (a^{2} b^{3} - b^{5}\right )} d \sin \left (d x + c\right ) +{\left (a^{3} b^{2} - a b^{4}\right )} d\right )}}, -\frac{{\left (a^{2} b - b^{3}\right )} d x \sin \left (d x + c\right ) +{\left (a^{3} - a b^{2}\right )} d x +{\left (a b \sin \left (d x + c\right ) + a^{2}\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (d x + c\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) +{\left (a^{2} b - b^{3}\right )} \cos \left (d x + c\right )}{{\left (a^{2} b^{3} - b^{5}\right )} d \sin \left (d x + c\right ) +{\left (a^{3} b^{2} - a b^{4}\right )} d}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.10802, size = 170, normalized size = 2.02 \begin{align*} \frac{\frac{2 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )} a}{\sqrt{a^{2} - b^{2}} b^{2}} - \frac{d x + c}{b^{2}} - \frac{2 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )} a b}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))*a/(sqrt(a^
2 - b^2)*b^2) - (d*x + c)/b^2 - 2*(b*tan(1/2*d*x + 1/2*c) + a)/((a*tan(1/2*d*x + 1/2*c)^2 + 2*b*tan(1/2*d*x +
1/2*c) + a)*a*b))/d