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

Optimal. Leaf size=70 \[ -\frac{2 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 d}+\frac{a x}{b^2}+\frac{\cos (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.115843, antiderivative size = 70, 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 = {2695, 2735, 2660, 618, 204} \[ -\frac{2 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{b^2 d}+\frac{a x}{b^2}+\frac{\cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2695

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 + p)), x] + Dist[(g^2*(p - 1))/(b*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && 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)} \, dx &=\frac{\cos (c+d x)}{b d}+\frac{\int \frac{b+a \sin (c+d x)}{a+b \sin (c+d x)} \, dx}{b}\\ &=\frac{a x}{b^2}+\frac{\cos (c+d x)}{b d}-\frac{\left (a^2-b^2\right ) \int \frac{1}{a+b \sin (c+d x)} \, dx}{b^2}\\ &=\frac{a x}{b^2}+\frac{\cos (c+d x)}{b d}-\frac{\left (2 \left (a^2-b^2\right )\right ) \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{a x}{b^2}+\frac{\cos (c+d x)}{b d}+\frac{\left (4 \left (a^2-b^2\right )\right ) \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{a x}{b^2}-\frac{2 \sqrt{a^2-b^2} \tan ^{-1}\left (\frac{b+a \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{b^2 d}+\frac{\cos (c+d x)}{b d}\\ \end{align*}

Mathematica [B]  time = 2.12276, size = 398, normalized size = 5.69 \[ \frac{b \cos (c+d x) \left (\sqrt{a+b} \left (2 \sqrt{-b^2} \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}} \sinh ^{-1}\left (\frac{\sqrt{a-b} \sqrt{-\frac{b (\sin (c+d x)+1)}{a-b}}}{\sqrt{2} \sqrt{b}}\right )+\sqrt{a-b} \sqrt{1-\sin (c+d x)} \left (\sqrt{-b} \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}} \sqrt{\frac{b (\sin (c+d x)+1)}{b-a}}+2 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{\frac{b (\sin (c+d x)+1)}{b-a}}}{\sqrt{-b} \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}}}\right )\right )\right )-2 \sqrt{-b} (b-a) \sqrt{1-\sin (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a-b} \sqrt{-\frac{b (\sin (c+d x)+1)}{a-b}}}{\sqrt{a+b} \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}}}\right )\right )}{(-b)^{5/2} d \sqrt{a-b} \sqrt{a+b} \sqrt{1-\sin (c+d x)} \sqrt{-\frac{b (\sin (c+d x)-1)}{a+b}} \sqrt{-\frac{b (\sin (c+d x)+1)}{a-b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.001, size = 142, normalized size = 2. \begin{align*} 2\,{\frac{1}{bd \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}+2\,{\frac{a\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{{b}^{2}d}}-2\,{\frac{{a}^{2}}{{b}^{2}d\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 ) }+2\,{\frac{1}{d\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)),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85215, size = 498, normalized size = 7.11 \begin{align*} \left [\frac{2 \, a d x + 2 \, b \cos \left (d x + c\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 \, b^{2} d}, \frac{a d x + b \cos \left (d x + c\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 )}{b^{2} 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)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.14904, size = 128, normalized size = 1.83 \begin{align*} \frac{\frac{{\left (d x + c\right )} a}{b^{2}} - \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 )} \sqrt{a^{2} - b^{2}}}{b^{2}} + \frac{2}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} 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)),x, algorithm="giac")

[Out]

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