3.200 \(\int \frac{A+C \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} \sqrt{a+a \cos (c+d x)}} \, dx\)

Optimal. Leaf size=133 \[ \frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{C \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{C \sin (c+d x) \sqrt{\cos (c+d x)}}{d \sqrt{a \cos (c+d x)+a}} \]

[Out]

-((C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(Sqrt[a]*d)) + (Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*
Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d) + (C*Sqrt[Cos[c + d*x]]*Sin[
c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.386999, antiderivative size = 133, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {3046, 2982, 2782, 205, 2774, 216} \[ \frac{\sqrt{2} (A+C) \tan ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{\cos (c+d x)} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{C \sin ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{C \sin (c+d x) \sqrt{\cos (c+d x)}}{d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]),x]

[Out]

-((C*ArcSin[(Sqrt[a]*Sin[c + d*x])/Sqrt[a + a*Cos[c + d*x]]])/(Sqrt[a]*d)) + (Sqrt[2]*(A + C)*ArcTan[(Sqrt[a]*
Sin[c + d*x])/(Sqrt[2]*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d) + (C*Sqrt[Cos[c + d*x]]*Sin[
c + d*x])/(d*Sqrt[a + a*Cos[c + d*x]])

Rule 3046

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*
sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n
+ 1))/(d*f*(m + n + 2)), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Simp
[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b
, c, d, e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^(-
1)] && NeQ[m + n + 2, 0]

Rule 2982

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin
[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[(A*b - a*B)/b, Int[1/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*
x]]), x], x] + Dist[B/b, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] /; FreeQ[{a, b, c, d, e
, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2782

Int[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> D
ist[(-2*a)/f, Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c
+ d*Sin[e + f*x]])], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 -
d^2, 0]

Rule 205

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

Rule 2774

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[-2/f, Su
bst[Int[1/Sqrt[1 - x^2/a], x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, d, e, f}, x]
&& EqQ[a^2 - b^2, 0] && EqQ[d, a/b]

Rule 216

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSin[(Rt[-b, 2]*x)/Sqrt[a]]/Rt[-b, 2], x] /; FreeQ[{a, b}
, x] && GtQ[a, 0] && NegQ[b]

Rubi steps

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

Mathematica [A]  time = 0.230397, size = 104, normalized size = 0.78 \[ \frac{\cos \left (\frac{1}{2} (c+d x)\right ) \left (2 (A+C) \tan ^{-1}\left (\frac{\sin \left (\frac{1}{2} (c+d x)\right )}{\sqrt{\cos (c+d x)}}\right )-\sqrt{2} C \sin ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )+2 C \sin \left (\frac{1}{2} (c+d x)\right ) \sqrt{\cos (c+d x)}\right )}{d \sqrt{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Cos[c + d*x]^2)/(Sqrt[Cos[c + d*x]]*Sqrt[a + a*Cos[c + d*x]]),x]

[Out]

(Cos[(c + d*x)/2]*(-(Sqrt[2]*C*ArcSin[Sqrt[2]*Sin[(c + d*x)/2]]) + 2*(A + C)*ArcTan[Sin[(c + d*x)/2]/Sqrt[Cos[
c + d*x]]] + 2*C*Sqrt[Cos[c + d*x]]*Sin[(c + d*x)/2]))/(d*Sqrt[a*(1 + Cos[c + d*x])])

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Maple [A]  time = 0.117, size = 178, normalized size = 1.3 \begin{align*} -{\frac{ \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{da \left ( \sin \left ( dx+c \right ) \right ) ^{4}} \left ( A\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sqrt{2}-C\sin \left ( dx+c \right ) \sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}+C\arcsin \left ({\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }} \right ) \sqrt{2}+C\arctan \left ({\frac{\sin \left ( dx+c \right ) }{\cos \left ( dx+c \right ) }\sqrt{{\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }}}} \right ) \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}\sqrt{a \left ( 1+\cos \left ( dx+c \right ) \right ) } \left ({\frac{\cos \left ( dx+c \right ) }{1+\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2),x)

[Out]

-1/d*(-1+cos(d*x+c))^2*(A*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)-C*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)+C*arcsin((-1+cos(d*x+c))/sin(d*x+c))*2^(1/2)+C*arctan(sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)/cos(d
*x+c)))*cos(d*x+c)^(3/2)*(a*(1+cos(d*x+c)))^(1/2)/a/sin(d*x+c)^4/(cos(d*x+c)/(1+cos(d*x+c)))^(3/2)

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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((A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 12.6548, size = 447, normalized size = 3.36 \begin{align*} \frac{\sqrt{a \cos \left (d x + c\right ) + a} C \sqrt{\cos \left (d x + c\right )} \sin \left (d x + c\right ) +{\left (C \cos \left (d x + c\right ) + C\right )} \sqrt{a} \arctan \left (\frac{\sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right ) - \frac{\sqrt{2}{\left ({\left (A + C\right )} a \cos \left (d x + c\right ) +{\left (A + C\right )} a\right )} \arctan \left (\frac{\sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{\sqrt{a} \sin \left (d x + c\right )}\right )}{\sqrt{a}}}{a d \cos \left (d x + c\right ) + a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

(sqrt(a*cos(d*x + c) + a)*C*sqrt(cos(d*x + c))*sin(d*x + c) + (C*cos(d*x + c) + C)*sqrt(a)*arctan(sqrt(a*cos(d
*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c))) - sqrt(2)*((A + C)*a*cos(d*x + c) + (A + C)*a)*arctan(
sqrt(2)*sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c) + a*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + C \cos ^{2}{\left (c + d x \right )}}{\sqrt{a \left (\cos{\left (c + d x \right )} + 1\right )} \sqrt{\cos{\left (c + d x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)**2)/cos(d*x+c)**(1/2)/(a+a*cos(d*x+c))**(1/2),x)

[Out]

Integral((A + C*cos(c + d*x)**2)/(sqrt(a*(cos(c + d*x) + 1))*sqrt(cos(c + d*x))), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + A}{\sqrt{a \cos \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)/(sqrt(a*cos(d*x + c) + a)*sqrt(cos(d*x + c))), x)