3.103 \(\int \frac{A+B \cos (c+d x)}{\sqrt{a+a \cos (c+d x)}} \, dx\)

Optimal. Leaf size=78 \[ \frac{\sqrt{2} (A-B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 B \sin (c+d x)}{d \sqrt{a \cos (c+d x)+a}} \]

[Out]

(Sqrt[2]*(A - B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d) + (2*B*Sin[c
+ d*x])/(d*Sqrt[a + a*Cos[c + d*x]])

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Rubi [A]  time = 0.0712777, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2751, 2649, 206} \[ \frac{\sqrt{2} (A-B) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x)}{\sqrt{2} \sqrt{a \cos (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 B \sin (c+d x)}{d \sqrt{a \cos (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2]*(A - B)*ArcTanh[(Sqrt[a]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Cos[c + d*x]])])/(Sqrt[a]*d) + (2*B*Sin[c
+ d*x])/(d*Sqrt[a + a*Cos[c + d*x]])

Rule 2751

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

Rule 2649

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

Rule 206

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

Rubi steps

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

Mathematica [A]  time = 0.0632729, size = 60, normalized size = 0.77 \[ \frac{2 \cos \left (\frac{1}{2} (c+d x)\right ) \left ((A-B) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 B \sin \left (\frac{1}{2} (c+d x)\right )\right )}{d \sqrt{a (\cos (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Cos[(c + d*x)/2]*((A - B)*ArcTanh[Sin[(c + d*x)/2]] + 2*B*Sin[(c + d*x)/2]))/(d*Sqrt[a*(1 + Cos[c + d*x])])

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Maple [B]  time = 1.862, size = 160, normalized size = 2.1 \begin{align*}{\frac{\sqrt{2}}{d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sqrt{a \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( A\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) a+2\,B\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}-B\ln \left ( 4\,{\frac{\sqrt{a}\sqrt{a \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}+a}{\cos \left ( 1/2\,dx+c/2 \right ) }} \right ) a \right ){a}^{-{\frac{3}{2}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{ \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

cos(1/2*d*x+1/2*c)*2^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)*(A*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*d*x+1
/2*c)^2)^(1/2)+a))*a+2*B*a^(1/2)*(a*sin(1/2*d*x+1/2*c)^2)^(1/2)-B*ln(4/cos(1/2*d*x+1/2*c)*(a^(1/2)*(a*sin(1/2*
d*x+1/2*c)^2)^(1/2)+a))*a)/a^(3/2)/sin(1/2*d*x+1/2*c)/(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/d

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

[Out]

Timed out

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Fricas [B]  time = 1.64399, size = 363, normalized size = 4.65 \begin{align*} \frac{4 \, \sqrt{a \cos \left (d x + c\right ) + a} B \sin \left (d x + c\right ) - \frac{\sqrt{2}{\left ({\left (A - B\right )} a \cos \left (d x + c\right ) +{\left (A - B\right )} a\right )} \log \left (-\frac{\cos \left (d x + c\right )^{2} + \frac{2 \, \sqrt{2} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{\sqrt{a}} - 2 \, \cos \left (d x + c\right ) - 3}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right )}{\sqrt{a}}}{2 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*sqrt(a*cos(d*x + c) + a)*B*sin(d*x + c) - sqrt(2)*((A - B)*a*cos(d*x + c) + (A - B)*a)*log(-(cos(d*x +
c)^2 + 2*sqrt(2)*sqrt(a*cos(d*x + c) + a)*sin(d*x + c)/sqrt(a) - 2*cos(d*x + c) - 3)/(cos(d*x + c)^2 + 2*cos(d
*x + c) + 1))/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 + B \cos{\left (c + d x \right )}}{\sqrt{a \left (\cos{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.6717, size = 119, normalized size = 1.53 \begin{align*} \frac{\frac{2 \, \sqrt{2} B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{\sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}} - \frac{{\left (\sqrt{2} A - \sqrt{2} B\right )} \log \left ({\left | -\sqrt{a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{a}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*sqrt(2)*B*tan(1/2*d*x + 1/2*c)/sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) - (sqrt(2)*A - sqrt(2)*B)*log(abs(-sqrt(a
)*tan(1/2*d*x + 1/2*c) + sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)))/sqrt(a))/d