3.310 \(\int \frac{A+B \sin (e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx\)

Optimal. Leaf size=79 \[ -\frac{\sqrt{2} (A-B) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} f}-\frac{2 B \cos (e+f x)}{f \sqrt{a \sin (e+f x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.0703238, antiderivative size = 79, 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} \[ -\frac{\sqrt{2} (A-B) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} f}-\frac{2 B \cos (e+f x)}{f \sqrt{a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sin[e + f*x])/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

-((Sqrt[2]*(A - B)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*f)) - (2*B*Cos
[e + f*x])/(f*Sqrt[a + a*Sin[e + f*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 \sin (e+f x)}{\sqrt{a+a \sin (e+f x)}} \, dx &=-\frac{2 B \cos (e+f x)}{f \sqrt{a+a \sin (e+f x)}}+(A-B) \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx\\ &=-\frac{2 B \cos (e+f x)}{f \sqrt{a+a \sin (e+f x)}}-\frac{(2 (A-B)) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{f}\\ &=-\frac{\sqrt{2} (A-B) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{\sqrt{a} f}-\frac{2 B \cos (e+f x)}{f \sqrt{a+a \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.219503, size = 106, normalized size = 1.34 \[ \frac{2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (B \left (\sin \left (\frac{1}{2} (e+f x)\right )-\cos \left (\frac{1}{2} (e+f x)\right )\right )+(1+i) (-1)^{3/4} (A-B) \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )\right )}{f \sqrt{a (\sin (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sin[e + f*x])/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

(2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((1 + I)*(-1)^(3/4)*(A - B)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[
(e + f*x)/4])] + B*(-Cos[(e + f*x)/2] + Sin[(e + f*x)/2])))/(f*Sqrt[a*(1 + Sin[e + f*x])])

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Maple [A]  time = 1.024, size = 128, normalized size = 1.6 \begin{align*} -{\frac{1+\sin \left ( fx+e \right ) }{af\cos \left ( fx+e \right ) }\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) } \left ( \sqrt{a}\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ) A-\sqrt{a}\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ) B+2\,\sqrt{a-a\sin \left ( fx+e \right ) }B \right ){\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2),x)

[Out]

-(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(a^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2)
)*A-a^(1/2)*2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*B+2*(a-a*sin(f*x+e))^(1/2)*B)/a/cos(f*
x+e)/(a+a*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sin \left (f x + e\right ) + A}{\sqrt{a \sin \left (f x + e\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sin(f*x + e) + A)/sqrt(a*sin(f*x + e) + a), x)

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Fricas [B]  time = 1.94569, size = 572, normalized size = 7.24 \begin{align*} -\frac{\frac{\sqrt{2}{\left ({\left (A - B\right )} a \cos \left (f x + e\right ) +{\left (A - B\right )} a \sin \left (f x + e\right ) +{\left (A - B\right )} a\right )} \log \left (-\frac{\cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac{2 \, \sqrt{2} \sqrt{a \sin \left (f x + e\right ) + a}{\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt{a}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} -{\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt{a}} + 4 \,{\left (B \cos \left (f x + e\right ) - B \sin \left (f x + e\right ) + B\right )} \sqrt{a \sin \left (f x + e\right ) + a}}{2 \,{\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/2*(sqrt(2)*((A - B)*a*cos(f*x + e) + (A - B)*a*sin(f*x + e) + (A - B)*a)*log(-(cos(f*x + e)^2 - (cos(f*x +
e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*(cos(f*x + e) - sin(f*x + e) + 1)/sqrt(a) + 3*cos(f*
x + e) + 2)/(cos(f*x + e)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2))/sqrt(a) + 4*(B*cos(f*x + e)
- B*sin(f*x + e) + B)*sqrt(a*sin(f*x + e) + a))/(a*f*cos(f*x + e) + a*f*sin(f*x + e) + a*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \sin{\left (e + f x \right )}}{\sqrt{a \left (\sin{\left (e + f x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))**(1/2),x)

[Out]

Integral((A + B*sin(e + f*x))/sqrt(a*(sin(e + f*x) + 1)), x)

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Giac [B]  time = 1.47835, size = 297, normalized size = 3.76 \begin{align*} \frac{2 \,{\left (\frac{\sqrt{2}{\left (A - B\right )} \arctan \left (-\frac{\sqrt{2}{\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a} + \sqrt{a}\right )}}{2 \, \sqrt{-a}}\right )}{\sqrt{-a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )} + \frac{\frac{B \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )} - \frac{B}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}}{\sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a}} - \frac{{\left (\sqrt{2} A a \arctan \left (\frac{\sqrt{a}}{\sqrt{-a}}\right ) - \sqrt{2} B a \arctan \left (\frac{\sqrt{a}}{\sqrt{-a}}\right ) - \sqrt{2} B \sqrt{-a} \sqrt{a}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}{\sqrt{-a} a}\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

2*(sqrt(2)*(A - B)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a) + sq
rt(a))/sqrt(-a))/(sqrt(-a)*sgn(tan(1/2*f*x + 1/2*e) + 1)) + (B*tan(1/2*f*x + 1/2*e)/sgn(tan(1/2*f*x + 1/2*e) +
1) - B/sgn(tan(1/2*f*x + 1/2*e) + 1))/sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a) - (sqrt(2)*A*a*arctan(sqrt(a)/sqrt(-
a)) - sqrt(2)*B*a*arctan(sqrt(a)/sqrt(-a)) - sqrt(2)*B*sqrt(-a)*sqrt(a))*sgn(tan(1/2*f*x + 1/2*e) + 1)/(sqrt(-
a)*a))/f