3.546 \(\int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx\)

Optimal. Leaf size=47 \[ -\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} f} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2649, 206} \[ -\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} f} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

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

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])

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]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a+a \sin (e+f x)}} \, dx &=-\frac {2 \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} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} f}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 73, normalized size = 1.55 \[ \frac {(2+2 i) (-1)^{3/4} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (e+f x)\right )-1\right )\right )}{f \sqrt {a (\sin (e+f x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + a*Sin[e + f*x]],x]

[Out]

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

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fricas [A]  time = 0.48, size = 167, normalized size = 3.55 \[ \left [\frac {\sqrt {2} \log \left (-\frac {\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) - \frac {2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{2 \, \sqrt {a} f}, \frac {\sqrt {2} \sqrt {-\frac {1}{a}} \arctan \left (\frac {\sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-\frac {1}{a}}}{\cos \left (f x + e\right )}\right )}{f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*sqrt(2)*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)*f), sqrt(2)*sqrt(-1/a)*arctan(sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(-1/a)/cos(f*x
+ e))/f]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)4*(1/8*sqrt(a)*ln(abs(sin(1/4*(2*f*x+2*exp(1)-pi))+1/sin(1/4*(2*f*x+2*exp(1)-pi))+2)
)/sqrt(2)/a/sign(cos(1/2*(f*x+exp(1))-1/4*pi))-1/8*sqrt(a)*ln(abs(sin(1/4*(2*f*x+2*exp(1)-pi))+1/sin(1/4*(2*f*
x+2*exp(1)-pi))-2))/sqrt(2)/a/sign(cos(1/2*(f*x+exp(1))-1/4*pi)))/f

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maple [A]  time = 0.52, size = 75, normalized size = 1.60 \[ -\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}\, \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a \sin \left (f x + e\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/sqrt(a*sin(f*x + e) + a), x)

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mupad [B]  time = 7.72, size = 49, normalized size = 1.04 \[ -\frac {\mathrm {F}\left (\frac {\pi }{4}-\frac {e}{2}-\frac {f\,x}{2}\middle |1\right )\,\sqrt {\frac {2\,\left (a+a\,\sin \left (e+f\,x\right )\right )}{a}}}{f\,\sqrt {a+a\,\sin \left (e+f\,x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + a*sin(e + f*x))^(1/2),x)

[Out]

-(ellipticF(pi/4 - e/2 - (f*x)/2, 1)*((2*(a + a*sin(e + f*x)))/a)^(1/2))/(f*(a + a*sin(e + f*x))^(1/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a \sin {\left (e + f x \right )} + a}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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