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

Optimal. Leaf size=37 \[ -\frac {2 \sqrt {a} \sin ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{f} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2774, 216} \[ -\frac {2 \sqrt {a} \sin ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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]

Rubi steps

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

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Mathematica [C]  time = 0.51, size = 164, normalized size = 4.43 \[ \frac {(1+i) e^{\frac {1}{2} i (e+f x)} \sqrt {-i e^{-i (e+f x)} \left (-1+e^{2 i (e+f x)}\right )} \sqrt {a (\sin (e+f x)+1)} \left (\tan ^{-1}\left (\sqrt {-1+e^{2 i (e+f x)}}\right )-i \tanh ^{-1}\left (\frac {e^{i (e+f x)}}{\sqrt {-1+e^{2 i (e+f x)}}}\right )\right )}{\sqrt {2} f \sqrt {-1+e^{2 i (e+f x)}} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.67, size = 330, normalized size = 8.92 \[ \left [\frac {\sqrt {-a} \log \left (\frac {128 \, a \cos \left (f x + e\right )^{5} - 128 \, a \cos \left (f x + e\right )^{4} - 416 \, a \cos \left (f x + e\right )^{3} + 128 \, a \cos \left (f x + e\right )^{2} - 8 \, {\left (16 \, \cos \left (f x + e\right )^{4} - 24 \, \cos \left (f x + e\right )^{3} - 66 \, \cos \left (f x + e\right )^{2} + {\left (16 \, \cos \left (f x + e\right )^{3} + 40 \, \cos \left (f x + e\right )^{2} - 26 \, \cos \left (f x + e\right ) - 51\right )} \sin \left (f x + e\right ) + 25 \, \cos \left (f x + e\right ) + 51\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-a} \sqrt {\sin \left (f x + e\right )} + 289 \, a \cos \left (f x + e\right ) + {\left (128 \, a \cos \left (f x + e\right )^{4} + 256 \, a \cos \left (f x + e\right )^{3} - 160 \, a \cos \left (f x + e\right )^{2} - 288 \, a \cos \left (f x + e\right ) + a\right )} \sin \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1}\right )}{4 \, f}, \frac {\sqrt {a} \arctan \left (\frac {{\left (8 \, \cos \left (f x + e\right )^{2} + 8 \, \sin \left (f x + e\right ) - 9\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} \sqrt {\sin \left (f x + e\right )}}{4 \, {\left (2 \, a \cos \left (f x + e\right )^{3} + a \cos \left (f x + e\right ) \sin \left (f x + e\right ) - 2 \, a \cos \left (f x + e\right )\right )}}\right )}{2 \, f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*sqrt(-a)*log((128*a*cos(f*x + e)^5 - 128*a*cos(f*x + e)^4 - 416*a*cos(f*x + e)^3 + 128*a*cos(f*x + e)^2 -
8*(16*cos(f*x + e)^4 - 24*cos(f*x + e)^3 - 66*cos(f*x + e)^2 + (16*cos(f*x + e)^3 + 40*cos(f*x + e)^2 - 26*co
s(f*x + e) - 51)*sin(f*x + e) + 25*cos(f*x + e) + 51)*sqrt(a*sin(f*x + e) + a)*sqrt(-a)*sqrt(sin(f*x + e)) + 2
89*a*cos(f*x + e) + (128*a*cos(f*x + e)^4 + 256*a*cos(f*x + e)^3 - 160*a*cos(f*x + e)^2 - 288*a*cos(f*x + e) +
a)*sin(f*x + e) + a)/(cos(f*x + e) + sin(f*x + e) + 1))/f, 1/2*sqrt(a)*arctan(1/4*(8*cos(f*x + e)^2 + 8*sin(f
*x + e) - 9)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*sqrt(sin(f*x + e))/(2*a*cos(f*x + e)^3 + a*cos(f*x + e)*sin(f*x
+ e) - 2*a*cos(f*x + e)))/f]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.32, size = 320, normalized size = 8.65 \[ \frac {\sqrt {-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (\sqrt {\sin }\left (f x +e \right )\right ) \left (\ln \left (-\frac {\sqrt {2}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sin \left (f x +e \right )+\sin \left (f x +e \right )-\cos \left (f x +e \right )+1}{\sqrt {2}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sin \left (f x +e \right )-\sin \left (f x +e \right )+\cos \left (f x +e \right )-1}\right )+4 \arctan \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {2}+1\right )+4 \arctan \left (\sqrt {-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {2}-1\right )+\ln \left (-\frac {\sqrt {2}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sin \left (f x +e \right )-\sin \left (f x +e \right )+\cos \left (f x +e \right )-1}{\sqrt {2}\, \sqrt {-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sin \left (f x +e \right )+\sin \left (f x +e \right )-\cos \left (f x +e \right )+1}\right )\right ) \sqrt {2}}{2 f \left (1-\cos \left (f x +e \right )+\sin \left (f x +e \right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/f*(-(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*(a*(1+sin(f*x+e)))^(1/2)*sin(f*x+e)^(1/2)*(ln(-(2^(1/2)*(-(-1+cos(f*
x+e))/sin(f*x+e))^(1/2)*sin(f*x+e)+sin(f*x+e)-cos(f*x+e)+1)/(2^(1/2)*(-(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*sin(f
*x+e)-sin(f*x+e)+cos(f*x+e)-1))+4*arctan((-(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*2^(1/2)+1)+4*arctan((-(-1+cos(f*x
+e))/sin(f*x+e))^(1/2)*2^(1/2)-1)+ln(-(2^(1/2)*(-(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*sin(f*x+e)-sin(f*x+e)+cos(f
*x+e)-1)/(2^(1/2)*(-(-1+cos(f*x+e))/sin(f*x+e))^(1/2)*sin(f*x+e)+sin(f*x+e)-cos(f*x+e)+1)))*2^(1/2)/(1-cos(f*x
+e)+sin(f*x+e))

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maxima [B]  time = 0.93, size = 210, normalized size = 5.68 \[ -\frac {2 \, \sqrt {2} \sqrt {a} \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )^{\frac {3}{2}} - 3 \, \sqrt {2} {\left (\sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}\right ) + \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}\right )}\right )\right )} \sqrt {a} + 6 \, \sqrt {2} \sqrt {a} \sqrt {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}} - \frac {2 \, {\left (\frac {3 \, \sqrt {2} \sqrt {a} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sqrt {2} \sqrt {a} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}}{\sqrt {\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}}}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(2*sqrt(2)*sqrt(a)*(sin(f*x + e)/(cos(f*x + e) + 1))^(3/2) - 3*sqrt(2)*(sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(
2) + 2*sqrt(sin(f*x + e)/(cos(f*x + e) + 1)))) + sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(sin(f*x + e)/(c
os(f*x + e) + 1)))))*sqrt(a) + 6*sqrt(2)*sqrt(a)*sqrt(sin(f*x + e)/(cos(f*x + e) + 1)) - 2*(3*sqrt(2)*sqrt(a)*
sin(f*x + e)/(cos(f*x + e) + 1) + sqrt(2)*sqrt(a)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)/sqrt(sin(f*x + e)/(cos(
f*x + e) + 1)))/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \[ \int \frac {\sqrt {a+a\,\sin \left (e+f\,x\right )}}{\sqrt {\sin \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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