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]

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

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Rubi [A]  time = 0.07, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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]

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)]

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*}

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Mathematica [C]  time = 0.22, 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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fricas [B]  time = 0.45, size = 210, normalized size = 2.66 \[ -\frac {\frac {\sqrt {2} {\left ({\left (A - B\right )} a \cos \left (f x + e\right ) + {\left (A - B\right )} a \sin \left (f x + e\right ) + {\left (A - B\right )} a\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {a}} + 4 \, {\left (B \cos \left (f x + e\right ) - B \sin \left (f x + e\right ) + B\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{2 \, {\left (a f \cos \left (f x + e\right ) + a f \sin \left (f x + e\right ) + a f\right )}} \]

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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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((A+B*sin(f*x+e))/(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)2/f*(2*sqrt(a*tan((f*x+exp(1))/2)^2+a)*(-1/2*B/sign(tan((f*x+exp(1))/2)+1)+1/2*B*tan
((f*x+exp(1))/2)/sign(tan((f*x+exp(1))/2)+1))/(a*tan((f*x+exp(1))/2)^2+a)+sqrt(2)*(A-B)*atan((-sqrt(a)*tan((f*
x+exp(1))/2)-sqrt(a)+sqrt(a*tan((f*x+exp(1))/2)^2+a))/sqrt(2)/sqrt(-a))/sqrt(-a)/sign(tan((f*x+exp(1))/2)+1)+(
-A*a*sqrt(2)*atan(sqrt(a)/sqrt(-a))+B*a*sqrt(2)*atan(sqrt(a)/sqrt(-a))+B*sqrt(-a)*sqrt(2)*sqrt(a))/a/sqrt(-a)*
sign(tan((f*x+exp(1))/2)+1))

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maple [A]  time = 1.29, size = 128, normalized size = 1.62 \[ -\frac {\left (1+\sin \left (f x +e \right )\right ) \sqrt {-a \left (\sin \left (f x +e \right )-1\right )}\, \left (\sqrt {a}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) A -\sqrt {a}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) B +2 \sqrt {a -a \sin \left (f x +e \right )}\, B \right )}{a \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(1+sin(f*x+e))*(-a*(sin(f*x+e)-1))^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sin \left (f x + e\right ) + A}{\sqrt {a \sin \left (f x + e\right ) + a}}\,{d x} \]

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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mupad [B]  time = 1.06, size = 151, normalized size = 1.91 \[ -\frac {A\,\mathrm {F}\left (\frac {\pi }{4}-\frac {e}{2}-\frac {f\,x}{2}\middle |1\right )\,\sqrt {\frac {2\,\left (a+a\,\sin \left (e+f\,x\right )\right )}{a}}}{f\,\sqrt {a+a\,\sin \left (e+f\,x\right )}}-\frac {B\,\left (4\,\mathrm {E}\left (\mathrm {asin}\left (\frac {\sqrt {2}\,\sqrt {1-\sin \left (e+f\,x\right )}}{2}\right )\middle |1\right )-2\,\mathrm {F}\left (\mathrm {asin}\left (\frac {\sqrt {2}\,\sqrt {1-\sin \left (e+f\,x\right )}}{2}\right )\middle |1\right )\right )\,\sqrt {{\cos \left (e+f\,x\right )}^2}\,\sqrt {\frac {a+a\,\sin \left (e+f\,x\right )}{2\,a}}}{f\,\cos \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (A*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)) - (
B*(4*ellipticE(asin((2^(1/2)*(1 - sin(e + f*x))^(1/2))/2), 1) - 2*ellipticF(asin((2^(1/2)*(1 - sin(e + f*x))^(
1/2))/2), 1))*(cos(e + f*x)^2)^(1/2)*((a + a*sin(e + f*x))/(2*a))^(1/2))/(f*cos(e + f*x)*(a + a*sin(e + f*x))^
(1/2))

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

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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