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

Optimal. Leaf size=77 \[ \frac {2 \sqrt {2} a \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {2736, 2679, 2649, 206} \[ \frac {2 \sqrt {2} a \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])/Sqrt[c - c*Sin[e + f*x]],x]

[Out]

(2*Sqrt[2]*a*ArcTanh[(Sqrt[c]*Cos[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sin[e + f*x]])])/(Sqrt[c]*f) - (2*a*Cos[e + f*
x])/(f*Sqrt[c - c*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 2679

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + p)), x] + Dist[(g^2*(p - 1))/(a*(m + p)), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[m, -1] && GtQ[p, 1] && (GtQ[m, -2] || EqQ[2*m + p + 1, 0] || (EqQ[m, -2] && IntegerQ[p])) && NeQ[m + p,
0] && IntegersQ[2*m, 2*p]

Rule 2736

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

Rubi steps

\begin {align*} \int \frac {a+a \sin (e+f x)}{\sqrt {c-c \sin (e+f x)}} \, dx &=(a c) \int \frac {\cos ^2(e+f x)}{(c-c \sin (e+f x))^{3/2}} \, dx\\ &=-\frac {2 a \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}+(2 a) \int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx\\ &=-\frac {2 a \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}-\frac {(4 a) \operatorname {Subst}\left (\int \frac {1}{2 c-x^2} \, dx,x,-\frac {c \cos (e+f x)}{\sqrt {c-c \sin (e+f x)}}\right )}{f}\\ &=\frac {2 \sqrt {2} a \tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 a \cos (e+f x)}{f \sqrt {c-c \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 0.65, size = 135, normalized size = 1.75 \[ -\frac {2 a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \left (\sqrt {c} (\sin (e+f x)+1)+\sqrt {2} \sqrt {-c (\sin (e+f x)+1)} \tan ^{-1}\left (\frac {\sqrt {-c (\sin (e+f x)+1)}}{\sqrt {2} \sqrt {c}}\right )\right )}{\sqrt {c} f \sqrt {c-c \sin (e+f x)} \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])/Sqrt[c - c*Sin[e + f*x]],x]

[Out]

(-2*a*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Sqrt[c]*(1 + Sin[e + f*x]) + Sqrt[2]*ArcTan[Sqrt[-(c*(1 + Sin[e +
f*x]))]/(Sqrt[2]*Sqrt[c])]*Sqrt[-(c*(1 + Sin[e + f*x]))]))/(Sqrt[c]*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*S
qrt[c - c*Sin[e + f*x]])

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fricas [B]  time = 0.45, size = 196, normalized size = 2.55 \[ \frac {\frac {\sqrt {2} {\left (a c \cos \left (f x + e\right ) - a c \sin \left (f x + e\right ) + a c\right )} \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right )}{\sqrt {c}} - 2 \, {\left (a \cos \left (f x + e\right ) + a \sin \left (f x + e\right ) + a\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}{c f \cos \left (f x + e\right ) - c f \sin \left (f x + e\right ) + c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(2)*(a*c*cos(f*x + e) - a*c*sin(f*x + e) + a*c)*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin(f*x + e) +
2*sqrt(2)*sqrt(-c*sin(f*x + e) + c)*(cos(f*x + e) + sin(f*x + e) + 1)/sqrt(c) + 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(c) - 2*(a*cos(f*x + e) + a*sin(f*x + e) + a)
*sqrt(-c*sin(f*x + e) + c))/(c*f*cos(f*x + e) - c*f*sin(f*x + e) + c*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+a*sin(f*x+e))/(c-c*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*(4*a*atan((-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c)+sqrt(c*tan((f*x+exp(1))/2)^2+c))
/sqrt(2)/sqrt(-c))/sqrt(2)/sqrt(-c)/sign(tan((f*x+exp(1))/2)-1)+2*sqrt(c*tan((f*x+exp(1))/2)^2+c)*(1/2*a/sign(
tan((f*x+exp(1))/2)-1)+1/2*a*tan((f*x+exp(1))/2)/sign(tan((f*x+exp(1))/2)-1))/(c*tan((f*x+exp(1))/2)^2+c)+(-4*
a*c*atan(sqrt(c)/sqrt(-c))-2*a*sqrt(-c)*sqrt(c))/c/sqrt(-c)/sqrt(2)*sign(tan((f*x+exp(1))/2)-1))

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maple [A]  time = 0.72, size = 94, normalized size = 1.22 \[ -\frac {2 \left (\sin \left (f x +e \right )-1\right ) \sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, a \left (\sqrt {c}\, \sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right )-\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\right )}{c \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+a\,\sin \left (e+f\,x\right )}{\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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