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

Optimal. Leaf size=83 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} a \sqrt {c} f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{a c f} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2736, 2675, 2649, 206} \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} a \sqrt {c} f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{a c f} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]]),x]

[Out]

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

Rule 2675

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p + 1)), x] + Dist[(a*(m + p + 1))/(g^2*(p + 1)), 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]
&& GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ[m + 1/2, 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 {1}{(a+a \sin (e+f x)) \sqrt {c-c \sin (e+f x)}} \, dx &=\frac {\int \sec ^2(e+f x) \sqrt {c-c \sin (e+f x)} \, dx}{a c}\\ &=-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{a c f}+\frac {\int \frac {1}{\sqrt {c-c \sin (e+f x)}} \, dx}{2 a}\\ &=-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{a c f}-\frac {\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 )}{a f}\\ &=\frac {\tanh ^{-1}\left (\frac {\sqrt {c} \cos (e+f x)}{\sqrt {2} \sqrt {c-c \sin (e+f x)}}\right )}{\sqrt {2} a \sqrt {c} f}-\frac {\sec (e+f x) \sqrt {c-c \sin (e+f x)}}{a c f}\\ \end {align*}

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Mathematica [C]  time = 0.33, size = 97, normalized size = 1.17 \[ -\frac {\cos (e+f x) \left (1+(1+i) \sqrt [4]{-1} \tan ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (\tan \left (\frac {1}{4} (e+f x)\right )+1\right )\right ) \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )\right )}{a f (\sin (e+f x)+1) \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])*Sqrt[c - c*Sin[e + f*x]]),x]

[Out]

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

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fricas [B]  time = 0.48, size = 154, normalized size = 1.86 \[ \frac {\sqrt {2} \sqrt {c} \cos \left (f x + e\right ) \log \left (-\frac {\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {-c \sin \left (f x + e\right ) + c} {\left (\cos \left (f x + e\right ) + \sin \left (f x + e\right ) + 1\right )}}{\sqrt {c}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) - 4 \, \sqrt {-c \sin \left (f x + e\right ) + c}}{4 \, a c f \cos \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(sqrt(2)*sqrt(c)*cos(f*x + e)*log(-(cos(f*x + e)^2 + (cos(f*x + e) - 2)*sin(f*x + e) + 2*sqrt(2)*sqrt(-c*s
in(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)) - 4*sqrt(-c*sin(f*x + e) + c))/(a*c*f*cos(f*x + e))

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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))/(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)2/f*((2*c*sqrt(2)*
atan(sqrt(c)/sqrt(-c))-2*c*atan(sqrt(c)/sqrt(-c))+sqrt(-c)*sqrt(c))/(2*a*c*sqrt(-c)*sqrt(2)-4*a*c*sqrt(-c))*si
gn(tan((f*x+exp(1))/2)-1)+2*(1/2*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c)+sqrt(c*tan((f*x+exp(1))/2)^2+c))/a/(-(-
sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*tan((f*x+exp(1))/2)^2+c))^2+2*sqrt(c)*(-sqrt(c)*tan((f*x+exp(1))/2)+sqrt(c*
tan((f*x+exp(1))/2)^2+c))+c)/sign(tan((f*x+exp(1))/2)-1)+1/2*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)/a/sign(tan((f*x+exp(1))/2)-1)))

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maple [A]  time = 0.96, size = 86, normalized size = 1.04 \[ \frac {\left (\sin \left (f x +e \right )-1\right ) \left (-\sqrt {2}\, \arctanh \left (\frac {\sqrt {c \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {2}}{2 \sqrt {c}}\right ) c \sqrt {c \left (1+\sin \left (f x +e \right )\right )}+2 c^{\frac {3}{2}}\right )}{2 a \,c^{\frac {3}{2}} \cos \left (f x +e \right ) \sqrt {c -c \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a*(sin(f*x+e)-1)*(-2^(1/2)*arctanh(1/2*(c*(1+sin(f*x+e)))^(1/2)*2^(1/2)/c^(1/2))*c*(c*(1+sin(f*x+e)))^(1/2
)+2*c^(3/2))/c^(3/2)/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 \[ \int \frac {1}{{\left (a \sin \left (f x + e\right ) + a\right )} \sqrt {-c \sin \left (f x + e\right ) + c}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(1/((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 \[ \int \frac {1}{\left (a+a\,\sin \left (e+f\,x\right )\right )\,\sqrt {c-c\,\sin \left (e+f\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(1/((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 \[ \frac {\int \frac {1}{\sqrt {- c \sin {\left (e + f x \right )} + c} \sin {\left (e + f x \right )} + \sqrt {- c \sin {\left (e + f x \right )} + c}}\, dx}{a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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