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

Optimal. Leaf size=102 \[ \frac {\sec (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)} \log (\sin (e+f x))}{c f}-\frac {a \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.46, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {2942, 2737, 2667, 31, 2948, 3475} \[ \frac {\sec (e+f x) \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)} \log (\sin (e+f x))}{c f}-\frac {a \cos (e+f x) \log (1-\sin (e+f x))}{f \sqrt {a \sin (e+f x)+a} \sqrt {c-c \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Cos[e + f*x]*Log[1 - Sin[e + f*x]])/(f*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])) + (Log[Sin[e +
f*x]]*Sec[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*Sqrt[c - c*Sin[e + f*x]])/(c*f)

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 2737

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(
a*c*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]]), Int[Cos[e + f*x]/(c + d*Sin[e + f*x]),
x], x] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 2942

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/(sin[(e_.) + (f_.)*(x_)]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x
_)]]), x_Symbol] :> -Dist[d/c, Int[Sqrt[a + b*Sin[e + f*x]]/Sqrt[c + d*Sin[e + f*x]], x], x] + Dist[1/c, Int[(
Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])/Sin[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ
[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[b*c + a*d, 0]

Rule 2948

Int[(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]])/sin[(e_.) + (f_.)*
(x_)], x_Symbol] :> Dist[(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])/Cos[e + f*x], Int[Cot[e + f*x], x
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && EqQ[c^2 - d^2, 0]

Rule 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [C]  time = 1.33, size = 144, normalized size = 1.41 \[ \frac {\sqrt {2} \left (e^{i (e+f x)}-i\right ) \sqrt {a (\sin (e+f x)+1)} \left (i \left (\log \left (1-e^{2 i (e+f x)}\right )-\log \left (1+e^{2 i (e+f x)}\right )\right )+2 \tan ^{-1}\left (e^{i (e+f x)}\right )\right )}{f \left (e^{i (e+f x)}+i\right ) \sqrt {i c e^{-i (e+f x)} \left (e^{i (e+f x)}-i\right )^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \sin \left (f x + e\right ) + a} \sqrt {-c \sin \left (f x + e\right ) + c}}{c \cos \left (f x + e\right )^{2} + c \sin \left (f x + e\right ) - c}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(a*sin(f*x + e) + a)*sqrt(-c*sin(f*x + e) + c)/(c*cos(f*x + e)^2 + c*sin(f*x + e) - c), x)

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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))^(1/2)/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: (8*pi/x/2)>(-8*pi/x/2)Unable to check sign: (8*pi/x/2)>(-8*pi/x/2)-sqrt(2*a)*sqrt(c)
*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*ln(abs(-4*(tan(1/2*(1/2*f*x+1/4*(2*exp(1)-pi)))^2+1/tan(1/2*(1/2*f*x+1/4*(
2*exp(1)-pi)))^2)+24))/sqrt(2)/c/f/sign(tan(1/2*(1/2*f*x+1/4*(2*exp(1)-pi)))^3+tan(1/2*(1/2*f*x+1/4*(2*exp(1)-
pi))))

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maple [A]  time = 0.47, size = 111, normalized size = 1.09 \[ \frac {\sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \left (-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )\right ) \left (2 \ln \left (-\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )\right )}{f \left (-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )\right ) \sqrt {-c \left (\sin \left (f x +e \right )-1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.43, size = 59, normalized size = 0.58 \[ \frac {\frac {2 \, \sqrt {a} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{\sqrt {c}} - \frac {\sqrt {a} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{\sqrt {c}}}{f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*sqrt(a)*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/sqrt(c) - sqrt(a)*log(sin(f*x + e)/(cos(f*x + e) + 1))/sqr
t(c))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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