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

Optimal. Leaf size=69 \[ \frac{2 \sec (e+f x) \sqrt{a \sin (e+f x)+a}}{c f}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{c f} \]

[Out]

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

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Rubi [A]  time = 0.306103, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {2934, 2773, 206, 2736, 2673} \[ \frac{2 \sec (e+f x) \sqrt{a \sin (e+f x)+a}}{c f}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2934

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/(sin[(e_.) + (f_.)*(x_)]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])
), x_Symbol] :> Dist[1/c, Int[Sqrt[a + b*Sin[e + f*x]]/Sin[e + f*x], x], x] - Dist[d/c, Int[Sqrt[a + b*Sin[e +
f*x]]/(c + d*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]

Rule 2773

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

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 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]))

Rule 2673

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 - 1))/(f*g*(m - 1)), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && Eq
Q[a^2 - b^2, 0] && EqQ[2*m + p - 1, 0] && NeQ[m, 1]

Rubi steps

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

Mathematica [B]  time = 0.357876, size = 157, normalized size = 2.28 \[ \frac{\sec (e+f x) \sqrt{a (\sin (e+f x)+1)} \left (\cos \left (\frac{1}{2} (e+f x)\right ) \left (\log \left (\sin \left (\frac{1}{2} (e+f x)\right )-\cos \left (\frac{1}{2} (e+f x)\right )+1\right )-\log \left (-\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )+1\right )\right )+\sin \left (\frac{1}{2} (e+f x)\right ) \left (\log \left (-\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )+1\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )-\cos \left (\frac{1}{2} (e+f x)\right )+1\right )\right )+2\right )}{c f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[e + f*x]*(2 + Cos[(e + f*x)/2]*(-Log[1 + Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] + Log[1 - Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2]]) + (Log[1 + Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[1 - Cos[(e + f*x)/2] + Sin[(e + f*x
)/2]])*Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])])/(c*f)

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Maple [A]  time = 0.82, size = 79, normalized size = 1.1 \begin{align*} -2\,{\frac{1+\sin \left ( fx+e \right ) }{\sqrt{a}c\cos \left ( fx+e \right ) \sqrt{a+a\sin \left ( fx+e \right ) }f} \left ({\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }}{\sqrt{a}}} \right ) a\sqrt{a-a\sin \left ( fx+e \right ) }-{a}^{3/2} \right ) } \end{align*}

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)),x)

[Out]

-2/a^(1/2)/c*(1+sin(f*x+e))*(arctanh((a-a*sin(f*x+e))^(1/2)/a^(1/2))*a*(a-a*sin(f*x+e))^(1/2)-a^(3/2))/cos(f*x
+e)/(a+a*sin(f*x+e))^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\int \frac{\sqrt{a \sin \left (f x + e\right ) + a}}{{\left (c \sin \left (f x + e\right ) - c\right )} \sin \left (f x + e\right )}\,{d x} \end{align*}

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)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.10138, size = 539, normalized size = 7.81 \begin{align*} \frac{\sqrt{a} \cos \left (f x + e\right ) \log \left (\frac{a \cos \left (f x + e\right )^{3} - 7 \, a \cos \left (f x + e\right )^{2} - 4 \,{\left (\cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right ) + 3\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) - 3\right )} \sqrt{a \sin \left (f x + e\right ) + a} \sqrt{a} - 9 \, a \cos \left (f x + e\right ) +{\left (a \cos \left (f x + e\right )^{2} + 8 \, a \cos \left (f x + e\right ) - a\right )} \sin \left (f x + e\right ) - a}{\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} +{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 1}\right ) + 4 \, \sqrt{a \sin \left (f x + e\right ) + a}}{2 \, c f \cos \left (f x + e\right )} \end{align*}

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)),x, algorithm="fricas")

[Out]

1/2*(sqrt(a)*cos(f*x + e)*log((a*cos(f*x + e)^3 - 7*a*cos(f*x + e)^2 - 4*(cos(f*x + e)^2 + (cos(f*x + e) + 3)*
sin(f*x + e) - 2*cos(f*x + e) - 3)*sqrt(a*sin(f*x + e) + a)*sqrt(a) - 9*a*cos(f*x + e) + (a*cos(f*x + e)^2 + 8
*a*cos(f*x + e) - a)*sin(f*x + e) - a)/(cos(f*x + e)^3 + cos(f*x + e)^2 + (cos(f*x + e)^2 - 1)*sin(f*x + e) -
cos(f*x + e) - 1)) + 4*sqrt(a*sin(f*x + e) + a))/(c*f*cos(f*x + e))

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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)),x)

[Out]

Timed out

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Giac [B]  time = 1.43416, size = 575, normalized size = 8.33 \begin{align*} \frac{\frac{2 \, a \arctan \left (-\frac{\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a}}{\sqrt{-a}}\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}{\sqrt{-a} c} - \frac{\sqrt{a} \log \left ({\left | -\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a} \right |}\right ) \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}{c} - \frac{{\left (2 \, \sqrt{2} a \arctan \left (\frac{\sqrt{2} \sqrt{a} + \sqrt{a}}{\sqrt{-a}}\right ) - \sqrt{2} \sqrt{-a} \sqrt{a} \log \left (\sqrt{2} \sqrt{a} + \sqrt{a}\right ) + 2 \, a \arctan \left (\frac{\sqrt{2} \sqrt{a} + \sqrt{a}}{\sqrt{-a}}\right ) - \sqrt{-a} \sqrt{a} \log \left (\sqrt{2} \sqrt{a} + \sqrt{a}\right ) - \sqrt{2} \sqrt{-a} \sqrt{a}\right )} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )}{\sqrt{2} \sqrt{-a} c + \sqrt{-a} c} + \frac{4 \,{\left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a}\right )} a \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right ) + a^{\frac{3}{2}} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 1\right )\right )}}{{\left ({\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a}\right )}^{2} - 2 \,{\left (\sqrt{a} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \sqrt{a \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + a}\right )} \sqrt{a} - a\right )} c}}{f} \end{align*}

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)),x, algorithm="giac")

[Out]

(2*a*arctan(-(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))/sqrt(-a))*sgn(tan(1/2*f*x + 1
/2*e) + 1)/(sqrt(-a)*c) - sqrt(a)*log(abs(-sqrt(a)*tan(1/2*f*x + 1/2*e) + sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a)))
*sgn(tan(1/2*f*x + 1/2*e) + 1)/c - (2*sqrt(2)*a*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - sqrt(2)*sqrt(-a
)*sqrt(a)*log(sqrt(2)*sqrt(a) + sqrt(a)) + 2*a*arctan((sqrt(2)*sqrt(a) + sqrt(a))/sqrt(-a)) - sqrt(-a)*sqrt(a)
*log(sqrt(2)*sqrt(a) + sqrt(a)) - sqrt(2)*sqrt(-a)*sqrt(a))*sgn(tan(1/2*f*x + 1/2*e) + 1)/(sqrt(2)*sqrt(-a)*c
+ sqrt(-a)*c) + 4*((sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*a*sgn(tan(1/2*f*x + 1/2
*e) + 1) + a^(3/2)*sgn(tan(1/2*f*x + 1/2*e) + 1))/(((sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e
)^2 + a))^2 - 2*(sqrt(a)*tan(1/2*f*x + 1/2*e) - sqrt(a*tan(1/2*f*x + 1/2*e)^2 + a))*sqrt(a) - a)*c))/f