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

Optimal. Leaf size=105 \[ \frac{2 \sqrt{a} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{c f \sqrt{c+d}}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{c f} \]

[Out]

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

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Rubi [A]  time = 0.288795, antiderivative size = 105, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.121, Rules used = {2934, 2773, 206, 208} \[ \frac{2 \sqrt{a} \sqrt{d} \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{c f \sqrt{c+d}}-\frac{2 \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{c f} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(c + d*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*Sqrt[a]*Sqrt[d]*ArcTanh[(Sqrt
[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(c*Sqrt[c + d]*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 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

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

Mathematica [C]  time = 5.29928, size = 746, normalized size = 7.1 \[ -\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) \sqrt{a (\sin (e+f x)+1)} \left (\sqrt{d} \left (\cos \left (\frac{e}{2}\right )+i \sin \left (\frac{e}{2}\right )\right ) \text{RootSum}\left [2 i \text{$\#$1}^2 c e^{i e}+\text{$\#$1}^4 d e^{2 i e}-d\& ,\frac{\text{$\#$1}^3 \left (-\sqrt{d}\right ) e^{i e} f x \sqrt{c+d}-2 i \text{$\#$1}^3 \sqrt{d} e^{i e} \sqrt{c+d} \log \left (-\text{$\#$1}+e^{\frac{i f x}{2}}\right )+\frac{(1-i) \text{$\#$1}^2 c f x}{\sqrt{e^{-i e}}}+\frac{(2+2 i) \text{$\#$1}^2 c \log \left (-\text{$\#$1}+e^{\frac{i f x}{2}}\right )}{\sqrt{e^{-i e}}}-i \text{$\#$1} \sqrt{d} f x \sqrt{c+d}+2 \text{$\#$1} \sqrt{d} \sqrt{c+d} \log \left (-\text{$\#$1}+e^{\frac{i f x}{2}}\right )-(2-2 i) d \sqrt{e^{-i e}} \log \left (-\text{$\#$1}+e^{\frac{i f x}{2}}\right )+(1+i) d \sqrt{e^{-i e}} f x}{\text{$\#$1}^2 (-c) e^{i e}-i d}\& \right ]+\sqrt{d} \left (\cos \left (\frac{e}{2}\right )+i \sin \left (\frac{e}{2}\right )\right ) \text{RootSum}\left [2 i \text{$\#$1}^2 c e^{i e}+\text{$\#$1}^4 d e^{2 i e}-d\& ,\frac{-i \text{$\#$1}^3 \sqrt{d} e^{i e} f x \sqrt{c+d}+2 \text{$\#$1}^3 \sqrt{d} e^{i e} \sqrt{c+d} \log \left (-\text{$\#$1}+e^{\frac{i f x}{2}}\right )-\frac{(1+i) \text{$\#$1}^2 c f x}{\sqrt{e^{-i e}}}+\frac{(2-2 i) \text{$\#$1}^2 c \log \left (-\text{$\#$1}+e^{\frac{i f x}{2}}\right )}{\sqrt{e^{-i e}}}+\text{$\#$1} \sqrt{d} f x \sqrt{c+d}+2 i \text{$\#$1} \sqrt{d} \sqrt{c+d} \log \left (-\text{$\#$1}+e^{\frac{i f x}{2}}\right )+(2+2 i) d \sqrt{e^{-i e}} \log \left (-\text{$\#$1}+e^{\frac{i f x}{2}}\right )+(1-i) d \sqrt{e^{-i e}} f x}{d-i \text{$\#$1}^2 c e^{i e}}\& \right ]+(4+4 i) \sqrt{c+d} \left (\log \left (-\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )+1\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )-\cos \left (\frac{1}{2} (e+f x)\right )+1\right )\right )\right )}{c f \sqrt{c+d} \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/8 + I/8)*((4 + 4*I)*Sqrt[c + d]*(Log[1 + Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - Log[1 - Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2]]) + Sqrt[d]*RootSum[-d + (2*I)*c*E^(I*e)*#1^2 + d*E^((2*I)*e)*#1^4 & , ((1 + I)*d*Sqrt[E^((
-I)*e)]*f*x - (2 - 2*I)*d*Sqrt[E^((-I)*e)]*Log[E^((I/2)*f*x) - #1] - I*Sqrt[d]*Sqrt[c + d]*f*x*#1 + 2*Sqrt[d]*
Sqrt[c + d]*Log[E^((I/2)*f*x) - #1]*#1 + ((1 - I)*c*f*x*#1^2)/Sqrt[E^((-I)*e)] + ((2 + 2*I)*c*Log[E^((I/2)*f*x
) - #1]*#1^2)/Sqrt[E^((-I)*e)] - Sqrt[d]*Sqrt[c + d]*E^(I*e)*f*x*#1^3 - (2*I)*Sqrt[d]*Sqrt[c + d]*E^(I*e)*Log[
E^((I/2)*f*x) - #1]*#1^3)/((-I)*d - c*E^(I*e)*#1^2) & ]*(Cos[e/2] + I*Sin[e/2]) + Sqrt[d]*RootSum[-d + (2*I)*c
*E^(I*e)*#1^2 + d*E^((2*I)*e)*#1^4 & , ((1 - I)*d*Sqrt[E^((-I)*e)]*f*x + (2 + 2*I)*d*Sqrt[E^((-I)*e)]*Log[E^((
I/2)*f*x) - #1] + Sqrt[d]*Sqrt[c + d]*f*x*#1 + (2*I)*Sqrt[d]*Sqrt[c + d]*Log[E^((I/2)*f*x) - #1]*#1 - ((1 + I)
*c*f*x*#1^2)/Sqrt[E^((-I)*e)] + ((2 - 2*I)*c*Log[E^((I/2)*f*x) - #1]*#1^2)/Sqrt[E^((-I)*e)] - I*Sqrt[d]*Sqrt[c
+ d]*E^(I*e)*f*x*#1^3 + 2*Sqrt[d]*Sqrt[c + d]*E^(I*e)*Log[E^((I/2)*f*x) - #1]*#1^3)/(d - I*c*E^(I*e)*#1^2) &
]*(Cos[e/2] + I*Sin[e/2]))*Sqrt[a*(1 + Sin[e + f*x])])/(c*Sqrt[c + d]*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]))

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Maple [A]  time = 1.32, size = 120, normalized size = 1.1 \begin{align*} 2\,{\frac{ \left ( 1+\sin \left ( fx+e \right ) \right ) \sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }}{\sqrt{a}c\sqrt{a \left ( c+d \right ) d}\cos \left ( fx+e \right ) \sqrt{a+a\sin \left ( fx+e \right ) }f} \left ( d{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }d}{\sqrt{a \left ( c+d \right ) d}}} \right ){a}^{3/2}-{\it Artanh} \left ({\frac{\sqrt{-a \left ( -1+\sin \left ( fx+e \right ) \right ) }}{\sqrt{a}}} \right ) a\sqrt{a \left ( c+d \right ) d} \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+d*sin(f*x+e)),x)

[Out]

2/a^(1/2)*(1+sin(f*x+e))*(-a*(-1+sin(f*x+e)))^(1/2)*(d*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))
*a^(3/2)-arctanh((-a*(-1+sin(f*x+e)))^(1/2)/a^(1/2))*a*(a*(c+d)*d)^(1/2))/c/(a*(c+d)*d)^(1/2)/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 (d \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+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 5.94278, size = 1953, normalized size = 18.6 \begin{align*} \text{result too large to display} \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+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

[1/2*(sqrt(a*d/(c + d))*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*cos(f*x + e)
^2 + 4*((c*d + d^2)*cos(f*x + e)^2 - c^2 - 4*c*d - 3*d^2 - (c^2 + 3*c*d + 2*d^2)*cos(f*x + e) + (c^2 + 4*c*d +
3*d^2 + (c*d + d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(a*d/(c + d)) - (a*c^2 + 8*a*c*d
+ 9*a*d^2)*cos(f*x + e) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*a*c*d + 4*a*d^2)*cos(f*x + e
))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 - (c^2 + d^2)*cos(f*x
+ e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))) + sqrt(a)*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)/(co
s(f*x + e)^3 + cos(f*x + e)^2 + (cos(f*x + e)^2 - 1)*sin(f*x + e) - cos(f*x + e) - 1)))/(c*f), 1/2*(2*sqrt(-a*
d/(c + d))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*sin(f*x + e) - c - 2*d)*sqrt(-a*d/(c + d))/(a*d*cos(f*x + e)
)) + sqrt(a)*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)))/(c*f)]

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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+d*sin(f*x+e)),x)

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \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+d*sin(f*x+e)),x, algorithm="giac")

[Out]

Exception raised: TypeError