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

Optimal. Leaf size=120 \[ \frac{\sec (e+f x) \sqrt{a \sin (e+f x)+a}}{a c f}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} c f}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{2} \sqrt{a} c f} \]

[Out]

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

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Rubi [A]  time = 0.442856, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.206, Rules used = {2940, 2736, 2673, 2985, 2649, 206, 2773} \[ \frac{\sec (e+f x) \sqrt{a \sin (e+f x)+a}}{a c f}-\frac{2 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{a} c f}+\frac{\tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{\sqrt{2} \sqrt{a} c f} \]

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

Rule 2940

Int[1/(sin[(e_.) + (f_.)*(x_)]*Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)
])), x_Symbol] :> Dist[d^2/(c*(b*c - a*d)), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] + Dist[1
/(c*(b*c - a*d)), Int[(b*c - a*d - b*d*Sin[e + f*x])/(Sin[e + f*x]*Sqrt[a + b*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 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]

Rule 2985

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

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

Mathematica [C]  time = 0.46554, size = 234, normalized size = 1.95 \[ \frac{\cos (e+f x) \left (\cos \left (\frac{1}{2} (e+f x)\right ) \left (\log \left (-\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )+1\right )-\log \left (\sin \left (\frac{1}{2} (e+f x)\right )-\cos \left (\frac{1}{2} (e+f x)\right )+1\right )\right )-\sin \left (\frac{1}{2} (e+f x)\right ) \log \left (-\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )+1\right )+\sin \left (\frac{1}{2} (e+f x)\right ) \log \left (\sin \left (\frac{1}{2} (e+f x)\right )-\cos \left (\frac{1}{2} (e+f x)\right )+1\right )+(1+i) (-1)^{3/4} \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )-1\right )}{c f (\sin (e+f x)-1) \sqrt{a (\sin (e+f x)+1)}} \]

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]

(Cos[e + f*x]*(-1 + 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]]) + (1 + I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(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] + Log[1 - Cos[(e + f
*x)/2] + Sin[(e + f*x)/2]]*Sin[(e + f*x)/2]))/(c*f*(-1 + Sin[e + f*x])*Sqrt[a*(1 + Sin[e + f*x])])

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Maple [A]  time = 1.125, size = 124, normalized size = 1. \begin{align*}{\frac{1+\sin \left ( fx+e \right ) }{2\,cf\cos \left ( fx+e \right ) } \left ( \sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}}{2}\sqrt{a-a\sin \left ( fx+e \right ) }{\frac{1}{\sqrt{a}}}} \right ){a}^{2}\sqrt{a-a\sin \left ( fx+e \right ) }+2\,{a}^{5/2}-4\,{\it Artanh} \left ({\frac{\sqrt{a-a\sin \left ( fx+e \right ) }}{\sqrt{a}}} \right ){a}^{2}\sqrt{a-a\sin \left ( fx+e \right ) } \right ){a}^{-{\frac{5}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( fx+e \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/c*(1+sin(f*x+e))*(2^(1/2)*arctanh(1/2*(a-a*sin(f*x+e))^(1/2)*2^(1/2)/a^(1/2))*a^2*(a-a*sin(f*x+e))^(1/2)+2
*a^(5/2)-4*arctanh((a-a*sin(f*x+e))^(1/2)/a^(1/2))*a^2*(a-a*sin(f*x+e))^(1/2))/a^(5/2)/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{1}{\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(1/sin(f*x+e)/(c-c*sin(f*x+e))/(a+a*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

-integrate(1/(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 = 2.24324, size = 892, normalized size = 7.43 \begin{align*} \frac{\sqrt{2} \sqrt{a} \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{a \sin \left (f x + e\right ) + a}{\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt{a}} + 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 ) + 2 \, \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}}{4 \, a c f \cos \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \frac{\int \frac{1}{\sqrt{a \sin{\left (e + f x \right )} + a} \sin ^{2}{\left (e + f x \right )} - \sqrt{a \sin{\left (e + f x \right )} + a} \sin{\left (e + f x \right )}}\, dx}{c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError