\(\int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))} \, dx\) [14]

Optimal result

Integrand size = 34, antiderivative size = 120 \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} c f}+\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {2} \sqrt {a} c f}+\frac {\sec (e+f x) \sqrt {a+a \sin (e+f x)}}{a c f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {3019, 2815, 2752, 3064, 2728, 212, 2852} \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} c f}+\frac {\text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {2} \sqrt {a} c f}+\frac {\sec (e+f x) \sqrt {a \sin (e+f x)+a}}{a c f} \]

[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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2728

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 2752

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[b*(g*C
os[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 2815

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 2852

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 3019

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 3064

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]

Rubi steps \begin{align*} \text {integral}& = \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 {\text {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 \text {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 \text {arctanh}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} c f}+\frac {\text {arctanh}\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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.32 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.62 \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))} \, dx=-\frac {\left (\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},\frac {1}{2} (1-\sin (e+f x))\right )-2 \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},1,\frac {1}{2},1-\sin (e+f x)\right )\right ) \sec (e+f x) \sqrt {a (1+\sin (e+f x))}}{a c f} \]

[In]

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

[Out]

-(((Hypergeometric2F1[-1/2, 1, 1/2, (1 - Sin[e + f*x])/2] - 2*Hypergeometric2F1[-1/2, 1, 1/2, 1 - Sin[e + f*x]
])*Sec[e + f*x]*Sqrt[a*(1 + Sin[e + f*x])])/(a*c*f))

Maple [A] (verified)

Time = 0.39 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.03

method	result	size
default	\(\frac {\left (1+\sin \left (f x +e \right )\right ) \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \sqrt {a -a \sin \left (f x +e \right )}+2 a^{\frac {5}{2}}-4 \,\operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (f x +e \right )}}{\sqrt {a}}\right ) a^{2} \sqrt {a -a \sin \left (f x +e \right )}\right )}{2 c \,a^{\frac {5}{2}} \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f}\)	\(124\)

[In]

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

[Out]

1/2*(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))/c/a^(5/2)/cos(f*x+e)/(a+a*sin(f*x
+e))^(1/2)/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 328 vs. \(2 (104) = 208\).

Time = 0.29 (sec) , antiderivative size = 328, normalized size of antiderivative = 2.73 \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))} \, dx=\frac {\sqrt {2} \sqrt {a} \cos \left (f x + e\right ) \log \left (-\frac {\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) - 2\right )} \sin \left (f x + e\right ) + \frac {2 \, \sqrt {2} \sqrt {a \sin \left (f x + e\right ) + a} {\left (\cos \left (f x + e\right ) - \sin \left (f x + e\right ) + 1\right )}}{\sqrt {a}} + 3 \, \cos \left (f x + e\right ) + 2}{\cos \left (f x + e\right )^{2} - {\left (\cos \left (f x + e\right ) + 2\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 2}\right ) + 2 \, \sqrt {a} \cos \left (f x + e\right ) \log \left (\frac {a \cos \left (f x + e\right )^{3} - 7 \, a \cos \left (f x + e\right )^{2} - 4 \, {\left (\cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right ) + 3\right )} \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right ) - 3\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {a} - 9 \, a \cos \left (f x + e\right ) + {\left (a \cos \left (f x + e\right )^{2} + 8 \, a \cos \left (f x + e\right ) - a\right )} \sin \left (f x + e\right ) - a}{\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} + {\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right ) - \cos \left (f x + e\right ) - 1}\right ) + 4 \, \sqrt {a \sin \left (f x + e\right ) + a}}{4 \, a c f \cos \left (f x + e\right )} \]

[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))

Sympy [F]

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

[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

Maxima [F]

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

[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)

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.18 \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))} \, dx=-\frac {\sqrt {2} {\left (\frac {2 \, \sqrt {2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) \right |}}\right )}{c} + \frac {\log \left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c} - \frac {\log \left (-\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1\right )}{c} + \frac {2}{c \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )}\right )}}{4 \, \sqrt {a} f \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \]

[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]

-1/4*sqrt(2)*(2*sqrt(2)*log(abs(-2*sqrt(2) + 4*sin(-1/4*pi + 1/2*f*x + 1/2*e))/abs(2*sqrt(2) + 4*sin(-1/4*pi +
1/2*f*x + 1/2*e)))/c + log(sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/c - log(-sin(-1/4*pi + 1/2*f*x + 1/2*e) + 1)/c
+ 2/(c*sin(-1/4*pi + 1/2*f*x + 1/2*e)))/(sqrt(a)*f*sgn(cos(-1/4*pi + 1/2*f*x + 1/2*e)))

Mupad [F(-1)]

Timed out. \[ \int \frac {\csc (e+f x)}{\sqrt {a+a \sin (e+f x)} (c-c \sin (e+f x))} \, dx=\int \frac {1}{\sin \left (e+f\,x\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\left (c-c\,\sin \left (e+f\,x\right )\right )} \,d x \]

[In]

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

[Out]

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