∫ csc(e+fx)________∘ a+a sin(e+fx) (c+d sin(e+fx)) dx

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

Optimal. Leaf size=165 \[ -\frac {2 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} c f (c-d) \sqrt {c+d}}+\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} f (c-d)}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} c f} \]

[Out]

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

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

*x+e))^(1/2))/c/(c-d)/f/a^(1/2)/(c+d)^(1/2)

________________________________________________________________________________________

Rubi [A] time = 0.46, antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {2940, 2773, 208, 2985, 2649, 206} \[ -\frac {2 d^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} c f (c-d) \sqrt {c+d}}+\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} f (c-d)}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a}}\right )}{\sqrt {a} c f} \]

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

+ f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*(c - d)*f) - (2*d^(3/2)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e +

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

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]

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 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 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 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]

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

________________________________________________________________________________________

Mathematica [C] time = 2.24, size = 331, normalized size = 2.01 \[ -\frac {\left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right ) \left (d^{3/2} \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (\sqrt {c+d}-\sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )+\sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right )\right )\right )-d^{3/2} \log \left (\sec ^2\left (\frac {1}{4} (e+f x)\right ) \left (\sqrt {c+d}+\sqrt {d} \sin \left (\frac {1}{2} (e+f x)\right )-\sqrt {d} \cos \left (\frac {1}{2} (e+f x)\right )\right )\right )+(2+2 i) (-1)^{3/4} c \sqrt {c+d} \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (e+f x)\right )-1\right )\right )+d \sqrt {c+d} \log \left (\sin \left (\frac {1}{2} (e+f x)\right )-\cos \left (\frac {1}{2} (e+f x)\right )+1\right )+(c-d) \sqrt {c+d} \log \left (-\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )+1\right )-c \sqrt {c+d} \log \left (\sin \left (\frac {1}{2} (e+f x)\right )-\cos \left (\frac {1}{2} (e+f x)\right )+1\right )\right )}{c f (c-d) \sqrt {c+d} \sqrt {a (\sin (e+f x)+1)}} \]

Antiderivative was successfully veriﬁed.

[In]

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

[Out]

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

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

]] + d*Sqrt[c + d]*Log[1 - Cos[(e + f*x)/2] + Sin[(e + f*x)/2]] + d^(3/2)*Log[Sec[(e + f*x)/4]^2*(Sqrt[c + d]

+ Sqrt[d]*Cos[(e + f*x)/2] - Sqrt[d]*Sin[(e + f*x)/2])] - d^(3/2)*Log[Sec[(e + f*x)/4]^2*(Sqrt[c + d] - Sqrt[d

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

*f*Sqrt[a*(1 + Sin[e + f*x])]))

________________________________________________________________________________________

fricas [B] time = 2.61, size = 1044, normalized size = 6.33 \[ \text {result too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/2*(a*d*sqrt(d/(a*c + a*d))*log((d^2*cos(f*x + e)^3 - (6*c*d + 7*d^2)*cos(f*x + e)^2 - c^2 - 2*c*d - d^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(d/(a*c + a*d)) - (c^2 + 8*c*d + 9*d^

2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 + 2*(3*c*d + 4*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(2)*sqrt(a)*c*log(-(cos(f*x + e)^2 - (cos

(f*x + e) - 2)*sin(f*x + e) - 2*sqrt(2)*sqrt(a*sin(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)) - sqrt(a)*(c - d)*lo

g((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)))/((a*c^2

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

qrt(-d/(a*c + a*d))/(d*cos(f*x + e))) + sqrt(2)*sqrt(a)*c*log(-(cos(f*x + e)^2 - (cos(f*x + e) - 2)*sin(f*x +

e) - 2*sqrt(2)*sqrt(a*sin(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)) - sqrt(a)*(c - d)*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)))/((a*c^2 - a*c*d)*f)]

________________________________________________________________________________________

giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab

le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (4*p

i/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unab

le to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*p

i/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (2*pi/t_nost

ep/2)>(-2*pi/t_nostep/2)Unable to check sign: (2*pi/t_nostep/2)>(-2*pi/t_nostep/2)Warning, integration of abs

or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(cos((f*t_nostep+exp(1)

)/2-pi/4))]Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(

-4*pi/t_nostep/2)Discontinuities at zeroes of cos((f*t_nostep+exp(1))/2-pi/4) were not checkedUnable to check

sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to

check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unab

le to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/

2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign