3.190 \(\int \frac{\csc (x)}{(a+b \sin (x))^2} \, dx\)

Optimal. Leaf size=93 \[ -\frac{2 b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2}}-\frac{b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}-\frac{\tanh ^{-1}(\cos (x))}{a^2} \]

[Out]

(-2*b*(2*a^2 - b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(3/2)) - ArcTanh[Cos[x]]/a^2 -
(b^2*Cos[x])/(a*(a^2 - b^2)*(a + b*Sin[x]))

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Rubi [A]  time = 0.179824, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.546, Rules used = {2802, 3001, 3770, 2660, 618, 204} \[ -\frac{2 b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2}}-\frac{b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}-\frac{\tanh ^{-1}(\cos (x))}{a^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]/(a + b*Sin[x])^2,x]

[Out]

(-2*b*(2*a^2 - b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2*(a^2 - b^2)^(3/2)) - ArcTanh[Cos[x]]/a^2 -
(b^2*Cos[x])/(a*(a^2 - b^2)*(a + b*Sin[x]))

Rule 2802

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 -
b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n
*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n
+ 3)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !
(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3001

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

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

Rubi steps

\begin{align*} \int \frac{\csc (x)}{(a+b \sin (x))^2} \, dx &=-\frac{b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac{\int \frac{\csc (x) \left (a^2-b^2-a b \sin (x)\right )}{a+b \sin (x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac{\int \csc (x) \, dx}{a^2}-\frac{\left (b \left (2 a^2-b^2\right )\right ) \int \frac{1}{a+b \sin (x)} \, dx}{a^2 \left (a^2-b^2\right )}\\ &=-\frac{\tanh ^{-1}(\cos (x))}{a^2}-\frac{b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}-\frac{\left (2 b \left (2 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac{x}{2}\right )\right )}{a^2 \left (a^2-b^2\right )}\\ &=-\frac{\tanh ^{-1}(\cos (x))}{a^2}-\frac{b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac{\left (4 b \left (2 a^2-b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac{x}{2}\right )\right )}{a^2 \left (a^2-b^2\right )}\\ &=-\frac{2 b \left (2 a^2-b^2\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2}}-\frac{\tanh ^{-1}(\cos (x))}{a^2}-\frac{b^2 \cos (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}\\ \end{align*}

Mathematica [A]  time = 0.239066, size = 99, normalized size = 1.06 \[ \frac{\frac{2 b \left (b^2-2 a^2\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-\frac{a b^2 \cos (x)}{(a-b) (a+b) (a+b \sin (x))}+\log \left (\sin \left (\frac{x}{2}\right )\right )-\log \left (\cos \left (\frac{x}{2}\right )\right )}{a^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]/(a + b*Sin[x])^2,x]

[Out]

((2*b*(-2*a^2 + b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) - Log[Cos[x/2]] + Log[Sin[x/2
]] - (a*b^2*Cos[x])/((a - b)*(a + b)*(a + b*Sin[x])))/a^2

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Maple [A]  time = 0.062, size = 174, normalized size = 1.9 \begin{align*} -2\,{\frac{{b}^{3}\tan \left ( x/2 \right ) }{{a}^{2} \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}-2\,{\frac{{b}^{2}}{a \left ( \left ( \tan \left ( x/2 \right ) \right ) ^{2}a+2\,\tan \left ( x/2 \right ) b+a \right ) \left ({a}^{2}-{b}^{2} \right ) }}-4\,{\frac{b}{ \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+2\,{\frac{{b}^{3}}{{a}^{2} \left ({a}^{2}-{b}^{2} \right ) ^{3/2}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+{\frac{1}{{a}^{2}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)/(a+b*sin(x))^2,x)

[Out]

-2/a^2*b^3/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)/(a^2-b^2)*tan(1/2*x)-2/a*b^2/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)/(a
^2-b^2)-4*b/(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))+2/a^2*b^3/(a^2-b^2)^(3/2)*arctan(
1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))+1/a^2*ln(tan(1/2*x))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a+b*sin(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.09561, size = 1184, normalized size = 12.73 \begin{align*} \left [-\frac{{\left (2 \, a^{3} b - a b^{3} +{\left (2 \, a^{2} b^{2} - b^{4}\right )} \sin \left (x\right )\right )} \sqrt{-a^{2} + b^{2}} \log \left (-\frac{{\left (2 \, a^{2} - b^{2}\right )} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2} - 2 \,{\left (a \cos \left (x\right ) \sin \left (x\right ) + b \cos \left (x\right )\right )} \sqrt{-a^{2} + b^{2}}}{b^{2} \cos \left (x\right )^{2} - 2 \, a b \sin \left (x\right ) - a^{2} - b^{2}}\right ) + 2 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) -{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right )}{2 \,{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4} +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \sin \left (x\right )\right )}}, \frac{2 \,{\left (2 \, a^{3} b - a b^{3} +{\left (2 \, a^{2} b^{2} - b^{4}\right )} \sin \left (x\right )\right )} \sqrt{a^{2} - b^{2}} \arctan \left (-\frac{a \sin \left (x\right ) + b}{\sqrt{a^{2} - b^{2}} \cos \left (x\right )}\right ) - 2 \,{\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (x\right ) -{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )} \log \left (\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right ) +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \sin \left (x\right )\right )} \log \left (-\frac{1}{2} \, \cos \left (x\right ) + \frac{1}{2}\right )}{2 \,{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4} +{\left (a^{6} b - 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} \sin \left (x\right )\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a+b*sin(x))^2,x, algorithm="fricas")

[Out]

[-1/2*((2*a^3*b - a*b^3 + (2*a^2*b^2 - b^4)*sin(x))*sqrt(-a^2 + b^2)*log(-((2*a^2 - b^2)*cos(x)^2 - 2*a*b*sin(
x) - a^2 - b^2 - 2*(a*cos(x)*sin(x) + b*cos(x))*sqrt(-a^2 + b^2))/(b^2*cos(x)^2 - 2*a*b*sin(x) - a^2 - b^2)) +
2*(a^3*b^2 - a*b^4)*cos(x) + (a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*sin(x))*log(1/2*cos(x) + 1/
2) - (a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b^5)*sin(x))*log(-1/2*cos(x) + 1/2))/(a^7 - 2*a^5*b^2 + a
^3*b^4 + (a^6*b - 2*a^4*b^3 + a^2*b^5)*sin(x)), 1/2*(2*(2*a^3*b - a*b^3 + (2*a^2*b^2 - b^4)*sin(x))*sqrt(a^2 -
b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x))) - 2*(a^3*b^2 - a*b^4)*cos(x) - (a^5 - 2*a^3*b^2 + a*b^4
+ (a^4*b - 2*a^2*b^3 + b^5)*sin(x))*log(1/2*cos(x) + 1/2) + (a^5 - 2*a^3*b^2 + a*b^4 + (a^4*b - 2*a^2*b^3 + b
^5)*sin(x))*log(-1/2*cos(x) + 1/2))/(a^7 - 2*a^5*b^2 + a^3*b^4 + (a^6*b - 2*a^4*b^3 + a^2*b^5)*sin(x))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc{\left (x \right )}}{\left (a + b \sin{\left (x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a+b*sin(x))**2,x)

[Out]

Integral(csc(x)/(a + b*sin(x))**2, x)

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Giac [A]  time = 1.60205, size = 181, normalized size = 1.95 \begin{align*} -\frac{2 \,{\left (2 \, a^{2} b - b^{3}\right )}{\left (\pi \left \lfloor \frac{x}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \arctan \left (\frac{a \tan \left (\frac{1}{2} \, x\right ) + b}{\sqrt{a^{2} - b^{2}}}\right )\right )}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt{a^{2} - b^{2}}} - \frac{2 \,{\left (b^{3} \tan \left (\frac{1}{2} \, x\right ) + a b^{2}\right )}}{{\left (a^{4} - a^{2} b^{2}\right )}{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\right )}} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right )}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)/(a+b*sin(x))^2,x, algorithm="giac")

[Out]

-2*(2*a^2*b - b^3)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))/((a^4 - a^2*
b^2)*sqrt(a^2 - b^2)) - 2*(b^3*tan(1/2*x) + a*b^2)/((a^4 - a^2*b^2)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)) + l
og(abs(tan(1/2*x)))/a^2