3.199 \(\int \frac{\csc (x)}{(a+b \sin (x))^3} \, dx\)
Optimal. Leaf size=145 \[ -\frac{b \left (-5 a^2 b^2+6 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{5/2}}-\frac{b^2 \left (5 a^2-2 b^2\right ) \cos (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac{b^2 \cos (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{\tanh ^{-1}(\cos (x))}{a^3} \]
[Out]
-((b*(6*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(5/2))) - ArcTanh[
Cos[x]]/a^3 - (b^2*Cos[x])/(2*a*(a^2 - b^2)*(a + b*Sin[x])^2) - (b^2*(5*a^2 - 2*b^2)*Cos[x])/(2*a^2*(a^2 - b^2
)^2*(a + b*Sin[x]))
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Rubi [A] time = 0.370485, antiderivative size = 145, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 11, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.636, Rules used =
{2802, 3055, 3001, 3770, 2660, 618, 204} \[ -\frac{b \left (-5 a^2 b^2+6 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{5/2}}-\frac{b^2 \left (5 a^2-2 b^2\right ) \cos (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac{b^2 \cos (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{\tanh ^{-1}(\cos (x))}{a^3} \]
Antiderivative was successfully verified.
[In]
Int[Csc[x]/(a + b*Sin[x])^3,x]
[Out]
-((b*(6*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(5/2))) - ArcTanh[
Cos[x]]/a^3 - (b^2*Cos[x])/(2*a*(a^2 - b^2)*(a + b*Sin[x])^2) - (b^2*(5*a^2 - 2*b^2)*Cos[x])/(2*a^2*(a^2 - b^2
)^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 3055
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*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] + Dis
t[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[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && Lt
Q[m, -1] && ((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))^3} \, dx &=-\frac{b^2 \cos (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}+\frac{\int \frac{\csc (x) \left (2 \left (a^2-b^2\right )-2 a b \sin (x)+b^2 \sin ^2(x)\right )}{(a+b \sin (x))^2} \, dx}{2 a \left (a^2-b^2\right )}\\ &=-\frac{b^2 \cos (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{b^2 \left (5 a^2-2 b^2\right ) \cos (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{\int \frac{\csc (x) \left (2 \left (a^2-b^2\right )^2-a b \left (4 a^2-b^2\right ) \sin (x)\right )}{a+b \sin (x)} \, dx}{2 a^2 \left (a^2-b^2\right )^2}\\ &=-\frac{b^2 \cos (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{b^2 \left (5 a^2-2 b^2\right ) \cos (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{\int \csc (x) \, dx}{a^3}-\frac{\left (b \left (6 a^4-5 a^2 b^2+2 b^4\right )\right ) \int \frac{1}{a+b \sin (x)} \, dx}{2 a^3 \left (a^2-b^2\right )^2}\\ &=-\frac{\tanh ^{-1}(\cos (x))}{a^3}-\frac{b^2 \cos (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{b^2 \left (5 a^2-2 b^2\right ) \cos (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}-\frac{\left (b \left (6 a^4-5 a^2 b^2+2 b^4\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^3 \left (a^2-b^2\right )^2}\\ &=-\frac{\tanh ^{-1}(\cos (x))}{a^3}-\frac{b^2 \cos (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{b^2 \left (5 a^2-2 b^2\right ) \cos (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}+\frac{\left (2 b \left (6 a^4-5 a^2 b^2+2 b^4\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^3 \left (a^2-b^2\right )^2}\\ &=-\frac{b \left (6 a^4-5 a^2 b^2+2 b^4\right ) \tan ^{-1}\left (\frac{b+a \tan \left (\frac{x}{2}\right )}{\sqrt{a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{5/2}}-\frac{\tanh ^{-1}(\cos (x))}{a^3}-\frac{b^2 \cos (x)}{2 a \left (a^2-b^2\right ) (a+b \sin (x))^2}-\frac{b^2 \left (5 a^2-2 b^2\right ) \cos (x)}{2 a^2 \left (a^2-b^2\right )^2 (a+b \sin (x))}\\ \end{align*}
Mathematica [A] time = 0.795476, size = 140, normalized size = 0.97 \[ -\frac{\frac{2 b \left (-5 a^2 b^2+6 a^4+2 b^4\right ) \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\left (a^2-b^2\right )^{5/2}}+\frac{a b^2 \cos (x) \left (b \left (5 a^2-2 b^2\right ) \sin (x)+6 a^3-3 a b^2\right )}{(a-b)^2 (a+b)^2 (a+b \sin (x))^2}-2 \log \left (\sin \left (\frac{x}{2}\right )\right )+2 \log \left (\cos \left (\frac{x}{2}\right )\right )}{2 a^3} \]
Antiderivative was successfully verified.
[In]
Integrate[Csc[x]/(a + b*Sin[x])^3,x]
[Out]
-((2*b*(6*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(5/2) + 2*Log[Cos[x/2
]] - 2*Log[Sin[x/2]] + (a*b^2*Cos[x]*(6*a^3 - 3*a*b^2 + b*(5*a^2 - 2*b^2)*Sin[x]))/((a - b)^2*(a + b)^2*(a + b
*Sin[x])^2))/(2*a^3)
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Maple [B] time = 0.075, size = 614, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(csc(x)/(a+b*sin(x))^3,x)
[Out]
-7*b^3/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^3+4/a^2*b^5/(tan(1/2*x)^2*a+2*tan(1/
2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^3-6*b^2*a/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*t
an(1/2*x)^2-9/a*b^4/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^2+6/a^3*b^6/(tan(1/2*x)
^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)^2-17*b^3/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^
2*b^2+b^4)*tan(1/2*x)+8/a^2*b^5/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*x)-6*a*b^2/(ta
n(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^2+b^4)+3/a*b^4/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)^2/(a^4-2*a^2*b^
2+b^4)-6*b*a/(a^4-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))+5/a*b^3/(a^4
-2*a^2*b^2+b^4)/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))-2/a^3*b^5/(a^4-2*a^2*b^2+b^4)
/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))+1/a^3*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))^3,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 8.20914, size = 2233, normalized size = 15.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(x)/(a+b*sin(x))^3,x, algorithm="fricas")
[Out]
[-1/4*(2*(5*a^5*b^3 - 7*a^3*b^5 + 2*a*b^7)*cos(x)*sin(x) + (6*a^6*b + a^4*b^3 - 3*a^2*b^5 + 2*b^7 - (6*a^4*b^3
- 5*a^2*b^5 + 2*b^7)*cos(x)^2 + 2*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6)*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)) + 6*(2*a^6*b^2 - 3*a^4*b^4 + a^2*b^6)*cos(x) + 2*(a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8 -
(a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(x)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*sin(x))*log(1/2*c
os(x) + 1/2) - 2*(a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8 - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(x)^2 + 2*(a^
7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*sin(x))*log(-1/2*cos(x) + 1/2))/(a^11 - 2*a^9*b^2 + 2*a^5*b^6 - a^3*b^8 -
(a^9*b^2 - 3*a^7*b^4 + 3*a^5*b^6 - a^3*b^8)*cos(x)^2 + 2*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*sin(x)),
-1/2*((5*a^5*b^3 - 7*a^3*b^5 + 2*a*b^7)*cos(x)*sin(x) - (6*a^6*b + a^4*b^3 - 3*a^2*b^5 + 2*b^7 - (6*a^4*b^3 -
5*a^2*b^5 + 2*b^7)*cos(x)^2 + 2*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) +
b)/(sqrt(a^2 - b^2)*cos(x))) + 3*(2*a^6*b^2 - 3*a^4*b^4 + a^2*b^6)*cos(x) + (a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8
- (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(x)^2 + 2*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*sin(x))*log(1/
2*cos(x) + 1/2) - (a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8 - (a^6*b^2 - 3*a^4*b^4 + 3*a^2*b^6 - b^8)*cos(x)^2 + 2*(a
^7*b - 3*a^5*b^3 + 3*a^3*b^5 - a*b^7)*sin(x))*log(-1/2*cos(x) + 1/2))/(a^11 - 2*a^9*b^2 + 2*a^5*b^6 - a^3*b^8
- (a^9*b^2 - 3*a^7*b^4 + 3*a^5*b^6 - a^3*b^8)*cos(x)^2 + 2*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*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 )^{3}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(x)/(a+b*sin(x))**3,x)
[Out]
Integral(csc(x)/(a + b*sin(x))**3, x)
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Giac [A] time = 1.62522, size = 332, normalized size = 2.29 \begin{align*} -\frac{{\left (6 \, a^{4} b - 5 \, a^{2} b^{3} + 2 \, b^{5}\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^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} \sqrt{a^{2} - b^{2}}} - \frac{7 \, a^{3} b^{3} \tan \left (\frac{1}{2} \, x\right )^{3} - 4 \, a b^{5} \tan \left (\frac{1}{2} \, x\right )^{3} + 6 \, a^{4} b^{2} \tan \left (\frac{1}{2} \, x\right )^{2} + 9 \, a^{2} b^{4} \tan \left (\frac{1}{2} \, x\right )^{2} - 6 \, b^{6} \tan \left (\frac{1}{2} \, x\right )^{2} + 17 \, a^{3} b^{3} \tan \left (\frac{1}{2} \, x\right ) - 8 \, a b^{5} \tan \left (\frac{1}{2} \, x\right ) + 6 \, a^{4} b^{2} - 3 \, a^{2} b^{4}}{{\left (a^{7} - 2 \, a^{5} b^{2} + a^{3} b^{4}\right )}{\left (a \tan \left (\frac{1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac{1}{2} \, x\right ) + a\right )}^{2}} + \frac{\log \left ({\left | \tan \left (\frac{1}{2} \, x\right ) \right |}\right )}{a^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(x)/(a+b*sin(x))^3,x, algorithm="giac")
[Out]
-(6*a^4*b - 5*a^2*b^3 + 2*b^5)*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))/
((a^7 - 2*a^5*b^2 + a^3*b^4)*sqrt(a^2 - b^2)) - (7*a^3*b^3*tan(1/2*x)^3 - 4*a*b^5*tan(1/2*x)^3 + 6*a^4*b^2*tan
(1/2*x)^2 + 9*a^2*b^4*tan(1/2*x)^2 - 6*b^6*tan(1/2*x)^2 + 17*a^3*b^3*tan(1/2*x) - 8*a*b^5*tan(1/2*x) + 6*a^4*b
^2 - 3*a^2*b^4)/((a^7 - 2*a^5*b^2 + a^3*b^4)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a)^2) + log(abs(tan(1/2*x)))/a
^3