3.192 \(\int \frac {\csc ^3(x)}{(a+b \sin (x))^2} \, dx\)

Optimal. Leaf size=168 \[ -\frac {\left (a^2-3 b^2\right ) \cot (x) \csc (x)}{2 a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x) \csc (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (x))}{2 a^4}-\frac {2 b^3 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{3/2}}+\frac {b \left (2 a^2-3 b^2\right ) \cot (x)}{a^3 \left (a^2-b^2\right )} \]

[Out]

-2*b^3*(4*a^2-3*b^2)*arctan((b+a*tan(1/2*x))/(a^2-b^2)^(1/2))/a^4/(a^2-b^2)^(3/2)-1/2*(a^2+6*b^2)*arctanh(cos(
x))/a^4+b*(2*a^2-3*b^2)*cot(x)/a^3/(a^2-b^2)-1/2*(a^2-3*b^2)*cot(x)*csc(x)/a^2/(a^2-b^2)-b^2*cot(x)*csc(x)/a/(
a^2-b^2)/(a+b*sin(x))

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Rubi [A]  time = 0.57, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {2802, 3055, 3001, 3770, 2660, 618, 204} \[ -\frac {2 b^3 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{3/2}}+\frac {b \left (2 a^2-3 b^2\right ) \cot (x)}{a^3 \left (a^2-b^2\right )}-\frac {\left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (x))}{2 a^4}-\frac {\left (a^2-3 b^2\right ) \cot (x) \csc (x)}{2 a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x) \csc (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))} \]

Antiderivative was successfully verified.

[In]

Int[Csc[x]^3/(a + b*Sin[x])^2,x]

[Out]

(-2*b^3*(4*a^2 - 3*b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^4*(a^2 - b^2)^(3/2)) - ((a^2 + 6*b^2)*Arc
Tanh[Cos[x]])/(2*a^4) + (b*(2*a^2 - 3*b^2)*Cot[x])/(a^3*(a^2 - b^2)) - ((a^2 - 3*b^2)*Cot[x]*Csc[x])/(2*a^2*(a
^2 - b^2)) - (b^2*Cot[x]*Csc[x])/(a*(a^2 - b^2)*(a + b*Sin[x]))

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])

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 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 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 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 3770

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

Rubi steps

\begin {align*} \int \frac {\csc ^3(x)}{(a+b \sin (x))^2} \, dx &=-\frac {b^2 \cot (x) \csc (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\int \frac {\csc ^3(x) \left (a^2-3 b^2-a b \sin (x)+2 b^2 \sin ^2(x)\right )}{a+b \sin (x)} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac {\left (a^2-3 b^2\right ) \cot (x) \csc (x)}{2 a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x) \csc (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\int \frac {\csc ^2(x) \left (-2 b \left (2 a^2-3 b^2\right )+a \left (a^2+b^2\right ) \sin (x)+b \left (a^2-3 b^2\right ) \sin ^2(x)\right )}{a+b \sin (x)} \, dx}{2 a^2 \left (a^2-b^2\right )}\\ &=\frac {b \left (2 a^2-3 b^2\right ) \cot (x)}{a^3 \left (a^2-b^2\right )}-\frac {\left (a^2-3 b^2\right ) \cot (x) \csc (x)}{2 a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x) \csc (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\int \frac {\csc (x) \left (a^4+5 a^2 b^2-6 b^4+a b \left (a^2-3 b^2\right ) \sin (x)\right )}{a+b \sin (x)} \, dx}{2 a^3 \left (a^2-b^2\right )}\\ &=\frac {b \left (2 a^2-3 b^2\right ) \cot (x)}{a^3 \left (a^2-b^2\right )}-\frac {\left (a^2-3 b^2\right ) \cot (x) \csc (x)}{2 a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x) \csc (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\left (b^3 \left (4 a^2-3 b^2\right )\right ) \int \frac {1}{a+b \sin (x)} \, dx}{a^4 \left (a^2-b^2\right )}+\frac {\left (a^2+6 b^2\right ) \int \csc (x) \, dx}{2 a^4}\\ &=-\frac {\left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (x))}{2 a^4}+\frac {b \left (2 a^2-3 b^2\right ) \cot (x)}{a^3 \left (a^2-b^2\right )}-\frac {\left (a^2-3 b^2\right ) \cot (x) \csc (x)}{2 a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x) \csc (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}-\frac {\left (2 b^3 \left (4 a^2-3 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^4 \left (a^2-b^2\right )}\\ &=-\frac {\left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (x))}{2 a^4}+\frac {b \left (2 a^2-3 b^2\right ) \cot (x)}{a^3 \left (a^2-b^2\right )}-\frac {\left (a^2-3 b^2\right ) \cot (x) \csc (x)}{2 a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x) \csc (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}+\frac {\left (4 b^3 \left (4 a^2-3 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^4 \left (a^2-b^2\right )}\\ &=-\frac {2 b^3 \left (4 a^2-3 b^2\right ) \tan ^{-1}\left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \left (a^2-b^2\right )^{3/2}}-\frac {\left (a^2+6 b^2\right ) \tanh ^{-1}(\cos (x))}{2 a^4}+\frac {b \left (2 a^2-3 b^2\right ) \cot (x)}{a^3 \left (a^2-b^2\right )}-\frac {\left (a^2-3 b^2\right ) \cot (x) \csc (x)}{2 a^2 \left (a^2-b^2\right )}-\frac {b^2 \cot (x) \csc (x)}{a \left (a^2-b^2\right ) (a+b \sin (x))}\\ \end {align*}

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Mathematica [A]  time = 1.01, size = 171, normalized size = 1.02 \[ \frac {4 \left (a^2+6 b^2\right ) \log \left (\sin \left (\frac {x}{2}\right )\right )-4 \left (a^2+6 b^2\right ) \log \left (\cos \left (\frac {x}{2}\right )\right )+\frac {16 b^3 \left (3 b^2-4 a^2\right ) \tan ^{-1}\left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2}}-a^2 \csc ^2\left (\frac {x}{2}\right )+a^2 \sec ^2\left (\frac {x}{2}\right )-\frac {8 a b^4 \cos (x)}{(a-b) (a+b) (a+b \sin (x))}-8 a b \tan \left (\frac {x}{2}\right )+8 a b \cot \left (\frac {x}{2}\right )}{8 a^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[x]^3/(a + b*Sin[x])^2,x]

[Out]

((16*b^3*(-4*a^2 + 3*b^2)*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/(a^2 - b^2)^(3/2) + 8*a*b*Cot[x/2] - a^2*C
sc[x/2]^2 - 4*(a^2 + 6*b^2)*Log[Cos[x/2]] + 4*(a^2 + 6*b^2)*Log[Sin[x/2]] + a^2*Sec[x/2]^2 - (8*a*b^4*Cos[x])/
((a - b)*(a + b)*(a + b*Sin[x])) - 8*a*b*Tan[x/2])/(8*a^4)

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fricas [B]  time = 1.27, size = 1174, normalized size = 6.99 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*sin(x))^2,x, algorithm="fricas")

[Out]

[-1/4*(4*(2*a^5*b^2 - 5*a^3*b^4 + 3*a*b^6)*cos(x)^3 - 6*(a^6*b - 2*a^4*b^3 + a^2*b^5)*cos(x)*sin(x) + 2*(4*a^3
*b^3 - 3*a*b^5 - (4*a^3*b^3 - 3*a*b^5)*cos(x)^2 + (4*a^2*b^4 - 3*b^6 - (4*a^2*b^4 - 3*b^6)*cos(x)^2)*sin(x))*s
qrt(-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^7 - 6*a^5*b^2 + 11*a^3*b^4 - 6*a*b^6)*cos(x) +
(a^7 + 4*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6 - (a^7 + 4*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6)*cos(x)^2 + (a^6*b + 4*a^4*b
^3 - 11*a^2*b^5 + 6*b^7 - (a^6*b + 4*a^4*b^3 - 11*a^2*b^5 + 6*b^7)*cos(x)^2)*sin(x))*log(1/2*cos(x) + 1/2) - (
a^7 + 4*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6 - (a^7 + 4*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6)*cos(x)^2 + (a^6*b + 4*a^4*b^
3 - 11*a^2*b^5 + 6*b^7 - (a^6*b + 4*a^4*b^3 - 11*a^2*b^5 + 6*b^7)*cos(x)^2)*sin(x))*log(-1/2*cos(x) + 1/2))/(a
^9 - 2*a^7*b^2 + a^5*b^4 - (a^9 - 2*a^7*b^2 + a^5*b^4)*cos(x)^2 + (a^8*b - 2*a^6*b^3 + a^4*b^5 - (a^8*b - 2*a^
6*b^3 + a^4*b^5)*cos(x)^2)*sin(x)), -1/4*(4*(2*a^5*b^2 - 5*a^3*b^4 + 3*a*b^6)*cos(x)^3 - 6*(a^6*b - 2*a^4*b^3
+ a^2*b^5)*cos(x)*sin(x) - 4*(4*a^3*b^3 - 3*a*b^5 - (4*a^3*b^3 - 3*a*b^5)*cos(x)^2 + (4*a^2*b^4 - 3*b^6 - (4*a
^2*b^4 - 3*b^6)*cos(x)^2)*sin(x))*sqrt(a^2 - b^2)*arctan(-(a*sin(x) + b)/(sqrt(a^2 - b^2)*cos(x))) + 2*(a^7 -
6*a^5*b^2 + 11*a^3*b^4 - 6*a*b^6)*cos(x) + (a^7 + 4*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6 - (a^7 + 4*a^5*b^2 - 11*a^3
*b^4 + 6*a*b^6)*cos(x)^2 + (a^6*b + 4*a^4*b^3 - 11*a^2*b^5 + 6*b^7 - (a^6*b + 4*a^4*b^3 - 11*a^2*b^5 + 6*b^7)*
cos(x)^2)*sin(x))*log(1/2*cos(x) + 1/2) - (a^7 + 4*a^5*b^2 - 11*a^3*b^4 + 6*a*b^6 - (a^7 + 4*a^5*b^2 - 11*a^3*
b^4 + 6*a*b^6)*cos(x)^2 + (a^6*b + 4*a^4*b^3 - 11*a^2*b^5 + 6*b^7 - (a^6*b + 4*a^4*b^3 - 11*a^2*b^5 + 6*b^7)*c
os(x)^2)*sin(x))*log(-1/2*cos(x) + 1/2))/(a^9 - 2*a^7*b^2 + a^5*b^4 - (a^9 - 2*a^7*b^2 + a^5*b^4)*cos(x)^2 + (
a^8*b - 2*a^6*b^3 + a^4*b^5 - (a^8*b - 2*a^6*b^3 + a^4*b^5)*cos(x)^2)*sin(x))]

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giac [A]  time = 0.18, size = 215, normalized size = 1.28 \[ -\frac {2 \, {\left (4 \, a^{2} b^{3} - 3 \, b^{5}\right )} {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} \sqrt {a^{2} - b^{2}}} - \frac {2 \, {\left (b^{5} \tan \left (\frac {1}{2} \, x\right ) + a b^{4}\right )}}{{\left (a^{6} - a^{4} b^{2}\right )} {\left (a \tan \left (\frac {1}{2} \, x\right )^{2} + 2 \, b \tan \left (\frac {1}{2} \, x\right ) + a\right )}} + \frac {{\left (a^{2} + 6 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{2 \, a^{4}} + \frac {a^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, x\right )}{8 \, a^{4}} - \frac {6 \, a^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 36 \, b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} - 8 \, a b \tan \left (\frac {1}{2} \, x\right ) + a^{2}}{8 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*sin(x))^2,x, algorithm="giac")

[Out]

-2*(4*a^2*b^3 - 3*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^6 -
a^4*b^2)*sqrt(a^2 - b^2)) - 2*(b^5*tan(1/2*x) + a*b^4)/((a^6 - a^4*b^2)*(a*tan(1/2*x)^2 + 2*b*tan(1/2*x) + a))
+ 1/2*(a^2 + 6*b^2)*log(abs(tan(1/2*x)))/a^4 + 1/8*(a^2*tan(1/2*x)^2 - 8*a*b*tan(1/2*x))/a^4 - 1/8*(6*a^2*tan
(1/2*x)^2 + 36*b^2*tan(1/2*x)^2 - 8*a*b*tan(1/2*x) + a^2)/(a^4*tan(1/2*x)^2)

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maple [A]  time = 0.15, size = 236, normalized size = 1.40 \[ \frac {\tan ^{2}\left (\frac {x}{2}\right )}{8 a^{2}}-\frac {\tan \left (\frac {x}{2}\right ) b}{a^{3}}-\frac {1}{8 a^{2} \tan \left (\frac {x}{2}\right )^{2}}+\frac {\ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 a^{2}}+\frac {3 \ln \left (\tan \left (\frac {x}{2}\right )\right ) b^{2}}{a^{4}}+\frac {b}{a^{3} \tan \left (\frac {x}{2}\right )}-\frac {2 b^{5} \tan \left (\frac {x}{2}\right )}{a^{4} \left (\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) a +2 \tan \left (\frac {x}{2}\right ) b +a \right ) \left (a^{2}-b^{2}\right )}-\frac {2 b^{4}}{a^{3} \left (\left (\tan ^{2}\left (\frac {x}{2}\right )\right ) a +2 \tan \left (\frac {x}{2}\right ) b +a \right ) \left (a^{2}-b^{2}\right )}-\frac {8 b^{3} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{2} \left (a^{2}-b^{2}\right )^{\frac {3}{2}}}+\frac {6 b^{5} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{4} \left (a^{2}-b^{2}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(x)^3/(a+b*sin(x))^2,x)

[Out]

1/8/a^2*tan(1/2*x)^2-1/a^3*tan(1/2*x)*b-1/8/a^2/tan(1/2*x)^2+1/2/a^2*ln(tan(1/2*x))+3/a^4*ln(tan(1/2*x))*b^2+b
/a^3/tan(1/2*x)-2/a^4*b^5/(tan(1/2*x)^2*a+2*tan(1/2*x)*b+a)/(a^2-b^2)*tan(1/2*x)-2/a^3*b^4/(tan(1/2*x)^2*a+2*t
an(1/2*x)*b+a)/(a^2-b^2)-8/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))+6/a^4*b^5/
(a^2-b^2)^(3/2)*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)^3/(a+b*sin(x))^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

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mupad [B]  time = 7.67, size = 1576, normalized size = 9.38 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(x)^3*(a + b*sin(x))^2),x)

[Out]

tan(x/2)^2/(8*a^2) - (a^2/2 - 3*a*b*tan(x/2) + (tan(x/2)^2*(a^4 + 32*b^4 - 17*a^2*b^2))/(2*(a^2 - b^2)) + (4*b
*tan(x/2)^3*(2*b^4 - a^4 + a^2*b^2))/(a*(a^2 - b^2)))/(4*a^4*tan(x/2)^2 + 4*a^4*tan(x/2)^4 + 8*a^3*b*tan(x/2)^
3) + (log(tan(x/2))*(a^2 + 6*b^2))/(2*a^4) - (b*tan(x/2))/a^3 + (b^3*atan(((b^3*(4*a^2 - 3*b^2)*(-(a + b)^3*(a
- b)^3)^(1/2)*((a^8*b - 12*a^4*b^5 + 13*a^6*b^3)/(a^8 - a^6*b^2) - (tan(x/2)*(a^10 - 24*a^2*b^8 + 56*a^4*b^6
- 35*a^6*b^4 + 2*a^8*b^2))/(a^9 + a^5*b^4 - 2*a^7*b^2) + (b^3*(4*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((2
*a^10*b - 2*a^8*b^3)/(a^8 - a^6*b^2) - (tan(x/2)*(6*a^12 - 8*a^6*b^6 + 22*a^8*b^4 - 20*a^10*b^2))/(a^9 + a^5*b
^4 - 2*a^7*b^2)))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))*1i)/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2) - (b^
3*(4*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((tan(x/2)*(a^10 - 24*a^2*b^8 + 56*a^4*b^6 - 35*a^6*b^4 + 2*a^8
*b^2))/(a^9 + a^5*b^4 - 2*a^7*b^2) - (a^8*b - 12*a^4*b^5 + 13*a^6*b^3)/(a^8 - a^6*b^2) + (b^3*(4*a^2 - 3*b^2)*
(-(a + b)^3*(a - b)^3)^(1/2)*((2*a^10*b - 2*a^8*b^3)/(a^8 - a^6*b^2) - (tan(x/2)*(6*a^12 - 8*a^6*b^6 + 22*a^8*
b^4 - 20*a^10*b^2))/(a^9 + a^5*b^4 - 2*a^7*b^2)))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2))*1i)/(a^10 - a^4*b^
6 + 3*a^6*b^4 - 3*a^8*b^2))/((2*(21*a^2*b^5 - 18*b^7 + 4*a^4*b^3))/(a^8 - a^6*b^2) + (2*tan(x/2)*(18*b^8 - 30*
a^2*b^6 + 8*a^4*b^4))/(a^9 + a^5*b^4 - 2*a^7*b^2) + (b^3*(4*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((a^8*b
- 12*a^4*b^5 + 13*a^6*b^3)/(a^8 - a^6*b^2) - (tan(x/2)*(a^10 - 24*a^2*b^8 + 56*a^4*b^6 - 35*a^6*b^4 + 2*a^8*b^
2))/(a^9 + a^5*b^4 - 2*a^7*b^2) + (b^3*(4*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*((2*a^10*b - 2*a^8*b^3)/(a
^8 - a^6*b^2) - (tan(x/2)*(6*a^12 - 8*a^6*b^6 + 22*a^8*b^4 - 20*a^10*b^2))/(a^9 + a^5*b^4 - 2*a^7*b^2)))/(a^10
- a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2) + (b^3*(4*a^2 - 3*b^2)*(-(a + b
)^3*(a - b)^3)^(1/2)*((tan(x/2)*(a^10 - 24*a^2*b^8 + 56*a^4*b^6 - 35*a^6*b^4 + 2*a^8*b^2))/(a^9 + a^5*b^4 - 2*
a^7*b^2) - (a^8*b - 12*a^4*b^5 + 13*a^6*b^3)/(a^8 - a^6*b^2) + (b^3*(4*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/
2)*((2*a^10*b - 2*a^8*b^3)/(a^8 - a^6*b^2) - (tan(x/2)*(6*a^12 - 8*a^6*b^6 + 22*a^8*b^4 - 20*a^10*b^2))/(a^9 +
a^5*b^4 - 2*a^7*b^2)))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)))*
(4*a^2 - 3*b^2)*(-(a + b)^3*(a - b)^3)^(1/2)*2i)/(a^10 - a^4*b^6 + 3*a^6*b^4 - 3*a^8*b^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{3}{\relax (x )}}{\left (a + b \sin {\relax (x )}\right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(x)**3/(a+b*sin(x))**2,x)

[Out]

Integral(csc(x)**3/(a + b*sin(x))**2, x)

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