\(\int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx\) [184]

Optimal result

Integrand size = 13, antiderivative size = 112 \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\frac {2 b^4 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \sqrt {a^2-b^2}}+\frac {b \left (a^2+2 b^2\right ) \text {arctanh}(\cos (x))}{2 a^4}-\frac {\left (2 a^2+3 b^2\right ) \cot (x)}{3 a^3}+\frac {b \cot (x) \csc (x)}{2 a^2}-\frac {\cot (x) \csc ^2(x)}{3 a} \]

[Out]

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

Rubi [A] (verified)

Time = 0.31 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.538, Rules used = {2881, 3134, 3080, 3855, 2739, 632, 210} \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\frac {b \cot (x) \csc (x)}{2 a^2}+\frac {2 b^4 \arctan \left (\frac {a \tan \left (\frac {x}{2}\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 \sqrt {a^2-b^2}}+\frac {b \left (a^2+2 b^2\right ) \text {arctanh}(\cos (x))}{2 a^4}-\frac {\left (2 a^2+3 b^2\right ) \cot (x)}{3 a^3}-\frac {\cot (x) \csc ^2(x)}{3 a} \]

[In]

Int[Csc[x]^4/(a + b*Sin[x]),x]

[Out]

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

Rule 210

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

Rule 632

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 2739

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 2881

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(-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 3080

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 3134

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] + D
ist[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] &&
LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n]
&&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3855

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

Rubi steps \begin{align*} \text {integral}& = -\frac {\cot (x) \csc ^2(x)}{3 a}+\frac {\int \frac {\csc ^3(x) \left (-3 b+2 a \sin (x)+2 b \sin ^2(x)\right )}{a+b \sin (x)} \, dx}{3 a} \\ & = \frac {b \cot (x) \csc (x)}{2 a^2}-\frac {\cot (x) \csc ^2(x)}{3 a}+\frac {\int \frac {\csc ^2(x) \left (2 \left (2 a^2+3 b^2\right )+a b \sin (x)-3 b^2 \sin ^2(x)\right )}{a+b \sin (x)} \, dx}{6 a^2} \\ & = -\frac {\left (2 a^2+3 b^2\right ) \cot (x)}{3 a^3}+\frac {b \cot (x) \csc (x)}{2 a^2}-\frac {\cot (x) \csc ^2(x)}{3 a}+\frac {\int \frac {\csc (x) \left (-3 b \left (a^2+2 b^2\right )-3 a b^2 \sin (x)\right )}{a+b \sin (x)} \, dx}{6 a^3} \\ & = -\frac {\left (2 a^2+3 b^2\right ) \cot (x)}{3 a^3}+\frac {b \cot (x) \csc (x)}{2 a^2}-\frac {\cot (x) \csc ^2(x)}{3 a}+\frac {b^4 \int \frac {1}{a+b \sin (x)} \, dx}{a^4}-\frac {\left (b \left (a^2+2 b^2\right )\right ) \int \csc (x) \, dx}{2 a^4} \\ & = \frac {b \left (a^2+2 b^2\right ) \text {arctanh}(\cos (x))}{2 a^4}-\frac {\left (2 a^2+3 b^2\right ) \cot (x)}{3 a^3}+\frac {b \cot (x) \csc (x)}{2 a^2}-\frac {\cot (x) \csc ^2(x)}{3 a}+\frac {\left (2 b^4\right ) \text {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {x}{2}\right )\right )}{a^4} \\ & = \frac {b \left (a^2+2 b^2\right ) \text {arctanh}(\cos (x))}{2 a^4}-\frac {\left (2 a^2+3 b^2\right ) \cot (x)}{3 a^3}+\frac {b \cot (x) \csc (x)}{2 a^2}-\frac {\cot (x) \csc ^2(x)}{3 a}-\frac {\left (4 b^4\right ) \text {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} \\ & = \frac {2 b^4 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{a^4 \sqrt {a^2-b^2}}+\frac {b \left (a^2+2 b^2\right ) \text {arctanh}(\cos (x))}{2 a^4}-\frac {\left (2 a^2+3 b^2\right ) \cot (x)}{3 a^3}+\frac {b \cot (x) \csc (x)}{2 a^2}-\frac {\cot (x) \csc ^2(x)}{3 a} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.12 \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\frac {\frac {24 b^4 \arctan \left (\frac {b+a \tan \left (\frac {x}{2}\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}+a \left (2 a^2+3 b^2\right ) \cos (3 x) \csc ^3(x)-3 a \cot (x) \csc (x) \left (-2 a b+\left (2 a^2+b^2\right ) \csc (x)\right )+6 b \left (a^2+2 b^2\right ) \left (\log \left (\cos \left (\frac {x}{2}\right )\right )-\log \left (\sin \left (\frac {x}{2}\right )\right )\right )}{12 a^4} \]

[In]

Integrate[Csc[x]^4/(a + b*Sin[x]),x]

[Out]

((24*b^4*ArcTan[(b + a*Tan[x/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2] + a*(2*a^2 + 3*b^2)*Cos[3*x]*Csc[x]^3 - 3*a
*Cot[x]*Csc[x]*(-2*a*b + (2*a^2 + b^2)*Csc[x]) + 6*b*(a^2 + 2*b^2)*(Log[Cos[x/2]] - Log[Sin[x/2]]))/(12*a^4)

Maple [A] (verified)

Time = 0.69 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.39

method	result	size
default	\(\frac {\frac {\left (\tan ^{3}\left (\frac {x}{2}\right )\right ) a^{2}}{3}-a b \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+3 a^{2} \tan \left (\frac {x}{2}\right )+4 b^{2} \tan \left (\frac {x}{2}\right )}{8 a^{3}}-\frac {1}{24 a \tan \left (\frac {x}{2}\right )^{3}}-\frac {3 a^{2}+4 b^{2}}{8 a^{3} \tan \left (\frac {x}{2}\right )}+\frac {b}{8 a^{2} \tan \left (\frac {x}{2}\right )^{2}}-\frac {b \left (a^{2}+2 b^{2}\right ) \ln \left (\tan \left (\frac {x}{2}\right )\right )}{2 a^{4}}+\frac {2 b^{4} \arctan \left (\frac {2 a \tan \left (\frac {x}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{a^{4} \sqrt {a^{2}-b^{2}}}\)	\(156\)
risch	\(-\frac {6 i b^{2} {\mathrm e}^{4 i x}+3 a b \,{\mathrm e}^{5 i x}-12 i a^{2} {\mathrm e}^{2 i x}-12 i b^{2} {\mathrm e}^{2 i x}+4 i a^{2}+6 i b^{2}-3 b a \,{\mathrm e}^{i x}}{3 a^{3} \left ({\mathrm e}^{2 i x}-1\right )^{3}}+\frac {b \ln \left ({\mathrm e}^{i x}+1\right )}{2 a^{2}}+\frac {b^{3} \ln \left ({\mathrm e}^{i x}+1\right )}{a^{4}}-\frac {b \ln \left ({\mathrm e}^{i x}-1\right )}{2 a^{2}}-\frac {b^{3} \ln \left ({\mathrm e}^{i x}-1\right )}{a^{4}}-\frac {b^{4} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}-a^{2}+b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, a^{4}}+\frac {b^{4} \ln \left ({\mathrm e}^{i x}+\frac {i a \sqrt {-a^{2}+b^{2}}+a^{2}-b^{2}}{b \sqrt {-a^{2}+b^{2}}}\right )}{\sqrt {-a^{2}+b^{2}}\, a^{4}}\)	\(268\)

[In]

int(csc(x)^4/(a+b*sin(x)),x,method=_RETURNVERBOSE)

[Out]

1/8/a^3*(1/3*tan(1/2*x)^3*a^2-a*b*tan(1/2*x)^2+3*a^2*tan(1/2*x)+4*b^2*tan(1/2*x))-1/24/a/tan(1/2*x)^3-1/8*(3*a
^2+4*b^2)/a^3/tan(1/2*x)+1/8/a^2*b/tan(1/2*x)^2-1/2/a^4*b*(a^2+2*b^2)*ln(tan(1/2*x))+2*b^4/a^4/(a^2-b^2)^(1/2)
*arctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 257 vs. \(2 (98) = 196\).

Time = 0.44 (sec) , antiderivative size = 577, normalized size of antiderivative = 5.15 \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\left [\frac {4 \, {\left (2 \, a^{5} + a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (x\right )^{3} + 6 \, {\left (b^{4} \cos \left (x\right )^{2} - b^{4}\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 ) \sin \left (x\right ) + 6 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (x\right ) \sin \left (x\right ) + 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5} - {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (x\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) - 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5} - {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) - 12 \, {\left (a^{5} - a b^{4}\right )} \cos \left (x\right )}{12 \, {\left (a^{6} - a^{4} b^{2} - {\left (a^{6} - a^{4} b^{2}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}, \frac {4 \, {\left (2 \, a^{5} + a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (x\right )^{3} + 12 \, {\left (b^{4} \cos \left (x\right )^{2} - b^{4}\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 ) \sin \left (x\right ) + 6 \, {\left (a^{4} b - a^{2} b^{3}\right )} \cos \left (x\right ) \sin \left (x\right ) + 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5} - {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (x\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) - 3 \, {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5} - {\left (a^{4} b + a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (x\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (x\right ) + \frac {1}{2}\right ) \sin \left (x\right ) - 12 \, {\left (a^{5} - a b^{4}\right )} \cos \left (x\right )}{12 \, {\left (a^{6} - a^{4} b^{2} - {\left (a^{6} - a^{4} b^{2}\right )} \cos \left (x\right )^{2}\right )} \sin \left (x\right )}\right ] \]

[In]

integrate(csc(x)^4/(a+b*sin(x)),x, algorithm="fricas")

[Out]

[1/12*(4*(2*a^5 + a^3*b^2 - 3*a*b^4)*cos(x)^3 + 6*(b^4*cos(x)^2 - b^4)*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))*sin(x) + 6*(a^4*b - a^2*b^3)*cos(x)*sin(x) + 3*(a^4*b + a^2*b^3 - 2*b^5 - (a^4*b + a^2*b^3 -
2*b^5)*cos(x)^2)*log(1/2*cos(x) + 1/2)*sin(x) - 3*(a^4*b + a^2*b^3 - 2*b^5 - (a^4*b + a^2*b^3 - 2*b^5)*cos(x)^
2)*log(-1/2*cos(x) + 1/2)*sin(x) - 12*(a^5 - a*b^4)*cos(x))/((a^6 - a^4*b^2 - (a^6 - a^4*b^2)*cos(x)^2)*sin(x)
), 1/12*(4*(2*a^5 + a^3*b^2 - 3*a*b^4)*cos(x)^3 + 12*(b^4*cos(x)^2 - b^4)*sqrt(a^2 - b^2)*arctan(-(a*sin(x) +
b)/(sqrt(a^2 - b^2)*cos(x)))*sin(x) + 6*(a^4*b - a^2*b^3)*cos(x)*sin(x) + 3*(a^4*b + a^2*b^3 - 2*b^5 - (a^4*b
+ a^2*b^3 - 2*b^5)*cos(x)^2)*log(1/2*cos(x) + 1/2)*sin(x) - 3*(a^4*b + a^2*b^3 - 2*b^5 - (a^4*b + a^2*b^3 - 2*
b^5)*cos(x)^2)*log(-1/2*cos(x) + 1/2)*sin(x) - 12*(a^5 - a*b^4)*cos(x))/((a^6 - a^4*b^2 - (a^6 - a^4*b^2)*cos(
x)^2)*sin(x))]

Sympy [F]

\[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\int \frac {\csc ^{4}{\left (x \right )}}{a + b \sin {\left (x \right )}}\, dx \]

[In]

integrate(csc(x)**4/(a+b*sin(x)),x)

[Out]

Integral(csc(x)**4/(a + b*sin(x)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(csc(x)^4/(a+b*sin(x)),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 de

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.73 \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\frac {2 \, {\left (\pi \left \lfloor \frac {x}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (a\right ) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, x\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{4}}{\sqrt {a^{2} - b^{2}} a^{4}} + \frac {a^{2} \tan \left (\frac {1}{2} \, x\right )^{3} - 3 \, a b \tan \left (\frac {1}{2} \, x\right )^{2} + 9 \, a^{2} \tan \left (\frac {1}{2} \, x\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, x\right )}{24 \, a^{3}} - \frac {{\left (a^{2} b + 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, x\right ) \right |}\right )}{2 \, a^{4}} + \frac {22 \, a^{2} b \tan \left (\frac {1}{2} \, x\right )^{3} + 44 \, b^{3} \tan \left (\frac {1}{2} \, x\right )^{3} - 9 \, a^{3} \tan \left (\frac {1}{2} \, x\right )^{2} - 12 \, a b^{2} \tan \left (\frac {1}{2} \, x\right )^{2} + 3 \, a^{2} b \tan \left (\frac {1}{2} \, x\right ) - a^{3}}{24 \, a^{4} \tan \left (\frac {1}{2} \, x\right )^{3}} \]

[In]

integrate(csc(x)^4/(a+b*sin(x)),x, algorithm="giac")

[Out]

2*(pi*floor(1/2*x/pi + 1/2)*sgn(a) + arctan((a*tan(1/2*x) + b)/sqrt(a^2 - b^2)))*b^4/(sqrt(a^2 - b^2)*a^4) + 1
/24*(a^2*tan(1/2*x)^3 - 3*a*b*tan(1/2*x)^2 + 9*a^2*tan(1/2*x) + 12*b^2*tan(1/2*x))/a^3 - 1/2*(a^2*b + 2*b^3)*l
og(abs(tan(1/2*x)))/a^4 + 1/24*(22*a^2*b*tan(1/2*x)^3 + 44*b^3*tan(1/2*x)^3 - 9*a^3*tan(1/2*x)^2 - 12*a*b^2*ta
n(1/2*x)^2 + 3*a^2*b*tan(1/2*x) - a^3)/(a^4*tan(1/2*x)^3)

Mupad [B] (verification not implemented)

Time = 6.95 (sec) , antiderivative size = 586, normalized size of antiderivative = 5.23 \[ \int \frac {\csc ^4(x)}{a+b \sin (x)} \, dx=\frac {a^5\,\left (\frac {\cos \left (3\,x\right )}{12}-\frac {\cos \left (x\right )}{4}\right )-a\,\left (\frac {b^4\,\cos \left (3\,x\right )}{8}-\frac {b^4\,\cos \left (x\right )}{8}\right )+a^4\,\left (\frac {b\,\sin \left (2\,x\right )}{8}-\frac {3\,b\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\sin \left (x\right )}{16}+\frac {b\,\sin \left (3\,x\right )\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{16}\right )-a^2\,\left (\frac {b^3\,\sin \left (2\,x\right )}{8}-\frac {b^3\,\sin \left (3\,x\right )\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{16}+\frac {3\,b^3\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\sin \left (x\right )}{16}\right )+a^3\,\left (\frac {b^2\,\cos \left (3\,x\right )}{24}+\frac {b^2\,\cos \left (x\right )}{8}\right )-\frac {b^5\,\sin \left (3\,x\right )\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )}{8}+\frac {3\,b^5\,\ln \left (\frac {\sin \left (\frac {x}{2}\right )}{\cos \left (\frac {x}{2}\right )}\right )\,\sin \left (x\right )}{8}+\frac {b^4\,\mathrm {atan}\left (\frac {-a^4\,\sin \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+b^4\,\sin \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,8{}\mathrm {i}+a\,b^3\,\cos \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,4{}\mathrm {i}+a^3\,b\,\cos \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}}{\cos \left (\frac {x}{2}\right )\,a^5+2\,\sin \left (\frac {x}{2}\right )\,a^4\,b+\cos \left (\frac {x}{2}\right )\,a^3\,b^2+4\,\sin \left (\frac {x}{2}\right )\,a^2\,b^3-4\,\cos \left (\frac {x}{2}\right )\,a\,b^4-8\,\sin \left (\frac {x}{2}\right )\,b^5}\right )\,\sin \left (3\,x\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}}{4}-\frac {b^4\,\mathrm {atan}\left (\frac {-a^4\,\sin \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}+b^4\,\sin \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,8{}\mathrm {i}+a\,b^3\,\cos \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,4{}\mathrm {i}+a^3\,b\,\cos \left (\frac {x}{2}\right )\,\sqrt {b^2-a^2}\,1{}\mathrm {i}}{\cos \left (\frac {x}{2}\right )\,a^5+2\,\sin \left (\frac {x}{2}\right )\,a^4\,b+\cos \left (\frac {x}{2}\right )\,a^3\,b^2+4\,\sin \left (\frac {x}{2}\right )\,a^2\,b^3-4\,\cos \left (\frac {x}{2}\right )\,a\,b^4-8\,\sin \left (\frac {x}{2}\right )\,b^5}\right )\,\sin \left (x\right )\,\sqrt {b^2-a^2}\,3{}\mathrm {i}}{4}}{\frac {3\,a^6\,\sin \left (x\right )}{8}-\frac {a^6\,\sin \left (3\,x\right )}{8}+\frac {a^4\,b^2\,\sin \left (3\,x\right )}{8}-\frac {3\,a^4\,b^2\,\sin \left (x\right )}{8}} \]

[In]

int(1/(sin(x)^4*(a + b*sin(x))),x)

[Out]

(a^5*(cos(3*x)/12 - cos(x)/4) - a*((b^4*cos(3*x))/8 - (b^4*cos(x))/8) + a^4*((b*sin(2*x))/8 - (3*b*log(sin(x/2
)/cos(x/2))*sin(x))/16 + (b*sin(3*x)*log(sin(x/2)/cos(x/2)))/16) - a^2*((b^3*sin(2*x))/8 - (b^3*sin(3*x)*log(s
in(x/2)/cos(x/2)))/16 + (3*b^3*log(sin(x/2)/cos(x/2))*sin(x))/16) + a^3*((b^2*cos(3*x))/24 + (b^2*cos(x))/8) -
(b^5*sin(3*x)*log(sin(x/2)/cos(x/2)))/8 + (3*b^5*log(sin(x/2)/cos(x/2))*sin(x))/8 + (b^4*atan((b^4*sin(x/2)*(
b^2 - a^2)^(1/2)*8i - a^4*sin(x/2)*(b^2 - a^2)^(1/2)*1i + a*b^3*cos(x/2)*(b^2 - a^2)^(1/2)*4i + a^3*b*cos(x/2)
*(b^2 - a^2)^(1/2)*1i)/(a^5*cos(x/2) - 8*b^5*sin(x/2) + a^3*b^2*cos(x/2) + 4*a^2*b^3*sin(x/2) - 4*a*b^4*cos(x/
2) + 2*a^4*b*sin(x/2)))*sin(3*x)*(b^2 - a^2)^(1/2)*1i)/4 - (b^4*atan((b^4*sin(x/2)*(b^2 - a^2)^(1/2)*8i - a^4*
sin(x/2)*(b^2 - a^2)^(1/2)*1i + a*b^3*cos(x/2)*(b^2 - a^2)^(1/2)*4i + a^3*b*cos(x/2)*(b^2 - a^2)^(1/2)*1i)/(a^
5*cos(x/2) - 8*b^5*sin(x/2) + a^3*b^2*cos(x/2) + 4*a^2*b^3*sin(x/2) - 4*a*b^4*cos(x/2) + 2*a^4*b*sin(x/2)))*si
n(x)*(b^2 - a^2)^(1/2)*3i)/4)/((3*a^6*sin(x))/8 - (a^6*sin(3*x))/8 + (a^4*b^2*sin(3*x))/8 - (3*a^4*b^2*sin(x))
/8)