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

Optimal. Leaf size=112 \[ \frac{2 b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 \sqrt{a^2-b^2}}-\frac{\left (2 a^2+3 b^2\right ) \cot (x)}{3 a^3}+\frac{b \left (a^2+2 b^2\right ) \tanh ^{-1}(\cos (x))}{2 a^4}+\frac{b \cot (x) \csc (x)}{2 a^2}-\frac{\cot (x) \csc ^2(x)}{3 a} \]

[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)

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Rubi [A]  time = 0.433336, antiderivative size = 112, normalized size of antiderivative = 1., 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} \[ \frac{2 b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{a^4 \sqrt{a^2-b^2}}-\frac{\left (2 a^2+3 b^2\right ) \cot (x)}{3 a^3}+\frac{b \left (a^2+2 b^2\right ) \tanh ^{-1}(\cos (x))}{2 a^4}+\frac{b \cot (x) \csc (x)}{2 a^2}-\frac{\cot (x) \csc ^2(x)}{3 a} \]

Antiderivative was successfully verified.

[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 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 ^4(x)}{a+b \sin (x)} \, dx &=-\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 ) \tanh ^{-1}(\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 ) \operatorname{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 ) \tanh ^{-1}(\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 ) \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}\\ &=\frac{2 b^4 \tan ^{-1}\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 ) \tanh ^{-1}(\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]  time = 1.54879, size = 125, normalized size = 1.12 \[ \frac{\frac{24 b^4 \tan ^{-1}\left (\frac{a \tan \left (\frac{x}{2}\right )+b}{\sqrt{a^2-b^2}}\right )}{\sqrt{a^2-b^2}}+a \left (2 a^2+3 b^2\right ) \cos (3 x) \csc ^3(x)-3 a \cot (x) \csc (x) \left (\left (2 a^2+b^2\right ) \csc (x)-2 a b\right )+6 b \left (a^2+2 b^2\right ) \left (\log \left (\cos \left (\frac{x}{2}\right )\right )-\log \left (\sin \left (\frac{x}{2}\right )\right )\right )}{12 a^4} \]

Antiderivative was successfully verified.

[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)

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Maple [A]  time = 0.051, size = 162, normalized size = 1.5 \begin{align*}{\frac{1}{24\,a} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{3}}-{\frac{b}{8\,{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{2}}+{\frac{3}{8\,a}\tan \left ({\frac{x}{2}} \right ) }+{\frac{{b}^{2}}{2\,{a}^{3}}\tan \left ({\frac{x}{2}} \right ) }+2\,{\frac{{b}^{4}}{{a}^{4}\sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( x/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }-{\frac{1}{24\,a} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-3}}-{\frac{3}{8\,a} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}-{\frac{{b}^{2}}{2\,{a}^{3}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-1}}+{\frac{b}{8\,{a}^{2}} \left ( \tan \left ({\frac{x}{2}} \right ) \right ) ^{-2}}-{\frac{b}{2\,{a}^{2}}\ln \left ( \tan \left ({\frac{x}{2}} \right ) \right ) }-{\frac{{b}^{3}}{{a}^{4}}\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)^4/(a+b*sin(x)),x)

[Out]

1/24/a*tan(1/2*x)^3-1/8/a^2*tan(1/2*x)^2*b+3/8/a*tan(1/2*x)+1/2/a^3*b^2*tan(1/2*x)+2/a^4*b^4/(a^2-b^2)^(1/2)*a
rctan(1/2*(2*a*tan(1/2*x)+2*b)/(a^2-b^2)^(1/2))-1/24/a/tan(1/2*x)^3-3/8/a/tan(1/2*x)-1/2/a^3/tan(1/2*x)*b^2+1/
8/a^2*b/tan(1/2*x)^2-1/2/a^2*b*ln(tan(1/2*x))-1/a^4*b^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)^4/(a+b*sin(x)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.05268, size = 1361, normalized size = 12.15 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{4}{\left (x \right )}}{a + b \sin{\left (x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(x)**4/(a + b*sin(x)), x)

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Giac [A]  time = 1.50635, size = 262, normalized size = 2.34 \begin{align*} \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}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)