3.50 \(\int \frac{1}{(a+b \csc (c+d x))^3} \, dx\)

Optimal. Leaf size=170 \[ \frac{b \left (-5 a^2 b^2+6 a^4+2 b^4\right ) \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{5/2}}-\frac{b^2 \left (5 a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \csc (c+d x))}-\frac{b^2 \cot (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \csc (c+d x))^2}+\frac{x}{a^3} \]

[Out]

x/a^3 + (b*(6*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTanh[(a + b*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(5/
2)*d) - (b^2*Cot[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Csc[c + d*x])^2) - (b^2*(5*a^2 - 2*b^2)*Cot[c + d*x])/(2*
a^2*(a^2 - b^2)^2*d*(a + b*Csc[c + d*x]))

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Rubi [A]  time = 0.318968, antiderivative size = 170, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {3785, 4060, 3919, 3831, 2660, 618, 206} \[ \frac{b \left (-5 a^2 b^2+6 a^4+2 b^4\right ) \tanh ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{5/2}}-\frac{b^2 \left (5 a^2-2 b^2\right ) \cot (c+d x)}{2 a^2 d \left (a^2-b^2\right )^2 (a+b \csc (c+d x))}-\frac{b^2 \cot (c+d x)}{2 a d \left (a^2-b^2\right ) (a+b \csc (c+d x))^2}+\frac{x}{a^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Csc[c + d*x])^(-3),x]

[Out]

x/a^3 + (b*(6*a^4 - 5*a^2*b^2 + 2*b^4)*ArcTanh[(a + b*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^3*(a^2 - b^2)^(5/
2)*d) - (b^2*Cot[c + d*x])/(2*a*(a^2 - b^2)*d*(a + b*Csc[c + d*x])^2) - (b^2*(5*a^2 - 2*b^2)*Cot[c + d*x])/(2*
a^2*(a^2 - b^2)^2*d*(a + b*Csc[c + d*x]))

Rule 3785

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n +
1))/(a*d*(n + 1)*(a^2 - b^2)), x] + Dist[1/(a*(n + 1)*(a^2 - b^2)), Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^
2 - b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 4060

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_))^(m_), x_Symbol] :> Simp[((A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(a*f*(m + 1
)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*(a^2 - b^2)*(m +
1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + 2)*Csc[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 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 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 206

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

Rubi steps

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

Mathematica [A]  time = 1.10303, size = 216, normalized size = 1.27 \[ \frac{\csc ^2(c+d x) (a \sin (c+d x)+b) \left (-\frac{3 a b^2 \left (2 a^2-b^2\right ) \cot (c+d x) (a \sin (c+d x)+b)}{(a-b)^2 (a+b)^2}-\frac{2 b \left (-5 a^2 b^2+6 a^4+2 b^4\right ) \csc (c+d x) (a \sin (c+d x)+b)^2 \tan ^{-1}\left (\frac{a+b \tan \left (\frac{1}{2} (c+d x)\right )}{\sqrt{b^2-a^2}}\right )}{\left (b^2-a^2\right )^{5/2}}+\frac{a b^3 \cot (c+d x)}{(a-b) (a+b)}+2 (c+d x) \csc (c+d x) (a \sin (c+d x)+b)^2\right )}{2 a^3 d (a+b \csc (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Csc[c + d*x])^(-3),x]

[Out]

(Csc[c + d*x]^2*(b + a*Sin[c + d*x])*((a*b^3*Cot[c + d*x])/((a - b)*(a + b)) - (3*a*b^2*(2*a^2 - b^2)*Cot[c +
d*x]*(b + a*Sin[c + d*x]))/((a - b)^2*(a + b)^2) + 2*(c + d*x)*Csc[c + d*x]*(b + a*Sin[c + d*x])^2 - (2*b*(6*a
^4 - 5*a^2*b^2 + 2*b^4)*ArcTan[(a + b*Tan[(c + d*x)/2])/Sqrt[-a^2 + b^2]]*Csc[c + d*x]*(b + a*Sin[c + d*x])^2)
/(-a^2 + b^2)^(5/2)))/(2*a^3*d*(a + b*Csc[c + d*x])^3)

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Maple [B]  time = 0.115, size = 796, normalized size = 4.7 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*csc(d*x+c))^3,x)

[Out]

2/d/a^3*arctan(tan(1/2*d*x+1/2*c))-4/d*a*b^2/(tan(1/2*d*x+1/2*c)^2*b+2*a*tan(1/2*d*x+1/2*c)+b)^2/(a^4-2*a^2*b^
2+b^4)*tan(1/2*d*x+1/2*c)^3+1/d/a*b^4/(tan(1/2*d*x+1/2*c)^2*b+2*a*tan(1/2*d*x+1/2*c)+b)^2/(a^4-2*a^2*b^2+b^4)*
tan(1/2*d*x+1/2*c)^3-10/d*a^2*b/(tan(1/2*d*x+1/2*c)^2*b+2*a*tan(1/2*d*x+1/2*c)+b)^2/(a^4-2*a^2*b^2+b^4)*tan(1/
2*d*x+1/2*c)^2-1/d*b^3/(tan(1/2*d*x+1/2*c)^2*b+2*a*tan(1/2*d*x+1/2*c)+b)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*d*x+1/2
*c)^2+2/d/a^2*b^5/(tan(1/2*d*x+1/2*c)^2*b+2*a*tan(1/2*d*x+1/2*c)+b)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)^2
-16/d*a*b^2/(tan(1/2*d*x+1/2*c)^2*b+2*a*tan(1/2*d*x+1/2*c)+b)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)+7/d/a*b
^4/(tan(1/2*d*x+1/2*c)^2*b+2*a*tan(1/2*d*x+1/2*c)+b)^2/(a^4-2*a^2*b^2+b^4)*tan(1/2*d*x+1/2*c)-5/d*b^3/(tan(1/2
*d*x+1/2*c)^2*b+2*a*tan(1/2*d*x+1/2*c)+b)^2/(a^4-2*a^2*b^2+b^4)+2/d/a^2*b^5/(tan(1/2*d*x+1/2*c)^2*b+2*a*tan(1/
2*d*x+1/2*c)+b)^2/(a^4-2*a^2*b^2+b^4)-6/d*a*b/(a^4-2*a^2*b^2+b^4)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*d*x
+1/2*c)+2*a)/(-a^2+b^2)^(1/2))+5/d/a*b^3/(a^4-2*a^2*b^2+b^4)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*d*x+1/2*
c)+2*a)/(-a^2+b^2)^(1/2))-2/d/a^3*b^5/(a^4-2*a^2*b^2+b^4)/(-a^2+b^2)^(1/2)*arctan(1/2*(2*b*tan(1/2*d*x+1/2*c)+
2*a)/(-a^2+b^2)^(1/2))

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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(1/(a+b*csc(d*x+c))^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 0.645056, size = 1994, normalized size = 11.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csc(d*x+c))^3,x, algorithm="fricas")

[Out]

[1/4*(4*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x*cos(d*x + c)^2 - 4*(a^8 - 2*a^6*b^2 + 2*a^2*b^6 - b^8)*d*x
- (6*a^6*b + a^4*b^3 - 3*a^2*b^5 + 2*b^7 - (6*a^6*b - 5*a^4*b^3 + 2*a^2*b^5)*cos(d*x + c)^2 + 2*(6*a^5*b^2 -
5*a^3*b^4 + 2*a*b^6)*sin(d*x + c))*sqrt(a^2 - b^2)*log(((a^2 - 2*b^2)*cos(d*x + c)^2 + 2*a*b*sin(d*x + c) + a^
2 + b^2 + 2*(b*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c))*sqrt(a^2 - b^2))/(a^2*cos(d*x + c)^2 - 2*a*b*sin(d*
x + c) - a^2 - b^2)) + 2*(5*a^5*b^3 - 7*a^3*b^5 + 2*a*b^7)*cos(d*x + c) - 2*(4*(a^7*b - 3*a^5*b^3 + 3*a^3*b^5
- a*b^7)*d*x - 3*(2*a^6*b^2 - 3*a^4*b^4 + a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^11 - 3*a^9*b^2 + 3*a^7*b^4
- a^5*b^6)*d*cos(d*x + c)^2 - 2*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*sin(d*x + c) - (a^11 - 2*a^9*b^2
+ 2*a^5*b^6 - a^3*b^8)*d), 1/2*(2*(a^8 - 3*a^6*b^2 + 3*a^4*b^4 - a^2*b^6)*d*x*cos(d*x + c)^2 - 2*(a^8 - 2*a^6*
b^2 + 2*a^2*b^6 - b^8)*d*x - (6*a^6*b + a^4*b^3 - 3*a^2*b^5 + 2*b^7 - (6*a^6*b - 5*a^4*b^3 + 2*a^2*b^5)*cos(d*
x + c)^2 + 2*(6*a^5*b^2 - 5*a^3*b^4 + 2*a*b^6)*sin(d*x + c))*sqrt(-a^2 + b^2)*arctan(-sqrt(-a^2 + b^2)*(b*sin(
d*x + c) + a)/((a^2 - b^2)*cos(d*x + c))) + (5*a^5*b^3 - 7*a^3*b^5 + 2*a*b^7)*cos(d*x + c) - (4*(a^7*b - 3*a^5
*b^3 + 3*a^3*b^5 - a*b^7)*d*x - 3*(2*a^6*b^2 - 3*a^4*b^4 + a^2*b^6)*cos(d*x + c))*sin(d*x + c))/((a^11 - 3*a^9
*b^2 + 3*a^7*b^4 - a^5*b^6)*d*cos(d*x + c)^2 - 2*(a^10*b - 3*a^8*b^3 + 3*a^6*b^5 - a^4*b^7)*d*sin(d*x + c) - (
a^11 - 2*a^9*b^2 + 2*a^5*b^6 - a^3*b^8)*d)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (a + b \csc{\left (c + d x \right )}\right )^{3}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csc(d*x+c))**3,x)

[Out]

Integral((a + b*csc(c + d*x))**(-3), x)

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Giac [A]  time = 1.18064, size = 401, normalized size = 2.36 \begin{align*} -\frac{\frac{{\left (6 \, a^{4} b - 5 \, a^{2} b^{3} + 2 \, b^{5}\right )}{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a}{\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{4 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 10 \, a^{4} b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a^{2} b^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 2 \, b^{5} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 16 \, a^{3} b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 7 \, a b^{4} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 5 \, a^{2} b^{3} - 2 \, b^{5}}{{\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )}{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b\right )}^{2}} - \frac{d x + c}{a^{3}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*csc(d*x+c))^3,x, algorithm="giac")

[Out]

-((6*a^4*b - 5*a^2*b^3 + 2*b^5)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(b) + arctan((b*tan(1/2*d*x + 1/2*c) + a)
/sqrt(-a^2 + b^2)))/((a^7 - 2*a^5*b^2 + a^3*b^4)*sqrt(-a^2 + b^2)) + (4*a^3*b^2*tan(1/2*d*x + 1/2*c)^3 - a*b^4
*tan(1/2*d*x + 1/2*c)^3 + 10*a^4*b*tan(1/2*d*x + 1/2*c)^2 + a^2*b^3*tan(1/2*d*x + 1/2*c)^2 - 2*b^5*tan(1/2*d*x
+ 1/2*c)^2 + 16*a^3*b^2*tan(1/2*d*x + 1/2*c) - 7*a*b^4*tan(1/2*d*x + 1/2*c) + 5*a^2*b^3 - 2*b^5)/((a^6 - 2*a^
4*b^2 + a^2*b^4)*(b*tan(1/2*d*x + 1/2*c)^2 + 2*a*tan(1/2*d*x + 1/2*c) + b)^2) - (d*x + c)/a^3)/d