3.38 \(\int \frac {1}{x^4 (a+b \csc ^{-1}(c x))} \, dx\)

Optimal. Leaf size=117 \[ -\frac {c^3 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )}{4 b}+\frac {c^3 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 a}{b}+3 \csc ^{-1}(c x)\right )}{4 b}-\frac {c^3 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )}{4 b}+\frac {c^3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \csc ^{-1}(c x)\right )}{4 b} \]

[Out]

-1/4*c^3*Ci(a/b+arccsc(c*x))*cos(a/b)/b+1/4*c^3*Ci(3*a/b+3*arccsc(c*x))*cos(3*a/b)/b-1/4*c^3*Si(a/b+arccsc(c*x
))*sin(a/b)/b+1/4*c^3*Si(3*a/b+3*arccsc(c*x))*sin(3*a/b)/b

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Rubi [A]  time = 0.23, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5223, 4406, 3303, 3299, 3302} \[ -\frac {c^3 \cos \left (\frac {a}{b}\right ) \text {CosIntegral}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )}{4 b}+\frac {c^3 \cos \left (\frac {3 a}{b}\right ) \text {CosIntegral}\left (\frac {3 a}{b}+3 \csc ^{-1}(c x)\right )}{4 b}-\frac {c^3 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )}{4 b}+\frac {c^3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \csc ^{-1}(c x)\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^4*(a + b*ArcCsc[c*x])),x]

[Out]

-(c^3*Cos[a/b]*CosIntegral[a/b + ArcCsc[c*x]])/(4*b) + (c^3*Cos[(3*a)/b]*CosIntegral[(3*a)/b + 3*ArcCsc[c*x]])
/(4*b) - (c^3*Sin[a/b]*SinIntegral[a/b + ArcCsc[c*x]])/(4*b) + (c^3*Sin[(3*a)/b]*SinIntegral[(3*a)/b + 3*ArcCs
c[c*x]])/(4*b)

Rule 3299

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[SinIntegral[e + f*x]/d, x] /; FreeQ[{c, d,
e, f}, x] && EqQ[d*e - c*f, 0]

Rule 3302

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Simp[CosIntegral[e - Pi/2 + f*x]/d, x] /; FreeQ
[{c, d, e, f}, x] && EqQ[d*(e - Pi/2) - c*f, 0]

Rule 3303

Int[sin[(e_.) + (f_.)*(x_)]/((c_.) + (d_.)*(x_)), x_Symbol] :> Dist[Cos[(d*e - c*f)/d], Int[Sin[(c*f)/d + f*x]
/(c + d*x), x], x] + Dist[Sin[(d*e - c*f)/d], Int[Cos[(c*f)/d + f*x]/(c + d*x), x], x] /; FreeQ[{c, d, e, f},
x] && NeQ[d*e - c*f, 0]

Rule 4406

Int[Cos[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Int[E
xpandTrigReduce[(c + d*x)^m, Sin[a + b*x]^n*Cos[a + b*x]^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0]
&& IGtQ[p, 0]

Rule 5223

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))^(n_)*(x_)^(m_.), x_Symbol] :> -Dist[(c^(m + 1))^(-1), Subst[Int[(a + b*
x)^n*Csc[x]^(m + 1)*Cot[x], x], x, ArcCsc[c*x]], x] /; FreeQ[{a, b, c}, x] && IntegerQ[n] && IntegerQ[m] && (G
tQ[n, 0] || LtQ[m, -1])

Rubi steps

\begin {align*} \int \frac {1}{x^4 \left (a+b \csc ^{-1}(c x)\right )} \, dx &=-\left (c^3 \operatorname {Subst}\left (\int \frac {\cos (x) \sin ^2(x)}{a+b x} \, dx,x,\csc ^{-1}(c x)\right )\right )\\ &=-\left (c^3 \operatorname {Subst}\left (\int \left (\frac {\cos (x)}{4 (a+b x)}-\frac {\cos (3 x)}{4 (a+b x)}\right ) \, dx,x,\csc ^{-1}(c x)\right )\right )\\ &=-\left (\frac {1}{4} c^3 \operatorname {Subst}\left (\int \frac {\cos (x)}{a+b x} \, dx,x,\csc ^{-1}(c x)\right )\right )+\frac {1}{4} c^3 \operatorname {Subst}\left (\int \frac {\cos (3 x)}{a+b x} \, dx,x,\csc ^{-1}(c x)\right )\\ &=-\left (\frac {1}{4} \left (c^3 \cos \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\csc ^{-1}(c x)\right )\right )+\frac {1}{4} \left (c^3 \cos \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cos \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\csc ^{-1}(c x)\right )-\frac {1}{4} \left (c^3 \sin \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\csc ^{-1}(c x)\right )+\frac {1}{4} \left (c^3 \sin \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sin \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\csc ^{-1}(c x)\right )\\ &=-\frac {c^3 \cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )}{4 b}+\frac {c^3 \cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (\frac {3 a}{b}+3 \csc ^{-1}(c x)\right )}{4 b}-\frac {c^3 \sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )}{4 b}+\frac {c^3 \sin \left (\frac {3 a}{b}\right ) \text {Si}\left (\frac {3 a}{b}+3 \csc ^{-1}(c x)\right )}{4 b}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 91, normalized size = 0.78 \[ -\frac {c^3 \left (\cos \left (\frac {a}{b}\right ) \text {Ci}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )-\cos \left (\frac {3 a}{b}\right ) \text {Ci}\left (3 \left (\frac {a}{b}+\csc ^{-1}(c x)\right )\right )+\sin \left (\frac {a}{b}\right ) \text {Si}\left (\frac {a}{b}+\csc ^{-1}(c x)\right )-\sin \left (\frac {3 a}{b}\right ) \text {Si}\left (3 \left (\frac {a}{b}+\csc ^{-1}(c x)\right )\right )\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^4*(a + b*ArcCsc[c*x])),x]

[Out]

-1/4*(c^3*(Cos[a/b]*CosIntegral[a/b + ArcCsc[c*x]] - Cos[(3*a)/b]*CosIntegral[3*(a/b + ArcCsc[c*x])] + Sin[a/b
]*SinIntegral[a/b + ArcCsc[c*x]] - Sin[(3*a)/b]*SinIntegral[3*(a/b + ArcCsc[c*x])]))/b

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fricas [F]  time = 0.41, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{b x^{4} \operatorname {arccsc}\left (c x\right ) + a x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arccsc(c*x)),x, algorithm="fricas")

[Out]

integral(1/(b*x^4*arccsc(c*x) + a*x^4), x)

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giac [A]  time = 0.15, size = 200, normalized size = 1.71 \[ \frac {1}{4} \, {\left (\frac {4 \, c^{2} \cos \left (\frac {a}{b}\right )^{3} \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (\frac {1}{c x}\right )\right )}{b} + \frac {4 \, c^{2} \cos \left (\frac {a}{b}\right )^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (\frac {1}{c x}\right )\right )}{b} - \frac {3 \, c^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (\frac {1}{c x}\right )\right )}{b} - \frac {c^{2} \cos \left (\frac {a}{b}\right ) \operatorname {Ci}\left (\frac {a}{b} + \arcsin \left (\frac {1}{c x}\right )\right )}{b} - \frac {c^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {3 \, a}{b} + 3 \, \arcsin \left (\frac {1}{c x}\right )\right )}{b} - \frac {c^{2} \sin \left (\frac {a}{b}\right ) \operatorname {Si}\left (\frac {a}{b} + \arcsin \left (\frac {1}{c x}\right )\right )}{b}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arccsc(c*x)),x, algorithm="giac")

[Out]

1/4*(4*c^2*cos(a/b)^3*cos_integral(3*a/b + 3*arcsin(1/(c*x)))/b + 4*c^2*cos(a/b)^2*sin(a/b)*sin_integral(3*a/b
+ 3*arcsin(1/(c*x)))/b - 3*c^2*cos(a/b)*cos_integral(3*a/b + 3*arcsin(1/(c*x)))/b - c^2*cos(a/b)*cos_integral
(a/b + arcsin(1/(c*x)))/b - c^2*sin(a/b)*sin_integral(3*a/b + 3*arcsin(1/(c*x)))/b - c^2*sin(a/b)*sin_integral
(a/b + arcsin(1/(c*x)))/b)*c

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maple [A]  time = 0.10, size = 102, normalized size = 0.87 \[ c^{3} \left (-\frac {\Si \left (\frac {a}{b}+\mathrm {arccsc}\left (c x \right )\right ) \sin \left (\frac {a}{b}\right )}{4 b}-\frac {\Ci \left (\frac {a}{b}+\mathrm {arccsc}\left (c x \right )\right ) \cos \left (\frac {a}{b}\right )}{4 b}+\frac {\Si \left (\frac {3 a}{b}+3 \,\mathrm {arccsc}\left (c x \right )\right ) \sin \left (\frac {3 a}{b}\right )}{4 b}+\frac {\Ci \left (\frac {3 a}{b}+3 \,\mathrm {arccsc}\left (c x \right )\right ) \cos \left (\frac {3 a}{b}\right )}{4 b}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^4/(a+b*arccsc(c*x)),x)

[Out]

c^3*(-1/4*Si(a/b+arccsc(c*x))*sin(a/b)/b-1/4*Ci(a/b+arccsc(c*x))*cos(a/b)/b+1/4*Si(3*a/b+3*arccsc(c*x))*sin(3*
a/b)/b+1/4*Ci(3*a/b+3*arccsc(c*x))*cos(3*a/b)/b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \operatorname {arccsc}\left (c x\right ) + a\right )} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^4/(a+b*arccsc(c*x)),x, algorithm="maxima")

[Out]

integrate(1/((b*arccsc(c*x) + a)*x^4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{x^4\,\left (a+b\,\mathrm {asin}\left (\frac {1}{c\,x}\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^4*(a + b*asin(1/(c*x)))),x)

[Out]

int(1/(x^4*(a + b*asin(1/(c*x)))), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{x^{4} \left (a + b \operatorname {acsc}{\left (c x \right )}\right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**4/(a+b*acsc(c*x)),x)

[Out]

Integral(1/(x**4*(a + b*acsc(c*x))), x)

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