3.22 \(\int \frac {(a+b \csc ^{-1}(c x))^2}{x^4} \, dx\)

Optimal. Leaf size=102 \[ -\frac {2 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \csc ^{-1}(c x)\right )}{9 x^2}-\frac {4}{9} b c^3 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \csc ^{-1}(c x)\right )-\frac {\left (a+b \csc ^{-1}(c x)\right )^2}{3 x^3}+\frac {4 b^2 c^2}{9 x}+\frac {2 b^2}{27 x^3} \]

[Out]

2/27*b^2/x^3+4/9*b^2*c^2/x-1/3*(a+b*arccsc(c*x))^2/x^3-4/9*b*c^3*(a+b*arccsc(c*x))*(1-1/c^2/x^2)^(1/2)-2/9*b*c
*(a+b*arccsc(c*x))*(1-1/c^2/x^2)^(1/2)/x^2

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Rubi [A]  time = 0.10, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5223, 4404, 3310, 3296, 2637} \[ -\frac {4}{9} b c^3 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \csc ^{-1}(c x)\right )-\frac {2 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \csc ^{-1}(c x)\right )}{9 x^2}-\frac {\left (a+b \csc ^{-1}(c x)\right )^2}{3 x^3}+\frac {4 b^2 c^2}{9 x}+\frac {2 b^2}{27 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcCsc[c*x])^2/x^4,x]

[Out]

(2*b^2)/(27*x^3) + (4*b^2*c^2)/(9*x) - (4*b*c^3*Sqrt[1 - 1/(c^2*x^2)]*(a + b*ArcCsc[c*x]))/9 - (2*b*c*Sqrt[1 -
1/(c^2*x^2)]*(a + b*ArcCsc[c*x]))/(9*x^2) - (a + b*ArcCsc[c*x])^2/(3*x^3)

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 3310

Int[((c_.) + (d_.)*(x_))*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(d*(b*Sin[e + f*x])^n)/(f^2*n
^2), x] + (Dist[(b^2*(n - 1))/n, Int[(c + d*x)*(b*Sin[e + f*x])^(n - 2), x], x] - Simp[(b*(c + d*x)*Cos[e + f*
x]*(b*Sin[e + f*x])^(n - 1))/(f*n), x]) /; FreeQ[{b, c, d, e, f}, x] && GtQ[n, 1]

Rule 4404

Int[Cos[(a_.) + (b_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.)*Sin[(a_.) + (b_.)*(x_)]^(n_.), x_Symbol] :> Simp[((c +
d*x)^m*Sin[a + b*x]^(n + 1))/(b*(n + 1)), x] - Dist[(d*m)/(b*(n + 1)), Int[(c + d*x)^(m - 1)*Sin[a + b*x]^(n +
1), x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && NeQ[n, -1]

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 {\left (a+b \csc ^{-1}(c x)\right )^2}{x^4} \, dx &=-\left (c^3 \operatorname {Subst}\left (\int (a+b x)^2 \cos (x) \sin ^2(x) \, dx,x,\csc ^{-1}(c x)\right )\right )\\ &=-\frac {\left (a+b \csc ^{-1}(c x)\right )^2}{3 x^3}+\frac {1}{3} \left (2 b c^3\right ) \operatorname {Subst}\left (\int (a+b x) \sin ^3(x) \, dx,x,\csc ^{-1}(c x)\right )\\ &=\frac {2 b^2}{27 x^3}-\frac {2 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \csc ^{-1}(c x)\right )}{9 x^2}-\frac {\left (a+b \csc ^{-1}(c x)\right )^2}{3 x^3}+\frac {1}{9} \left (4 b c^3\right ) \operatorname {Subst}\left (\int (a+b x) \sin (x) \, dx,x,\csc ^{-1}(c x)\right )\\ &=\frac {2 b^2}{27 x^3}-\frac {4}{9} b c^3 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \csc ^{-1}(c x)\right )-\frac {2 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \csc ^{-1}(c x)\right )}{9 x^2}-\frac {\left (a+b \csc ^{-1}(c x)\right )^2}{3 x^3}+\frac {1}{9} \left (4 b^2 c^3\right ) \operatorname {Subst}\left (\int \cos (x) \, dx,x,\csc ^{-1}(c x)\right )\\ &=\frac {2 b^2}{27 x^3}+\frac {4 b^2 c^2}{9 x}-\frac {4}{9} b c^3 \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \csc ^{-1}(c x)\right )-\frac {2 b c \sqrt {1-\frac {1}{c^2 x^2}} \left (a+b \csc ^{-1}(c x)\right )}{9 x^2}-\frac {\left (a+b \csc ^{-1}(c x)\right )^2}{3 x^3}\\ \end {align*}

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Mathematica [A]  time = 0.24, size = 108, normalized size = 1.06 \[ -\frac {9 a^2+6 a b c x \sqrt {1-\frac {1}{c^2 x^2}} \left (2 c^2 x^2+1\right )+6 b \csc ^{-1}(c x) \left (3 a+b c x \sqrt {1-\frac {1}{c^2 x^2}} \left (2 c^2 x^2+1\right )\right )-2 b^2 \left (6 c^2 x^2+1\right )+9 b^2 \csc ^{-1}(c x)^2}{27 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcCsc[c*x])^2/x^4,x]

[Out]

-1/27*(9*a^2 + 6*a*b*c*Sqrt[1 - 1/(c^2*x^2)]*x*(1 + 2*c^2*x^2) - 2*b^2*(1 + 6*c^2*x^2) + 6*b*(3*a + b*c*Sqrt[1
- 1/(c^2*x^2)]*x*(1 + 2*c^2*x^2))*ArcCsc[c*x] + 9*b^2*ArcCsc[c*x]^2)/x^3

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fricas [A]  time = 0.97, size = 93, normalized size = 0.91 \[ \frac {12 \, b^{2} c^{2} x^{2} - 9 \, b^{2} \operatorname {arccsc}\left (c x\right )^{2} - 18 \, a b \operatorname {arccsc}\left (c x\right ) - 9 \, a^{2} + 2 \, b^{2} - 6 \, {\left (2 \, a b c^{2} x^{2} + a b + {\left (2 \, b^{2} c^{2} x^{2} + b^{2}\right )} \operatorname {arccsc}\left (c x\right )\right )} \sqrt {c^{2} x^{2} - 1}}{27 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsc(c*x))^2/x^4,x, algorithm="fricas")

[Out]

1/27*(12*b^2*c^2*x^2 - 9*b^2*arccsc(c*x)^2 - 18*a*b*arccsc(c*x) - 9*a^2 + 2*b^2 - 6*(2*a*b*c^2*x^2 + a*b + (2*
b^2*c^2*x^2 + b^2)*arccsc(c*x))*sqrt(c^2*x^2 - 1))/x^3

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giac [B]  time = 0.17, size = 224, normalized size = 2.20 \[ \frac {1}{27} \, {\left (6 \, b^{2} c^{2} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} \arcsin \left (\frac {1}{c x}\right ) + 6 \, a b c^{2} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 18 \, b^{2} c^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} \arcsin \left (\frac {1}{c x}\right ) - \frac {9 \, b^{2} c {\left (\frac {1}{c^{2} x^{2}} - 1\right )} \arcsin \left (\frac {1}{c x}\right )^{2}}{x} - 18 \, a b c^{2} \sqrt {-\frac {1}{c^{2} x^{2}} + 1} - \frac {18 \, a b c {\left (\frac {1}{c^{2} x^{2}} - 1\right )} \arcsin \left (\frac {1}{c x}\right )}{x} - \frac {9 \, b^{2} c \arcsin \left (\frac {1}{c x}\right )^{2}}{x} + \frac {2 \, b^{2} c {\left (\frac {1}{c^{2} x^{2}} - 1\right )}}{x} - \frac {18 \, a b c \arcsin \left (\frac {1}{c x}\right )}{x} + \frac {14 \, b^{2} c}{x} - \frac {9 \, a^{2}}{c x^{3}}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsc(c*x))^2/x^4,x, algorithm="giac")

[Out]

1/27*(6*b^2*c^2*(-1/(c^2*x^2) + 1)^(3/2)*arcsin(1/(c*x)) + 6*a*b*c^2*(-1/(c^2*x^2) + 1)^(3/2) - 18*b^2*c^2*sqr
t(-1/(c^2*x^2) + 1)*arcsin(1/(c*x)) - 9*b^2*c*(1/(c^2*x^2) - 1)*arcsin(1/(c*x))^2/x - 18*a*b*c^2*sqrt(-1/(c^2*
x^2) + 1) - 18*a*b*c*(1/(c^2*x^2) - 1)*arcsin(1/(c*x))/x - 9*b^2*c*arcsin(1/(c*x))^2/x + 2*b^2*c*(1/(c^2*x^2)
- 1)/x - 18*a*b*c*arcsin(1/(c*x))/x + 14*b^2*c/x - 9*a^2/(c*x^3))*c

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maple [A]  time = 0.57, size = 154, normalized size = 1.51 \[ c^{3} \left (-\frac {a^{2}}{3 c^{3} x^{3}}+b^{2} \left (-\frac {\mathrm {arccsc}\left (c x \right )^{2}}{3 c^{3} x^{3}}-\frac {2 \,\mathrm {arccsc}\left (c x \right ) \left (2 c^{2} x^{2}+1\right ) \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}}{9 c^{2} x^{2}}+\frac {2}{27 c^{3} x^{3}}+\frac {4}{9 c x}\right )+2 a b \left (-\frac {\mathrm {arccsc}\left (c x \right )}{3 c^{3} x^{3}}-\frac {\left (c^{2} x^{2}-1\right ) \left (2 c^{2} x^{2}+1\right )}{9 \sqrt {\frac {c^{2} x^{2}-1}{c^{2} x^{2}}}\, c^{4} x^{4}}\right )\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arccsc(c*x))^2/x^4,x)

[Out]

c^3*(-1/3*a^2/c^3/x^3+b^2*(-1/3/c^3/x^3*arccsc(c*x)^2-2/9*arccsc(c*x)*(2*c^2*x^2+1)/c^2/x^2*((c^2*x^2-1)/c^2/x
^2)^(1/2)+2/27/c^3/x^3+4/9/c/x)+2*a*b*(-1/3/c^3/x^3*arccsc(c*x)-1/9*(c^2*x^2-1)*(2*c^2*x^2+1)/((c^2*x^2-1)/c^2
/x^2)^(1/2)/c^4/x^4))

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maxima [B]  time = 1.95, size = 197, normalized size = 1.93 \[ \frac {2}{9} \, a b {\left (\frac {c^{4} {\left (-\frac {1}{c^{2} x^{2}} + 1\right )}^{\frac {3}{2}} - 3 \, c^{4} \sqrt {-\frac {1}{c^{2} x^{2}} + 1}}{c} - \frac {3 \, \operatorname {arccsc}\left (c x\right )}{x^{3}}\right )} - \frac {b^{2} \operatorname {arccsc}\left (c x\right )^{2}}{3 \, x^{3}} - \frac {a^{2}}{3 \, x^{3}} - \frac {2 \, {\left (6 \, c^{5} x^{4} \arctan \left (1, \sqrt {c x + 1} \sqrt {c x - 1}\right ) - 3 \, c^{3} x^{2} \arctan \left (1, \sqrt {c x + 1} \sqrt {c x - 1}\right ) - {\left (6 \, c^{3} x^{2} + c\right )} \sqrt {c x + 1} \sqrt {c x - 1} - 3 \, c \arctan \left (1, \sqrt {c x + 1} \sqrt {c x - 1}\right )\right )} b^{2}}{27 \, \sqrt {c x + 1} \sqrt {c x - 1} c x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arccsc(c*x))^2/x^4,x, algorithm="maxima")

[Out]

2/9*a*b*((c^4*(-1/(c^2*x^2) + 1)^(3/2) - 3*c^4*sqrt(-1/(c^2*x^2) + 1))/c - 3*arccsc(c*x)/x^3) - 1/3*b^2*arccsc
(c*x)^2/x^3 - 1/3*a^2/x^3 - 2/27*(6*c^5*x^4*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1)) - 3*c^3*x^2*arctan2(1, sqr
t(c*x + 1)*sqrt(c*x - 1)) - (6*c^3*x^2 + c)*sqrt(c*x + 1)*sqrt(c*x - 1) - 3*c*arctan2(1, sqrt(c*x + 1)*sqrt(c*
x - 1)))*b^2/(sqrt(c*x + 1)*sqrt(c*x - 1)*c*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,\mathrm {asin}\left (\frac {1}{c\,x}\right )\right )}^2}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*asin(1/(c*x)))^2/x^4,x)

[Out]

int((a + b*asin(1/(c*x)))^2/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {acsc}{\left (c x \right )}\right )^{2}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*acsc(c*x))**2/x**4,x)

[Out]

Integral((a + b*acsc(c*x))**2/x**4, x)

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