3.79 \(\int \frac{(d+e x^2) (a+b \csc ^{-1}(c x))}{x^2} \, dx\)

Optimal. Leaf size=87 \[ -\frac{d \left (a+b \csc ^{-1}(c x)\right )}{x}+e x \left (a+b \csc ^{-1}(c x)\right )-\frac{b c d \sqrt{c^2 x^2-1}}{\sqrt{c^2 x^2}}+\frac{b e x \tanh ^{-1}\left (\frac{c x}{\sqrt{c^2 x^2-1}}\right )}{\sqrt{c^2 x^2}} \]

[Out]

-((b*c*d*Sqrt[-1 + c^2*x^2])/Sqrt[c^2*x^2]) - (d*(a + b*ArcCsc[c*x]))/x + e*x*(a + b*ArcCsc[c*x]) + (b*e*x*Arc
Tanh[(c*x)/Sqrt[-1 + c^2*x^2]])/Sqrt[c^2*x^2]

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Rubi [A]  time = 0.0645933, antiderivative size = 87, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {14, 5239, 451, 217, 206} \[ -\frac{d \left (a+b \csc ^{-1}(c x)\right )}{x}+e x \left (a+b \csc ^{-1}(c x)\right )-\frac{b c d \sqrt{c^2 x^2-1}}{\sqrt{c^2 x^2}}+\frac{b e x \tanh ^{-1}\left (\frac{c x}{\sqrt{c^2 x^2-1}}\right )}{\sqrt{c^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)*(a + b*ArcCsc[c*x]))/x^2,x]

[Out]

-((b*c*d*Sqrt[-1 + c^2*x^2])/Sqrt[c^2*x^2]) - (d*(a + b*ArcCsc[c*x]))/x + e*x*(a + b*ArcCsc[c*x]) + (b*e*x*Arc
Tanh[(c*x)/Sqrt[-1 + c^2*x^2]])/Sqrt[c^2*x^2]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 5239

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Dist[a + b*ArcCsc[c*x], u, x] + Dist[(b*c*x)/Sqrt[c^2*x^2], Int[SimplifyI
ntegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] &&  !(ILtQ
[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (I
LtQ[(m + 2*p + 1)/2, 0] &&  !ILtQ[(m - 1)/2, 0]))

Rule 451

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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{\left (d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right )}{x^2} \, dx &=-\frac{d \left (a+b \csc ^{-1}(c x)\right )}{x}+e x \left (a+b \csc ^{-1}(c x)\right )+\frac{(b c x) \int \frac{-d+e x^2}{x^2 \sqrt{-1+c^2 x^2}} \, dx}{\sqrt{c^2 x^2}}\\ &=-\frac{b c d \sqrt{-1+c^2 x^2}}{\sqrt{c^2 x^2}}-\frac{d \left (a+b \csc ^{-1}(c x)\right )}{x}+e x \left (a+b \csc ^{-1}(c x)\right )+\frac{(b c e x) \int \frac{1}{\sqrt{-1+c^2 x^2}} \, dx}{\sqrt{c^2 x^2}}\\ &=-\frac{b c d \sqrt{-1+c^2 x^2}}{\sqrt{c^2 x^2}}-\frac{d \left (a+b \csc ^{-1}(c x)\right )}{x}+e x \left (a+b \csc ^{-1}(c x)\right )+\frac{(b c e x) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^2} \, dx,x,\frac{x}{\sqrt{-1+c^2 x^2}}\right )}{\sqrt{c^2 x^2}}\\ &=-\frac{b c d \sqrt{-1+c^2 x^2}}{\sqrt{c^2 x^2}}-\frac{d \left (a+b \csc ^{-1}(c x)\right )}{x}+e x \left (a+b \csc ^{-1}(c x)\right )+\frac{b e x \tanh ^{-1}\left (\frac{c x}{\sqrt{-1+c^2 x^2}}\right )}{\sqrt{c^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.119431, size = 104, normalized size = 1.2 \[ -\frac{a d}{x}+a e x-b c d \sqrt{\frac{c^2 x^2-1}{c^2 x^2}}+\frac{b e x \sqrt{1-\frac{1}{c^2 x^2}} \tanh ^{-1}\left (\frac{c x}{\sqrt{c^2 x^2-1}}\right )}{\sqrt{c^2 x^2-1}}-\frac{b d \csc ^{-1}(c x)}{x}+b e x \csc ^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*d)/x) + a*e*x - b*c*d*Sqrt[(-1 + c^2*x^2)/(c^2*x^2)] - (b*d*ArcCsc[c*x])/x + b*e*x*ArcCsc[c*x] + (b*e*Sqr
t[1 - 1/(c^2*x^2)]*x*ArcTanh[(c*x)/Sqrt[-1 + c^2*x^2]])/Sqrt[-1 + c^2*x^2]

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Maple [A]  time = 0.186, size = 136, normalized size = 1.6 \begin{align*} aex-{\frac{ad}{x}}+b{\rm arccsc} \left (cx\right )ex-{\frac{b{\rm arccsc} \left (cx\right )d}{x}}-{bcd{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{bd}{c{x}^{2}}{\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}}+{\frac{be}{{c}^{2}x}\sqrt{{c}^{2}{x}^{2}-1}\ln \left ( cx+\sqrt{{c}^{2}{x}^{2}-1} \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}-1}{{c}^{2}{x}^{2}}}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*e*x-a*d/x+b*arccsc(c*x)*e*x-b*arccsc(c*x)*d/x-c*b/((c^2*x^2-1)/c^2/x^2)^(1/2)*d+b/c/x^2/((c^2*x^2-1)/c^2/x^2
)^(1/2)*d+b/c^2*(c^2*x^2-1)^(1/2)/((c^2*x^2-1)/c^2/x^2)^(1/2)/x*e*ln(c*x+(c^2*x^2-1)^(1/2))

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Maxima [A]  time = 0.988959, size = 120, normalized size = 1.38 \begin{align*} -{\left (c \sqrt{-\frac{1}{c^{2} x^{2}} + 1} + \frac{\operatorname{arccsc}\left (c x\right )}{x}\right )} b d + a e x + \frac{{\left (2 \, c x \operatorname{arccsc}\left (c x\right ) + \log \left (\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right ) - \log \left (-\sqrt{-\frac{1}{c^{2} x^{2}} + 1} + 1\right )\right )} b e}{2 \, c} - \frac{a d}{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(c*sqrt(-1/(c^2*x^2) + 1) + arccsc(c*x)/x)*b*d + a*e*x + 1/2*(2*c*x*arccsc(c*x) + log(sqrt(-1/(c^2*x^2) + 1)
+ 1) - log(-sqrt(-1/(c^2*x^2) + 1) + 1))*b*e/c - a*d/x

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Fricas [A]  time = 2.90754, size = 288, normalized size = 3.31 \begin{align*} -\frac{b c^{2} d x - a c e x^{2} + b e x \log \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) + \sqrt{c^{2} x^{2} - 1} b c d + a c d - 2 \,{\left (b c d - b c e\right )} x \arctan \left (-c x + \sqrt{c^{2} x^{2} - 1}\right ) -{\left (b c e x^{2} - b c d +{\left (b c d - b c e\right )} x\right )} \operatorname{arccsc}\left (c x\right )}{c x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(b*c^2*d*x - a*c*e*x^2 + b*e*x*log(-c*x + sqrt(c^2*x^2 - 1)) + sqrt(c^2*x^2 - 1)*b*c*d + a*c*d - 2*(b*c*d - b
*c*e)*x*arctan(-c*x + sqrt(c^2*x^2 - 1)) - (b*c*e*x^2 - b*c*d + (b*c*d - b*c*e)*x)*arccsc(c*x))/(c*x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b \operatorname{acsc}{\left (c x \right )}\right ) \left (d + e x^{2}\right )}{x^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}{\left (b \operatorname{arccsc}\left (c x\right ) + a\right )}}{x^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((e*x^2 + d)*(b*arccsc(c*x) + a)/x^2, x)