3.158 \(\int \frac {x (a+b \csc ^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=138 \[ -\frac {a+b \csc ^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}+\frac {b c x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{3 d^{3/2} e \sqrt {c^2 x^2}}+\frac {b c x \sqrt {c^2 x^2-1}}{3 d \sqrt {c^2 x^2} \left (c^2 d+e\right ) \sqrt {d+e x^2}} \]

[Out]

1/3*(-a-b*arccsc(c*x))/e/(e*x^2+d)^(3/2)+1/3*b*c*x*arctan((e*x^2+d)^(1/2)/d^(1/2)/(c^2*x^2-1)^(1/2))/d^(3/2)/e
/(c^2*x^2)^(1/2)+1/3*b*c*x*(c^2*x^2-1)^(1/2)/d/(c^2*d+e)/(c^2*x^2)^(1/2)/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {5237, 446, 96, 93, 204} \[ -\frac {a+b \csc ^{-1}(c x)}{3 e \left (d+e x^2\right )^{3/2}}+\frac {b c x \tan ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{3 d^{3/2} e \sqrt {c^2 x^2}}+\frac {b c x \sqrt {c^2 x^2-1}}{3 d \sqrt {c^2 x^2} \left (c^2 d+e\right ) \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*ArcCsc[c*x]))/(d + e*x^2)^(5/2),x]

[Out]

(b*c*x*Sqrt[-1 + c^2*x^2])/(3*d*(c^2*d + e)*Sqrt[c^2*x^2]*Sqrt[d + e*x^2]) - (a + b*ArcCsc[c*x])/(3*e*(d + e*x
^2)^(3/2)) + (b*c*x*ArcTan[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[-1 + c^2*x^2])])/(3*d^(3/2)*e*Sqrt[c^2*x^2])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

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])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5237

Int[((a_.) + ArcCsc[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x^2)^(p +
1)*(a + b*ArcCsc[c*x]))/(2*e*(p + 1)), x] + Dist[(b*c*x)/(2*e*(p + 1)*Sqrt[c^2*x^2]), Int[(d + e*x^2)^(p + 1)/
(x*Sqrt[c^2*x^2 - 1]), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.21, size = 157, normalized size = 1.14 \[ \frac {-a d \left (c^2 d+e\right )+b c e x \sqrt {1-\frac {1}{c^2 x^2}} \left (d+e x^2\right )-b d \left (c^2 d+e\right ) \csc ^{-1}(c x)}{3 d e \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}-\frac {b c x \sqrt {1-\frac {1}{c^2 x^2}} \tan ^{-1}\left (\frac {\sqrt {d} \sqrt {c^2 x^2-1}}{\sqrt {d+e x^2}}\right )}{3 d^{3/2} e \sqrt {c^2 x^2-1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(a + b*ArcCsc[c*x]))/(d + e*x^2)^(5/2),x]

[Out]

(-(a*d*(c^2*d + e)) + b*c*e*Sqrt[1 - 1/(c^2*x^2)]*x*(d + e*x^2) - b*d*(c^2*d + e)*ArcCsc[c*x])/(3*d*e*(c^2*d +
e)*(d + e*x^2)^(3/2)) - (b*c*Sqrt[1 - 1/(c^2*x^2)]*x*ArcTan[(Sqrt[d]*Sqrt[-1 + c^2*x^2])/Sqrt[d + e*x^2]])/(3
*d^(3/2)*e*Sqrt[-1 + c^2*x^2])

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fricas [B]  time = 1.06, size = 573, normalized size = 4.15 \[ \left [-\frac {{\left (b c^{2} d^{3} + {\left (b c^{2} d e^{2} + b e^{3}\right )} x^{4} + b d^{2} e + 2 \, {\left (b c^{2} d^{2} e + b d e^{2}\right )} x^{2}\right )} \sqrt {-d} \log \left (\frac {{\left (c^{4} d^{2} - 6 \, c^{2} d e + e^{2}\right )} x^{4} - 8 \, {\left (c^{2} d^{2} - d e\right )} x^{2} + 4 \, \sqrt {c^{2} x^{2} - 1} {\left ({\left (c^{2} d - e\right )} x^{2} - 2 \, d\right )} \sqrt {e x^{2} + d} \sqrt {-d} + 8 \, d^{2}}{x^{4}}\right ) + 4 \, {\left (a c^{2} d^{3} + a d^{2} e + {\left (b c^{2} d^{3} + b d^{2} e\right )} \operatorname {arccsc}\left (c x\right ) - {\left (b d e^{2} x^{2} + b d^{2} e\right )} \sqrt {c^{2} x^{2} - 1}\right )} \sqrt {e x^{2} + d}}{12 \, {\left (c^{2} d^{5} e + d^{4} e^{2} + {\left (c^{2} d^{3} e^{3} + d^{2} e^{4}\right )} x^{4} + 2 \, {\left (c^{2} d^{4} e^{2} + d^{3} e^{3}\right )} x^{2}\right )}}, \frac {{\left (b c^{2} d^{3} + {\left (b c^{2} d e^{2} + b e^{3}\right )} x^{4} + b d^{2} e + 2 \, {\left (b c^{2} d^{2} e + b d e^{2}\right )} x^{2}\right )} \sqrt {d} \arctan \left (-\frac {\sqrt {c^{2} x^{2} - 1} {\left ({\left (c^{2} d - e\right )} x^{2} - 2 \, d\right )} \sqrt {e x^{2} + d} \sqrt {d}}{2 \, {\left (c^{2} d e x^{4} + {\left (c^{2} d^{2} - d e\right )} x^{2} - d^{2}\right )}}\right ) - 2 \, {\left (a c^{2} d^{3} + a d^{2} e + {\left (b c^{2} d^{3} + b d^{2} e\right )} \operatorname {arccsc}\left (c x\right ) - {\left (b d e^{2} x^{2} + b d^{2} e\right )} \sqrt {c^{2} x^{2} - 1}\right )} \sqrt {e x^{2} + d}}{6 \, {\left (c^{2} d^{5} e + d^{4} e^{2} + {\left (c^{2} d^{3} e^{3} + d^{2} e^{4}\right )} x^{4} + 2 \, {\left (c^{2} d^{4} e^{2} + d^{3} e^{3}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arccsc(c*x))/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*((b*c^2*d^3 + (b*c^2*d*e^2 + b*e^3)*x^4 + b*d^2*e + 2*(b*c^2*d^2*e + b*d*e^2)*x^2)*sqrt(-d)*log(((c^4*d
^2 - 6*c^2*d*e + e^2)*x^4 - 8*(c^2*d^2 - d*e)*x^2 + 4*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d
)*sqrt(-d) + 8*d^2)/x^4) + 4*(a*c^2*d^3 + a*d^2*e + (b*c^2*d^3 + b*d^2*e)*arccsc(c*x) - (b*d*e^2*x^2 + b*d^2*e
)*sqrt(c^2*x^2 - 1))*sqrt(e*x^2 + d))/(c^2*d^5*e + d^4*e^2 + (c^2*d^3*e^3 + d^2*e^4)*x^4 + 2*(c^2*d^4*e^2 + d^
3*e^3)*x^2), 1/6*((b*c^2*d^3 + (b*c^2*d*e^2 + b*e^3)*x^4 + b*d^2*e + 2*(b*c^2*d^2*e + b*d*e^2)*x^2)*sqrt(d)*ar
ctan(-1/2*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(d)/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2
- d^2)) - 2*(a*c^2*d^3 + a*d^2*e + (b*c^2*d^3 + b*d^2*e)*arccsc(c*x) - (b*d*e^2*x^2 + b*d^2*e)*sqrt(c^2*x^2 -
1))*sqrt(e*x^2 + d))/(c^2*d^5*e + d^4*e^2 + (c^2*d^3*e^3 + d^2*e^4)*x^4 + 2*(c^2*d^4*e^2 + d^3*e^3)*x^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arccsc}\left (c x\right ) + a\right )} x}{{\left (e x^{2} + d\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arccsc(c*x))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccsc(c*x) + a)*x/(e*x^2 + d)^(5/2), x)

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maple [F]  time = 4.94, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a +b \,\mathrm {arccsc}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arccsc(c*x))/(e*x^2+d)^(5/2),x)

[Out]

int(x*(a+b*arccsc(c*x))/(e*x^2+d)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ b \int \frac {x \arctan \left (1, \sqrt {c x + 1} \sqrt {c x - 1}\right )}{{\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt {e x^{2} + d}}\,{d x} - \frac {a}{3 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arccsc(c*x))/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

b*integrate(x*arctan2(1, sqrt(c*x + 1)*sqrt(c*x - 1))/((e^2*x^4 + 2*d*e*x^2 + d^2)*sqrt(e*x^2 + d)), x) - 1/3*
a/((e*x^2 + d)^(3/2)*e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x\,\left (a+b\,\mathrm {asin}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*asin(1/(c*x))))/(d + e*x^2)^(5/2),x)

[Out]

int((x*(a + b*asin(1/(c*x))))/(d + e*x^2)^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x \left (a + b \operatorname {acsc}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*acsc(c*x))/(e*x**2+d)**(5/2),x)

[Out]

Integral(x*(a + b*acsc(c*x))/(d + e*x**2)**(5/2), x)

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