3.2.48 \(\int \frac {x^3 (a+b \csc ^{-1}(c x))}{(d+e x^2)^{3/2}} \, dx\) [148]

Optimal. Leaf size=156 \[ \frac {d \left (a+b \csc ^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} \left (a+b \csc ^{-1}(c x)\right )}{e^2}-\frac {2 b c \sqrt {d} x \text {ArcTan}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1+c^2 x^2}}\right )}{e^2 \sqrt {c^2 x^2}}+\frac {b x \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {-1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{e^{3/2} \sqrt {c^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 11, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.478, Rules used = {272, 45, 5347, 12, 587, 163, 65, 223, 212, 95, 210} \begin {gather*} \frac {\sqrt {d+e x^2} \left (a+b \csc ^{-1}(c x)\right )}{e^2}+\frac {d \left (a+b \csc ^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}-\frac {2 b c \sqrt {d} x \text {ArcTan}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{e^2 \sqrt {c^2 x^2}}+\frac {b x \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{e^{3/2} \sqrt {c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcCsc[c*x]))/(d + e*x^2)^(3/2),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 95

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 163

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 210

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

Rule 212

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

Rule 223

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 272

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

Rule 587

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

Rule 5347

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]))

Rubi steps

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

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Mathematica [A]
time = 0.14, size = 146, normalized size = 0.94 \begin {gather*} \frac {\left (2 d+e x^2\right ) \left (a+b \csc ^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}+\frac {b \sqrt {1-\frac {1}{c^2 x^2}} x \left (2 c \sqrt {d} \text {ArcTan}\left (\frac {\sqrt {d} \sqrt {-1+c^2 x^2}}{\sqrt {d+e x^2}}\right )+\sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {-1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )\right )}{e^2 \sqrt {-1+c^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcCsc[c*x]))/(d + e*x^2)^(3/2),x]

[Out]

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

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \left (a +b \,\mathrm {arccsc}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arccsc(c*x))/(e*x^2+d)^(3/2),x)

[Out]

int(x^3*(a+b*arccsc(c*x))/(e*x^2+d)^(3/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccsc(c*x))/(e*x^2+d)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (131) = 262\).
time = 0.45, size = 572, normalized size = 3.67 \begin {gather*} \left [\frac {{\left (b x^{2} e + b d\right )} e^{\frac {1}{2}} \log \left (c^{4} d^{2} + 4 \, {\left (c^{3} d + {\left (2 \, c^{3} x^{2} - c\right )} e\right )} \sqrt {c^{2} x^{2} - 1} \sqrt {x^{2} e + d} e^{\frac {1}{2}} + {\left (8 \, c^{4} x^{4} - 8 \, c^{2} x^{2} + 1\right )} e^{2} + 2 \, {\left (4 \, c^{4} d x^{2} - 3 \, c^{2} d\right )} e\right ) + 2 \, {\left (b c x^{2} e + b c d\right )} \sqrt {-d} \log \left (\frac {c^{4} d^{2} x^{4} - 8 \, c^{2} d^{2} x^{2} + x^{4} e^{2} + 4 \, {\left (c^{2} d x^{2} - x^{2} e - 2 \, d\right )} \sqrt {c^{2} x^{2} - 1} \sqrt {x^{2} e + d} \sqrt {-d} + 8 \, d^{2} - 2 \, {\left (3 \, c^{2} d x^{4} - 4 \, d x^{2}\right )} e}{x^{4}}\right ) + 4 \, {\left (a c x^{2} e + 2 \, a c d + {\left (b c x^{2} e + 2 \, b c d\right )} \operatorname {arccsc}\left (c x\right )\right )} \sqrt {x^{2} e + d}}{4 \, {\left (c x^{2} e^{3} + c d e^{2}\right )}}, \frac {{\left (b x^{2} e + b d\right )} e^{\frac {1}{2}} \log \left (c^{4} d^{2} + 4 \, {\left (c^{3} d + {\left (2 \, c^{3} x^{2} - c\right )} e\right )} \sqrt {c^{2} x^{2} - 1} \sqrt {x^{2} e + d} e^{\frac {1}{2}} + {\left (8 \, c^{4} x^{4} - 8 \, c^{2} x^{2} + 1\right )} e^{2} + 2 \, {\left (4 \, c^{4} d x^{2} - 3 \, c^{2} d\right )} e\right ) - 4 \, {\left (b c x^{2} e + b c d\right )} \sqrt {d} \arctan \left (-\frac {{\left (c^{2} d x^{2} - x^{2} e - 2 \, d\right )} \sqrt {c^{2} x^{2} - 1} \sqrt {x^{2} e + d} \sqrt {d}}{2 \, {\left (c^{2} d^{2} x^{2} - d^{2} + {\left (c^{2} d x^{4} - d x^{2}\right )} e\right )}}\right ) + 4 \, {\left (a c x^{2} e + 2 \, a c d + {\left (b c x^{2} e + 2 \, b c d\right )} \operatorname {arccsc}\left (c x\right )\right )} \sqrt {x^{2} e + d}}{4 \, {\left (c x^{2} e^{3} + c d e^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccsc(c*x))/(e*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

[1/4*((b*x^2*e + b*d)*e^(1/2)*log(c^4*d^2 + 4*(c^3*d + (2*c^3*x^2 - c)*e)*sqrt(c^2*x^2 - 1)*sqrt(x^2*e + d)*e^
(1/2) + (8*c^4*x^4 - 8*c^2*x^2 + 1)*e^2 + 2*(4*c^4*d*x^2 - 3*c^2*d)*e) + 2*(b*c*x^2*e + b*c*d)*sqrt(-d)*log((c
^4*d^2*x^4 - 8*c^2*d^2*x^2 + x^4*e^2 + 4*(c^2*d*x^2 - x^2*e - 2*d)*sqrt(c^2*x^2 - 1)*sqrt(x^2*e + d)*sqrt(-d)
+ 8*d^2 - 2*(3*c^2*d*x^4 - 4*d*x^2)*e)/x^4) + 4*(a*c*x^2*e + 2*a*c*d + (b*c*x^2*e + 2*b*c*d)*arccsc(c*x))*sqrt
(x^2*e + d))/(c*x^2*e^3 + c*d*e^2), 1/4*((b*x^2*e + b*d)*e^(1/2)*log(c^4*d^2 + 4*(c^3*d + (2*c^3*x^2 - c)*e)*s
qrt(c^2*x^2 - 1)*sqrt(x^2*e + d)*e^(1/2) + (8*c^4*x^4 - 8*c^2*x^2 + 1)*e^2 + 2*(4*c^4*d*x^2 - 3*c^2*d)*e) - 4*
(b*c*x^2*e + b*c*d)*sqrt(d)*arctan(-1/2*(c^2*d*x^2 - x^2*e - 2*d)*sqrt(c^2*x^2 - 1)*sqrt(x^2*e + d)*sqrt(d)/(c
^2*d^2*x^2 - d^2 + (c^2*d*x^4 - d*x^2)*e)) + 4*(a*c*x^2*e + 2*a*c*d + (b*c*x^2*e + 2*b*c*d)*arccsc(c*x))*sqrt(
x^2*e + d))/(c*x^2*e^3 + c*d*e^2)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (a + b \operatorname {acsc}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*acsc(c*x))/(e*x**2+d)**(3/2),x)

[Out]

Integral(x**3*(a + b*acsc(c*x))/(d + e*x**2)**(3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arccsc(c*x))/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

integrate((b*arccsc(c*x) + a)*x^3/(e*x^2 + d)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^3\,\left (a+b\,\mathrm {asin}\left (\frac {1}{c\,x}\right )\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*asin(1/(c*x))))/(d + e*x^2)^(3/2),x)

[Out]

int((x^3*(a + b*asin(1/(c*x))))/(d + e*x^2)^(3/2), x)

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