3.2.57 \(\int \frac {x^3 (a+b \csc ^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\) [157]

Optimal. Leaf size=163 \[ -\frac {b c x \sqrt {-1+c^2 x^2}}{3 e \left (c^2 d+e\right ) \sqrt {c^2 x^2} \sqrt {d+e x^2}}+\frac {d \left (a+b \csc ^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \csc ^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {2 b c x \text {ArcTan}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1+c^2 x^2}}\right )}{3 \sqrt {d} e^2 \sqrt {c^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.348, Rules used = {272, 45, 5347, 12, 587, 157, 95, 210} \begin {gather*} -\frac {a+b \csc ^{-1}(c x)}{e^2 \sqrt {d+e x^2}}+\frac {d \left (a+b \csc ^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}+\frac {2 b c x \text {ArcTan}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{3 \sqrt {d} e^2 \sqrt {c^2 x^2}}-\frac {b c x \sqrt {c^2 x^2-1}}{3 e \sqrt {c^2 x^2} \left (c^2 d+e\right ) \sqrt {d+e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcCsc[c*x]))/(d + e*x^2)^(5/2),x]

[Out]

-1/3*(b*c*x*Sqrt[-1 + c^2*x^2])/(e*(c^2*d + e)*Sqrt[c^2*x^2]*Sqrt[d + e*x^2]) + (d*(a + b*ArcCsc[c*x]))/(3*e^2
*(d + e*x^2)^(3/2)) - (a + b*ArcCsc[c*x])/(e^2*Sqrt[d + e*x^2]) + (2*b*c*x*ArcTan[Sqrt[d + e*x^2]/(Sqrt[d]*Sqr
t[-1 + c^2*x^2])])/(3*Sqrt[d]*e^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 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 157

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(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[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*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

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

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Mathematica [A]
time = 0.16, size = 173, normalized size = 1.06 \begin {gather*} \frac {-b c e \sqrt {1-\frac {1}{c^2 x^2}} x \left (d+e x^2\right )-a \left (c^2 d+e\right ) \left (2 d+3 e x^2\right )-b \left (c^2 d+e\right ) \left (2 d+3 e x^2\right ) \csc ^{-1}(c x)}{3 e^2 \left (c^2 d+e\right ) \left (d+e x^2\right )^{3/2}}-\frac {2 b c \sqrt {1-\frac {1}{c^2 x^2}} x \text {ArcTan}\left (\frac {\sqrt {d} \sqrt {-1+c^2 x^2}}{\sqrt {d+e x^2}}\right )}{3 \sqrt {d} 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)^(5/2),x]

[Out]

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

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \left (a +b \,\mathrm {arccsc}\left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arccsc(c*x))/(e*x^2+d)^(5/2),x)

[Out]

int(x^3*(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 \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)^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 332 vs. \(2 (138) = 276\).
time = 0.48, size = 690, normalized size = 4.23 \begin {gather*} \left [-\frac {{\left (b c^{2} d^{3} + b x^{4} e^{3} + {\left (b c^{2} d x^{4} + 2 \, b d x^{2}\right )} e^{2} + {\left (2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} e\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 ) + 2 \, {\left (2 \, a c^{2} d^{3} + 3 \, a d x^{2} e^{2} + {\left (2 \, b c^{2} d^{3} + 3 \, b d x^{2} e^{2} + {\left (3 \, b c^{2} d^{2} x^{2} + 2 \, b d^{2}\right )} e\right )} \operatorname {arccsc}\left (c x\right ) + {\left (3 \, a c^{2} d^{2} x^{2} + 2 \, a d^{2}\right )} e + {\left (b d x^{2} e^{2} + b d^{2} e\right )} \sqrt {c^{2} x^{2} - 1}\right )} \sqrt {x^{2} e + d}}{6 \, {\left (c^{2} d^{4} e^{2} + d x^{4} e^{5} + {\left (c^{2} d^{2} x^{4} + 2 \, d^{2} x^{2}\right )} e^{4} + {\left (2 \, c^{2} d^{3} x^{2} + d^{3}\right )} e^{3}\right )}}, \frac {{\left (b c^{2} d^{3} + b x^{4} e^{3} + {\left (b c^{2} d x^{4} + 2 \, b d x^{2}\right )} e^{2} + {\left (2 \, b c^{2} d^{2} x^{2} + b d^{2}\right )} e\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 ) - {\left (2 \, a c^{2} d^{3} + 3 \, a d x^{2} e^{2} + {\left (2 \, b c^{2} d^{3} + 3 \, b d x^{2} e^{2} + {\left (3 \, b c^{2} d^{2} x^{2} + 2 \, b d^{2}\right )} e\right )} \operatorname {arccsc}\left (c x\right ) + {\left (3 \, a c^{2} d^{2} x^{2} + 2 \, a d^{2}\right )} e + {\left (b d x^{2} e^{2} + b d^{2} e\right )} \sqrt {c^{2} x^{2} - 1}\right )} \sqrt {x^{2} e + d}}{3 \, {\left (c^{2} d^{4} e^{2} + d x^{4} e^{5} + {\left (c^{2} d^{2} x^{4} + 2 \, d^{2} x^{2}\right )} e^{4} + {\left (2 \, c^{2} d^{3} x^{2} + d^{3}\right )} e^{3}\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)^(5/2),x, algorithm="fricas")

[Out]

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

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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 {5}{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)**(5/2),x)

[Out]

Integral(x**3*(a + b*acsc(c*x))/(d + e*x**2)**(5/2), 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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arccsc(c*x) + a)*x^3/(e*x^2 + d)^(5/2), 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 )}^{5/2}} \,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)^(5/2),x)

[Out]

int((x^3*(a + b*asin(1/(c*x))))/(d + e*x^2)^(5/2), x)

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