3.1219 \(\int \frac {x^2 (a+b \tan ^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=109 \[ \frac {x^3 \left (a+b \tan ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {b c}{3 e \left (c^2 d-e\right ) \sqrt {d+e x^2}}-\frac {b \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d \left (c^2 d-e\right )^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {264, 4976, 446, 78, 63, 208} \[ \frac {x^3 \left (a+b \tan ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac {b c}{3 e \left (c^2 d-e\right ) \sqrt {d+e x^2}}-\frac {b \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d \left (c^2 d-e\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c)/(3*(c^2*d - e)*e*Sqrt[d + e*x^2]) + (x^3*(a + b*ArcTan[c*x]))/(3*d*(d + e*x^2)^(3/2)) - (b*ArcTanh[(c*Sq
rt[d + e*x^2])/Sqrt[c^2*d - e]])/(3*d*(c^2*d - e)^(3/2))

Rule 63

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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 208

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

Rule 264

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

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 4976

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

Rubi steps

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

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Mathematica [C]  time = 1.11, size = 252, normalized size = 2.31 \[ -\frac {-\frac {2 \left (a x \left (c^2 d-e\right )+b c d\right )}{e \left (c^2 d-e\right ) \sqrt {d+e x^2}}+\frac {2 a d x}{e \left (d+e x^2\right )^{3/2}}+\frac {b \log \left (\frac {12 c d \sqrt {c^2 d-e} \left (\sqrt {c^2 d-e} \sqrt {d+e x^2}+c d-i e x\right )}{b (c x+i)}\right )}{\left (c^2 d-e\right )^{3/2}}+\frac {b \log \left (\frac {12 c d \sqrt {c^2 d-e} \left (\sqrt {c^2 d-e} \sqrt {d+e x^2}+c d+i e x\right )}{b (c x-i)}\right )}{\left (c^2 d-e\right )^{3/2}}-\frac {2 b x^3 \tan ^{-1}(c x)}{\left (d+e x^2\right )^{3/2}}}{6 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/6*((2*a*d*x)/(e*(d + e*x^2)^(3/2)) - (2*(b*c*d + a*(c^2*d - e)*x))/((c^2*d - e)*e*Sqrt[d + e*x^2]) - (2*b*x
^3*ArcTan[c*x])/(d + e*x^2)^(3/2) + (b*Log[(12*c*d*Sqrt[c^2*d - e]*(c*d - I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x
^2]))/(b*(I + c*x))])/(c^2*d - e)^(3/2) + (b*Log[(12*c*d*Sqrt[c^2*d - e]*(c*d + I*e*x + Sqrt[c^2*d - e]*Sqrt[d
+ e*x^2]))/(b*(-I + c*x))])/(c^2*d - e)^(3/2))/d

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fricas [B]  time = 0.76, size = 676, normalized size = 6.20 \[ \left [-\frac {{\left (b e^{3} x^{4} + 2 \, b d e^{2} x^{2} + b d^{2} e\right )} \sqrt {c^{2} d - e} \log \left (\frac {c^{4} e^{2} x^{4} + 8 \, c^{4} d^{2} - 8 \, c^{2} d e + 2 \, {\left (4 \, c^{4} d e - 3 \, c^{2} e^{2}\right )} x^{2} + 4 \, {\left (c^{3} e x^{2} + 2 \, c^{3} d - c e\right )} \sqrt {c^{2} d - e} \sqrt {e x^{2} + d} + e^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) - 4 \, {\left (b c^{3} d^{3} - b c d^{2} e + {\left (b c^{4} d^{2} e - 2 \, b c^{2} d e^{2} + b e^{3}\right )} x^{3} \arctan \left (c x\right ) + {\left (a c^{4} d^{2} e - 2 \, a c^{2} d e^{2} + a e^{3}\right )} x^{3} + {\left (b c^{3} d^{2} e - b c d e^{2}\right )} x^{2}\right )} \sqrt {e x^{2} + d}}{12 \, {\left (c^{4} d^{5} e - 2 \, c^{2} d^{4} e^{2} + d^{3} e^{3} + {\left (c^{4} d^{3} e^{3} - 2 \, c^{2} d^{2} e^{4} + d e^{5}\right )} x^{4} + 2 \, {\left (c^{4} d^{4} e^{2} - 2 \, c^{2} d^{3} e^{3} + d^{2} e^{4}\right )} x^{2}\right )}}, -\frac {{\left (b e^{3} x^{4} + 2 \, b d e^{2} x^{2} + b d^{2} e\right )} \sqrt {-c^{2} d + e} \arctan \left (-\frac {{\left (c^{2} e x^{2} + 2 \, c^{2} d - e\right )} \sqrt {-c^{2} d + e} \sqrt {e x^{2} + d}}{2 \, {\left (c^{3} d^{2} - c d e + {\left (c^{3} d e - c e^{2}\right )} x^{2}\right )}}\right ) - 2 \, {\left (b c^{3} d^{3} - b c d^{2} e + {\left (b c^{4} d^{2} e - 2 \, b c^{2} d e^{2} + b e^{3}\right )} x^{3} \arctan \left (c x\right ) + {\left (a c^{4} d^{2} e - 2 \, a c^{2} d e^{2} + a e^{3}\right )} x^{3} + {\left (b c^{3} d^{2} e - b c d e^{2}\right )} x^{2}\right )} \sqrt {e x^{2} + d}}{6 \, {\left (c^{4} d^{5} e - 2 \, c^{2} d^{4} e^{2} + d^{3} e^{3} + {\left (c^{4} d^{3} e^{3} - 2 \, c^{2} d^{2} e^{4} + d e^{5}\right )} x^{4} + 2 \, {\left (c^{4} d^{4} e^{2} - 2 \, c^{2} d^{3} e^{3} + d^{2} e^{4}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [F]  time = 1.10, size = 0, normalized size = 0.00 \[ \int \frac {x^{2} \left (a +b \arctan \left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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