3.1166 \(\int \frac {x^3 (a+b \tan ^{-1}(c x))}{(d+e x^2)^3} \, dx\)

Optimal. Leaf size=130 \[ \frac {x^4 \left (a+b \tan ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac {b c \left (c^2 d-3 e\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 \sqrt {d} e^{3/2} \left (c^2 d-e\right )^2}+\frac {b c x}{8 e \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac {b \tan ^{-1}(c x)}{4 d \left (c^2 d-e\right )^2} \]

[Out]

1/8*b*c*x/(c^2*d-e)/e/(e*x^2+d)-1/4*b*arctan(c*x)/d/(c^2*d-e)^2+1/4*x^4*(a+b*arctan(c*x))/d/(e*x^2+d)^2-1/8*b*
c*(c^2*d-3*e)*arctan(x*e^(1/2)/d^(1/2))/(c^2*d-e)^2/e^(3/2)/d^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {264, 4976, 12, 470, 522, 205} \[ \frac {x^4 \left (a+b \tan ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac {b c \left (c^2 d-3 e\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 \sqrt {d} e^{3/2} \left (c^2 d-e\right )^2}+\frac {b c x}{8 e \left (c^2 d-e\right ) \left (d+e x^2\right )}-\frac {b \tan ^{-1}(c x)}{4 d \left (c^2 d-e\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*x)/(8*(c^2*d - e)*e*(d + e*x^2)) - (b*ArcTan[c*x])/(4*d*(c^2*d - e)^2) + (x^4*(a + b*ArcTan[c*x]))/(4*d*(
d + e*x^2)^2) - (b*c*(c^2*d - 3*e)*ArcTan[(Sqrt[e]*x)/Sqrt[d]])/(8*Sqrt[d]*(c^2*d - e)^2*e^(3/2))

Rule 12

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[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 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

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^3 \left (a+b \tan ^{-1}(c x)\right )}{\left (d+e x^2\right )^3} \, dx &=\frac {x^4 \left (a+b \tan ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-(b c) \int \frac {x^4}{4 \left (d+c^2 d x^2\right ) \left (d+e x^2\right )^2} \, dx\\ &=\frac {x^4 \left (a+b \tan ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac {1}{4} (b c) \int \frac {x^4}{\left (d+c^2 d x^2\right ) \left (d+e x^2\right )^2} \, dx\\ &=\frac {b c x}{8 \left (c^2 d-e\right ) e \left (d+e x^2\right )}+\frac {x^4 \left (a+b \tan ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac {(b c) \int \frac {d^2+d \left (c^2 d-2 e\right ) x^2}{\left (d+c^2 d x^2\right ) \left (d+e x^2\right )} \, dx}{8 d \left (c^2 d-e\right ) e}\\ &=\frac {b c x}{8 \left (c^2 d-e\right ) e \left (d+e x^2\right )}+\frac {x^4 \left (a+b \tan ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac {(b c) \int \frac {1}{d+c^2 d x^2} \, dx}{4 \left (c^2 d-e\right )^2}-\frac {\left (b c \left (c^2 d-3 e\right )\right ) \int \frac {1}{d+e x^2} \, dx}{8 \left (c^2 d-e\right )^2 e}\\ &=\frac {b c x}{8 \left (c^2 d-e\right ) e \left (d+e x^2\right )}-\frac {b \tan ^{-1}(c x)}{4 d \left (c^2 d-e\right )^2}+\frac {x^4 \left (a+b \tan ^{-1}(c x)\right )}{4 d \left (d+e x^2\right )^2}-\frac {b c \left (c^2 d-3 e\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{8 \sqrt {d} \left (c^2 d-e\right )^2 e^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 3.76, size = 158, normalized size = 1.22 \[ \frac {\frac {-4 a c^2 d+4 a e+b c e x}{\left (c^2 d-e\right ) \left (d+e x^2\right )}+\frac {2 a d}{\left (d+e x^2\right )^2}+\frac {2 b c^2 \left (c^2 d-2 e\right ) \tan ^{-1}(c x)}{\left (e-c^2 d\right )^2}-\frac {b c \sqrt {e} \left (c^2 d-3 e\right ) \tan ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d}}\right )}{\sqrt {d} \left (e-c^2 d\right )^2}-\frac {2 b \tan ^{-1}(c x) \left (d+2 e x^2\right )}{\left (d+e x^2\right )^2}}{8 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*d)/(d + e*x^2)^2 + (-4*a*c^2*d + 4*a*e + b*c*e*x)/((c^2*d - e)*(d + e*x^2)) + (2*b*c^2*(c^2*d - 2*e)*Arc
Tan[c*x])/(-(c^2*d) + e)^2 - (2*b*(d + 2*e*x^2)*ArcTan[c*x])/(d + e*x^2)^2 - (b*c*(c^2*d - 3*e)*Sqrt[e]*ArcTan
[(Sqrt[e]*x)/Sqrt[d]])/(Sqrt[d]*(-(c^2*d) + e)^2))/(8*e^2)

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fricas [B]  time = 0.59, size = 697, normalized size = 5.36 \[ \left [-\frac {4 \, a c^{4} d^{4} - 8 \, a c^{2} d^{3} e + 4 \, a d^{2} e^{2} - 2 \, {\left (b c^{3} d^{2} e^{2} - b c d e^{3}\right )} x^{3} + 8 \, {\left (a c^{4} d^{3} e - 2 \, a c^{2} d^{2} e^{2} + a d e^{3}\right )} x^{2} - {\left (b c^{3} d^{3} - 3 \, b c d^{2} e + {\left (b c^{3} d e^{2} - 3 \, b c e^{3}\right )} x^{4} + 2 \, {\left (b c^{3} d^{2} e - 3 \, b c d e^{2}\right )} x^{2}\right )} \sqrt {-d e} \log \left (\frac {e x^{2} - 2 \, \sqrt {-d e} x - d}{e x^{2} + d}\right ) - 2 \, {\left (b c^{3} d^{3} e - b c d^{2} e^{2}\right )} x + 4 \, {\left (2 \, b d e^{3} x^{2} + b d^{2} e^{2} - {\left (b c^{4} d^{2} e^{2} - 2 \, b c^{2} d e^{3}\right )} x^{4}\right )} \arctan \left (c x\right )}{16 \, {\left (c^{4} d^{5} e^{2} - 2 \, c^{2} d^{4} e^{3} + d^{3} e^{4} + {\left (c^{4} d^{3} e^{4} - 2 \, c^{2} d^{2} e^{5} + d e^{6}\right )} x^{4} + 2 \, {\left (c^{4} d^{4} e^{3} - 2 \, c^{2} d^{3} e^{4} + d^{2} e^{5}\right )} x^{2}\right )}}, -\frac {2 \, a c^{4} d^{4} - 4 \, a c^{2} d^{3} e + 2 \, a d^{2} e^{2} - {\left (b c^{3} d^{2} e^{2} - b c d e^{3}\right )} x^{3} + 4 \, {\left (a c^{4} d^{3} e - 2 \, a c^{2} d^{2} e^{2} + a d e^{3}\right )} x^{2} + {\left (b c^{3} d^{3} - 3 \, b c d^{2} e + {\left (b c^{3} d e^{2} - 3 \, b c e^{3}\right )} x^{4} + 2 \, {\left (b c^{3} d^{2} e - 3 \, b c d e^{2}\right )} x^{2}\right )} \sqrt {d e} \arctan \left (\frac {\sqrt {d e} x}{d}\right ) - {\left (b c^{3} d^{3} e - b c d^{2} e^{2}\right )} x + 2 \, {\left (2 \, b d e^{3} x^{2} + b d^{2} e^{2} - {\left (b c^{4} d^{2} e^{2} - 2 \, b c^{2} d e^{3}\right )} x^{4}\right )} \arctan \left (c x\right )}{8 \, {\left (c^{4} d^{5} e^{2} - 2 \, c^{2} d^{4} e^{3} + d^{3} e^{4} + {\left (c^{4} d^{3} e^{4} - 2 \, c^{2} d^{2} e^{5} + d e^{6}\right )} x^{4} + 2 \, {\left (c^{4} d^{4} e^{3} - 2 \, c^{2} d^{3} e^{4} + d^{2} e^{5}\right )} x^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/16*(4*a*c^4*d^4 - 8*a*c^2*d^3*e + 4*a*d^2*e^2 - 2*(b*c^3*d^2*e^2 - b*c*d*e^3)*x^3 + 8*(a*c^4*d^3*e - 2*a*c
^2*d^2*e^2 + a*d*e^3)*x^2 - (b*c^3*d^3 - 3*b*c*d^2*e + (b*c^3*d*e^2 - 3*b*c*e^3)*x^4 + 2*(b*c^3*d^2*e - 3*b*c*
d*e^2)*x^2)*sqrt(-d*e)*log((e*x^2 - 2*sqrt(-d*e)*x - d)/(e*x^2 + d)) - 2*(b*c^3*d^3*e - b*c*d^2*e^2)*x + 4*(2*
b*d*e^3*x^2 + b*d^2*e^2 - (b*c^4*d^2*e^2 - 2*b*c^2*d*e^3)*x^4)*arctan(c*x))/(c^4*d^5*e^2 - 2*c^2*d^4*e^3 + d^3
*e^4 + (c^4*d^3*e^4 - 2*c^2*d^2*e^5 + d*e^6)*x^4 + 2*(c^4*d^4*e^3 - 2*c^2*d^3*e^4 + d^2*e^5)*x^2), -1/8*(2*a*c
^4*d^4 - 4*a*c^2*d^3*e + 2*a*d^2*e^2 - (b*c^3*d^2*e^2 - b*c*d*e^3)*x^3 + 4*(a*c^4*d^3*e - 2*a*c^2*d^2*e^2 + a*
d*e^3)*x^2 + (b*c^3*d^3 - 3*b*c*d^2*e + (b*c^3*d*e^2 - 3*b*c*e^3)*x^4 + 2*(b*c^3*d^2*e - 3*b*c*d*e^2)*x^2)*sqr
t(d*e)*arctan(sqrt(d*e)*x/d) - (b*c^3*d^3*e - b*c*d^2*e^2)*x + 2*(2*b*d*e^3*x^2 + b*d^2*e^2 - (b*c^4*d^2*e^2 -
2*b*c^2*d*e^3)*x^4)*arctan(c*x))/(c^4*d^5*e^2 - 2*c^2*d^4*e^3 + d^3*e^4 + (c^4*d^3*e^4 - 2*c^2*d^2*e^5 + d*e^
6)*x^4 + 2*(c^4*d^4*e^3 - 2*c^2*d^3*e^4 + d^2*e^5)*x^2)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.05, size = 297, normalized size = 2.28 \[ \frac {c^{4} a d}{4 e^{2} \left (c^{2} e \,x^{2}+c^{2} d \right )^{2}}-\frac {c^{2} a}{2 e^{2} \left (c^{2} e \,x^{2}+c^{2} d \right )}+\frac {c^{4} b \arctan \left (c x \right ) d}{4 e^{2} \left (c^{2} e \,x^{2}+c^{2} d \right )^{2}}-\frac {c^{2} b \arctan \left (c x \right )}{2 e^{2} \left (c^{2} e \,x^{2}+c^{2} d \right )}+\frac {c^{5} b d x}{8 e \left (c^{2} d -e \right )^{2} \left (c^{2} e \,x^{2}+c^{2} d \right )}-\frac {c^{3} b x}{8 \left (c^{2} d -e \right )^{2} \left (c^{2} e \,x^{2}+c^{2} d \right )}-\frac {c^{3} b d \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 e \left (c^{2} d -e \right )^{2} \sqrt {d e}}+\frac {3 c b \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{8 \left (c^{2} d -e \right )^{2} \sqrt {d e}}+\frac {c^{4} b d \arctan \left (c x \right )}{4 e^{2} \left (c^{2} d -e \right )^{2}}-\frac {c^{2} b \arctan \left (c x \right )}{2 e \left (c^{2} d -e \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*c^4*a/e^2*d/(c^2*e*x^2+c^2*d)^2-1/2*c^2*a/e^2/(c^2*e*x^2+c^2*d)+1/4*c^4*b*arctan(c*x)/e^2*d/(c^2*e*x^2+c^2
*d)^2-1/2*c^2*b*arctan(c*x)/e^2/(c^2*e*x^2+c^2*d)+1/8*c^5*b/e*d/(c^2*d-e)^2*x/(c^2*e*x^2+c^2*d)-1/8*c^3*b/(c^2
*d-e)^2*x/(c^2*e*x^2+c^2*d)-1/8*c^3*b/e*d/(c^2*d-e)^2/(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))+3/8*c*b/(c^2*d-e)^2/
(d*e)^(1/2)*arctan(e*x/(d*e)^(1/2))+1/4*c^4*b/e^2*d/(c^2*d-e)^2*arctan(c*x)-1/2*c^2*b/e/(c^2*d-e)^2*arctan(c*x
)

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maxima [A]  time = 0.42, size = 216, normalized size = 1.66 \[ -\frac {1}{8} \, {\left (c {\left (\frac {{\left (c^{2} d - 3 \, e\right )} \arctan \left (\frac {e x}{\sqrt {d e}}\right )}{{\left (c^{4} d^{2} e - 2 \, c^{2} d e^{2} + e^{3}\right )} \sqrt {d e}} - \frac {x}{c^{2} d^{2} e - d e^{2} + {\left (c^{2} d e^{2} - e^{3}\right )} x^{2}} - \frac {2 \, {\left (c^{4} d - 2 \, c^{2} e\right )} \arctan \left (c x\right )}{{\left (c^{4} d^{2} e^{2} - 2 \, c^{2} d e^{3} + e^{4}\right )} c}\right )} + \frac {2 \, {\left (2 \, e x^{2} + d\right )} \arctan \left (c x\right )}{e^{4} x^{4} + 2 \, d e^{3} x^{2} + d^{2} e^{2}}\right )} b - \frac {{\left (2 \, e x^{2} + d\right )} a}{4 \, {\left (e^{4} x^{4} + 2 \, d e^{3} x^{2} + d^{2} e^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(c*((c^2*d - 3*e)*arctan(e*x/sqrt(d*e))/((c^4*d^2*e - 2*c^2*d*e^2 + e^3)*sqrt(d*e)) - x/(c^2*d^2*e - d*e^
2 + (c^2*d*e^2 - e^3)*x^2) - 2*(c^4*d - 2*c^2*e)*arctan(c*x)/((c^4*d^2*e^2 - 2*c^2*d*e^3 + e^4)*c)) + 2*(2*e*x
^2 + d)*arctan(c*x)/(e^4*x^4 + 2*d*e^3*x^2 + d^2*e^2))*b - 1/4*(2*e*x^2 + d)*a/(e^4*x^4 + 2*d*e^3*x^2 + d^2*e^
2)

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mupad [B]  time = 3.30, size = 273, normalized size = 2.10 \[ \frac {b\,c^4\,d\,\mathrm {atan}\left (c\,x\right )}{4\,e^2\,{\left (e-c^2\,d\right )}^2}-\frac {a\,d}{4\,e^2\,{\left (e\,x^2+d\right )}^2}-\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{4\,e^2\,{\left (e\,x^2+d\right )}^2}-\frac {b\,c\,x^3}{8\,\left (e-c^2\,d\right )\,{\left (e\,x^2+d\right )}^2}-\frac {b\,c^2\,\mathrm {atan}\left (c\,x\right )}{2\,e\,{\left (e-c^2\,d\right )}^2}-\frac {b\,x^2\,\mathrm {atan}\left (c\,x\right )}{2\,e\,{\left (e\,x^2+d\right )}^2}-\frac {b\,c^3\,\mathrm {atan}\left (\frac {x\,\sqrt {-d\,e^3}\,1{}\mathrm {i}}{d\,e}\right )\,\sqrt {-d\,e^3}\,1{}\mathrm {i}}{8\,e^3\,{\left (e-c^2\,d\right )}^2}-\frac {a\,x^2}{2\,e\,{\left (e\,x^2+d\right )}^2}-\frac {b\,c\,d\,x}{8\,e\,\left (e-c^2\,d\right )\,{\left (e\,x^2+d\right )}^2}+\frac {b\,c\,\mathrm {atan}\left (\frac {x\,\sqrt {-d\,e^3}\,1{}\mathrm {i}}{d\,e}\right )\,\sqrt {-d\,e^3}\,3{}\mathrm {i}}{8\,d\,e^2\,{\left (e-c^2\,d\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*c^4*d*atan(c*x))/(4*e^2*(e - c^2*d)^2) - (a*d)/(4*e^2*(d + e*x^2)^2) - (b*d*atan(c*x))/(4*e^2*(d + e*x^2)^2
) - (b*c*x^3)/(8*(e - c^2*d)*(d + e*x^2)^2) - (b*c^2*atan(c*x))/(2*e*(e - c^2*d)^2) - (b*x^2*atan(c*x))/(2*e*(
d + e*x^2)^2) - (b*c^3*atan((x*(-d*e^3)^(1/2)*1i)/(d*e))*(-d*e^3)^(1/2)*1i)/(8*e^3*(e - c^2*d)^2) - (a*x^2)/(2
*e*(d + e*x^2)^2) - (b*c*d*x)/(8*e*(e - c^2*d)*(d + e*x^2)^2) + (b*c*atan((x*(-d*e^3)^(1/2)*1i)/(d*e))*(-d*e^3
)^(1/2)*3i)/(8*d*e^2*(e - c^2*d)^2)

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

[Out]

Timed out

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