3.1144 \(\int \frac {(d+e x^2)^3 (a+b \tan ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=158 \[ -\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{3} b c d^2 \log (x) \left (c^2 d-9 e\right )+\frac {b \left (c^2 d+e\right ) \left (c^4 d^2-10 c^2 d e+e^2\right ) \log \left (c^2 x^2+1\right )}{6 c^3}-\frac {b c d^3}{6 x^2}-\frac {b e^3 x^2}{6 c} \]

[Out]

-1/6*b*c*d^3/x^2-1/6*b*e^3*x^2/c-1/3*d^3*(a+b*arctan(c*x))/x^3-3*d^2*e*(a+b*arctan(c*x))/x+3*d*e^2*x*(a+b*arct
an(c*x))+1/3*e^3*x^3*(a+b*arctan(c*x))-1/3*b*c*d^2*(c^2*d-9*e)*ln(x)+1/6*b*(c^2*d+e)*(c^4*d^2-10*c^2*d*e+e^2)*
ln(c^2*x^2+1)/c^3

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Rubi [A]  time = 0.26, antiderivative size = 158, 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 = {270, 4976, 12, 1799, 1620} \[ -\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {b \left (c^2 d+e\right ) \left (c^4 d^2-10 c^2 d e+e^2\right ) \log \left (c^2 x^2+1\right )}{6 c^3}-\frac {1}{3} b c d^2 \log (x) \left (c^2 d-9 e\right )-\frac {b c d^3}{6 x^2}-\frac {b e^3 x^2}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^4,x]

[Out]

-(b*c*d^3)/(6*x^2) - (b*e^3*x^2)/(6*c) - (d^3*(a + b*ArcTan[c*x]))/(3*x^3) - (3*d^2*e*(a + b*ArcTan[c*x]))/x +
3*d*e^2*x*(a + b*ArcTan[c*x]) + (e^3*x^3*(a + b*ArcTan[c*x]))/3 - (b*c*d^2*(c^2*d - 9*e)*Log[x])/3 + (b*(c^2*
d + e)*(c^4*d^2 - 10*c^2*d*e + e^2)*Log[1 + c^2*x^2])/(6*c^3)

Rule 12

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

Rule 270

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

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 1799

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]

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 {\left (d+e x^2\right )^3 \left (a+b \tan ^{-1}(c x)\right )}{x^4} \, dx &=-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac {-d^3-9 d^2 e x^2+9 d e^2 x^4+e^3 x^6}{3 x^3 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{3} (b c) \int \frac {-d^3-9 d^2 e x^2+9 d e^2 x^4+e^3 x^6}{x^3 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{6} (b c) \operatorname {Subst}\left (\int \frac {-d^3-9 d^2 e x+9 d e^2 x^2+e^3 x^3}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{6} (b c) \operatorname {Subst}\left (\int \left (\frac {e^3}{c^2}-\frac {d^3}{x^2}+\frac {d^2 \left (c^2 d-9 e\right )}{x}+\frac {\left (c^2 d+e\right ) \left (-c^4 d^2+10 c^2 d e-e^2\right )}{c^2 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac {b c d^3}{6 x^2}-\frac {b e^3 x^2}{6 c}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{3} e^3 x^3 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{3} b c d^2 \left (c^2 d-9 e\right ) \log (x)+\frac {b \left (c^2 d+e\right ) \left (c^4 d^2-10 c^2 d e+e^2\right ) \log \left (1+c^2 x^2\right )}{6 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 166, normalized size = 1.05 \[ \frac {1}{6} \left (-\frac {2 a d^3}{x^3}-\frac {18 a d^2 e}{x}+18 a d e^2 x+2 a e^3 x^3-2 b c d^2 \log (x) \left (c^2 d-9 e\right )+\frac {b \left (c^6 d^3-9 c^4 d^2 e-9 c^2 d e^2+e^3\right ) \log \left (c^2 x^2+1\right )}{c^3}-\frac {b c d^3}{x^2}+\frac {2 b \tan ^{-1}(c x) \left (-d^3-9 d^2 e x^2+9 d e^2 x^4+e^3 x^6\right )}{x^3}-\frac {b e^3 x^2}{c}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^4,x]

[Out]

((-2*a*d^3)/x^3 - (b*c*d^3)/x^2 - (18*a*d^2*e)/x + 18*a*d*e^2*x - (b*e^3*x^2)/c + 2*a*e^3*x^3 + (2*b*(-d^3 - 9
*d^2*e*x^2 + 9*d*e^2*x^4 + e^3*x^6)*ArcTan[c*x])/x^3 - 2*b*c*d^2*(c^2*d - 9*e)*Log[x] + (b*(c^6*d^3 - 9*c^4*d^
2*e - 9*c^2*d*e^2 + e^3)*Log[1 + c^2*x^2])/c^3)/6

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fricas [A]  time = 0.46, size = 205, normalized size = 1.30 \[ \frac {2 \, a c^{3} e^{3} x^{6} + 18 \, a c^{3} d e^{2} x^{4} - b c^{2} e^{3} x^{5} - b c^{4} d^{3} x - 18 \, a c^{3} d^{2} e x^{2} - 2 \, a c^{3} d^{3} + {\left (b c^{6} d^{3} - 9 \, b c^{4} d^{2} e - 9 \, b c^{2} d e^{2} + b e^{3}\right )} x^{3} \log \left (c^{2} x^{2} + 1\right ) - 2 \, {\left (b c^{6} d^{3} - 9 \, b c^{4} d^{2} e\right )} x^{3} \log \relax (x) + 2 \, {\left (b c^{3} e^{3} x^{6} + 9 \, b c^{3} d e^{2} x^{4} - 9 \, b c^{3} d^{2} e x^{2} - b c^{3} d^{3}\right )} \arctan \left (c x\right )}{6 \, c^{3} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arctan(c*x))/x^4,x, algorithm="fricas")

[Out]

1/6*(2*a*c^3*e^3*x^6 + 18*a*c^3*d*e^2*x^4 - b*c^2*e^3*x^5 - b*c^4*d^3*x - 18*a*c^3*d^2*e*x^2 - 2*a*c^3*d^3 + (
b*c^6*d^3 - 9*b*c^4*d^2*e - 9*b*c^2*d*e^2 + b*e^3)*x^3*log(c^2*x^2 + 1) - 2*(b*c^6*d^3 - 9*b*c^4*d^2*e)*x^3*lo
g(x) + 2*(b*c^3*e^3*x^6 + 9*b*c^3*d*e^2*x^4 - 9*b*c^3*d^2*e*x^2 - b*c^3*d^3)*arctan(c*x))/(c^3*x^3)

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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((e*x^2+d)^3*(a+b*arctan(c*x))/x^4,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.05, size = 213, normalized size = 1.35 \[ \frac {a \,x^{3} e^{3}}{3}+3 a d \,e^{2} x -\frac {a \,d^{3}}{3 x^{3}}-\frac {3 a \,d^{2} e}{x}+\frac {b \arctan \left (c x \right ) x^{3} e^{3}}{3}+3 b \arctan \left (c x \right ) d \,e^{2} x -\frac {b \arctan \left (c x \right ) d^{3}}{3 x^{3}}-\frac {3 b \arctan \left (c x \right ) d^{2} e}{x}-\frac {b \,e^{3} x^{2}}{6 c}-\frac {c^{3} b \,d^{3} \ln \left (c x \right )}{3}+3 c b \ln \left (c x \right ) d^{2} e -\frac {b c \,d^{3}}{6 x^{2}}+\frac {b \,c^{3} d^{3} \ln \left (c^{2} x^{2}+1\right )}{6}-\frac {3 c b \ln \left (c^{2} x^{2}+1\right ) d^{2} e}{2}-\frac {3 b \ln \left (c^{2} x^{2}+1\right ) d \,e^{2}}{2 c}+\frac {b \ln \left (c^{2} x^{2}+1\right ) e^{3}}{6 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^3*(a+b*arctan(c*x))/x^4,x)

[Out]

1/3*a*x^3*e^3+3*a*d*e^2*x-1/3*a*d^3/x^3-3*a*d^2*e/x+1/3*b*arctan(c*x)*x^3*e^3+3*b*arctan(c*x)*d*e^2*x-1/3*b*ar
ctan(c*x)*d^3/x^3-3*b*arctan(c*x)*d^2*e/x-1/6*b*e^3*x^2/c-1/3*c^3*b*d^3*ln(c*x)+3*c*b*ln(c*x)*d^2*e-1/6*b*c*d^
3/x^2+1/6*b*c^3*d^3*ln(c^2*x^2+1)-3/2*c*b*ln(c^2*x^2+1)*d^2*e-3/2/c*b*ln(c^2*x^2+1)*d*e^2+1/6/c^3*b*ln(c^2*x^2
+1)*e^3

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maxima [A]  time = 0.34, size = 193, normalized size = 1.22 \[ \frac {1}{3} \, a e^{3} x^{3} + \frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d^{3} - \frac {3}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b d^{2} e + \frac {1}{6} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b e^{3} + 3 \, a d e^{2} x + \frac {3 \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d e^{2}}{2 \, c} - \frac {3 \, a d^{2} e}{x} - \frac {a d^{3}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arctan(c*x))/x^4,x, algorithm="maxima")

[Out]

1/3*a*e^3*x^3 + 1/6*((c^2*log(c^2*x^2 + 1) - c^2*log(x^2) - 1/x^2)*c - 2*arctan(c*x)/x^3)*b*d^3 - 3/2*(c*(log(
c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*d^2*e + 1/6*(2*x^3*arctan(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c
^4))*b*e^3 + 3*a*d*e^2*x + 3/2*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*d*e^2/c - 3*a*d^2*e/x - 1/3*a*d^3/x^3

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mupad [B]  time = 0.64, size = 203, normalized size = 1.28 \[ \frac {a\,e^3\,x^3}{3}-\ln \relax (x)\,\left (\frac {b\,c^3\,d^3}{3}-3\,b\,c\,d^2\,e\right )-\frac {\frac {b\,c^2\,d^3\,x}{2}+a\,c\,d^3+9\,a\,e\,c\,d^2\,x^2}{3\,c\,x^3}-x\,\left (\frac {a\,e^3}{c^2}-\frac {a\,e^2\,\left (3\,d\,c^2+e\right )}{c^2}\right )+\frac {\ln \left (c^2\,x^2+1\right )\,\left (b\,c^6\,d^3-9\,b\,c^4\,d^2\,e-9\,b\,c^2\,d\,e^2+b\,e^3\right )}{6\,c^3}-\frac {\mathrm {atan}\left (c\,x\right )\,\left (\frac {b\,d^3}{3}+3\,b\,d^2\,e\,x^2-3\,b\,d\,e^2\,x^4-\frac {b\,e^3\,x^6}{3}\right )}{x^3}-\frac {b\,e^3\,x^2}{6\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + e*x^2)^3)/x^4,x)

[Out]

(a*e^3*x^3)/3 - log(x)*((b*c^3*d^3)/3 - 3*b*c*d^2*e) - (a*c*d^3 + (b*c^2*d^3*x)/2 + 9*a*c*d^2*e*x^2)/(3*c*x^3)
- x*((a*e^3)/c^2 - (a*e^2*(e + 3*c^2*d))/c^2) + (log(c^2*x^2 + 1)*(b*e^3 + b*c^6*d^3 - 9*b*c^2*d*e^2 - 9*b*c^
4*d^2*e))/(6*c^3) - (atan(c*x)*((b*d^3)/3 - (b*e^3*x^6)/3 + 3*b*d^2*e*x^2 - 3*b*d*e^2*x^4))/x^3 - (b*e^3*x^2)/
(6*c)

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sympy [A]  time = 3.30, size = 272, normalized size = 1.72 \[ \begin {cases} - \frac {a d^{3}}{3 x^{3}} - \frac {3 a d^{2} e}{x} + 3 a d e^{2} x + \frac {a e^{3} x^{3}}{3} - \frac {b c^{3} d^{3} \log {\relax (x )}}{3} + \frac {b c^{3} d^{3} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6} - \frac {b c d^{3}}{6 x^{2}} + 3 b c d^{2} e \log {\relax (x )} - \frac {3 b c d^{2} e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b d^{3} \operatorname {atan}{\left (c x \right )}}{3 x^{3}} - \frac {3 b d^{2} e \operatorname {atan}{\left (c x \right )}}{x} + 3 b d e^{2} x \operatorname {atan}{\left (c x \right )} + \frac {b e^{3} x^{3} \operatorname {atan}{\left (c x \right )}}{3} - \frac {3 b d e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} - \frac {b e^{3} x^{2}}{6 c} + \frac {b e^{3} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6 c^{3}} & \text {for}\: c \neq 0 \\a \left (- \frac {d^{3}}{3 x^{3}} - \frac {3 d^{2} e}{x} + 3 d e^{2} x + \frac {e^{3} x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**3*(a+b*atan(c*x))/x**4,x)

[Out]

Piecewise((-a*d**3/(3*x**3) - 3*a*d**2*e/x + 3*a*d*e**2*x + a*e**3*x**3/3 - b*c**3*d**3*log(x)/3 + b*c**3*d**3
*log(x**2 + c**(-2))/6 - b*c*d**3/(6*x**2) + 3*b*c*d**2*e*log(x) - 3*b*c*d**2*e*log(x**2 + c**(-2))/2 - b*d**3
*atan(c*x)/(3*x**3) - 3*b*d**2*e*atan(c*x)/x + 3*b*d*e**2*x*atan(c*x) + b*e**3*x**3*atan(c*x)/3 - 3*b*d*e**2*l
og(x**2 + c**(-2))/(2*c) - b*e**3*x**2/(6*c) + b*e**3*log(x**2 + c**(-2))/(6*c**3), Ne(c, 0)), (a*(-d**3/(3*x*
*3) - 3*d**2*e/x + 3*d*e**2*x + e**3*x**3/3), True))

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