3.1142 \(\int \frac {(d+e x^2)^3 (a+b \tan ^{-1}(c x))}{x^2} \, dx\)

Optimal. Leaf size=160 \[ -\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d^2 e x \left (a+b \tan ^{-1}(c x)\right )+d e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{5} e^3 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {b e^2 x^2 \left (5 c^2 d-e\right )}{10 c^3}-\frac {b \left (5 c^6 d^3+15 c^4 d^2 e-5 c^2 d e^2+e^3\right ) \log \left (c^2 x^2+1\right )}{10 c^5}+b c d^3 \log (x)-\frac {b e^3 x^4}{20 c} \]

[Out]

-1/10*b*(5*c^2*d-e)*e^2*x^2/c^3-1/20*b*e^3*x^4/c-d^3*(a+b*arctan(c*x))/x+3*d^2*e*x*(a+b*arctan(c*x))+d*e^2*x^3
*(a+b*arctan(c*x))+1/5*e^3*x^5*(a+b*arctan(c*x))+b*c*d^3*ln(x)-1/10*b*(5*c^6*d^3+15*c^4*d^2*e-5*c^2*d*e^2+e^3)
*ln(c^2*x^2+1)/c^5

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Rubi [A]  time = 0.26, antiderivative size = 160, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {270, 4976, 1799, 1620} \[ 3 d^2 e x \left (a+b \tan ^{-1}(c x)\right )-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+d e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{5} e^3 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {b \left (15 c^4 d^2 e+5 c^6 d^3-5 c^2 d e^2+e^3\right ) \log \left (c^2 x^2+1\right )}{10 c^5}-\frac {b e^2 x^2 \left (5 c^2 d-e\right )}{10 c^3}+b c d^3 \log (x)-\frac {b e^3 x^4}{20 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*(5*c^2*d - e)*e^2*x^2)/(10*c^3) - (b*e^3*x^4)/(20*c) - (d^3*(a + b*ArcTan[c*x]))/x + 3*d^2*e*x*(a + b*ArcT
an[c*x]) + d*e^2*x^3*(a + b*ArcTan[c*x]) + (e^3*x^5*(a + b*ArcTan[c*x]))/5 + b*c*d^3*Log[x] - (b*(5*c^6*d^3 +
15*c^4*d^2*e - 5*c^2*d*e^2 + e^3)*Log[1 + c^2*x^2])/(10*c^5)

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^2} \, dx &=-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d^2 e x \left (a+b \tan ^{-1}(c x)\right )+d e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{5} e^3 x^5 \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac {-d^3+3 d^2 e x^2+d e^2 x^4+\frac {e^3 x^6}{5}}{x \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d^2 e x \left (a+b \tan ^{-1}(c x)\right )+d e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{5} e^3 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{2} (b c) \operatorname {Subst}\left (\int \frac {-d^3+3 d^2 e x+d e^2 x^2+\frac {e^3 x^3}{5}}{x \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d^2 e x \left (a+b \tan ^{-1}(c x)\right )+d e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{5} e^3 x^5 \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{2} (b c) \operatorname {Subst}\left (\int \left (\frac {\left (5 c^2 d-e\right ) e^2}{5 c^4}-\frac {d^3}{x}+\frac {e^3 x}{5 c^2}+\frac {5 c^6 d^3+15 c^4 d^2 e-5 c^2 d e^2+e^3}{5 c^4 \left (1+c^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac {b \left (5 c^2 d-e\right ) e^2 x^2}{10 c^3}-\frac {b e^3 x^4}{20 c}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{x}+3 d^2 e x \left (a+b \tan ^{-1}(c x)\right )+d e^2 x^3 \left (a+b \tan ^{-1}(c x)\right )+\frac {1}{5} e^3 x^5 \left (a+b \tan ^{-1}(c x)\right )+b c d^3 \log (x)-\frac {b \left (5 c^6 d^3+15 c^4 d^2 e-5 c^2 d e^2+e^3\right ) \log \left (1+c^2 x^2\right )}{10 c^5}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 169, normalized size = 1.06 \[ \frac {1}{20} \left (-\frac {20 a d^3}{x}+60 a d^2 e x+20 a d e^2 x^3+4 a e^3 x^5+\frac {2 b e^2 x^2 \left (e-5 c^2 d\right )}{c^3}-\frac {2 b \left (5 c^6 d^3+15 c^4 d^2 e-5 c^2 d e^2+e^3\right ) \log \left (c^2 x^2+1\right )}{c^5}+20 b c d^3 \log (x)+\frac {4 b \tan ^{-1}(c x) \left (-5 d^3+15 d^2 e x^2+5 d e^2 x^4+e^3 x^6\right )}{x}-\frac {b e^3 x^4}{c}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-20*a*d^3)/x + 60*a*d^2*e*x + (2*b*e^2*(-5*c^2*d + e)*x^2)/c^3 + 20*a*d*e^2*x^3 - (b*e^3*x^4)/c + 4*a*e^3*x^
5 + (4*b*(-5*d^3 + 15*d^2*e*x^2 + 5*d*e^2*x^4 + e^3*x^6)*ArcTan[c*x])/x + 20*b*c*d^3*Log[x] - (2*b*(5*c^6*d^3
+ 15*c^4*d^2*e - 5*c^2*d*e^2 + e^3)*Log[1 + c^2*x^2])/c^5)/20

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fricas [A]  time = 0.63, size = 206, normalized size = 1.29 \[ \frac {4 \, a c^{5} e^{3} x^{6} + 20 \, a c^{5} d e^{2} x^{4} - b c^{4} e^{3} x^{5} + 20 \, b c^{6} d^{3} x \log \relax (x) + 60 \, a c^{5} d^{2} e x^{2} - 20 \, a c^{5} d^{3} - 2 \, {\left (5 \, b c^{4} d e^{2} - b c^{2} e^{3}\right )} x^{3} - 2 \, {\left (5 \, b c^{6} d^{3} + 15 \, b c^{4} d^{2} e - 5 \, b c^{2} d e^{2} + b e^{3}\right )} x \log \left (c^{2} x^{2} + 1\right ) + 4 \, {\left (b c^{5} e^{3} x^{6} + 5 \, b c^{5} d e^{2} x^{4} + 15 \, b c^{5} d^{2} e x^{2} - 5 \, b c^{5} d^{3}\right )} \arctan \left (c x\right )}{20 \, c^{5} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*(4*a*c^5*e^3*x^6 + 20*a*c^5*d*e^2*x^4 - b*c^4*e^3*x^5 + 20*b*c^6*d^3*x*log(x) + 60*a*c^5*d^2*e*x^2 - 20*a
*c^5*d^3 - 2*(5*b*c^4*d*e^2 - b*c^2*e^3)*x^3 - 2*(5*b*c^6*d^3 + 15*b*c^4*d^2*e - 5*b*c^2*d*e^2 + b*e^3)*x*log(
c^2*x^2 + 1) + 4*(b*c^5*e^3*x^6 + 5*b*c^5*d*e^2*x^4 + 15*b*c^5*d^2*e*x^2 - 5*b*c^5*d^3)*arctan(c*x))/(c^5*x)

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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^2,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.05, size = 211, normalized size = 1.32 \[ \frac {a \,x^{5} e^{3}}{5}+a \,x^{3} d \,e^{2}+3 a \,d^{2} e x -\frac {a \,d^{3}}{x}+\frac {b \arctan \left (c x \right ) x^{5} e^{3}}{5}+b \arctan \left (c x \right ) x^{3} d \,e^{2}+3 b \arctan \left (c x \right ) d^{2} e x -\frac {b \arctan \left (c x \right ) d^{3}}{x}-\frac {b \,e^{3} x^{4}}{20 c}-\frac {b d \,e^{2} x^{2}}{2 c}+\frac {b \,x^{2} e^{3}}{10 c^{3}}+c b \,d^{3} \ln \left (c x \right )-\frac {b c \,d^{3} \ln \left (c^{2} x^{2}+1\right )}{2}-\frac {3 b \ln \left (c^{2} x^{2}+1\right ) d^{2} e}{2 c}+\frac {b \ln \left (c^{2} x^{2}+1\right ) d \,e^{2}}{2 c^{3}}-\frac {b \ln \left (c^{2} x^{2}+1\right ) e^{3}}{10 c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*x^5*e^3+a*x^3*d*e^2+3*a*d^2*e*x-a*d^3/x+1/5*b*arctan(c*x)*x^5*e^3+b*arctan(c*x)*x^3*d*e^2+3*b*arctan(c*x
)*d^2*e*x-b*arctan(c*x)*d^3/x-1/20*b*e^3*x^4/c-1/2*b*d*e^2*x^2/c+1/10*b/c^3*x^2*e^3+c*b*d^3*ln(c*x)-1/2*b*c*d^
3*ln(c^2*x^2+1)-3/2*b/c*ln(c^2*x^2+1)*d^2*e+1/2*b/c^3*ln(c^2*x^2+1)*d*e^2-1/10*b/c^5*ln(c^2*x^2+1)*e^3

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maxima [A]  time = 0.33, size = 197, normalized size = 1.23 \[ \frac {1}{5} \, a e^{3} x^{5} + a d e^{2} x^{3} - \frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b d^{3} + \frac {1}{2} \, {\left (2 \, x^{3} \arctan \left (c x\right ) - c {\left (\frac {x^{2}}{c^{2}} - \frac {\log \left (c^{2} x^{2} + 1\right )}{c^{4}}\right )}\right )} b d e^{2} + \frac {1}{20} \, {\left (4 \, x^{5} \arctan \left (c x\right ) - c {\left (\frac {c^{2} x^{4} - 2 \, x^{2}}{c^{4}} + \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{6}}\right )}\right )} b e^{3} + 3 \, a d^{2} e x + \frac {3 \, {\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b d^{2} e}{2 \, c} - \frac {a d^{3}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*e^3*x^5 + a*d*e^2*x^3 - 1/2*(c*(log(c^2*x^2 + 1) - log(x^2)) + 2*arctan(c*x)/x)*b*d^3 + 1/2*(2*x^3*arcta
n(c*x) - c*(x^2/c^2 - log(c^2*x^2 + 1)/c^4))*b*d*e^2 + 1/20*(4*x^5*arctan(c*x) - c*((c^2*x^4 - 2*x^2)/c^4 + 2*
log(c^2*x^2 + 1)/c^6))*b*e^3 + 3*a*d^2*e*x + 3/2*(2*c*x*arctan(c*x) - log(c^2*x^2 + 1))*b*d^2*e/c - a*d^3/x

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mupad [B]  time = 0.64, size = 236, normalized size = 1.48 \[ x\,\left (\frac {\frac {a\,e^3}{c^2}-\frac {a\,e^2\,\left (3\,d\,c^2+e\right )}{c^2}}{c^2}+\frac {3\,a\,d\,e\,\left (d\,c^2+e\right )}{c^2}\right )-x^3\,\left (\frac {a\,e^3}{3\,c^2}-\frac {a\,e^2\,\left (3\,d\,c^2+e\right )}{3\,c^2}\right )+x^2\,\left (\frac {b\,e^3}{10\,c^3}-\frac {b\,d\,e^2}{2\,c}\right )-\frac {a\,d^3}{x}+\frac {a\,e^3\,x^5}{5}-\frac {\ln \left (c^2\,x^2+1\right )\,\left (5\,b\,c^6\,d^3+15\,b\,c^4\,d^2\,e-5\,b\,c^2\,d\,e^2+b\,e^3\right )}{10\,c^5}+\frac {\mathrm {atan}\left (c\,x\right )\,\left (-b\,d^3+3\,b\,d^2\,e\,x^2+b\,d\,e^2\,x^4+\frac {b\,e^3\,x^6}{5}\right )}{x}-\frac {b\,e^3\,x^4}{20\,c}+b\,c\,d^3\,\ln \relax (x) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*(((a*e^3)/c^2 - (a*e^2*(e + 3*c^2*d))/c^2)/c^2 + (3*a*d*e*(e + c^2*d))/c^2) - x^3*((a*e^3)/(3*c^2) - (a*e^2*
(e + 3*c^2*d))/(3*c^2)) + x^2*((b*e^3)/(10*c^3) - (b*d*e^2)/(2*c)) - (a*d^3)/x + (a*e^3*x^5)/5 - (log(c^2*x^2
+ 1)*(b*e^3 + 5*b*c^6*d^3 - 5*b*c^2*d*e^2 + 15*b*c^4*d^2*e))/(10*c^5) + (atan(c*x)*((b*e^3*x^6)/5 - b*d^3 + 3*
b*d^2*e*x^2 + b*d*e^2*x^4))/x - (b*e^3*x^4)/(20*c) + b*c*d^3*log(x)

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sympy [A]  time = 3.29, size = 258, normalized size = 1.61 \[ \begin {cases} - \frac {a d^{3}}{x} + 3 a d^{2} e x + a d e^{2} x^{3} + \frac {a e^{3} x^{5}}{5} + b c d^{3} \log {\relax (x )} - \frac {b c d^{3} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b d^{3} \operatorname {atan}{\left (c x \right )}}{x} + 3 b d^{2} e x \operatorname {atan}{\left (c x \right )} + b d e^{2} x^{3} \operatorname {atan}{\left (c x \right )} + \frac {b e^{3} x^{5} \operatorname {atan}{\left (c x \right )}}{5} - \frac {3 b d^{2} e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} - \frac {b d e^{2} x^{2}}{2 c} - \frac {b e^{3} x^{4}}{20 c} + \frac {b d e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c^{3}} + \frac {b e^{3} x^{2}}{10 c^{3}} - \frac {b e^{3} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10 c^{5}} & \text {for}\: c \neq 0 \\a \left (- \frac {d^{3}}{x} + 3 d^{2} e x + d e^{2} x^{3} + \frac {e^{3} x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-a*d**3/x + 3*a*d**2*e*x + a*d*e**2*x**3 + a*e**3*x**5/5 + b*c*d**3*log(x) - b*c*d**3*log(x**2 + c*
*(-2))/2 - b*d**3*atan(c*x)/x + 3*b*d**2*e*x*atan(c*x) + b*d*e**2*x**3*atan(c*x) + b*e**3*x**5*atan(c*x)/5 - 3
*b*d**2*e*log(x**2 + c**(-2))/(2*c) - b*d*e**2*x**2/(2*c) - b*e**3*x**4/(20*c) + b*d*e**2*log(x**2 + c**(-2))/
(2*c**3) + b*e**3*x**2/(10*c**3) - b*e**3*log(x**2 + c**(-2))/(10*c**5), Ne(c, 0)), (a*(-d**3/x + 3*d**2*e*x +
d*e**2*x**3 + e**3*x**5/5), True))

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