3.1132 \(\int \frac {(d+e x^2)^2 (a+b \tan ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=115 \[ -\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{3} b c d \log (x) \left (c^2 d-6 e\right )+\frac {b \left (c^4 d^2-6 c^2 d e-3 e^2\right ) \log \left (c^2 x^2+1\right )}{6 c}-\frac {b c d^2}{6 x^2} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 115, 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, 1251, 893} \[ -\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \tan ^{-1}(c x)\right )+\frac {b \left (c^4 d^2-6 c^2 d e-3 e^2\right ) \log \left (c^2 x^2+1\right )}{6 c}-\frac {1}{3} b c d \log (x) \left (c^2 d-6 e\right )-\frac {b c d^2}{6 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 0]))

Rule 1251

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2,
Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] &&
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 )^2 \left (a+b \tan ^{-1}(c x)\right )}{x^4} \, dx &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \tan ^{-1}(c x)\right )-(b c) \int \frac {-d^2-6 d e x^2+3 e^2 x^4}{3 x^3 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{3} (b c) \int \frac {-d^2-6 d e x^2+3 e^2 x^4}{x^3 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{6} (b c) \operatorname {Subst}\left (\int \frac {-d^2-6 d e x+3 e^2 x^2}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{6} (b c) \operatorname {Subst}\left (\int \left (-\frac {d^2}{x^2}+\frac {d \left (c^2 d-6 e\right )}{x}+\frac {-c^4 d^2+6 c^2 d e+3 e^2}{1+c^2 x}\right ) \, dx,x,x^2\right )\\ &=-\frac {b c d^2}{6 x^2}-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{x}+e^2 x \left (a+b \tan ^{-1}(c x)\right )-\frac {1}{3} b c d \left (c^2 d-6 e\right ) \log (x)+\frac {b \left (c^4 d^2-6 c^2 d e-3 e^2\right ) \log \left (1+c^2 x^2\right )}{6 c}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 119, normalized size = 1.03 \[ \frac {1}{6} \left (-\frac {2 a d^2}{x^3}-\frac {12 a d e}{x}+6 a e^2 x-2 b c d \log (x) \left (c^2 d-6 e\right )+\frac {b \left (c^4 d^2-6 c^2 d e-3 e^2\right ) \log \left (c^2 x^2+1\right )}{c}-\frac {2 b \tan ^{-1}(c x) \left (d^2+6 d e x^2-3 e^2 x^4\right )}{x^3}-\frac {b c d^2}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*a*d^2)/x^3 - (b*c*d^2)/x^2 - (12*a*d*e)/x + 6*a*e^2*x - (2*b*(d^2 + 6*d*e*x^2 - 3*e^2*x^4)*ArcTan[c*x])/x
^3 - 2*b*c*d*(c^2*d - 6*e)*Log[x] + (b*(c^4*d^2 - 6*c^2*d*e - 3*e^2)*Log[1 + c^2*x^2])/c)/6

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fricas [A]  time = 0.43, size = 139, normalized size = 1.21 \[ \frac {6 \, a c e^{2} x^{4} - b c^{2} d^{2} x - 12 \, a c d e x^{2} + {\left (b c^{4} d^{2} - 6 \, b c^{2} d e - 3 \, b e^{2}\right )} x^{3} \log \left (c^{2} x^{2} + 1\right ) - 2 \, {\left (b c^{4} d^{2} - 6 \, b c^{2} d e\right )} x^{3} \log \relax (x) - 2 \, a c d^{2} + 2 \, {\left (3 \, b c e^{2} x^{4} - 6 \, b c d e x^{2} - b c d^{2}\right )} \arctan \left (c x\right )}{6 \, c x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(6*a*c*e^2*x^4 - b*c^2*d^2*x - 12*a*c*d*e*x^2 + (b*c^4*d^2 - 6*b*c^2*d*e - 3*b*e^2)*x^3*log(c^2*x^2 + 1) -
2*(b*c^4*d^2 - 6*b*c^2*d*e)*x^3*log(x) - 2*a*c*d^2 + 2*(3*b*c*e^2*x^4 - 6*b*c*d*e*x^2 - b*c*d^2)*arctan(c*x))
/(c*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)^2*(a+b*arctan(c*x))/x^4,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.05, size = 147, normalized size = 1.28 \[ a x \,e^{2}-\frac {a \,d^{2}}{3 x^{3}}-\frac {2 a e d}{x}+b \arctan \left (c x \right ) x \,e^{2}-\frac {b \arctan \left (c x \right ) d^{2}}{3 x^{3}}-\frac {2 b \arctan \left (c x \right ) e d}{x}-\frac {c^{3} b \,d^{2} \ln \left (c x \right )}{3}+2 c b \ln \left (c x \right ) d e -\frac {b c \,d^{2}}{6 x^{2}}+\frac {c^{3} b \ln \left (c^{2} x^{2}+1\right ) d^{2}}{6}-c b \ln \left (c^{2} x^{2}+1\right ) e d -\frac {b \ln \left (c^{2} x^{2}+1\right ) e^{2}}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 135, normalized size = 1.17 \[ \frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d^{2} - {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b d e + a e^{2} x + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b e^{2}}{2 \, c} - \frac {2 \, a d e}{x} - \frac {a d^{2}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.68, size = 142, normalized size = 1.23 \[ a\,e^2\,x-\frac {a\,d^2}{3\,x^3}+\frac {b\,c^3\,d^2\,\ln \left (c^2\,x^2+1\right )}{6}-\frac {b\,e^2\,\ln \left (c^2\,x^2+1\right )}{2\,c}-\frac {b\,c^3\,d^2\,\ln \relax (x)}{3}-\frac {2\,a\,d\,e}{x}+b\,e^2\,x\,\mathrm {atan}\left (c\,x\right )-\frac {b\,c\,d^2}{6\,x^2}-\frac {b\,d^2\,\mathrm {atan}\left (c\,x\right )}{3\,x^3}-b\,c\,d\,e\,\ln \left (c^2\,x^2+1\right )+2\,b\,c\,d\,e\,\ln \relax (x)-\frac {2\,b\,d\,e\,\mathrm {atan}\left (c\,x\right )}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.87, size = 180, normalized size = 1.57 \[ \begin {cases} - \frac {a d^{2}}{3 x^{3}} - \frac {2 a d e}{x} + a e^{2} x - \frac {b c^{3} d^{2} \log {\relax (x )}}{3} + \frac {b c^{3} d^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6} - \frac {b c d^{2}}{6 x^{2}} + 2 b c d e \log {\relax (x )} - b c d e \log {\left (x^{2} + \frac {1}{c^{2}} \right )} - \frac {b d^{2} \operatorname {atan}{\left (c x \right )}}{3 x^{3}} - \frac {2 b d e \operatorname {atan}{\left (c x \right )}}{x} + b e^{2} x \operatorname {atan}{\left (c x \right )} - \frac {b e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} & \text {for}\: c \neq 0 \\a \left (- \frac {d^{2}}{3 x^{3}} - \frac {2 d e}{x} + e^{2} x\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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