3.1121 \(\int \frac {(d+e x^2) (a+b \tan ^{-1}(c x))}{x^4} \, dx\)

Optimal. Leaf size=83 \[ -\frac {d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {1}{6} b c \left (c^2 d-3 e\right ) \log \left (c^2 x^2+1\right )-\frac {1}{3} b c \log (x) \left (c^2 d-3 e\right )-\frac {b c d}{6 x^2} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {14, 4976, 12, 446, 77} \[ -\frac {d \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {e \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {1}{6} b c \left (c^2 d-3 e\right ) \log \left (c^2 x^2+1\right )-\frac {1}{3} b c \log (x) \left (c^2 d-3 e\right )-\frac {b c d}{6 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

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

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

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Mathematica [A]  time = 0.04, size = 98, normalized size = 1.18 \[ -\frac {a d}{3 x^3}-\frac {a e}{x}+\frac {1}{6} b c d \left (c^2 \log \left (c^2 x^2+1\right )-2 c^2 \log (x)-\frac {1}{x^2}\right )-\frac {1}{2} b c e \log \left (c^2 x^2+1\right )-\frac {b d \tan ^{-1}(c x)}{3 x^3}+b c e \log (x)-\frac {b e \tan ^{-1}(c x)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 85, normalized size = 1.02 \[ \frac {{\left (b c^{3} d - 3 \, b c e\right )} x^{3} \log \left (c^{2} x^{2} + 1\right ) - 2 \, {\left (b c^{3} d - 3 \, b c e\right )} x^{3} \log \relax (x) - b c d x - 6 \, a e x^{2} - 2 \, a d - 2 \, {\left (3 \, b e x^{2} + b d\right )} \arctan \left (c x\right )}{6 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

sage0*x

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maple [A]  time = 0.06, size = 97, normalized size = 1.17 \[ -\frac {a d}{3 x^{3}}-\frac {a e}{x}-\frac {b \arctan \left (c x \right ) d}{3 x^{3}}-\frac {b \arctan \left (c x \right ) e}{x}-\frac {c^{3} b d \ln \left (c x \right )}{3}+c b \ln \left (c x \right ) e -\frac {b c d}{6 x^{2}}+\frac {c^{3} b \ln \left (c^{2} x^{2}+1\right ) d}{6}-\frac {c b \ln \left (c^{2} x^{2}+1\right ) e}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 93, normalized size = 1.12 \[ \frac {1}{6} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d - \frac {1}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b e - \frac {a e}{x} - \frac {a d}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 0.55, size = 92, normalized size = 1.11 \[ b\,c\,e\,\ln \relax (x)-\frac {a\,e}{x}-\frac {b\,c\,e\,\ln \left (c^2\,x^2+1\right )}{2}-\frac {b\,c\,d}{6\,x^2}-\frac {a\,d}{3\,x^3}-\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{3\,x^3}-\frac {b\,e\,\mathrm {atan}\left (c\,x\right )}{x}+\frac {b\,c^3\,d\,\ln \left (c^2\,x^2+1\right )}{6}-\frac {b\,c^3\,d\,\ln \relax (x)}{3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.29, size = 116, normalized size = 1.40 \[ \begin {cases} - \frac {a d}{3 x^{3}} - \frac {a e}{x} - \frac {b c^{3} d \log {\relax (x )}}{3} + \frac {b c^{3} d \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{6} - \frac {b c d}{6 x^{2}} + b c e \log {\relax (x )} - \frac {b c e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b d \operatorname {atan}{\left (c x \right )}}{3 x^{3}} - \frac {b e \operatorname {atan}{\left (c x \right )}}{x} & \text {for}\: c \neq 0 \\a \left (- \frac {d}{3 x^{3}} - \frac {e}{x}\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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