3.12.19 \(\int \frac {(d+e x^2) (a+b \text {ArcTan}(c x))}{x^2} \, dx\) [1119]

Optimal. Leaf size=57 \[ -\frac {d (a+b \text {ArcTan}(c x))}{x}+e x (a+b \text {ArcTan}(c x))+b c d \log (x)-\frac {b \left (c^2 d+e\right ) \log \left (1+c^2 x^2\right )}{2 c} \]

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {14, 5096, 457, 78} \begin {gather*} -\frac {d (a+b \text {ArcTan}(c x))}{x}+e x (a+b \text {ArcTan}(c x))-\frac {b \left (c^2 d+e\right ) \log \left (c^2 x^2+1\right )}{2 c}+b c d \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 78

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 457

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 5096

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

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Mathematica [A]
time = 0.01, size = 73, normalized size = 1.28 \begin {gather*} -\frac {a d}{x}+a e x-\frac {b d \text {ArcTan}(c x)}{x}+b e x \text {ArcTan}(c x)+b c d \log (x)-\frac {1}{2} b c d \log \left (1+c^2 x^2\right )-\frac {b e \log \left (1+c^2 x^2\right )}{2 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.10, size = 84, normalized size = 1.47

method	result	size
derivativedivides	\(c \left (\frac {a \left (e c x -\frac {d c}{x}\right )}{c^{2}}+\frac {b \arctan \left (c x \right ) e x}{c}-\frac {b \arctan \left (c x \right ) d}{c x}-\frac {b \ln \left (c^{2} x^{2}+1\right ) d}{2}-\frac {b \ln \left (c^{2} x^{2}+1\right ) e}{2 c^{2}}+b d \ln \left (c x \right )\right )\)	\(84\)
default	\(c \left (\frac {a \left (e c x -\frac {d c}{x}\right )}{c^{2}}+\frac {b \arctan \left (c x \right ) e x}{c}-\frac {b \arctan \left (c x \right ) d}{c x}-\frac {b \ln \left (c^{2} x^{2}+1\right ) d}{2}-\frac {b \ln \left (c^{2} x^{2}+1\right ) e}{2 c^{2}}+b d \ln \left (c x \right )\right )\)	\(84\)
risch	\(\frac {i b \left (-e \,x^{2}+d \right ) \ln \left (i c x +1\right )}{2 x}+\frac {i b c e \,x^{2} \ln \left (-i c x +1\right )+2 b \,c^{2} d \ln \left (x \right ) x -\ln \left (c^{2} x^{2}+1\right ) b \,c^{2} d x -i b c d \ln \left (-i c x +1\right )+2 a e \,x^{2} c -\ln \left (c^{2} x^{2}+1\right ) b e x -2 a d c}{2 c x}\)	\(121\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.25, size = 75, normalized size = 1.32 \begin {gather*} -\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 + a x e + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b e}{2 \, c} - \frac {a d}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.05, size = 78, normalized size = 1.37 \begin {gather*} \frac {2 \, b c^{2} d x \log \left (x\right ) + 2 \, a c x^{2} e - 2 \, a c d + 2 \, {\left (b c x^{2} e - b c d\right )} \arctan \left (c x\right ) - {\left (b c^{2} d x + b x e\right )} \log \left (c^{2} x^{2} + 1\right )}{2 \, c x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.36, size = 80, normalized size = 1.40 \begin {gather*} \begin {cases} - \frac {a d}{x} + a e x + b c d \log {\left (x \right )} - \frac {b c d \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b d \operatorname {atan}{\left (c x \right )}}{x} + b e x \operatorname {atan}{\left (c x \right )} - \frac {b e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} & \text {for}\: c \neq 0 \\a \left (- \frac {d}{x} + e x\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [B]
time = 0.23, size = 69, normalized size = 1.21 \begin {gather*} a\,e\,x-\frac {a\,d}{x}+b\,e\,x\,\mathrm {atan}\left (c\,x\right )-\frac {b\,c\,d\,\ln \left (c^2\,x^2+1\right )}{2}+b\,c\,d\,\ln \left (x\right )-\frac {b\,d\,\mathrm {atan}\left (c\,x\right )}{x}-\frac {b\,e\,\ln \left (c^2\,x^2+1\right )}{2\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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