3.12.34 \(\int \frac {(d+e x^2)^2 (a+b \text {ArcTan}(c x))}{x^6} \, dx\) [1134]

Optimal. Leaf size=150 \[ -\frac {b c d^2}{20 x^4}+\frac {b c d \left (3 c^2 d-10 e\right )}{30 x^2}-\frac {d^2 (a+b \text {ArcTan}(c x))}{5 x^5}-\frac {2 d e (a+b \text {ArcTan}(c x))}{3 x^3}-\frac {e^2 (a+b \text {ArcTan}(c x))}{x}+\frac {1}{15} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log (x)-\frac {1}{30} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log \left (1+c^2 x^2\right ) \]

[Out]

-1/20*b*c*d^2/x^4+1/30*b*c*d*(3*c^2*d-10*e)/x^2-1/5*d^2*(a+b*arctan(c*x))/x^5-2/3*d*e*(a+b*arctan(c*x))/x^3-e^
2*(a+b*arctan(c*x))/x+1/15*b*c*(3*c^4*d^2-10*c^2*d*e+15*e^2)*ln(x)-1/30*b*c*(3*c^4*d^2-10*c^2*d*e+15*e^2)*ln(c
^2*x^2+1)

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Rubi [A]
time = 0.13, antiderivative size = 150, 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 = {276, 5096, 12, 1265, 907} \begin {gather*} -\frac {d^2 (a+b \text {ArcTan}(c x))}{5 x^5}-\frac {2 d e (a+b \text {ArcTan}(c x))}{3 x^3}-\frac {e^2 (a+b \text {ArcTan}(c x))}{x}+\frac {b c d \left (3 c^2 d-10 e\right )}{30 x^2}-\frac {1}{30} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log \left (c^2 x^2+1\right )+\frac {1}{15} b c \log (x) \left (3 c^4 d^2-10 c^2 d e+15 e^2\right )-\frac {b c d^2}{20 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^2*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

-1/20*(b*c*d^2)/x^4 + (b*c*d*(3*c^2*d - 10*e))/(30*x^2) - (d^2*(a + b*ArcTan[c*x]))/(5*x^5) - (2*d*e*(a + b*Ar
cTan[c*x]))/(3*x^3) - (e^2*(a + b*ArcTan[c*x]))/x + (b*c*(3*c^4*d^2 - 10*c^2*d*e + 15*e^2)*Log[x])/15 - (b*c*(
3*c^4*d^2 - 10*c^2*d*e + 15*e^2)*Log[1 + c^2*x^2])/30

Rule 12

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

Rule 276

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 907

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 1265

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 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 )^2 \left (a+b \tan ^{-1}(c x)\right )}{x^6} \, dx &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-(b c) \int \frac {-3 d^2-10 d e x^2-15 e^2 x^4}{15 x^5 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {1}{15} (b c) \int \frac {-3 d^2-10 d e x^2-15 e^2 x^4}{x^5 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {1}{30} (b c) \text {Subst}\left (\int \frac {-3 d^2-10 d e x-15 e^2 x^2}{x^3 \left (1+c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}-\frac {1}{30} (b c) \text {Subst}\left (\int \left (-\frac {3 d^2}{x^3}+\frac {d \left (3 c^2 d-10 e\right )}{x^2}+\frac {-3 c^4 d^2+10 c^2 d e-15 e^2}{x}+\frac {3 c^6 d^2-10 c^4 d e+15 c^2 e^2}{1+c^2 x}\right ) \, dx,x,x^2\right )\\ &=-\frac {b c d^2}{20 x^4}+\frac {b c d \left (3 c^2 d-10 e\right )}{30 x^2}-\frac {d^2 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {2 d e \left (a+b \tan ^{-1}(c x)\right )}{3 x^3}-\frac {e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+\frac {1}{15} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log (x)-\frac {1}{30} b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) \log \left (1+c^2 x^2\right )\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 153, normalized size = 1.02 \begin {gather*} -\frac {12 a d^2+3 b c d^2 x+40 a d e x^2-2 b c d \left (3 c^2 d-10 e\right ) x^3+60 a e^2 x^4+4 b \left (3 d^2+10 d e x^2+15 e^2 x^4\right ) \text {ArcTan}(c x)-4 b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) x^5 \log (x)+2 b c \left (3 c^4 d^2-10 c^2 d e+15 e^2\right ) x^5 \log \left (1+c^2 x^2\right )}{60 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^2*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

-1/60*(12*a*d^2 + 3*b*c*d^2*x + 40*a*d*e*x^2 - 2*b*c*d*(3*c^2*d - 10*e)*x^3 + 60*a*e^2*x^4 + 4*b*(3*d^2 + 10*d
*e*x^2 + 15*e^2*x^4)*ArcTan[c*x] - 4*b*c*(3*c^4*d^2 - 10*c^2*d*e + 15*e^2)*x^5*Log[x] + 2*b*c*(3*c^4*d^2 - 10*
c^2*d*e + 15*e^2)*x^5*Log[1 + c^2*x^2])/x^5

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Maple [A]
time = 0.20, size = 213, normalized size = 1.42

method	result	size
derivativedivides	\(c^{5} \left (\frac {a \left (-\frac {2 d e}{3 c \,x^{3}}-\frac {d^{2}}{5 c \,x^{5}}-\frac {e^{2}}{c x}\right )}{c^{4}}-\frac {2 b \arctan \left (c x \right ) d e}{3 c^{5} x^{3}}-\frac {b \arctan \left (c x \right ) d^{2}}{5 c^{5} x^{5}}-\frac {b \arctan \left (c x \right ) e^{2}}{c^{5} x}-\frac {b \ln \left (c^{2} x^{2}+1\right ) d^{2}}{10}+\frac {b \ln \left (c^{2} x^{2}+1\right ) d e}{3 c^{2}}-\frac {b \ln \left (c^{2} x^{2}+1\right ) e^{2}}{2 c^{4}}+\frac {b \,d^{2} \ln \left (c x \right )}{5}-\frac {2 b \ln \left (c x \right ) d e}{3 c^{2}}+\frac {b \ln \left (c x \right ) e^{2}}{c^{4}}+\frac {b \,d^{2}}{10 c^{2} x^{2}}-\frac {b d e}{3 c^{4} x^{2}}-\frac {b \,d^{2}}{20 c^{4} x^{4}}\right )\)	\(213\)
default	\(c^{5} \left (\frac {a \left (-\frac {2 d e}{3 c \,x^{3}}-\frac {d^{2}}{5 c \,x^{5}}-\frac {e^{2}}{c x}\right )}{c^{4}}-\frac {2 b \arctan \left (c x \right ) d e}{3 c^{5} x^{3}}-\frac {b \arctan \left (c x \right ) d^{2}}{5 c^{5} x^{5}}-\frac {b \arctan \left (c x \right ) e^{2}}{c^{5} x}-\frac {b \ln \left (c^{2} x^{2}+1\right ) d^{2}}{10}+\frac {b \ln \left (c^{2} x^{2}+1\right ) d e}{3 c^{2}}-\frac {b \ln \left (c^{2} x^{2}+1\right ) e^{2}}{2 c^{4}}+\frac {b \,d^{2} \ln \left (c x \right )}{5}-\frac {2 b \ln \left (c x \right ) d e}{3 c^{2}}+\frac {b \ln \left (c x \right ) e^{2}}{c^{4}}+\frac {b \,d^{2}}{10 c^{2} x^{2}}-\frac {b d e}{3 c^{4} x^{2}}-\frac {b \,d^{2}}{20 c^{4} x^{4}}\right )\)	\(213\)
risch	\(\frac {i b \left (15 e^{2} x^{4}+10 d e \,x^{2}+3 d^{2}\right ) \ln \left (i c x +1\right )}{30 x^{5}}-\frac {-12 \ln \left (x \right ) b \,c^{5} d^{2} x^{5}+6 \ln \left (-c^{2} x^{2}-1\right ) b \,c^{5} d^{2} x^{5}+40 \ln \left (x \right ) b \,c^{3} d e \,x^{5}-20 \ln \left (-c^{2} x^{2}-1\right ) b \,c^{3} d e \,x^{5}-60 \ln \left (x \right ) b c \,e^{2} x^{5}+30 \ln \left (-c^{2} x^{2}-1\right ) b c \,e^{2} x^{5}+30 i b \,e^{2} x^{4} \ln \left (-i c x +1\right )-6 b \,d^{2} c^{3} x^{3}+20 i b d e \,x^{2} \ln \left (-i c x +1\right )+60 a \,e^{2} x^{4}+20 b c d e \,x^{3}+6 i b \,d^{2} \ln \left (-i c x +1\right )+40 a d e \,x^{2}+3 b c \,d^{2} x +12 d^{2} a}{60 x^{5}}\)	\(251\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^2*(a+b*arctan(c*x))/x^6,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.25, size = 166, normalized size = 1.11 \begin {gather*} -\frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d^{2} + \frac {1}{3} \, {\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 e - \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^{2} - \frac {a e^{2}}{x} - \frac {2 \, a d e}{3 \, x^{3}} - \frac {a d^{2}}{5 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arctan(c*x))/x^6,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.55, size = 173, normalized size = 1.15 \begin {gather*} \frac {6 \, b c^{3} d^{2} x^{3} - 60 \, a x^{4} e^{2} - 3 \, b c d^{2} x - 12 \, a d^{2} - 4 \, {\left (15 \, b x^{4} e^{2} + 10 \, b d x^{2} e + 3 \, b d^{2}\right )} \arctan \left (c x\right ) - 20 \, {\left (b c d x^{3} + 2 \, a d x^{2}\right )} e - 2 \, {\left (3 \, b c^{5} d^{2} x^{5} - 10 \, b c^{3} d x^{5} e + 15 \, b c x^{5} e^{2}\right )} \log \left (c^{2} x^{2} + 1\right ) + 4 \, {\left (3 \, b c^{5} d^{2} x^{5} - 10 \, b c^{3} d x^{5} e + 15 \, b c x^{5} e^{2}\right )} \log \left (x\right )}{60 \, x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^2*(a+b*arctan(c*x))/x^6,x, algorithm="fricas")

[Out]

1/60*(6*b*c^3*d^2*x^3 - 60*a*x^4*e^2 - 3*b*c*d^2*x - 12*a*d^2 - 4*(15*b*x^4*e^2 + 10*b*d*x^2*e + 3*b*d^2)*arct
an(c*x) - 20*(b*c*d*x^3 + 2*a*d*x^2)*e - 2*(3*b*c^5*d^2*x^5 - 10*b*c^3*d*x^5*e + 15*b*c*x^5*e^2)*log(c^2*x^2 +
1) + 4*(3*b*c^5*d^2*x^5 - 10*b*c^3*d*x^5*e + 15*b*c*x^5*e^2)*log(x))/x^5

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Sympy [A]
time = 0.64, size = 235, normalized size = 1.57 \begin {gather*} \begin {cases} - \frac {a d^{2}}{5 x^{5}} - \frac {2 a d e}{3 x^{3}} - \frac {a e^{2}}{x} + \frac {b c^{5} d^{2} \log {\left (x \right )}}{5} - \frac {b c^{5} d^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10} + \frac {b c^{3} d^{2}}{10 x^{2}} - \frac {2 b c^{3} d e \log {\left (x \right )}}{3} + \frac {b c^{3} d e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{3} - \frac {b c d^{2}}{20 x^{4}} - \frac {b c d e}{3 x^{2}} + b c e^{2} \log {\left (x \right )} - \frac {b c e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b d^{2} \operatorname {atan}{\left (c x \right )}}{5 x^{5}} - \frac {2 b d e \operatorname {atan}{\left (c x \right )}}{3 x^{3}} - \frac {b e^{2} \operatorname {atan}{\left (c x \right )}}{x} & \text {for}\: c \neq 0 \\a \left (- \frac {d^{2}}{5 x^{5}} - \frac {2 d e}{3 x^{3}} - \frac {e^{2}}{x}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**2*(a+b*atan(c*x))/x**6,x)

[Out]

Piecewise((-a*d**2/(5*x**5) - 2*a*d*e/(3*x**3) - a*e**2/x + b*c**5*d**2*log(x)/5 - b*c**5*d**2*log(x**2 + c**(
-2))/10 + b*c**3*d**2/(10*x**2) - 2*b*c**3*d*e*log(x)/3 + b*c**3*d*e*log(x**2 + c**(-2))/3 - b*c*d**2/(20*x**4
) - b*c*d*e/(3*x**2) + b*c*e**2*log(x) - b*c*e**2*log(x**2 + c**(-2))/2 - b*d**2*atan(c*x)/(5*x**5) - 2*b*d*e*
atan(c*x)/(3*x**3) - b*e**2*atan(c*x)/x, Ne(c, 0)), (a*(-d**2/(5*x**5) - 2*d*e/(3*x**3) - e**2/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)^2*(a+b*arctan(c*x))/x^6,x, algorithm="giac")

[Out]

sage0*x

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Mupad [B]
time = 0.49, size = 179, normalized size = 1.19 \begin {gather*} \frac {b\,c^3\,d^2}{10\,x^2}-\frac {a\,e^2}{x}-\frac {b\,c^5\,d^2\,\ln \left (c^2\,x^2+1\right )}{10}-\frac {a\,d^2}{5\,x^5}+\frac {b\,c^5\,d^2\,\ln \left (x\right )}{5}-\frac {2\,a\,d\,e}{3\,x^3}-\frac {b\,c\,e^2\,\ln \left (c^2\,x^2+1\right )}{2}-\frac {b\,c\,d^2}{20\,x^4}+b\,c\,e^2\,\ln \left (x\right )-\frac {b\,d^2\,\mathrm {atan}\left (c\,x\right )}{5\,x^5}-\frac {b\,e^2\,\mathrm {atan}\left (c\,x\right )}{x}+\frac {b\,c^3\,d\,e\,\ln \left (c^2\,x^2+1\right )}{3}-\frac {2\,b\,c^3\,d\,e\,\ln \left (x\right )}{3}-\frac {b\,c\,d\,e}{3\,x^2}-\frac {2\,b\,d\,e\,\mathrm {atan}\left (c\,x\right )}{3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + e*x^2)^2)/x^6,x)

[Out]

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

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