3.1146 \(\int \frac {(d+e x^2)^3 (a+b \tan ^{-1}(c x))}{x^6} \, dx\)

Optimal. Leaf size=177 \[ -\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac {3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+e^3 x \left (a+b \tan ^{-1}(c x)\right )+\frac {b c d^2 \left (c^2 d-5 e\right )}{10 x^2}+\frac {1}{5} b c d \log (x) \left (c^4 d^2-5 c^2 d e+15 e^2\right )-\frac {b \left (c^6 d^3-5 c^4 d^2 e+15 c^2 d e^2+5 e^3\right ) \log \left (c^2 x^2+1\right )}{10 c}-\frac {b c d^3}{20 x^4} \]

[Out]

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

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Rubi [A]  time = 0.29, antiderivative size = 177, 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, 1799, 1620} \[ -\frac {d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^3}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{5 x^5}-\frac {3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x}+e^3 x \left (a+b \tan ^{-1}(c x)\right )-\frac {b \left (-5 c^4 d^2 e+c^6 d^3+15 c^2 d e^2+5 e^3\right ) \log \left (c^2 x^2+1\right )}{10 c}+\frac {1}{5} b c d \log (x) \left (c^4 d^2-5 c^2 d e+15 e^2\right )+\frac {b c d^2 \left (c^2 d-5 e\right )}{10 x^2}-\frac {b c d^3}{20 x^4} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

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

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Mathematica [A]  time = 0.19, size = 184, normalized size = 1.04 \[ \frac {1}{20} \left (-\frac {4 a d^3}{x^5}-\frac {20 a d^2 e}{x^3}-\frac {60 a d e^2}{x}+20 a e^3 x+\frac {2 b c d^2 \left (c^2 d-5 e\right )}{x^2}+4 b c d \log (x) \left (c^4 d^2-5 c^2 d e+15 e^2\right )-\frac {2 b \left (c^6 d^3-5 c^4 d^2 e+15 c^2 d e^2+5 e^3\right ) \log \left (c^2 x^2+1\right )}{c}-\frac {b c d^3}{x^4}-\frac {4 b \tan ^{-1}(c x) \left (d^3+5 d^2 e x^2+15 d e^2 x^4-5 e^3 x^6\right )}{x^5}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^6,x]

[Out]

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

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fricas [A]  time = 0.48, size = 214, normalized size = 1.21 \[ \frac {20 \, a c e^{3} x^{6} - 60 \, a c d e^{2} x^{4} - b c^{2} d^{3} x - 20 \, a c d^{2} e x^{2} - 2 \, {\left (b c^{6} d^{3} - 5 \, b c^{4} d^{2} e + 15 \, b c^{2} d e^{2} + 5 \, b e^{3}\right )} x^{5} \log \left (c^{2} x^{2} + 1\right ) + 4 \, {\left (b c^{6} d^{3} - 5 \, b c^{4} d^{2} e + 15 \, b c^{2} d e^{2}\right )} x^{5} \log \relax (x) - 4 \, a c d^{3} + 2 \, {\left (b c^{4} d^{3} - 5 \, b c^{2} d^{2} e\right )} x^{3} + 4 \, {\left (5 \, b c e^{3} x^{6} - 15 \, b c d e^{2} x^{4} - 5 \, b c d^{2} e x^{2} - b c d^{3}\right )} \arctan \left (c x\right )}{20 \, c x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arctan(c*x))/x^6,x, algorithm="fricas")

[Out]

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

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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^6,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.05, size = 236, normalized size = 1.33 \[ a \,e^{3} x -\frac {a \,d^{2} e}{x^{3}}-\frac {3 a d \,e^{2}}{x}-\frac {a \,d^{3}}{5 x^{5}}+b \arctan \left (c x \right ) e^{3} x -\frac {b \arctan \left (c x \right ) d^{2} e}{x^{3}}-\frac {3 b \arctan \left (c x \right ) d \,e^{2}}{x}-\frac {b \arctan \left (c x \right ) d^{3}}{5 x^{5}}+\frac {b \,c^{3} d^{3}}{10 x^{2}}-\frac {c b \,d^{2} e}{2 x^{2}}-\frac {b c \,d^{3}}{20 x^{4}}+\frac {c^{5} b \,d^{3} \ln \left (c x \right )}{5}-c^{3} b \ln \left (c x \right ) d^{2} e +3 c b \ln \left (c x \right ) d \,e^{2}-\frac {c^{5} b \ln \left (c^{2} x^{2}+1\right ) d^{3}}{10}+\frac {c^{3} b \ln \left (c^{2} x^{2}+1\right ) d^{2} e}{2}-\frac {3 c b \ln \left (c^{2} x^{2}+1\right ) d \,e^{2}}{2}-\frac {b \ln \left (c^{2} x^{2}+1\right ) e^{3}}{2 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^3*(a+b*arctan(c*x))/x^6,x)

[Out]

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

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maxima [A]  time = 0.33, size = 208, normalized size = 1.18 \[ -\frac {1}{20} \, {\left ({\left (2 \, c^{4} \log \left (c^{2} x^{2} + 1\right ) - 2 \, c^{4} \log \left (x^{2}\right ) - \frac {2 \, c^{2} x^{2} - 1}{x^{4}}\right )} c + \frac {4 \, \arctan \left (c x\right )}{x^{5}}\right )} b d^{3} + \frac {1}{2} \, {\left ({\left (c^{2} \log \left (c^{2} x^{2} + 1\right ) - c^{2} \log \left (x^{2}\right ) - \frac {1}{x^{2}}\right )} c - \frac {2 \, \arctan \left (c x\right )}{x^{3}}\right )} b d^{2} e - \frac {3}{2} \, {\left (c {\left (\log \left (c^{2} x^{2} + 1\right ) - \log \left (x^{2}\right )\right )} + \frac {2 \, \arctan \left (c x\right )}{x}\right )} b d e^{2} + a e^{3} x + \frac {{\left (2 \, c x \arctan \left (c x\right ) - \log \left (c^{2} x^{2} + 1\right )\right )} b e^{3}}{2 \, c} - \frac {3 \, a d e^{2}}{x} - \frac {a d^{2} e}{x^{3}} - \frac {a d^{3}}{5 \, x^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 0.64, size = 194, normalized size = 1.10 \[ \ln \relax (x)\,\left (\frac {b\,c^5\,d^3}{5}-b\,c^3\,d^2\,e+3\,b\,c\,d\,e^2\right )-\frac {a\,d^3-x^3\,\left (\frac {b\,c^3\,d^3}{2}-\frac {5\,b\,c\,d^2\,e}{2}\right )+\frac {b\,c\,d^3\,x}{4}+5\,a\,d^2\,e\,x^2+15\,a\,d\,e^2\,x^4}{5\,x^5}-\frac {\ln \left (c^2\,x^2+1\right )\,\left (b\,c^6\,d^3-5\,b\,c^4\,d^2\,e+15\,b\,c^2\,d\,e^2+5\,b\,e^3\right )}{10\,c}-\frac {\mathrm {atan}\left (c\,x\right )\,\left (\frac {b\,d^3}{5}+b\,d^2\,e\,x^2+3\,b\,d\,e^2\,x^4-b\,e^3\,x^6\right )}{x^5}+a\,e^3\,x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + e*x^2)^3)/x^6,x)

[Out]

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

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sympy [A]  time = 3.38, size = 289, normalized size = 1.63 \[ \begin {cases} - \frac {a d^{3}}{5 x^{5}} - \frac {a d^{2} e}{x^{3}} - \frac {3 a d e^{2}}{x} + a e^{3} x + \frac {b c^{5} d^{3} \log {\relax (x )}}{5} - \frac {b c^{5} d^{3} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{10} + \frac {b c^{3} d^{3}}{10 x^{2}} - b c^{3} d^{2} e \log {\relax (x )} + \frac {b c^{3} d^{2} e \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b c d^{3}}{20 x^{4}} - \frac {b c d^{2} e}{2 x^{2}} + 3 b c d e^{2} \log {\relax (x )} - \frac {3 b c d e^{2} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2} - \frac {b d^{3} \operatorname {atan}{\left (c x \right )}}{5 x^{5}} - \frac {b d^{2} e \operatorname {atan}{\left (c x \right )}}{x^{3}} - \frac {3 b d e^{2} \operatorname {atan}{\left (c x \right )}}{x} + b e^{3} x \operatorname {atan}{\left (c x \right )} - \frac {b e^{3} \log {\left (x^{2} + \frac {1}{c^{2}} \right )}}{2 c} & \text {for}\: c \neq 0 \\a \left (- \frac {d^{3}}{5 x^{5}} - \frac {d^{2} e}{x^{3}} - \frac {3 d e^{2}}{x} + e^{3} x\right ) & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**3*(a+b*atan(c*x))/x**6,x)

[Out]

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

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