3.1149 \(\int \frac {(d+e x^2)^3 (a+b \tan ^{-1}(c x))}{x^9} \, dx\)

Optimal. Leaf size=152 \[ -\frac {\left (d+e x^2\right )^4 \left (a+b \tan ^{-1}(c x)\right )}{8 d x^8}+\frac {b c d^2 \left (c^2 d-4 e\right )}{40 x^5}+\frac {b \left (c^2 d-e\right )^4 \tan ^{-1}(c x)}{8 d}-\frac {b c d \left (c^4 d^2-4 c^2 d e+6 e^2\right )}{24 x^3}+\frac {b c \left (c^2 d-2 e\right ) \left (c^4 d^2-2 c^2 d e+2 e^2\right )}{8 x}-\frac {b c d^3}{56 x^7} \]

[Out]

-1/56*b*c*d^3/x^7+1/40*b*c*d^2*(c^2*d-4*e)/x^5-1/24*b*c*d*(c^4*d^2-4*c^2*d*e+6*e^2)/x^3+1/8*b*c*(c^2*d-2*e)*(c
^4*d^2-2*c^2*d*e+2*e^2)/x+1/8*b*(c^2*d-e)^4*arctan(c*x)/d-1/8*(e*x^2+d)^4*(a+b*arctan(c*x))/d/x^8

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Rubi [A]  time = 0.20, antiderivative size = 152, 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 = {264, 4976, 12, 461, 203} \[ -\frac {\left (d+e x^2\right )^4 \left (a+b \tan ^{-1}(c x)\right )}{8 d x^8}-\frac {b c d \left (c^4 d^2-4 c^2 d e+6 e^2\right )}{24 x^3}+\frac {b c \left (c^2 d-2 e\right ) \left (c^4 d^2-2 c^2 d e+2 e^2\right )}{8 x}+\frac {b c d^2 \left (c^2 d-4 e\right )}{40 x^5}+\frac {b \left (c^2 d-e\right )^4 \tan ^{-1}(c x)}{8 d}-\frac {b c d^3}{56 x^7} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^9,x]

[Out]

-(b*c*d^3)/(56*x^7) + (b*c*d^2*(c^2*d - 4*e))/(40*x^5) - (b*c*d*(c^4*d^2 - 4*c^2*d*e + 6*e^2))/(24*x^3) + (b*c
*(c^2*d - 2*e)*(c^4*d^2 - 2*c^2*d*e + 2*e^2))/(8*x) + (b*(c^2*d - e)^4*ArcTan[c*x])/(8*d) - ((d + e*x^2)^4*(a
+ b*ArcTan[c*x]))/(8*d*x^8)

Rule 12

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

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 264

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a
*c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] && EqQ[(m + 1)/n + p + 1, 0] && NeQ[m, -1]

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

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^9} \, dx &=-\frac {\left (d+e x^2\right )^4 \left (a+b \tan ^{-1}(c x)\right )}{8 d x^8}-(b c) \int \frac {\left (d+e x^2\right )^4}{8 x^8 \left (-d-c^2 d x^2\right )} \, dx\\ &=-\frac {\left (d+e x^2\right )^4 \left (a+b \tan ^{-1}(c x)\right )}{8 d x^8}-\frac {1}{8} (b c) \int \frac {\left (d+e x^2\right )^4}{x^8 \left (-d-c^2 d x^2\right )} \, dx\\ &=-\frac {\left (d+e x^2\right )^4 \left (a+b \tan ^{-1}(c x)\right )}{8 d x^8}-\frac {1}{8} (b c) \int \left (-\frac {d^3}{x^8}+\frac {d^2 \left (c^2 d-4 e\right )}{x^6}-\frac {d \left (c^4 d^2-4 c^2 d e+6 e^2\right )}{x^4}+\frac {\left (c^2 d-2 e\right ) \left (c^4 d^2-2 c^2 d e+2 e^2\right )}{x^2}-\frac {\left (c^2 d-e\right )^4}{d \left (1+c^2 x^2\right )}\right ) \, dx\\ &=-\frac {b c d^3}{56 x^7}+\frac {b c d^2 \left (c^2 d-4 e\right )}{40 x^5}-\frac {b c d \left (c^4 d^2-4 c^2 d e+6 e^2\right )}{24 x^3}+\frac {b c \left (c^2 d-2 e\right ) \left (c^4 d^2-2 c^2 d e+2 e^2\right )}{8 x}-\frac {\left (d+e x^2\right )^4 \left (a+b \tan ^{-1}(c x)\right )}{8 d x^8}+\frac {\left (b c \left (c^2 d-e\right )^4\right ) \int \frac {1}{1+c^2 x^2} \, dx}{8 d}\\ &=-\frac {b c d^3}{56 x^7}+\frac {b c d^2 \left (c^2 d-4 e\right )}{40 x^5}-\frac {b c d \left (c^4 d^2-4 c^2 d e+6 e^2\right )}{24 x^3}+\frac {b c \left (c^2 d-2 e\right ) \left (c^4 d^2-2 c^2 d e+2 e^2\right )}{8 x}+\frac {b \left (c^2 d-e\right )^4 \tan ^{-1}(c x)}{8 d}-\frac {\left (d+e x^2\right )^4 \left (a+b \tan ^{-1}(c x)\right )}{8 d x^8}\\ \end {align*}

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Mathematica [C]  time = 0.19, size = 154, normalized size = 1.01 \[ -\frac {35 \left (\left (d^3+4 d^2 e x^2+6 d e^2 x^4+4 e^3 x^6\right ) \left (a+b \tan ^{-1}(c x)\right )+2 b c d e^2 x^5 \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};-c^2 x^2\right )+4 b c e^3 x^7 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};-c^2 x^2\right )\right )+5 b c d^3 x \, _2F_1\left (-\frac {7}{2},1;-\frac {5}{2};-c^2 x^2\right )+28 b c d^2 e x^3 \, _2F_1\left (-\frac {5}{2},1;-\frac {3}{2};-c^2 x^2\right )}{280 x^8} \]

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^9,x]

[Out]

-1/280*(5*b*c*d^3*x*Hypergeometric2F1[-7/2, 1, -5/2, -(c^2*x^2)] + 28*b*c*d^2*e*x^3*Hypergeometric2F1[-5/2, 1,
-3/2, -(c^2*x^2)] + 35*((d^3 + 4*d^2*e*x^2 + 6*d*e^2*x^4 + 4*e^3*x^6)*(a + b*ArcTan[c*x]) + 2*b*c*d*e^2*x^5*H
ypergeometric2F1[-3/2, 1, -1/2, -(c^2*x^2)] + 4*b*c*e^3*x^7*Hypergeometric2F1[-1/2, 1, 1/2, -(c^2*x^2)]))/x^8

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fricas [A]  time = 0.43, size = 228, normalized size = 1.50 \[ -\frac {420 \, a e^{3} x^{6} + 630 \, a d e^{2} x^{4} - 105 \, {\left (b c^{7} d^{3} - 4 \, b c^{5} d^{2} e + 6 \, b c^{3} d e^{2} - 4 \, b c e^{3}\right )} x^{7} + 15 \, b c d^{3} x + 420 \, a d^{2} e x^{2} + 35 \, {\left (b c^{5} d^{3} - 4 \, b c^{3} d^{2} e + 6 \, b c d e^{2}\right )} x^{5} + 105 \, a d^{3} - 21 \, {\left (b c^{3} d^{3} - 4 \, b c d^{2} e\right )} x^{3} + 105 \, {\left (4 \, b e^{3} x^{6} - {\left (b c^{8} d^{3} - 4 \, b c^{6} d^{2} e + 6 \, b c^{4} d e^{2} - 4 \, b c^{2} e^{3}\right )} x^{8} + 6 \, b d e^{2} x^{4} + 4 \, b d^{2} e x^{2} + b d^{3}\right )} \arctan \left (c x\right )}{840 \, x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arctan(c*x))/x^9,x, algorithm="fricas")

[Out]

-1/840*(420*a*e^3*x^6 + 630*a*d*e^2*x^4 - 105*(b*c^7*d^3 - 4*b*c^5*d^2*e + 6*b*c^3*d*e^2 - 4*b*c*e^3)*x^7 + 15
*b*c*d^3*x + 420*a*d^2*e*x^2 + 35*(b*c^5*d^3 - 4*b*c^3*d^2*e + 6*b*c*d*e^2)*x^5 + 105*a*d^3 - 21*(b*c^3*d^3 -
4*b*c*d^2*e)*x^3 + 105*(4*b*e^3*x^6 - (b*c^8*d^3 - 4*b*c^6*d^2*e + 6*b*c^4*d*e^2 - 4*b*c^2*e^3)*x^8 + 6*b*d*e^
2*x^4 + 4*b*d^2*e*x^2 + b*d^3)*arctan(c*x))/x^8

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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^9,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.05, size = 265, normalized size = 1.74 \[ -\frac {a \,d^{3}}{8 x^{8}}-\frac {3 a d \,e^{2}}{4 x^{4}}-\frac {a \,d^{2} e}{2 x^{6}}-\frac {a \,e^{3}}{2 x^{2}}-\frac {b \arctan \left (c x \right ) d^{3}}{8 x^{8}}-\frac {3 b \arctan \left (c x \right ) d \,e^{2}}{4 x^{4}}-\frac {b \arctan \left (c x \right ) d^{2} e}{2 x^{6}}-\frac {b \arctan \left (c x \right ) e^{3}}{2 x^{2}}+\frac {c^{7} b \,d^{3}}{8 x}-\frac {c^{5} b \,d^{2} e}{2 x}+\frac {3 c^{3} b d \,e^{2}}{4 x}-\frac {c b \,e^{3}}{2 x}+\frac {c^{3} b \,d^{3}}{40 x^{5}}-\frac {c b \,d^{2} e}{10 x^{5}}-\frac {b c \,d^{3}}{56 x^{7}}-\frac {c^{5} b \,d^{3}}{24 x^{3}}+\frac {c^{3} b \,d^{2} e}{6 x^{3}}-\frac {c b d \,e^{2}}{4 x^{3}}+\frac {c^{8} b \arctan \left (c x \right ) d^{3}}{8}-\frac {c^{6} b \arctan \left (c x \right ) d^{2} e}{2}+\frac {3 c^{4} b \arctan \left (c x \right ) d \,e^{2}}{4}-\frac {c^{2} b \arctan \left (c x \right ) e^{3}}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^3*(a+b*arctan(c*x))/x^9,x)

[Out]

-1/8*a*d^3/x^8-3/4*a*d*e^2/x^4-1/2*a*d^2*e/x^6-1/2*a*e^3/x^2-1/8*b*arctan(c*x)*d^3/x^8-3/4*b*arctan(c*x)*d*e^2
/x^4-1/2*b*arctan(c*x)*d^2*e/x^6-1/2*b*arctan(c*x)*e^3/x^2+1/8*c^7*b*d^3/x-1/2*c^5*b*d^2*e/x+3/4*c^3*b*d*e^2/x
-1/2*c*b*e^3/x+1/40*c^3*b*d^3/x^5-1/10*c*b*d^2*e/x^5-1/56*b*c*d^3/x^7-1/24*c^5*b*d^3/x^3+1/6*c^3*b*d^2*e/x^3-1
/4*c*b*d*e^2/x^3+1/8*c^8*b*arctan(c*x)*d^3-1/2*c^6*b*arctan(c*x)*d^2*e+3/4*c^4*b*arctan(c*x)*d*e^2-1/2*c^2*b*a
rctan(c*x)*e^3

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maxima [A]  time = 0.43, size = 218, normalized size = 1.43 \[ \frac {1}{840} \, {\left ({\left (105 \, c^{7} \arctan \left (c x\right ) + \frac {105 \, c^{6} x^{6} - 35 \, c^{4} x^{4} + 21 \, c^{2} x^{2} - 15}{x^{7}}\right )} c - \frac {105 \, \arctan \left (c x\right )}{x^{8}}\right )} b d^{3} - \frac {1}{30} \, {\left ({\left (15 \, c^{5} \arctan \left (c x\right ) + \frac {15 \, c^{4} x^{4} - 5 \, c^{2} x^{2} + 3}{x^{5}}\right )} c + \frac {15 \, \arctan \left (c x\right )}{x^{6}}\right )} b d^{2} e + \frac {1}{4} \, {\left ({\left (3 \, c^{3} \arctan \left (c x\right ) + \frac {3 \, c^{2} x^{2} - 1}{x^{3}}\right )} c - \frac {3 \, \arctan \left (c x\right )}{x^{4}}\right )} b d e^{2} - \frac {1}{2} \, {\left ({\left (c \arctan \left (c x\right ) + \frac {1}{x}\right )} c + \frac {\arctan \left (c x\right )}{x^{2}}\right )} b e^{3} - \frac {a e^{3}}{2 \, x^{2}} - \frac {3 \, a d e^{2}}{4 \, x^{4}} - \frac {a d^{2} e}{2 \, x^{6}} - \frac {a d^{3}}{8 \, x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arctan(c*x))/x^9,x, algorithm="maxima")

[Out]

1/840*((105*c^7*arctan(c*x) + (105*c^6*x^6 - 35*c^4*x^4 + 21*c^2*x^2 - 15)/x^7)*c - 105*arctan(c*x)/x^8)*b*d^3
- 1/30*((15*c^5*arctan(c*x) + (15*c^4*x^4 - 5*c^2*x^2 + 3)/x^5)*c + 15*arctan(c*x)/x^6)*b*d^2*e + 1/4*((3*c^3
*arctan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*b*d*e^2 - 1/2*((c*arctan(c*x) + 1/x)*c + arctan(c*x
)/x^2)*b*e^3 - 1/2*a*e^3/x^2 - 3/4*a*d*e^2/x^4 - 1/2*a*d^2*e/x^6 - 1/8*a*d^3/x^8

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mupad [B]  time = 0.63, size = 301, normalized size = 1.98 \[ \frac {b\,c^2\,\mathrm {atan}\left (\frac {b\,c^2\,x\,\left (2\,e-c^2\,d\right )\,\left (c^4\,d^2-2\,c^2\,d\,e+2\,e^2\right )}{b\,c^7\,d^3-4\,b\,c^5\,d^2\,e+6\,b\,c^3\,d\,e^2-4\,b\,c\,e^3}\right )\,\left (2\,e-c^2\,d\right )\,\left (c^4\,d^2-2\,c^2\,d\,e+2\,e^2\right )}{8}-\frac {\mathrm {atan}\left (c\,x\right )\,\left (\frac {b\,d^3}{8}+\frac {b\,d^2\,e\,x^2}{2}+\frac {3\,b\,d\,e^2\,x^4}{4}+\frac {b\,e^3\,x^6}{2}\right )}{x^8}-\frac {a\,d^3-x^3\,\left (\frac {b\,c^3\,d^3}{5}-\frac {4\,b\,c\,d^2\,e}{5}\right )-x^7\,\left (b\,c^7\,d^3-4\,b\,c^5\,d^2\,e+6\,b\,c^3\,d\,e^2-4\,b\,c\,e^3\right )+x^5\,\left (\frac {b\,c^5\,d^3}{3}-\frac {4\,b\,c^3\,d^2\,e}{3}+2\,b\,c\,d\,e^2\right )+4\,a\,e^3\,x^6+\frac {b\,c\,d^3\,x}{7}+4\,a\,d^2\,e\,x^2+6\,a\,d\,e^2\,x^4}{8\,x^8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + e*x^2)^3)/x^9,x)

[Out]

(b*c^2*atan((b*c^2*x*(2*e - c^2*d)*(2*e^2 + c^4*d^2 - 2*c^2*d*e))/(b*c^7*d^3 - 4*b*c*e^3 + 6*b*c^3*d*e^2 - 4*b
*c^5*d^2*e))*(2*e - c^2*d)*(2*e^2 + c^4*d^2 - 2*c^2*d*e))/8 - (atan(c*x)*((b*d^3)/8 + (b*e^3*x^6)/2 + (b*d^2*e
*x^2)/2 + (3*b*d*e^2*x^4)/4))/x^8 - (a*d^3 - x^3*((b*c^3*d^3)/5 - (4*b*c*d^2*e)/5) - x^7*(b*c^7*d^3 - 4*b*c*e^
3 + 6*b*c^3*d*e^2 - 4*b*c^5*d^2*e) + x^5*((b*c^5*d^3)/3 + 2*b*c*d*e^2 - (4*b*c^3*d^2*e)/3) + 4*a*e^3*x^6 + (b*
c*d^3*x)/7 + 4*a*d^2*e*x^2 + 6*a*d*e^2*x^4)/(8*x^8)

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sympy [B]  time = 3.16, size = 309, normalized size = 2.03 \[ - \frac {a d^{3}}{8 x^{8}} - \frac {a d^{2} e}{2 x^{6}} - \frac {3 a d e^{2}}{4 x^{4}} - \frac {a e^{3}}{2 x^{2}} + \frac {b c^{8} d^{3} \operatorname {atan}{\left (c x \right )}}{8} + \frac {b c^{7} d^{3}}{8 x} - \frac {b c^{6} d^{2} e \operatorname {atan}{\left (c x \right )}}{2} - \frac {b c^{5} d^{3}}{24 x^{3}} - \frac {b c^{5} d^{2} e}{2 x} + \frac {3 b c^{4} d e^{2} \operatorname {atan}{\left (c x \right )}}{4} + \frac {b c^{3} d^{3}}{40 x^{5}} + \frac {b c^{3} d^{2} e}{6 x^{3}} + \frac {3 b c^{3} d e^{2}}{4 x} - \frac {b c^{2} e^{3} \operatorname {atan}{\left (c x \right )}}{2} - \frac {b c d^{3}}{56 x^{7}} - \frac {b c d^{2} e}{10 x^{5}} - \frac {b c d e^{2}}{4 x^{3}} - \frac {b c e^{3}}{2 x} - \frac {b d^{3} \operatorname {atan}{\left (c x \right )}}{8 x^{8}} - \frac {b d^{2} e \operatorname {atan}{\left (c x \right )}}{2 x^{6}} - \frac {3 b d e^{2} \operatorname {atan}{\left (c x \right )}}{4 x^{4}} - \frac {b e^{3} \operatorname {atan}{\left (c x \right )}}{2 x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**3*(a+b*atan(c*x))/x**9,x)

[Out]

-a*d**3/(8*x**8) - a*d**2*e/(2*x**6) - 3*a*d*e**2/(4*x**4) - a*e**3/(2*x**2) + b*c**8*d**3*atan(c*x)/8 + b*c**
7*d**3/(8*x) - b*c**6*d**2*e*atan(c*x)/2 - b*c**5*d**3/(24*x**3) - b*c**5*d**2*e/(2*x) + 3*b*c**4*d*e**2*atan(
c*x)/4 + b*c**3*d**3/(40*x**5) + b*c**3*d**2*e/(6*x**3) + 3*b*c**3*d*e**2/(4*x) - b*c**2*e**3*atan(c*x)/2 - b*
c*d**3/(56*x**7) - b*c*d**2*e/(10*x**5) - b*c*d*e**2/(4*x**3) - b*c*e**3/(2*x) - b*d**3*atan(c*x)/(8*x**8) - b
*d**2*e*atan(c*x)/(2*x**6) - 3*b*d*e**2*atan(c*x)/(4*x**4) - b*e**3*atan(c*x)/(2*x**2)

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