3.12.47 \(\int \frac {(d+e x^2)^3 (a+b \text {ArcTan}(c x))}{x^7} \, dx\) [1147]

Optimal. Leaf size=228 \[ -\frac {b c d^3}{30 x^5}+\frac {b c^3 d^3}{18 x^3}-\frac {b c d^2 e}{4 x^3}-\frac {b c^5 d^3}{6 x}+\frac {3 b c^3 d^2 e}{4 x}-\frac {3 b c d e^2}{2 x}-\frac {1}{6} b c^6 d^3 \text {ArcTan}(c x)+\frac {3}{4} b c^4 d^2 e \text {ArcTan}(c x)-\frac {3}{2} b c^2 d e^2 \text {ArcTan}(c x)-\frac {d^3 (a+b \text {ArcTan}(c x))}{6 x^6}-\frac {3 d^2 e (a+b \text {ArcTan}(c x))}{4 x^4}-\frac {3 d e^2 (a+b \text {ArcTan}(c x))}{2 x^2}+a e^3 \log (x)+\frac {1}{2} i b e^3 \text {PolyLog}(2,-i c x)-\frac {1}{2} i b e^3 \text {PolyLog}(2,i c x) \]

[Out]

-1/30*b*c*d^3/x^5+1/18*b*c^3*d^3/x^3-1/4*b*c*d^2*e/x^3-1/6*b*c^5*d^3/x+3/4*b*c^3*d^2*e/x-3/2*b*c*d*e^2/x-1/6*b
*c^6*d^3*arctan(c*x)+3/4*b*c^4*d^2*e*arctan(c*x)-3/2*b*c^2*d*e^2*arctan(c*x)-1/6*d^3*(a+b*arctan(c*x))/x^6-3/4
*d^2*e*(a+b*arctan(c*x))/x^4-3/2*d*e^2*(a+b*arctan(c*x))/x^2+a*e^3*ln(x)+1/2*I*b*e^3*polylog(2,-I*c*x)-1/2*I*b
*e^3*polylog(2,I*c*x)

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Rubi [A]
time = 0.15, antiderivative size = 228, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {5100, 4946, 331, 209, 4940, 2438} \begin {gather*} -\frac {d^3 (a+b \text {ArcTan}(c x))}{6 x^6}-\frac {3 d^2 e (a+b \text {ArcTan}(c x))}{4 x^4}-\frac {3 d e^2 (a+b \text {ArcTan}(c x))}{2 x^2}+a e^3 \log (x)-\frac {1}{6} b c^6 d^3 \text {ArcTan}(c x)+\frac {3}{4} b c^4 d^2 e \text {ArcTan}(c x)-\frac {3}{2} b c^2 d e^2 \text {ArcTan}(c x)-\frac {b c^5 d^3}{6 x}+\frac {b c^3 d^3}{18 x^3}+\frac {3 b c^3 d^2 e}{4 x}-\frac {b c d^3}{30 x^5}-\frac {b c d^2 e}{4 x^3}-\frac {3 b c d e^2}{2 x}+\frac {1}{2} i b e^3 \text {Li}_2(-i c x)-\frac {1}{2} i b e^3 \text {Li}_2(i c x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^7,x]

[Out]

-1/30*(b*c*d^3)/x^5 + (b*c^3*d^3)/(18*x^3) - (b*c*d^2*e)/(4*x^3) - (b*c^5*d^3)/(6*x) + (3*b*c^3*d^2*e)/(4*x) -
(3*b*c*d*e^2)/(2*x) - (b*c^6*d^3*ArcTan[c*x])/6 + (3*b*c^4*d^2*e*ArcTan[c*x])/4 - (3*b*c^2*d*e^2*ArcTan[c*x])
/2 - (d^3*(a + b*ArcTan[c*x]))/(6*x^6) - (3*d^2*e*(a + b*ArcTan[c*x]))/(4*x^4) - (3*d*e^2*(a + b*ArcTan[c*x]))
/(2*x^2) + a*e^3*Log[x] + (I/2)*b*e^3*PolyLog[2, (-I)*c*x] - (I/2)*b*e^3*PolyLog[2, I*c*x]

Rule 209

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

Rule 331

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] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 4940

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[I*(b/2), Int[Log[1 - I*c*x
]/x, x], x] - Dist[I*(b/2), Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

Rule 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[c*x^n])^(p - 1)/(1 + c^2*x^(2*n))),
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1]

Rule 5100

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With
[{u = ExpandIntegrand[(a + b*ArcTan[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b,
c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && ((EqQ[p, 1] && GtQ[q, 0]) || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {\left (d+e x^2\right )^3 \left (a+b \tan ^{-1}(c x)\right )}{x^7} \, dx &=\int \left (\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{x^7}+\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{x^5}+\frac {3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{x^3}+\frac {e^3 \left (a+b \tan ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d^3 \int \frac {a+b \tan ^{-1}(c x)}{x^7} \, dx+\left (3 d^2 e\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^5} \, dx+\left (3 d e^2\right ) \int \frac {a+b \tan ^{-1}(c x)}{x^3} \, dx+e^3 \int \frac {a+b \tan ^{-1}(c x)}{x} \, dx\\ &=-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac {3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+a e^3 \log (x)+\frac {1}{6} \left (b c d^3\right ) \int \frac {1}{x^6 \left (1+c^2 x^2\right )} \, dx+\frac {1}{4} \left (3 b c d^2 e\right ) \int \frac {1}{x^4 \left (1+c^2 x^2\right )} \, dx+\frac {1}{2} \left (3 b c d e^2\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx+\frac {1}{2} \left (i b e^3\right ) \int \frac {\log (1-i c x)}{x} \, dx-\frac {1}{2} \left (i b e^3\right ) \int \frac {\log (1+i c x)}{x} \, dx\\ &=-\frac {b c d^3}{30 x^5}-\frac {b c d^2 e}{4 x^3}-\frac {3 b c d e^2}{2 x}-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac {3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+a e^3 \log (x)+\frac {1}{2} i b e^3 \text {Li}_2(-i c x)-\frac {1}{2} i b e^3 \text {Li}_2(i c x)-\frac {1}{6} \left (b c^3 d^3\right ) \int \frac {1}{x^4 \left (1+c^2 x^2\right )} \, dx-\frac {1}{4} \left (3 b c^3 d^2 e\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx-\frac {1}{2} \left (3 b c^3 d e^2\right ) \int \frac {1}{1+c^2 x^2} \, dx\\ &=-\frac {b c d^3}{30 x^5}+\frac {b c^3 d^3}{18 x^3}-\frac {b c d^2 e}{4 x^3}+\frac {3 b c^3 d^2 e}{4 x}-\frac {3 b c d e^2}{2 x}-\frac {3}{2} b c^2 d e^2 \tan ^{-1}(c x)-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac {3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+a e^3 \log (x)+\frac {1}{2} i b e^3 \text {Li}_2(-i c x)-\frac {1}{2} i b e^3 \text {Li}_2(i c x)+\frac {1}{6} \left (b c^5 d^3\right ) \int \frac {1}{x^2 \left (1+c^2 x^2\right )} \, dx+\frac {1}{4} \left (3 b c^5 d^2 e\right ) \int \frac {1}{1+c^2 x^2} \, dx\\ &=-\frac {b c d^3}{30 x^5}+\frac {b c^3 d^3}{18 x^3}-\frac {b c d^2 e}{4 x^3}-\frac {b c^5 d^3}{6 x}+\frac {3 b c^3 d^2 e}{4 x}-\frac {3 b c d e^2}{2 x}+\frac {3}{4} b c^4 d^2 e \tan ^{-1}(c x)-\frac {3}{2} b c^2 d e^2 \tan ^{-1}(c x)-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac {3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+a e^3 \log (x)+\frac {1}{2} i b e^3 \text {Li}_2(-i c x)-\frac {1}{2} i b e^3 \text {Li}_2(i c x)-\frac {1}{6} \left (b c^7 d^3\right ) \int \frac {1}{1+c^2 x^2} \, dx\\ &=-\frac {b c d^3}{30 x^5}+\frac {b c^3 d^3}{18 x^3}-\frac {b c d^2 e}{4 x^3}-\frac {b c^5 d^3}{6 x}+\frac {3 b c^3 d^2 e}{4 x}-\frac {3 b c d e^2}{2 x}-\frac {1}{6} b c^6 d^3 \tan ^{-1}(c x)+\frac {3}{4} b c^4 d^2 e \tan ^{-1}(c x)-\frac {3}{2} b c^2 d e^2 \tan ^{-1}(c x)-\frac {d^3 \left (a+b \tan ^{-1}(c x)\right )}{6 x^6}-\frac {3 d^2 e \left (a+b \tan ^{-1}(c x)\right )}{4 x^4}-\frac {3 d e^2 \left (a+b \tan ^{-1}(c x)\right )}{2 x^2}+a e^3 \log (x)+\frac {1}{2} i b e^3 \text {Li}_2(-i c x)-\frac {1}{2} i b e^3 \text {Li}_2(i c x)\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 190, normalized size = 0.83 \begin {gather*} -\frac {a d^3}{6 x^6}-\frac {3 a d^2 e}{4 x^4}-\frac {3 a d e^2}{2 x^2}+\frac {b d^2 e \left (c x \left (-1+3 c^2 x^2\right )+3 \left (-1+c^4 x^4\right ) \text {ArcTan}(c x)\right )}{4 x^4}-\frac {b d^3 \left (c x \left (3-5 c^2 x^2+15 c^4 x^4\right )+15 \left (1+c^6 x^6\right ) \text {ArcTan}(c x)\right )}{90 x^6}-\frac {3 b d e^2 (\text {ArcTan}(c x)+c x (1+c x \text {ArcTan}(c x)))}{2 x^2}+a e^3 \log (x)+\frac {1}{2} i b e^3 (\text {PolyLog}(2,-i c x)-\text {PolyLog}(2,i c x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + e*x^2)^3*(a + b*ArcTan[c*x]))/x^7,x]

[Out]

-1/6*(a*d^3)/x^6 - (3*a*d^2*e)/(4*x^4) - (3*a*d*e^2)/(2*x^2) + (b*d^2*e*(c*x*(-1 + 3*c^2*x^2) + 3*(-1 + c^4*x^
4)*ArcTan[c*x]))/(4*x^4) - (b*d^3*(c*x*(3 - 5*c^2*x^2 + 15*c^4*x^4) + 15*(1 + c^6*x^6)*ArcTan[c*x]))/(90*x^6)
- (3*b*d*e^2*(ArcTan[c*x] + c*x*(1 + c*x*ArcTan[c*x])))/(2*x^2) + a*e^3*Log[x] + (I/2)*b*e^3*(PolyLog[2, (-I)*
c*x] - PolyLog[2, I*c*x])

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Maple [A]
time = 0.18, size = 315, normalized size = 1.38

method	result	size
derivativedivides	\(c^{6} \left (-\frac {3 a \,d^{2} e}{4 c^{6} x^{4}}-\frac {a \,d^{3}}{6 c^{6} x^{6}}-\frac {3 a d \,e^{2}}{2 c^{6} x^{2}}+\frac {a \,e^{3} \ln \left (c x \right )}{c^{6}}-\frac {3 b \arctan \left (c x \right ) d^{2} e}{4 c^{6} x^{4}}-\frac {b \arctan \left (c x \right ) d^{3}}{6 c^{6} x^{6}}-\frac {3 b \arctan \left (c x \right ) d \,e^{2}}{2 c^{6} x^{2}}+\frac {b \arctan \left (c x \right ) e^{3} \ln \left (c x \right )}{c^{6}}+\frac {i b \,e^{3} \dilog \left (i c x +1\right )}{2 c^{6}}-\frac {i b \,e^{3} \dilog \left (-i c x +1\right )}{2 c^{6}}+\frac {i b \,e^{3} \ln \left (c x \right ) \ln \left (i c x +1\right )}{2 c^{6}}-\frac {i b \,e^{3} \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2 c^{6}}-\frac {b \,d^{3} \arctan \left (c x \right )}{6}+\frac {3 b \,d^{2} e \arctan \left (c x \right )}{4 c^{2}}-\frac {3 b d \,e^{2} \arctan \left (c x \right )}{2 c^{4}}-\frac {b \,d^{3}}{6 c x}+\frac {3 b \,d^{2} e}{4 c^{3} x}-\frac {3 b d \,e^{2}}{2 c^{5} x}-\frac {b \,d^{3}}{30 c^{5} x^{5}}+\frac {b \,d^{3}}{18 c^{3} x^{3}}-\frac {b \,d^{2} e}{4 c^{5} x^{3}}\right )\)	\(315\)
default	\(c^{6} \left (-\frac {3 a \,d^{2} e}{4 c^{6} x^{4}}-\frac {a \,d^{3}}{6 c^{6} x^{6}}-\frac {3 a d \,e^{2}}{2 c^{6} x^{2}}+\frac {a \,e^{3} \ln \left (c x \right )}{c^{6}}-\frac {3 b \arctan \left (c x \right ) d^{2} e}{4 c^{6} x^{4}}-\frac {b \arctan \left (c x \right ) d^{3}}{6 c^{6} x^{6}}-\frac {3 b \arctan \left (c x \right ) d \,e^{2}}{2 c^{6} x^{2}}+\frac {b \arctan \left (c x \right ) e^{3} \ln \left (c x \right )}{c^{6}}+\frac {i b \,e^{3} \dilog \left (i c x +1\right )}{2 c^{6}}-\frac {i b \,e^{3} \dilog \left (-i c x +1\right )}{2 c^{6}}+\frac {i b \,e^{3} \ln \left (c x \right ) \ln \left (i c x +1\right )}{2 c^{6}}-\frac {i b \,e^{3} \ln \left (c x \right ) \ln \left (-i c x +1\right )}{2 c^{6}}-\frac {b \,d^{3} \arctan \left (c x \right )}{6}+\frac {3 b \,d^{2} e \arctan \left (c x \right )}{4 c^{2}}-\frac {3 b d \,e^{2} \arctan \left (c x \right )}{2 c^{4}}-\frac {b \,d^{3}}{6 c x}+\frac {3 b \,d^{2} e}{4 c^{3} x}-\frac {3 b d \,e^{2}}{2 c^{5} x}-\frac {b \,d^{3}}{30 c^{5} x^{5}}+\frac {b \,d^{3}}{18 c^{3} x^{3}}-\frac {b \,d^{2} e}{4 c^{5} x^{3}}\right )\)	\(315\)
risch	\(a \,e^{3} \ln \left (-i c x \right )+\frac {3 i b \,e^{2} d \ln \left (i c x +1\right )}{4 x^{2}}-\frac {3 i b \,c^{2} e^{2} d \ln \left (i c x \right )}{4}-\frac {a \,d^{3}}{6 x^{6}}-\frac {i b \,d^{3} \ln \left (-i c x +1\right )}{12 x^{6}}+\frac {i c^{6} b \,d^{3} \ln \left (-i c x \right )}{12}-\frac {i b \,c^{6} d^{3} \ln \left (i c x \right )}{12}+\frac {i b \,d^{3} \ln \left (i c x +1\right )}{12 x^{6}}-\frac {3 i b \,d^{2} e \ln \left (-i c x +1\right )}{8 x^{4}}-\frac {b c \,d^{2} e}{4 x^{3}}+\frac {3 b \,c^{3} d^{2} e}{4 x}-\frac {3 b c d \,e^{2}}{2 x}+\frac {3 b \,c^{4} d^{2} e \arctan \left (c x \right )}{4}-\frac {3 b \,c^{2} d \,e^{2} \arctan \left (c x \right )}{2}-\frac {b \,c^{6} d^{3} \arctan \left (c x \right )}{6}+\frac {i b \,e^{3} \dilog \left (i c x +1\right )}{2}-\frac {i b \,e^{3} \dilog \left (-i c x +1\right )}{2}-\frac {3 a \,d^{2} e}{4 x^{4}}-\frac {3 a d \,e^{2}}{2 x^{2}}-\frac {b c \,d^{3}}{30 x^{5}}+\frac {b \,c^{3} d^{3}}{18 x^{3}}-\frac {b \,c^{5} d^{3}}{6 x}-\frac {3 i c^{4} b \,d^{2} e \ln \left (-i c x \right )}{8}-\frac {3 i b d \,e^{2} \ln \left (-i c x +1\right )}{4 x^{2}}+\frac {3 i c^{2} b d \,e^{2} \ln \left (-i c x \right )}{4}+\frac {3 i b \,d^{2} e \ln \left (i c x +1\right )}{8 x^{4}}+\frac {3 i b \,c^{4} d^{2} e \ln \left (i c x \right )}{8}\)	\(394\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)^3*(a+b*arctan(c*x))/x^7,x,method=_RETURNVERBOSE)

[Out]

c^6*(-3/4*a/c^6*d^2*e/x^4-1/6*a*d^3/c^6/x^6-3/2*a/c^6*d*e^2/x^2+a/c^6*e^3*ln(c*x)-3/4*b/c^6*arctan(c*x)*d^2*e/
x^4-1/6*b*arctan(c*x)*d^3/c^6/x^6-3/2*b/c^6*arctan(c*x)*d*e^2/x^2+b/c^6*arctan(c*x)*e^3*ln(c*x)+1/2*I*b/c^6*e^
3*ln(c*x)*ln(1+I*c*x)-1/2*I*b/c^6*e^3*dilog(1-I*c*x)+1/2*I*b/c^6*e^3*dilog(1+I*c*x)-1/2*I*b/c^6*e^3*ln(c*x)*ln
(1-I*c*x)-1/6*b*d^3*arctan(c*x)+3/4*b*d^2*e*arctan(c*x)/c^2-3/2*b*d*e^2*arctan(c*x)/c^4-1/6*b*d^3/c/x+3/4*b/c^
3*d^2*e/x-3/2*b/c^5*d*e^2/x-1/30*b*d^3/c^5/x^5+1/18*b*d^3/c^3/x^3-1/4*b/c^5*d^2*e/x^3)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)^3*(a+b*arctan(c*x))/x^7,x, algorithm="maxima")

[Out]

-1/90*((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^3 + 1/4*((3*c^3*arc
tan(c*x) + (3*c^2*x^2 - 1)/x^3)*c - 3*arctan(c*x)/x^4)*b*d^2*e - 3/2*((c*arctan(c*x) + 1/x)*c + arctan(c*x)/x^
2)*b*d*e^2 + b*e^3*integrate(arctan(c*x)/x, x) + a*e^3*log(x) - 3/2*a*d*e^2/x^2 - 3/4*a*d^2*e/x^4 - 1/6*a*d^3/
x^6

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Fricas [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)^3*(a+b*arctan(c*x))/x^7,x, algorithm="fricas")

[Out]

integral((a*x^6*e^3 + 3*a*d*x^4*e^2 + 3*a*d^2*x^2*e + a*d^3 + (b*x^6*e^3 + 3*b*d*x^4*e^2 + 3*b*d^2*x^2*e + b*d
^3)*arctan(c*x))/x^7, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {atan}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{3}}{x^{7}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)**3*(a+b*atan(c*x))/x**7,x)

[Out]

Integral((a + b*atan(c*x))*(d + e*x**2)**3/x**7, x)

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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)^3*(a+b*arctan(c*x))/x^7,x, algorithm="giac")

[Out]

sage0*x

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Mupad [B]
time = 0.84, size = 261, normalized size = 1.14 \begin {gather*} \left \{\begin {array}{cl} a\,e^3\,\ln \left (x\right )-\frac {\frac {a\,d^3}{6}+\frac {3\,a\,d^2\,e\,x^2}{4}+\frac {3\,a\,d\,e^2\,x^4}{2}}{x^6} & \text {\ if\ \ }c=0\\ a\,e^3\,\ln \left (x\right )-\frac {\frac {a\,d^3}{6}+\frac {3\,a\,d^2\,e\,x^2}{4}+\frac {3\,a\,d\,e^2\,x^4}{2}}{x^6}-3\,b\,d^2\,e\,\left (\frac {\mathrm {atan}\left (c\,x\right )}{4\,x^4}+\frac {\frac {\frac {c^2}{3}-c^4\,x^2}{x^3}-c^5\,\mathrm {atan}\left (c\,x\right )}{4\,c}\right )-\frac {b\,d^3\,\left (\frac {c^6\,x^4-\frac {c^4\,x^2}{3}+\frac {c^2}{5}}{x^5}+c^7\,\mathrm {atan}\left (c\,x\right )\right )}{6\,c}-3\,b\,d\,e^2\,\left (\frac {c^3\,\mathrm {atan}\left (c\,x\right )+\frac {c^2}{x}}{2\,c}+\frac {\mathrm {atan}\left (c\,x\right )}{2\,x^2}\right )-\frac {b\,d^3\,\mathrm {atan}\left (c\,x\right )}{6\,x^6}-\frac {b\,e^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1-c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2}+\frac {b\,e^3\,{\mathrm {Li}}_{\mathrm {2}}\left (1+c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2} & \text {\ if\ \ }c\neq 0 \end {array}\right . \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atan(c*x))*(d + e*x^2)^3)/x^7,x)

[Out]

piecewise(c == 0, - ((a*d^3)/6 + (3*a*d^2*e*x^2)/4 + (3*a*d*e^2*x^4)/2)/x^6 + a*e^3*log(x), c ~= 0, - ((a*d^3)
/6 + (3*a*d^2*e*x^2)/4 + (3*a*d*e^2*x^4)/2)/x^6 + a*e^3*log(x) - (b*e^3*dilog(- c*x*1i + 1)*1i)/2 + (b*e^3*dil
og(c*x*1i + 1)*1i)/2 - 3*b*d^2*e*(atan(c*x)/(4*x^4) + ((c^2/3 - c^4*x^2)/x^3 - c^5*atan(c*x))/(4*c)) - (b*d^3*
((c^2/5 - (c^4*x^2)/3 + c^6*x^4)/x^5 + c^7*atan(c*x)))/(6*c) - 3*b*d*e^2*((c^3*atan(c*x) + c^2/x)/(2*c) + atan
(c*x)/(2*x^2)) - (b*d^3*atan(c*x))/(6*x^6))

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