3.12.92 \(\int \frac {(d+e x^2)^{3/2} (a+b \text {ArcTan}(c x))}{x^6} \, dx\) [1192]

Optimal. Leaf size=178 \[ \frac {b c \left (4 c^2 d-7 e\right ) \sqrt {d+e x^2}}{40 x^2}-\frac {b c \left (d+e x^2\right )^{3/2}}{20 x^4}-\frac {\left (d+e x^2\right )^{5/2} (a+b \text {ArcTan}(c x))}{5 d x^5}-\frac {b c \left (8 c^4 d^2-20 c^2 d e+15 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{40 \sqrt {d}}+\frac {b \left (c^2 d-e\right )^{5/2} \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{5 d} \]

[Out]

-1/20*b*c*(e*x^2+d)^(3/2)/x^4-1/5*(e*x^2+d)^(5/2)*(a+b*arctan(c*x))/d/x^5+1/5*b*(c^2*d-e)^(5/2)*arctanh(c*(e*x
^2+d)^(1/2)/(c^2*d-e)^(1/2))/d-1/40*b*c*(8*c^4*d^2-20*c^2*d*e+15*e^2)*arctanh((e*x^2+d)^(1/2)/d^(1/2))/d^(1/2)
+1/40*b*c*(4*c^2*d-7*e)*(e*x^2+d)^(1/2)/x^2

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Rubi [A]
time = 0.24, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {270, 5096, 12, 457, 100, 154, 162, 65, 214} \begin {gather*} -\frac {\left (d+e x^2\right )^{5/2} (a+b \text {ArcTan}(c x))}{5 d x^5}+\frac {b c \left (4 c^2 d-7 e\right ) \sqrt {d+e x^2}}{40 x^2}+\frac {b \left (c^2 d-e\right )^{5/2} \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{5 d}-\frac {b c \left (8 c^4 d^2-20 c^2 d e+15 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{40 \sqrt {d}}-\frac {b c \left (d+e x^2\right )^{3/2}}{20 x^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c*(4*c^2*d - 7*e)*Sqrt[d + e*x^2])/(40*x^2) - (b*c*(d + e*x^2)^(3/2))/(20*x^4) - ((d + e*x^2)^(5/2)*(a + b*
ArcTan[c*x]))/(5*d*x^5) - (b*c*(8*c^4*d^2 - 20*c^2*d*e + 15*e^2)*ArcTanh[Sqrt[d + e*x^2]/Sqrt[d]])/(40*Sqrt[d]
) + (b*(c^2*d - e)^(5/2)*ArcTanh[(c*Sqrt[d + e*x^2])/Sqrt[c^2*d - e]])/(5*d)

Rule 12

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

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 154

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && ILtQ[m, -1] && GtQ[n, 0]

Rule 162

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 214

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

Rule 270

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 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 )^{3/2} \left (a+b \tan ^{-1}(c x)\right )}{x^6} \, dx &=-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 d x^5}-(b c) \int \frac {\left (d+e x^2\right )^{5/2}}{5 x^5 \left (-d-c^2 d x^2\right )} \, dx\\ &=-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 d x^5}-\frac {1}{5} (b c) \int \frac {\left (d+e x^2\right )^{5/2}}{x^5 \left (-d-c^2 d x^2\right )} \, dx\\ &=-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 d x^5}-\frac {1}{10} (b c) \text {Subst}\left (\int \frac {(d+e x)^{5/2}}{x^3 \left (-d-c^2 d x\right )} \, dx,x,x^2\right )\\ &=-\frac {b c \left (d+e x^2\right )^{3/2}}{20 x^4}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 d x^5}-\frac {(b c) \text {Subst}\left (\int \frac {\sqrt {d+e x} \left (-\frac {1}{2} d^2 \left (4 c^2 d-7 e\right )-\frac {1}{2} d \left (c^2 d-4 e\right ) e x\right )}{x^2 \left (-d-c^2 d x\right )} \, dx,x,x^2\right )}{20 d}\\ &=\frac {b c \left (4 c^2 d-7 e\right ) \sqrt {d+e x^2}}{40 x^2}-\frac {b c \left (d+e x^2\right )^{3/2}}{20 x^4}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 d x^5}+\frac {(b c) \text {Subst}\left (\int \frac {-\frac {1}{4} d^3 \left (8 c^4 d^2-20 c^2 d e+15 e^2\right )-\frac {1}{4} d^2 e \left (4 c^4 d^2-9 c^2 d e+8 e^2\right ) x}{x \left (-d-c^2 d x\right ) \sqrt {d+e x}} \, dx,x,x^2\right )}{20 d^2}\\ &=\frac {b c \left (4 c^2 d-7 e\right ) \sqrt {d+e x^2}}{40 x^2}-\frac {b c \left (d+e x^2\right )^{3/2}}{20 x^4}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 d x^5}+\frac {1}{10} \left (b c \left (c^2 d-e\right )^3\right ) \text {Subst}\left (\int \frac {1}{\left (-d-c^2 d x\right ) \sqrt {d+e x}} \, dx,x,x^2\right )+\frac {1}{80} \left (b c \left (8 c^4 d^2-20 c^2 d e+15 e^2\right )\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )\\ &=\frac {b c \left (4 c^2 d-7 e\right ) \sqrt {d+e x^2}}{40 x^2}-\frac {b c \left (d+e x^2\right )^{3/2}}{20 x^4}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 d x^5}+\frac {\left (b c \left (c^2 d-e\right )^3\right ) \text {Subst}\left (\int \frac {1}{-d+\frac {c^2 d^2}{e}-\frac {c^2 d x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{5 e}+\frac {\left (b c \left (8 c^4 d^2-20 c^2 d e+15 e^2\right )\right ) \text {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{40 e}\\ &=\frac {b c \left (4 c^2 d-7 e\right ) \sqrt {d+e x^2}}{40 x^2}-\frac {b c \left (d+e x^2\right )^{3/2}}{20 x^4}-\frac {\left (d+e x^2\right )^{5/2} \left (a+b \tan ^{-1}(c x)\right )}{5 d x^5}-\frac {b c \left (8 c^4 d^2-20 c^2 d e+15 e^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{40 \sqrt {d}}+\frac {b \left (c^2 d-e\right )^{5/2} \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{5 d}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.34, size = 334, normalized size = 1.88 \begin {gather*} \frac {-\sqrt {d+e x^2} \left (8 a \left (d+e x^2\right )^2+b c d x \left (9 e x^2+d \left (2-4 c^2 x^2\right )\right )\right )-8 b \left (d+e x^2\right )^{5/2} \text {ArcTan}(c x)+b c \sqrt {d} \left (8 c^4 d^2-20 c^2 d e+15 e^2\right ) x^5 \log (x)-b c \sqrt {d} \left (8 c^4 d^2-20 c^2 d e+15 e^2\right ) x^5 \log \left (d+\sqrt {d} \sqrt {d+e x^2}\right )+4 b \left (c^2 d-e\right )^{5/2} x^5 \log \left (-\frac {20 c d \left (c d-i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \left (c^2 d-e\right )^{7/2} (i+c x)}\right )+4 b \left (c^2 d-e\right )^{5/2} x^5 \log \left (-\frac {20 c d \left (c d+i e x+\sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \left (c^2 d-e\right )^{7/2} (-i+c x)}\right )}{40 d x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(Sqrt[d + e*x^2]*(8*a*(d + e*x^2)^2 + b*c*d*x*(9*e*x^2 + d*(2 - 4*c^2*x^2)))) - 8*b*(d + e*x^2)^(5/2)*ArcTan
[c*x] + b*c*Sqrt[d]*(8*c^4*d^2 - 20*c^2*d*e + 15*e^2)*x^5*Log[x] - b*c*Sqrt[d]*(8*c^4*d^2 - 20*c^2*d*e + 15*e^
2)*x^5*Log[d + Sqrt[d]*Sqrt[d + e*x^2]] + 4*b*(c^2*d - e)^(5/2)*x^5*Log[(-20*c*d*(c*d - I*e*x + Sqrt[c^2*d - e
]*Sqrt[d + e*x^2]))/(b*(c^2*d - e)^(7/2)*(I + c*x))] + 4*b*(c^2*d - e)^(5/2)*x^5*Log[(-20*c*d*(c*d + I*e*x + S
qrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(c^2*d - e)^(7/2)*(-I + c*x))])/(40*d*x^5)

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Maple [F]
time = 0.13, size = 0, normalized size = 0.00 \[\int \frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \arctan \left (c x \right )\right )}{x^{6}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(c^2*d-%e>0)', see `assume?` fo
r more detai

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Fricas [A]
time = 2.58, size = 1227, normalized size = 6.89 \begin {gather*} \left [\frac {4 \, {\left (b c^{4} d^{2} x^{5} - 2 \, b c^{2} d x^{5} e + b x^{5} e^{2}\right )} \sqrt {c^{2} d - e} \log \left (\frac {8 \, c^{4} d^{2} + 4 \, {\left (2 \, c^{3} d + {\left (c^{3} x^{2} - c\right )} e\right )} \sqrt {c^{2} d - e} \sqrt {x^{2} e + d} + {\left (c^{4} x^{4} - 6 \, c^{2} x^{2} + 1\right )} e^{2} + 8 \, {\left (c^{4} d x^{2} - c^{2} d\right )} e}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) + {\left (8 \, b c^{5} d^{2} x^{5} - 20 \, b c^{3} d x^{5} e + 15 \, b c x^{5} e^{2}\right )} \sqrt {d} \log \left (-\frac {x^{2} e - 2 \, \sqrt {x^{2} e + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (4 \, b c^{3} d^{2} x^{3} - 8 \, a x^{4} e^{2} - 2 \, b c d^{2} x - 8 \, a d^{2} - 8 \, {\left (b x^{4} e^{2} + 2 \, b d x^{2} e + b d^{2}\right )} \arctan \left (c x\right ) - {\left (9 \, b c d x^{3} + 16 \, a d x^{2}\right )} e\right )} \sqrt {x^{2} e + d}}{80 \, d x^{5}}, \frac {8 \, {\left (b c^{4} d^{2} x^{5} - 2 \, b c^{2} d x^{5} e + b x^{5} e^{2}\right )} \sqrt {-c^{2} d + e} \arctan \left (-\frac {{\left (2 \, c^{2} d + {\left (c^{2} x^{2} - 1\right )} e\right )} \sqrt {-c^{2} d + e} \sqrt {x^{2} e + d}}{2 \, {\left (c^{3} d^{2} - c x^{2} e^{2} + {\left (c^{3} d x^{2} - c d\right )} e\right )}}\right ) + {\left (8 \, b c^{5} d^{2} x^{5} - 20 \, b c^{3} d x^{5} e + 15 \, b c x^{5} e^{2}\right )} \sqrt {d} \log \left (-\frac {x^{2} e - 2 \, \sqrt {x^{2} e + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (4 \, b c^{3} d^{2} x^{3} - 8 \, a x^{4} e^{2} - 2 \, b c d^{2} x - 8 \, a d^{2} - 8 \, {\left (b x^{4} e^{2} + 2 \, b d x^{2} e + b d^{2}\right )} \arctan \left (c x\right ) - {\left (9 \, b c d x^{3} + 16 \, a d x^{2}\right )} e\right )} \sqrt {x^{2} e + d}}{80 \, d x^{5}}, \frac {{\left (8 \, b c^{5} d^{2} x^{5} - 20 \, b c^{3} d x^{5} e + 15 \, b c x^{5} e^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) + 2 \, {\left (b c^{4} d^{2} x^{5} - 2 \, b c^{2} d x^{5} e + b x^{5} e^{2}\right )} \sqrt {c^{2} d - e} \log \left (\frac {8 \, c^{4} d^{2} + 4 \, {\left (2 \, c^{3} d + {\left (c^{3} x^{2} - c\right )} e\right )} \sqrt {c^{2} d - e} \sqrt {x^{2} e + d} + {\left (c^{4} x^{4} - 6 \, c^{2} x^{2} + 1\right )} e^{2} + 8 \, {\left (c^{4} d x^{2} - c^{2} d\right )} e}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) + {\left (4 \, b c^{3} d^{2} x^{3} - 8 \, a x^{4} e^{2} - 2 \, b c d^{2} x - 8 \, a d^{2} - 8 \, {\left (b x^{4} e^{2} + 2 \, b d x^{2} e + b d^{2}\right )} \arctan \left (c x\right ) - {\left (9 \, b c d x^{3} + 16 \, a d x^{2}\right )} e\right )} \sqrt {x^{2} e + d}}{40 \, d x^{5}}, \frac {4 \, {\left (b c^{4} d^{2} x^{5} - 2 \, b c^{2} d x^{5} e + b x^{5} e^{2}\right )} \sqrt {-c^{2} d + e} \arctan \left (-\frac {{\left (2 \, c^{2} d + {\left (c^{2} x^{2} - 1\right )} e\right )} \sqrt {-c^{2} d + e} \sqrt {x^{2} e + d}}{2 \, {\left (c^{3} d^{2} - c x^{2} e^{2} + {\left (c^{3} d x^{2} - c d\right )} e\right )}}\right ) + {\left (8 \, b c^{5} d^{2} x^{5} - 20 \, b c^{3} d x^{5} e + 15 \, b c x^{5} e^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d}}{\sqrt {x^{2} e + d}}\right ) + {\left (4 \, b c^{3} d^{2} x^{3} - 8 \, a x^{4} e^{2} - 2 \, b c d^{2} x - 8 \, a d^{2} - 8 \, {\left (b x^{4} e^{2} + 2 \, b d x^{2} e + b d^{2}\right )} \arctan \left (c x\right ) - {\left (9 \, b c d x^{3} + 16 \, a d x^{2}\right )} e\right )} \sqrt {x^{2} e + d}}{40 \, d x^{5}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/80*(4*(b*c^4*d^2*x^5 - 2*b*c^2*d*x^5*e + b*x^5*e^2)*sqrt(c^2*d - e)*log((8*c^4*d^2 + 4*(2*c^3*d + (c^3*x^2
- c)*e)*sqrt(c^2*d - e)*sqrt(x^2*e + d) + (c^4*x^4 - 6*c^2*x^2 + 1)*e^2 + 8*(c^4*d*x^2 - c^2*d)*e)/(c^4*x^4 +
2*c^2*x^2 + 1)) + (8*b*c^5*d^2*x^5 - 20*b*c^3*d*x^5*e + 15*b*c*x^5*e^2)*sqrt(d)*log(-(x^2*e - 2*sqrt(x^2*e + d
)*sqrt(d) + 2*d)/x^2) + 2*(4*b*c^3*d^2*x^3 - 8*a*x^4*e^2 - 2*b*c*d^2*x - 8*a*d^2 - 8*(b*x^4*e^2 + 2*b*d*x^2*e
+ b*d^2)*arctan(c*x) - (9*b*c*d*x^3 + 16*a*d*x^2)*e)*sqrt(x^2*e + d))/(d*x^5), 1/80*(8*(b*c^4*d^2*x^5 - 2*b*c^
2*d*x^5*e + b*x^5*e^2)*sqrt(-c^2*d + e)*arctan(-1/2*(2*c^2*d + (c^2*x^2 - 1)*e)*sqrt(-c^2*d + e)*sqrt(x^2*e +
d)/(c^3*d^2 - c*x^2*e^2 + (c^3*d*x^2 - c*d)*e)) + (8*b*c^5*d^2*x^5 - 20*b*c^3*d*x^5*e + 15*b*c*x^5*e^2)*sqrt(d
)*log(-(x^2*e - 2*sqrt(x^2*e + d)*sqrt(d) + 2*d)/x^2) + 2*(4*b*c^3*d^2*x^3 - 8*a*x^4*e^2 - 2*b*c*d^2*x - 8*a*d
^2 - 8*(b*x^4*e^2 + 2*b*d*x^2*e + b*d^2)*arctan(c*x) - (9*b*c*d*x^3 + 16*a*d*x^2)*e)*sqrt(x^2*e + d))/(d*x^5),
1/40*((8*b*c^5*d^2*x^5 - 20*b*c^3*d*x^5*e + 15*b*c*x^5*e^2)*sqrt(-d)*arctan(sqrt(-d)/sqrt(x^2*e + d)) + 2*(b*
c^4*d^2*x^5 - 2*b*c^2*d*x^5*e + b*x^5*e^2)*sqrt(c^2*d - e)*log((8*c^4*d^2 + 4*(2*c^3*d + (c^3*x^2 - c)*e)*sqrt
(c^2*d - e)*sqrt(x^2*e + d) + (c^4*x^4 - 6*c^2*x^2 + 1)*e^2 + 8*(c^4*d*x^2 - c^2*d)*e)/(c^4*x^4 + 2*c^2*x^2 +
1)) + (4*b*c^3*d^2*x^3 - 8*a*x^4*e^2 - 2*b*c*d^2*x - 8*a*d^2 - 8*(b*x^4*e^2 + 2*b*d*x^2*e + b*d^2)*arctan(c*x)
- (9*b*c*d*x^3 + 16*a*d*x^2)*e)*sqrt(x^2*e + d))/(d*x^5), 1/40*(4*(b*c^4*d^2*x^5 - 2*b*c^2*d*x^5*e + b*x^5*e^
2)*sqrt(-c^2*d + e)*arctan(-1/2*(2*c^2*d + (c^2*x^2 - 1)*e)*sqrt(-c^2*d + e)*sqrt(x^2*e + d)/(c^3*d^2 - c*x^2*
e^2 + (c^3*d*x^2 - c*d)*e)) + (8*b*c^5*d^2*x^5 - 20*b*c^3*d*x^5*e + 15*b*c*x^5*e^2)*sqrt(-d)*arctan(sqrt(-d)/s
qrt(x^2*e + d)) + (4*b*c^3*d^2*x^3 - 8*a*x^4*e^2 - 2*b*c*d^2*x - 8*a*d^2 - 8*(b*x^4*e^2 + 2*b*d*x^2*e + b*d^2)
*arctan(c*x) - (9*b*c*d*x^3 + 16*a*d*x^2)*e)*sqrt(x^2*e + d))/(d*x^5)]

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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 )^{\frac {3}{2}}}{x^{6}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {atan}\left (c\,x\right )\right )\,{\left (e\,x^2+d\right )}^{3/2}}{x^6} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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