3.1223 \(\int \frac {a+b \tan ^{-1}(c x)}{x^2 (d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=274 \[ -\frac {8 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {4 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \tan ^{-1}(c x)}{d x \left (d+e x^2\right )^{3/2}}-\frac {b \left (3 c^4 d^2-12 c^2 d e+8 e^2\right )}{3 c d^3 \left (c^2 d-e\right ) \sqrt {d+e x^2}}+\frac {b \left (3 c^4 d^2-12 c^2 d e+8 e^2\right ) \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d^3 \left (c^2 d-e\right )^{3/2}}-\frac {b c \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}-\frac {8 b e}{3 c d^3 \sqrt {d+e x^2}}+\frac {b c}{d^2 \sqrt {d+e x^2}} \]

[Out]

(-a-b*arctan(c*x))/d/x/(e*x^2+d)^(3/2)-4/3*e*x*(a+b*arctan(c*x))/d^2/(e*x^2+d)^(3/2)-b*c*arctanh((e*x^2+d)^(1/
2)/d^(1/2))/d^(5/2)+1/3*b*(3*c^4*d^2-12*c^2*d*e+8*e^2)*arctanh(c*(e*x^2+d)^(1/2)/(c^2*d-e)^(1/2))/d^3/(c^2*d-e
)^(3/2)+b*c/d^2/(e*x^2+d)^(1/2)-8/3*b*e/c/d^3/(e*x^2+d)^(1/2)-1/3*b*(3*c^4*d^2-12*c^2*d*e+8*e^2)/c/d^3/(c^2*d-
e)/(e*x^2+d)^(1/2)-8/3*e*x*(a+b*arctan(c*x))/d^3/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.93, antiderivative size = 274, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 12, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {271, 192, 191, 4976, 12, 6725, 266, 51, 63, 208, 261, 444} \[ -\frac {8 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {4 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {a+b \tan ^{-1}(c x)}{d x \left (d+e x^2\right )^{3/2}}-\frac {b \left (3 c^4 d^2-12 c^2 d e+8 e^2\right )}{3 c d^3 \left (c^2 d-e\right ) \sqrt {d+e x^2}}+\frac {b \left (3 c^4 d^2-12 c^2 d e+8 e^2\right ) \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d^3 \left (c^2 d-e\right )^{3/2}}+\frac {b c}{d^2 \sqrt {d+e x^2}}-\frac {8 b e}{3 c d^3 \sqrt {d+e x^2}}-\frac {b c \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c*x])/(x^2*(d + e*x^2)^(5/2)),x]

[Out]

(b*c)/(d^2*Sqrt[d + e*x^2]) - (8*b*e)/(3*c*d^3*Sqrt[d + e*x^2]) - (b*(3*c^4*d^2 - 12*c^2*d*e + 8*e^2))/(3*c*d^
3*(c^2*d - e)*Sqrt[d + e*x^2]) - (a + b*ArcTan[c*x])/(d*x*(d + e*x^2)^(3/2)) - (4*e*x*(a + b*ArcTan[c*x]))/(3*
d^2*(d + e*x^2)^(3/2)) - (8*e*x*(a + b*ArcTan[c*x]))/(3*d^3*Sqrt[d + e*x^2]) - (b*c*ArcTanh[Sqrt[d + e*x^2]/Sq
rt[d]])/d^(5/2) + (b*(3*c^4*d^2 - 12*c^2*d*e + 8*e^2)*ArcTanh[(c*Sqrt[d + e*x^2])/Sqrt[c^2*d - e]])/(3*d^3*(c^
2*d - e)^(3/2))

Rule 12

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

Rule 51

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

Rule 63

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 191

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

Rule 192

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

Rule 208

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

Rule 261

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

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 271

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

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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]))

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {a+b \tan ^{-1}(c x)}{x^2 \left (d+e x^2\right )^{5/2}} \, dx &=-\frac {a+b \tan ^{-1}(c x)}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^3 \sqrt {d+e x^2}}-(b c) \int \frac {-3 d^2-12 d e x^2-8 e^2 x^4}{3 d^3 x \left (1+c^2 x^2\right ) \left (d+e x^2\right )^{3/2}} \, dx\\ &=-\frac {a+b \tan ^{-1}(c x)}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {(b c) \int \frac {-3 d^2-12 d e x^2-8 e^2 x^4}{x \left (1+c^2 x^2\right ) \left (d+e x^2\right )^{3/2}} \, dx}{3 d^3}\\ &=-\frac {a+b \tan ^{-1}(c x)}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {(b c) \int \left (-\frac {3 d^2}{x \left (d+e x^2\right )^{3/2}}-\frac {8 e^2 x}{c^2 \left (d+e x^2\right )^{3/2}}+\frac {\left (3 c^4 d^2-12 c^2 d e+8 e^2\right ) x}{c^2 \left (1+c^2 x^2\right ) \left (d+e x^2\right )^{3/2}}\right ) \, dx}{3 d^3}\\ &=-\frac {a+b \tan ^{-1}(c x)}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {(b c) \int \frac {1}{x \left (d+e x^2\right )^{3/2}} \, dx}{d}+\frac {\left (8 b e^2\right ) \int \frac {x}{\left (d+e x^2\right )^{3/2}} \, dx}{3 c d^3}-\frac {\left (b \left (3 c^4 d^2-12 c^2 d e+8 e^2\right )\right ) \int \frac {x}{\left (1+c^2 x^2\right ) \left (d+e x^2\right )^{3/2}} \, dx}{3 c d^3}\\ &=-\frac {8 b e}{3 c d^3 \sqrt {d+e x^2}}-\frac {a+b \tan ^{-1}(c x)}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x (d+e x)^{3/2}} \, dx,x,x^2\right )}{2 d}-\frac {\left (b \left (3 c^4 d^2-12 c^2 d e+8 e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+c^2 x\right ) (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 c d^3}\\ &=\frac {b c}{d^2 \sqrt {d+e x^2}}-\frac {8 b e}{3 c d^3 \sqrt {d+e x^2}}-\frac {b \left (3 c^4 d^2-12 c^2 d e+8 e^2\right )}{3 c d^3 \left (c^2 d-e\right ) \sqrt {d+e x^2}}-\frac {a+b \tan ^{-1}(c x)}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{2 d^2}-\frac {\left (b c \left (3 c^4 d^2-12 c^2 d e+8 e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+c^2 x\right ) \sqrt {d+e x}} \, dx,x,x^2\right )}{6 d^3 \left (c^2 d-e\right )}\\ &=\frac {b c}{d^2 \sqrt {d+e x^2}}-\frac {8 b e}{3 c d^3 \sqrt {d+e x^2}}-\frac {b \left (3 c^4 d^2-12 c^2 d e+8 e^2\right )}{3 c d^3 \left (c^2 d-e\right ) \sqrt {d+e x^2}}-\frac {a+b \tan ^{-1}(c x)}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^3 \sqrt {d+e x^2}}+\frac {(b c) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{d^2 e}-\frac {\left (b c \left (3 c^4 d^2-12 c^2 d e+8 e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {c^2 d}{e}+\frac {c^2 x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{3 d^3 \left (c^2 d-e\right ) e}\\ &=\frac {b c}{d^2 \sqrt {d+e x^2}}-\frac {8 b e}{3 c d^3 \sqrt {d+e x^2}}-\frac {b \left (3 c^4 d^2-12 c^2 d e+8 e^2\right )}{3 c d^3 \left (c^2 d-e\right ) \sqrt {d+e x^2}}-\frac {a+b \tan ^{-1}(c x)}{d x \left (d+e x^2\right )^{3/2}}-\frac {4 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \left (d+e x^2\right )^{3/2}}-\frac {8 e x \left (a+b \tan ^{-1}(c x)\right )}{3 d^3 \sqrt {d+e x^2}}-\frac {b c \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{d^{5/2}}+\frac {b \left (3 c^4 d^2-12 c^2 d e+8 e^2\right ) \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d^3 \left (c^2 d-e\right )^{3/2}}\\ \end {align*}

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Mathematica [C]  time = 1.18, size = 418, normalized size = 1.53 \[ \frac {\frac {2 e \left (5 a x \left (e-c^2 d\right )+b c d\right )}{\left (c^2 d-e\right ) \sqrt {d+e x^2}}-\frac {6 a \sqrt {d+e x^2}}{x}-\frac {2 a d e x}{\left (d+e x^2\right )^{3/2}}+\frac {b \left (3 c^4 d^2-12 c^2 d e+8 e^2\right ) \log \left (-\frac {12 c d^3 \sqrt {c^2 d-e} \left (\sqrt {c^2 d-e} \sqrt {d+e x^2}+c d-i e x\right )}{b (c x+i) \left (3 c^4 d^2-12 c^2 d e+8 e^2\right )}\right )}{\left (c^2 d-e\right )^{3/2}}+\frac {b \left (3 c^4 d^2-12 c^2 d e+8 e^2\right ) \log \left (-\frac {12 c d^3 \sqrt {c^2 d-e} \left (\sqrt {c^2 d-e} \sqrt {d+e x^2}+c d+i e x\right )}{b (c x-i) \left (3 c^4 d^2-12 c^2 d e+8 e^2\right )}\right )}{\left (c^2 d-e\right )^{3/2}}-\frac {2 b \tan ^{-1}(c x) \left (3 d^2+12 d e x^2+8 e^2 x^4\right )}{x \left (d+e x^2\right )^{3/2}}-6 b c \sqrt {d} \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right )+6 b c \sqrt {d} \log (x)}{6 d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTan[c*x])/(x^2*(d + e*x^2)^(5/2)),x]

[Out]

((-2*a*d*e*x)/(d + e*x^2)^(3/2) + (2*e*(b*c*d + 5*a*(-(c^2*d) + e)*x))/((c^2*d - e)*Sqrt[d + e*x^2]) - (6*a*Sq
rt[d + e*x^2])/x - (2*b*(3*d^2 + 12*d*e*x^2 + 8*e^2*x^4)*ArcTan[c*x])/(x*(d + e*x^2)^(3/2)) + 6*b*c*Sqrt[d]*Lo
g[x] - 6*b*c*Sqrt[d]*Log[d + Sqrt[d]*Sqrt[d + e*x^2]] + (b*(3*c^4*d^2 - 12*c^2*d*e + 8*e^2)*Log[(-12*c*d^3*Sqr
t[c^2*d - e]*(c*d - I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(3*c^4*d^2 - 12*c^2*d*e + 8*e^2)*(I + c*x))])
/(c^2*d - e)^(3/2) + (b*(3*c^4*d^2 - 12*c^2*d*e + 8*e^2)*Log[(-12*c*d^3*Sqrt[c^2*d - e]*(c*d + I*e*x + Sqrt[c^
2*d - e]*Sqrt[d + e*x^2]))/(b*(3*c^4*d^2 - 12*c^2*d*e + 8*e^2)*(-I + c*x))])/(c^2*d - e)^(3/2))/(6*d^3)

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fricas [B]  time = 2.14, size = 2714, normalized size = 9.91 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x^2/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*(((3*b*c^4*d^2*e^2 - 12*b*c^2*d*e^3 + 8*b*e^4)*x^5 + 2*(3*b*c^4*d^3*e - 12*b*c^2*d^2*e^2 + 8*b*d*e^3)*x
^3 + (3*b*c^4*d^4 - 12*b*c^2*d^3*e + 8*b*d^2*e^2)*x)*sqrt(c^2*d - e)*log((c^4*e^2*x^4 + 8*c^4*d^2 - 8*c^2*d*e
+ 2*(4*c^4*d*e - 3*c^2*e^2)*x^2 - 4*(c^3*e*x^2 + 2*c^3*d - c*e)*sqrt(c^2*d - e)*sqrt(e*x^2 + d) + e^2)/(c^4*x^
4 + 2*c^2*x^2 + 1)) - 6*((b*c^5*d^2*e^2 - 2*b*c^3*d*e^3 + b*c*e^4)*x^5 + 2*(b*c^5*d^3*e - 2*b*c^3*d^2*e^2 + b*
c*d*e^3)*x^3 + (b*c^5*d^4 - 2*b*c^3*d^3*e + b*c*d^2*e^2)*x)*sqrt(d)*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) +
2*d)/x^2) + 4*(3*a*c^4*d^4 - 6*a*c^2*d^3*e + 3*a*d^2*e^2 + 8*(a*c^4*d^2*e^2 - 2*a*c^2*d*e^3 + a*e^4)*x^4 - (b*
c^3*d^2*e^2 - b*c*d*e^3)*x^3 + 12*(a*c^4*d^3*e - 2*a*c^2*d^2*e^2 + a*d*e^3)*x^2 - (b*c^3*d^3*e - b*c*d^2*e^2)*
x + (3*b*c^4*d^4 - 6*b*c^2*d^3*e + 3*b*d^2*e^2 + 8*(b*c^4*d^2*e^2 - 2*b*c^2*d*e^3 + b*e^4)*x^4 + 12*(b*c^4*d^3
*e - 2*b*c^2*d^2*e^2 + b*d*e^3)*x^2)*arctan(c*x))*sqrt(e*x^2 + d))/((c^4*d^5*e^2 - 2*c^2*d^4*e^3 + d^3*e^4)*x^
5 + 2*(c^4*d^6*e - 2*c^2*d^5*e^2 + d^4*e^3)*x^3 + (c^4*d^7 - 2*c^2*d^6*e + d^5*e^2)*x), 1/6*(((3*b*c^4*d^2*e^2
- 12*b*c^2*d*e^3 + 8*b*e^4)*x^5 + 2*(3*b*c^4*d^3*e - 12*b*c^2*d^2*e^2 + 8*b*d*e^3)*x^3 + (3*b*c^4*d^4 - 12*b*
c^2*d^3*e + 8*b*d^2*e^2)*x)*sqrt(-c^2*d + e)*arctan(-1/2*(c^2*e*x^2 + 2*c^2*d - e)*sqrt(-c^2*d + e)*sqrt(e*x^2
+ d)/(c^3*d^2 - c*d*e + (c^3*d*e - c*e^2)*x^2)) + 3*((b*c^5*d^2*e^2 - 2*b*c^3*d*e^3 + b*c*e^4)*x^5 + 2*(b*c^5
*d^3*e - 2*b*c^3*d^2*e^2 + b*c*d*e^3)*x^3 + (b*c^5*d^4 - 2*b*c^3*d^3*e + b*c*d^2*e^2)*x)*sqrt(d)*log(-(e*x^2 -
2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) - 2*(3*a*c^4*d^4 - 6*a*c^2*d^3*e + 3*a*d^2*e^2 + 8*(a*c^4*d^2*e^2 - 2*a
*c^2*d*e^3 + a*e^4)*x^4 - (b*c^3*d^2*e^2 - b*c*d*e^3)*x^3 + 12*(a*c^4*d^3*e - 2*a*c^2*d^2*e^2 + a*d*e^3)*x^2 -
(b*c^3*d^3*e - b*c*d^2*e^2)*x + (3*b*c^4*d^4 - 6*b*c^2*d^3*e + 3*b*d^2*e^2 + 8*(b*c^4*d^2*e^2 - 2*b*c^2*d*e^3
+ b*e^4)*x^4 + 12*(b*c^4*d^3*e - 2*b*c^2*d^2*e^2 + b*d*e^3)*x^2)*arctan(c*x))*sqrt(e*x^2 + d))/((c^4*d^5*e^2
- 2*c^2*d^4*e^3 + d^3*e^4)*x^5 + 2*(c^4*d^6*e - 2*c^2*d^5*e^2 + d^4*e^3)*x^3 + (c^4*d^7 - 2*c^2*d^6*e + d^5*e^
2)*x), 1/12*(12*((b*c^5*d^2*e^2 - 2*b*c^3*d*e^3 + b*c*e^4)*x^5 + 2*(b*c^5*d^3*e - 2*b*c^3*d^2*e^2 + b*c*d*e^3)
*x^3 + (b*c^5*d^4 - 2*b*c^3*d^3*e + b*c*d^2*e^2)*x)*sqrt(-d)*arctan(sqrt(-d)/sqrt(e*x^2 + d)) - ((3*b*c^4*d^2*
e^2 - 12*b*c^2*d*e^3 + 8*b*e^4)*x^5 + 2*(3*b*c^4*d^3*e - 12*b*c^2*d^2*e^2 + 8*b*d*e^3)*x^3 + (3*b*c^4*d^4 - 12
*b*c^2*d^3*e + 8*b*d^2*e^2)*x)*sqrt(c^2*d - e)*log((c^4*e^2*x^4 + 8*c^4*d^2 - 8*c^2*d*e + 2*(4*c^4*d*e - 3*c^2
*e^2)*x^2 - 4*(c^3*e*x^2 + 2*c^3*d - c*e)*sqrt(c^2*d - e)*sqrt(e*x^2 + d) + e^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) -
4*(3*a*c^4*d^4 - 6*a*c^2*d^3*e + 3*a*d^2*e^2 + 8*(a*c^4*d^2*e^2 - 2*a*c^2*d*e^3 + a*e^4)*x^4 - (b*c^3*d^2*e^2
- b*c*d*e^3)*x^3 + 12*(a*c^4*d^3*e - 2*a*c^2*d^2*e^2 + a*d*e^3)*x^2 - (b*c^3*d^3*e - b*c*d^2*e^2)*x + (3*b*c^4
*d^4 - 6*b*c^2*d^3*e + 3*b*d^2*e^2 + 8*(b*c^4*d^2*e^2 - 2*b*c^2*d*e^3 + b*e^4)*x^4 + 12*(b*c^4*d^3*e - 2*b*c^2
*d^2*e^2 + b*d*e^3)*x^2)*arctan(c*x))*sqrt(e*x^2 + d))/((c^4*d^5*e^2 - 2*c^2*d^4*e^3 + d^3*e^4)*x^5 + 2*(c^4*d
^6*e - 2*c^2*d^5*e^2 + d^4*e^3)*x^3 + (c^4*d^7 - 2*c^2*d^6*e + d^5*e^2)*x), 1/6*(((3*b*c^4*d^2*e^2 - 12*b*c^2*
d*e^3 + 8*b*e^4)*x^5 + 2*(3*b*c^4*d^3*e - 12*b*c^2*d^2*e^2 + 8*b*d*e^3)*x^3 + (3*b*c^4*d^4 - 12*b*c^2*d^3*e +
8*b*d^2*e^2)*x)*sqrt(-c^2*d + e)*arctan(-1/2*(c^2*e*x^2 + 2*c^2*d - e)*sqrt(-c^2*d + e)*sqrt(e*x^2 + d)/(c^3*d
^2 - c*d*e + (c^3*d*e - c*e^2)*x^2)) + 6*((b*c^5*d^2*e^2 - 2*b*c^3*d*e^3 + b*c*e^4)*x^5 + 2*(b*c^5*d^3*e - 2*b
*c^3*d^2*e^2 + b*c*d*e^3)*x^3 + (b*c^5*d^4 - 2*b*c^3*d^3*e + b*c*d^2*e^2)*x)*sqrt(-d)*arctan(sqrt(-d)/sqrt(e*x
^2 + d)) - 2*(3*a*c^4*d^4 - 6*a*c^2*d^3*e + 3*a*d^2*e^2 + 8*(a*c^4*d^2*e^2 - 2*a*c^2*d*e^3 + a*e^4)*x^4 - (b*c
^3*d^2*e^2 - b*c*d*e^3)*x^3 + 12*(a*c^4*d^3*e - 2*a*c^2*d^2*e^2 + a*d*e^3)*x^2 - (b*c^3*d^3*e - b*c*d^2*e^2)*x
+ (3*b*c^4*d^4 - 6*b*c^2*d^3*e + 3*b*d^2*e^2 + 8*(b*c^4*d^2*e^2 - 2*b*c^2*d*e^3 + b*e^4)*x^4 + 12*(b*c^4*d^3*
e - 2*b*c^2*d^2*e^2 + b*d*e^3)*x^2)*arctan(c*x))*sqrt(e*x^2 + d))/((c^4*d^5*e^2 - 2*c^2*d^4*e^3 + d^3*e^4)*x^5
+ 2*(c^4*d^6*e - 2*c^2*d^5*e^2 + d^4*e^3)*x^3 + (c^4*d^7 - 2*c^2*d^6*e + d^5*e^2)*x)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x^2/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [F]  time = 1.04, size = 0, normalized size = 0.00 \[ \int \frac {a +b \arctan \left (c x \right )}{x^{2} \left (e \,x^{2}+d \right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(c*x))/x^2/(e*x^2+d)^(5/2),x)

[Out]

int((a+b*arctan(c*x))/x^2/(e*x^2+d)^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {1}{3} \, a {\left (\frac {8 \, e x}{\sqrt {e x^{2} + d} d^{3}} + \frac {4 \, e x}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2}} + \frac {3}{{\left (e x^{2} + d\right )}^{\frac {3}{2}} d x}\right )} + 2 \, b \int \frac {\arctan \left (c x\right )}{2 \, {\left (e^{2} x^{6} + 2 \, d e x^{4} + d^{2} x^{2}\right )} \sqrt {e x^{2} + d}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x^2/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

-1/3*a*(8*e*x/(sqrt(e*x^2 + d)*d^3) + 4*e*x/((e*x^2 + d)^(3/2)*d^2) + 3/((e*x^2 + d)^(3/2)*d*x)) + 2*b*integra
te(1/2*arctan(c*x)/((e^2*x^6 + 2*d*e*x^4 + d^2*x^2)*sqrt(e*x^2 + d)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x^2\,{\left (e\,x^2+d\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atan(c*x))/(x^2*(d + e*x^2)^(5/2)),x)

[Out]

int((a + b*atan(c*x))/(x^2*(d + e*x^2)^(5/2)), x)

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sympy [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((a+b*atan(c*x))/x**2/(e*x**2+d)**(5/2),x)

[Out]

Timed out

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