3.3.23 \(\int \frac {(c+a^2 c x^2)^{5/2} \text {ArcTan}(a x)}{x^4} \, dx\) [223]

Optimal. Leaf size=372 \[ -\frac {1}{2} a^3 c^2 \sqrt {c+a^2 c x^2}-\frac {a c^2 \sqrt {c+a^2 c x^2}}{6 x^2}-\frac {2 a^2 c^2 \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)}{x}+\frac {1}{2} a^4 c^2 x \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)-\frac {c \left (c+a^2 c x^2\right )^{3/2} \text {ArcTan}(a x)}{3 x^3}-\frac {5 i a^3 c^3 \sqrt {1+a^2 x^2} \text {ArcTan}(a x) \text {ArcTan}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {13}{6} a^3 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )+\frac {5 i a^3 c^3 \sqrt {1+a^2 x^2} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}}-\frac {5 i a^3 c^3 \sqrt {1+a^2 x^2} \text {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}} \]

[Out]

-1/3*c*(a^2*c*x^2+c)^(3/2)*arctan(a*x)/x^3-13/6*a^3*c^(5/2)*arctanh((a^2*c*x^2+c)^(1/2)/c^(1/2))-5*I*a^3*c^3*a
rctan(a*x)*arctan((1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)+5/2*I*a^3*c^3*polylog
(2,-I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)-5/2*I*a^3*c^3*polylog(2,I*(1+I*a*
x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/(a^2*c*x^2+c)^(1/2)-1/2*a^3*c^2*(a^2*c*x^2+c)^(1/2)-1/6*a*c^2*(a^2
*c*x^2+c)^(1/2)/x^2-2*a^2*c^2*arctan(a*x)*(a^2*c*x^2+c)^(1/2)/x+1/2*a^4*c^2*x*arctan(a*x)*(a^2*c*x^2+c)^(1/2)

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Rubi [A]
time = 0.70, antiderivative size = 372, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 9, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.409, Rules used = {5070, 5064, 272, 43, 65, 214, 5010, 5006, 4998} \begin {gather*} -\frac {2 a^2 c^2 \text {ArcTan}(a x) \sqrt {a^2 c x^2+c}}{x}-\frac {c \text {ArcTan}(a x) \left (a^2 c x^2+c\right )^{3/2}}{3 x^3}-\frac {a c^2 \sqrt {a^2 c x^2+c}}{6 x^2}+\frac {1}{2} a^4 c^2 x \text {ArcTan}(a x) \sqrt {a^2 c x^2+c}-\frac {5 i a^3 c^3 \sqrt {a^2 x^2+1} \text {ArcTan}(a x) \text {ArcTan}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {13}{6} a^3 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {a^2 c x^2+c}}{\sqrt {c}}\right )+\frac {5 i a^3 c^3 \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{2 \sqrt {a^2 c x^2+c}}-\frac {5 i a^3 c^3 \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{2 \sqrt {a^2 c x^2+c}}-\frac {1}{2} a^3 c^2 \sqrt {a^2 c x^2+c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((c + a^2*c*x^2)^(5/2)*ArcTan[a*x])/x^4,x]

[Out]

-1/2*(a^3*c^2*Sqrt[c + a^2*c*x^2]) - (a*c^2*Sqrt[c + a^2*c*x^2])/(6*x^2) - (2*a^2*c^2*Sqrt[c + a^2*c*x^2]*ArcT
an[a*x])/x + (a^4*c^2*x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/2 - (c*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/(3*x^3) - (
(5*I)*a^3*c^3*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/Sqrt[c + a^2*c*x^2] - (13
*a^3*c^(5/2)*ArcTanh[Sqrt[c + a^2*c*x^2]/Sqrt[c]])/6 + (((5*I)/2)*a^3*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, ((-I)*S
qrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/Sqrt[c + a^2*c*x^2] - (((5*I)/2)*a^3*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqr
t[1 + I*a*x])/Sqrt[1 - I*a*x]])/Sqrt[c + a^2*c*x^2]

Rule 43

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

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 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 272

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 4998

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp[(-b)*((d + e*x^2)^q/(2*c
*q*(2*q + 1))), x] + (Dist[2*d*(q/(2*q + 1)), Int[(d + e*x^2)^(q - 1)*(a + b*ArcTan[c*x]), x], x] + Simp[x*(d
+ e*x^2)^q*((a + b*ArcTan[c*x])/(2*q + 1)), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[q, 0]

Rule 5006

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[-2*I*(a + b*ArcTan[c*x])*(
ArcTan[Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x]]/(c*Sqrt[d])), x] + (Simp[I*b*(PolyLog[2, (-I)*(Sqrt[1 + I*c*x]/Sqrt[1
- I*c*x])]/(c*Sqrt[d])), x] - Simp[I*b*(PolyLog[2, I*(Sqrt[1 + I*c*x]/Sqrt[1 - I*c*x])]/(c*Sqrt[d])), x]) /; F
reeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[d, 0]

Rule 5010

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + c^2*x^2]/Sq
rt[d + e*x^2], Int[(a + b*ArcTan[c*x])^p/Sqrt[1 + c^2*x^2], x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*
d] && IGtQ[p, 0] &&  !GtQ[d, 0]

Rule 5064

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Simp
[(f*x)^(m + 1)*(d + e*x^2)^(q + 1)*((a + b*ArcTan[c*x])^p/(d*f*(m + 1))), x] - Dist[b*c*(p/(f*(m + 1))), Int[(
f*x)^(m + 1)*(d + e*x^2)^q*(a + b*ArcTan[c*x])^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && EqQ[e,
c^2*d] && EqQ[m + 2*q + 3, 0] && GtQ[p, 0] && NeQ[m, -1]

Rule 5070

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

Rubi steps

\begin {align*} \int \frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{x^4} \, dx &=c \int \frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{x^4} \, dx+\left (a^2 c\right ) \int \frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{x^2} \, dx\\ &=c^2 \int \frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{x^4} \, dx+2 \left (\left (a^2 c^2\right ) \int \frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{x^2} \, dx\right )+\left (a^4 c^2\right ) \int \sqrt {c+a^2 c x^2} \tan ^{-1}(a x) \, dx\\ &=-\frac {1}{2} a^3 c^2 \sqrt {c+a^2 c x^2}+\frac {1}{2} a^4 c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)-\frac {c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 x^3}+\frac {1}{3} \left (a c^2\right ) \int \frac {\sqrt {c+a^2 c x^2}}{x^3} \, dx+\frac {1}{2} \left (a^4 c^3\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx+2 \left (\left (a^2 c^3\right ) \int \frac {\tan ^{-1}(a x)}{x^2 \sqrt {c+a^2 c x^2}} \, dx+\left (a^4 c^3\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx\right )\\ &=-\frac {1}{2} a^3 c^2 \sqrt {c+a^2 c x^2}+\frac {1}{2} a^4 c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)-\frac {c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 x^3}+\frac {1}{6} \left (a c^2\right ) \text {Subst}\left (\int \frac {\sqrt {c+a^2 c x}}{x^2} \, dx,x,x^2\right )+\frac {\left (a^4 c^3 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{2 \sqrt {c+a^2 c x^2}}+2 \left (-\frac {a^2 c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{x}+\left (a^3 c^3\right ) \int \frac {1}{x \sqrt {c+a^2 c x^2}} \, dx+\frac {\left (a^4 c^3 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\right )\\ &=-\frac {1}{2} a^3 c^2 \sqrt {c+a^2 c x^2}-\frac {a c^2 \sqrt {c+a^2 c x^2}}{6 x^2}+\frac {1}{2} a^4 c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)-\frac {c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 x^3}-\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}}-\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}}+\frac {1}{12} \left (a^3 c^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+a^2 c x}} \, dx,x,x^2\right )+2 \left (-\frac {a^2 c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{x}-\frac {2 i a^3 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {1}{2} \left (a^3 c^3\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {c+a^2 c x}} \, dx,x,x^2\right )\right )\\ &=-\frac {1}{2} a^3 c^2 \sqrt {c+a^2 c x^2}-\frac {a c^2 \sqrt {c+a^2 c x^2}}{6 x^2}+\frac {1}{2} a^4 c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)-\frac {c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 x^3}-\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}}-\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}}+\frac {1}{6} \left (a c^2\right ) \text {Subst}\left (\int \frac {1}{-\frac {1}{a^2}+\frac {x^2}{a^2 c}} \, dx,x,\sqrt {c+a^2 c x^2}\right )+2 \left (-\frac {a^2 c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{x}-\frac {2 i a^3 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}+\left (a c^2\right ) \text {Subst}\left (\int \frac {1}{-\frac {1}{a^2}+\frac {x^2}{a^2 c}} \, dx,x,\sqrt {c+a^2 c x^2}\right )\right )\\ &=-\frac {1}{2} a^3 c^2 \sqrt {c+a^2 c x^2}-\frac {a c^2 \sqrt {c+a^2 c x^2}}{6 x^2}+\frac {1}{2} a^4 c^2 x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)-\frac {c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{3 x^3}-\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {1}{6} a^3 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )+\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}}-\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{2 \sqrt {c+a^2 c x^2}}+2 \left (-\frac {a^2 c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{x}-\frac {2 i a^3 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-a^3 c^{5/2} \tanh ^{-1}\left (\frac {\sqrt {c+a^2 c x^2}}{\sqrt {c}}\right )+\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {i a^3 c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.75, size = 313, normalized size = 0.84 \begin {gather*} \frac {c^2 \sqrt {c+a^2 c x^2} \left (-a x \sqrt {1+a^2 x^2}-3 a^3 x^3 \sqrt {1+a^2 x^2}-2 \sqrt {1+a^2 x^2} \text {ArcTan}(a x)-14 a^2 x^2 \sqrt {1+a^2 x^2} \text {ArcTan}(a x)+3 a^4 x^4 \sqrt {1+a^2 x^2} \text {ArcTan}(a x)-a^3 x^3 \tanh ^{-1}\left (\sqrt {1+a^2 x^2}\right )+15 a^3 x^3 \text {ArcTan}(a x) \log \left (1-i e^{i \text {ArcTan}(a x)}\right )-15 a^3 x^3 \text {ArcTan}(a x) \log \left (1+i e^{i \text {ArcTan}(a x)}\right )-12 a^3 x^3 \log \left (\cos \left (\frac {1}{2} \text {ArcTan}(a x)\right )\right )+12 a^3 x^3 \log \left (\sin \left (\frac {1}{2} \text {ArcTan}(a x)\right )\right )+15 i a^3 x^3 \text {PolyLog}\left (2,-i e^{i \text {ArcTan}(a x)}\right )-15 i a^3 x^3 \text {PolyLog}\left (2,i e^{i \text {ArcTan}(a x)}\right )\right )}{6 x^3 \sqrt {1+a^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((c + a^2*c*x^2)^(5/2)*ArcTan[a*x])/x^4,x]

[Out]

(c^2*Sqrt[c + a^2*c*x^2]*(-(a*x*Sqrt[1 + a^2*x^2]) - 3*a^3*x^3*Sqrt[1 + a^2*x^2] - 2*Sqrt[1 + a^2*x^2]*ArcTan[
a*x] - 14*a^2*x^2*Sqrt[1 + a^2*x^2]*ArcTan[a*x] + 3*a^4*x^4*Sqrt[1 + a^2*x^2]*ArcTan[a*x] - a^3*x^3*ArcTanh[Sq
rt[1 + a^2*x^2]] + 15*a^3*x^3*ArcTan[a*x]*Log[1 - I*E^(I*ArcTan[a*x])] - 15*a^3*x^3*ArcTan[a*x]*Log[1 + I*E^(I
*ArcTan[a*x])] - 12*a^3*x^3*Log[Cos[ArcTan[a*x]/2]] + 12*a^3*x^3*Log[Sin[ArcTan[a*x]/2]] + (15*I)*a^3*x^3*Poly
Log[2, (-I)*E^(I*ArcTan[a*x])] - (15*I)*a^3*x^3*PolyLog[2, I*E^(I*ArcTan[a*x])]))/(6*x^3*Sqrt[1 + a^2*x^2])

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Maple [A]
time = 0.52, size = 270, normalized size = 0.73

method	result	size
default	\(\frac {c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (3 \arctan \left (a x \right ) a^{4} x^{4}-3 a^{3} x^{3}-14 \arctan \left (a x \right ) a^{2} x^{2}-a x -2 \arctan \left (a x \right )\right )}{6 x^{3}}-\frac {i a^{3} c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (15 i \arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-15 i \arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+13 i \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-1\right )-13 i \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+15 \dilog \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-15 \dilog \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{6 \sqrt {a^{2} x^{2}+1}}\)	\(270\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2*c*x^2+c)^(5/2)*arctan(a*x)/x^4,x,method=_RETURNVERBOSE)

[Out]

1/6*c^2*(c*(a*x-I)*(I+a*x))^(1/2)*(3*arctan(a*x)*a^4*x^4-3*a^3*x^3-14*arctan(a*x)*a^2*x^2-a*x-2*arctan(a*x))/x
^3-1/6*I*a^3*c^2*(c*(a*x-I)*(I+a*x))^(1/2)*(15*I*arctan(a*x)*ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-15*I*arctan(a
*x)*ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+13*I*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-1)-13*I*ln(1+(1+I*a*x)/(a^2*x^2+1)
^(1/2))+15*dilog(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-15*dilog(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2)))/(a^2*x^2+1)^(1/2)

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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((a^2*c*x^2+c)^(5/2)*arctan(a*x)/x^4,x, algorithm="maxima")

[Out]

integrate((a^2*c*x^2 + c)^(5/2)*arctan(a*x)/x^4, x)

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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((a^2*c*x^2+c)^(5/2)*arctan(a*x)/x^4,x, algorithm="fricas")

[Out]

integral((a^4*c^2*x^4 + 2*a^2*c^2*x^2 + c^2)*sqrt(a^2*c*x^2 + c)*arctan(a*x)/x^4, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {5}{2}} \operatorname {atan}{\left (a x \right )}}{x^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a**2*c*x**2+c)**(5/2)*atan(a*x)/x**4,x)

[Out]

Integral((c*(a**2*x**2 + 1))**(5/2)*atan(a*x)/x**4, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a^2*c*x^2+c)^(5/2)*arctan(a*x)/x^4,x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^{5/2}}{x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((atan(a*x)*(c + a^2*c*x^2)^(5/2))/x^4,x)

[Out]

int((atan(a*x)*(c + a^2*c*x^2)^(5/2))/x^4, x)

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