3.4.7 \(\int x^3 \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)^2 \, dx\) [307]

Optimal. Leaf size=385 \[ -\frac {11 \sqrt {c+a^2 c x^2}}{60 a^4}+\frac {\left (c+a^2 c x^2\right )^{3/2}}{30 a^4 c}+\frac {x \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)}{12 a^3}-\frac {x^3 \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)}{10 a}-\frac {2 \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)^2}{15 a^4}+\frac {x^2 \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)^2}{15 a^2}+\frac {1}{5} x^4 \sqrt {c+a^2 c x^2} \text {ArcTan}(a x)^2-\frac {11 i c \sqrt {1+a^2 x^2} \text {ArcTan}(a x) \text {ArcTan}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{30 a^4 \sqrt {c+a^2 c x^2}}+\frac {11 i c \sqrt {1+a^2 x^2} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{60 a^4 \sqrt {c+a^2 c x^2}}-\frac {11 i c \sqrt {1+a^2 x^2} \text {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{60 a^4 \sqrt {c+a^2 c x^2}} \]

[Out]

1/30*(a^2*c*x^2+c)^(3/2)/a^4/c-11/30*I*c*arctan(a*x)*arctan((1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)
/a^4/(a^2*c*x^2+c)^(1/2)+11/60*I*c*polylog(2,-I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a^4/(a^2*c*
x^2+c)^(1/2)-11/60*I*c*polylog(2,I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a^4/(a^2*c*x^2+c)^(1/2)-
11/60*(a^2*c*x^2+c)^(1/2)/a^4+1/12*x*arctan(a*x)*(a^2*c*x^2+c)^(1/2)/a^3-1/10*x^3*arctan(a*x)*(a^2*c*x^2+c)^(1
/2)/a-2/15*arctan(a*x)^2*(a^2*c*x^2+c)^(1/2)/a^4+1/15*x^2*arctan(a*x)^2*(a^2*c*x^2+c)^(1/2)/a^2+1/5*x^4*arctan
(a*x)^2*(a^2*c*x^2+c)^(1/2)

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Rubi [A]
time = 0.99, antiderivative size = 385, normalized size of antiderivative = 1.00, number of steps used = 26, number of rules used = 8, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5070, 5072, 267, 5010, 5006, 5050, 272, 45} \begin {gather*} \frac {x^2 \text {ArcTan}(a x)^2 \sqrt {a^2 c x^2+c}}{15 a^2}+\frac {1}{5} x^4 \text {ArcTan}(a x)^2 \sqrt {a^2 c x^2+c}-\frac {x^3 \text {ArcTan}(a x) \sqrt {a^2 c x^2+c}}{10 a}-\frac {2 \text {ArcTan}(a x)^2 \sqrt {a^2 c x^2+c}}{15 a^4}-\frac {11 i c \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 )}{30 a^4 \sqrt {a^2 c x^2+c}}+\frac {11 i c \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{60 a^4 \sqrt {a^2 c x^2+c}}-\frac {11 i c \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{60 a^4 \sqrt {a^2 c x^2+c}}+\frac {\left (a^2 c x^2+c\right )^{3/2}}{30 a^4 c}-\frac {11 \sqrt {a^2 c x^2+c}}{60 a^4}+\frac {x \text {ArcTan}(a x) \sqrt {a^2 c x^2+c}}{12 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2,x]

[Out]

(-11*Sqrt[c + a^2*c*x^2])/(60*a^4) + (c + a^2*c*x^2)^(3/2)/(30*a^4*c) + (x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(1
2*a^3) - (x^3*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(10*a) - (2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(15*a^4) + (x^2*
Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(15*a^2) + (x^4*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/5 - (((11*I)/30)*c*Sqrt[
1 + a^2*x^2]*ArcTan[a*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(a^4*Sqrt[c + a^2*c*x^2]) + (((11*I)/60)*c*S
qrt[1 + a^2*x^2]*PolyLog[2, ((-I)*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^4*Sqrt[c + a^2*c*x^2]) - (((11*I)/60)*
c*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^4*Sqrt[c + a^2*c*x^2])

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 267

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 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 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 5050

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

Rule 5072

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

Rubi steps

\begin {align*} \int x^3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2 \, dx &=c \int \frac {x^3 \tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx+\left (a^2 c\right ) \int \frac {x^5 \tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx\\ &=\frac {x^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 a^2}+\frac {1}{5} x^4 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2-\frac {1}{5} (4 c) \int \frac {x^3 \tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx-\frac {(2 c) \int \frac {x \tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx}{3 a^2}-\frac {(2 c) \int \frac {x^2 \tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{3 a}-\frac {1}{5} (2 a c) \int \frac {x^4 \tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 a^3}-\frac {x^3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{10 a}-\frac {2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 a^4}+\frac {x^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^2}+\frac {1}{5} x^4 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2+\frac {1}{10} c \int \frac {x^3}{\sqrt {c+a^2 c x^2}} \, dx+\frac {c \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{3 a^3}+\frac {(4 c) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{3 a^3}+\frac {c \int \frac {x}{\sqrt {c+a^2 c x^2}} \, dx}{3 a^2}+\frac {(8 c) \int \frac {x \tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx}{15 a^2}+\frac {(3 c) \int \frac {x^2 \tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{10 a}+\frac {(8 c) \int \frac {x^2 \tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{15 a}\\ &=\frac {\sqrt {c+a^2 c x^2}}{3 a^4}+\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{12 a^3}-\frac {x^3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{10 a}-\frac {2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^4}+\frac {x^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^2}+\frac {1}{5} x^4 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2+\frac {1}{20} c \text {Subst}\left (\int \frac {x}{\sqrt {c+a^2 c x}} \, dx,x,x^2\right )-\frac {(3 c) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{20 a^3}-\frac {(4 c) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{15 a^3}-\frac {(16 c) \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{15 a^3}-\frac {(3 c) \int \frac {x}{\sqrt {c+a^2 c x^2}} \, dx}{20 a^2}-\frac {(4 c) \int \frac {x}{\sqrt {c+a^2 c x^2}} \, dx}{15 a^2}+\frac {\left (c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{3 a^3 \sqrt {c+a^2 c x^2}}+\frac {\left (4 c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{3 a^3 \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2}}{12 a^4}+\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{12 a^3}-\frac {x^3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{10 a}-\frac {2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^4}+\frac {x^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^2}+\frac {1}{5} x^4 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2-\frac {10 i c \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 )}{3 a^4 \sqrt {c+a^2 c x^2}}+\frac {5 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{3 a^4 \sqrt {c+a^2 c x^2}}-\frac {5 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{3 a^4 \sqrt {c+a^2 c x^2}}+\frac {1}{20} c \text {Subst}\left (\int \left (-\frac {1}{a^2 \sqrt {c+a^2 c x}}+\frac {\sqrt {c+a^2 c x}}{a^2 c}\right ) \, dx,x,x^2\right )-\frac {\left (3 c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{20 a^3 \sqrt {c+a^2 c x^2}}-\frac {\left (4 c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{15 a^3 \sqrt {c+a^2 c x^2}}-\frac {\left (16 c \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{15 a^3 \sqrt {c+a^2 c x^2}}\\ &=-\frac {11 \sqrt {c+a^2 c x^2}}{60 a^4}+\frac {\left (c+a^2 c x^2\right )^{3/2}}{30 a^4 c}+\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{12 a^3}-\frac {x^3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{10 a}-\frac {2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^4}+\frac {x^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{15 a^2}+\frac {1}{5} x^4 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2-\frac {11 i c \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 )}{30 a^4 \sqrt {c+a^2 c x^2}}+\frac {11 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{60 a^4 \sqrt {c+a^2 c x^2}}-\frac {11 i c \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{60 a^4 \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.84, size = 360, normalized size = 0.94 \begin {gather*} -\frac {\left (1+a^2 x^2\right )^2 \sqrt {c \left (1+a^2 x^2\right )} \left (50-32 \text {ArcTan}(a x)^2+72 \cos (2 \text {ArcTan}(a x))+160 \text {ArcTan}(a x)^2 \cos (2 \text {ArcTan}(a x))+22 \cos (4 \text {ArcTan}(a x))-\frac {110 \text {ArcTan}(a x) \log \left (1-i e^{i \text {ArcTan}(a x)}\right )}{\sqrt {1+a^2 x^2}}-55 \text {ArcTan}(a x) \cos (3 \text {ArcTan}(a x)) \log \left (1-i e^{i \text {ArcTan}(a x)}\right )-11 \text {ArcTan}(a x) \cos (5 \text {ArcTan}(a x)) \log \left (1-i e^{i \text {ArcTan}(a x)}\right )+\frac {110 \text {ArcTan}(a x) \log \left (1+i e^{i \text {ArcTan}(a x)}\right )}{\sqrt {1+a^2 x^2}}+55 \text {ArcTan}(a x) \cos (3 \text {ArcTan}(a x)) \log \left (1+i e^{i \text {ArcTan}(a x)}\right )+11 \text {ArcTan}(a x) \cos (5 \text {ArcTan}(a x)) \log \left (1+i e^{i \text {ArcTan}(a x)}\right )-\frac {176 i \text {PolyLog}\left (2,-i e^{i \text {ArcTan}(a x)}\right )}{\left (1+a^2 x^2\right )^{5/2}}+\frac {176 i \text {PolyLog}\left (2,i e^{i \text {ArcTan}(a x)}\right )}{\left (1+a^2 x^2\right )^{5/2}}+4 \text {ArcTan}(a x) \sin (2 \text {ArcTan}(a x))-22 \text {ArcTan}(a x) \sin (4 \text {ArcTan}(a x))\right )}{960 a^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2,x]

[Out]

-1/960*((1 + a^2*x^2)^2*Sqrt[c*(1 + a^2*x^2)]*(50 - 32*ArcTan[a*x]^2 + 72*Cos[2*ArcTan[a*x]] + 160*ArcTan[a*x]
^2*Cos[2*ArcTan[a*x]] + 22*Cos[4*ArcTan[a*x]] - (110*ArcTan[a*x]*Log[1 - I*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^
2] - 55*ArcTan[a*x]*Cos[3*ArcTan[a*x]]*Log[1 - I*E^(I*ArcTan[a*x])] - 11*ArcTan[a*x]*Cos[5*ArcTan[a*x]]*Log[1
- I*E^(I*ArcTan[a*x])] + (110*ArcTan[a*x]*Log[1 + I*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] + 55*ArcTan[a*x]*Cos
[3*ArcTan[a*x]]*Log[1 + I*E^(I*ArcTan[a*x])] + 11*ArcTan[a*x]*Cos[5*ArcTan[a*x]]*Log[1 + I*E^(I*ArcTan[a*x])]
- ((176*I)*PolyLog[2, (-I)*E^(I*ArcTan[a*x])])/(1 + a^2*x^2)^(5/2) + ((176*I)*PolyLog[2, I*E^(I*ArcTan[a*x])])
/(1 + a^2*x^2)^(5/2) + 4*ArcTan[a*x]*Sin[2*ArcTan[a*x]] - 22*ArcTan[a*x]*Sin[4*ArcTan[a*x]]))/a^4

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Maple [A]
time = 1.71, size = 235, normalized size = 0.61

method	result	size
default	\(\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (12 \arctan \left (a x \right )^{2} a^{4} x^{4}-6 \arctan \left (a x \right ) a^{3} x^{3}+4 \arctan \left (a x \right )^{2} a^{2} x^{2}+2 a^{2} x^{2}+5 \arctan \left (a x \right ) a x -8 \arctan \left (a x \right )^{2}-9\right )}{60 a^{4}}-\frac {11 \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (\arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \dilog \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+i \dilog \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{60 a^{4} \sqrt {a^{2} x^{2}+1}}\)	\(235\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctan(a*x)^2*(a^2*c*x^2+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/60/a^4*(c*(a*x-I)*(I+a*x))^(1/2)*(12*arctan(a*x)^2*a^4*x^4-6*arctan(a*x)*a^3*x^3+4*arctan(a*x)^2*a^2*x^2+2*a
^2*x^2+5*arctan(a*x)*a*x-8*arctan(a*x)^2-9)-11/60*(c*(a*x-I)*(I+a*x))^(1/2)*(arctan(a*x)*ln(1+I*(1+I*a*x)/(a^2
*x^2+1)^(1/2))-arctan(a*x)*ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-I*dilog(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+I*dilo
g(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2)))/a^4/(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(x^3*arctan(a*x)^2*(a^2*c*x^2+c)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a^2*c*x^2 + c)*x^3*arctan(a*x)^2, 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(x^3*arctan(a*x)^2*(a^2*c*x^2+c)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(a^2*c*x^2 + c)*x^3*arctan(a*x)^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{3} \sqrt {c \left (a^{2} x^{2} + 1\right )} \operatorname {atan}^{2}{\left (a x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atan(a*x)**2*(a**2*c*x**2+c)**(1/2),x)

[Out]

Integral(x**3*sqrt(c*(a**2*x**2 + 1))*atan(a*x)**2, 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(x^3*arctan(a*x)^2*(a^2*c*x^2+c)^(1/2),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 x^3\,{\mathrm {atan}\left (a\,x\right )}^2\,\sqrt {c\,a^2\,x^2+c} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*atan(a*x)^2*(c + a^2*c*x^2)^(1/2),x)

[Out]

int(x^3*atan(a*x)^2*(c + a^2*c*x^2)^(1/2), x)

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