3.209 \(\int x^2 (c+a^2 c x^2)^{3/2} \tan ^{-1}(a x) \, dx\)

Optimal. Leaf size=357 \[ \frac {c x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{16 a^2}+\frac {1}{6} a^2 c x^5 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)+\frac {7}{24} c x^3 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)-\frac {i c^2 \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{16 a^3 \sqrt {a^2 c x^2+c}}+\frac {i c^2 \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{16 a^3 \sqrt {a^2 c x^2+c}}+\frac {i c^2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a^3 \sqrt {a^2 c x^2+c}}-\frac {\left (a^2 c x^2+c\right )^{5/2}}{30 a^3 c}+\frac {\left (a^2 c x^2+c\right )^{3/2}}{72 a^3}+\frac {c \sqrt {a^2 c x^2+c}}{16 a^3} \]

[Out]

1/72*(a^2*c*x^2+c)^(3/2)/a^3-1/30*(a^2*c*x^2+c)^(5/2)/a^3/c+1/8*I*c^2*arctan(a*x)*arctan((1+I*a*x)^(1/2)/(1-I*
a*x)^(1/2))*(a^2*x^2+1)^(1/2)/a^3/(a^2*c*x^2+c)^(1/2)-1/16*I*c^2*polylog(2,-I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))
*(a^2*x^2+1)^(1/2)/a^3/(a^2*c*x^2+c)^(1/2)+1/16*I*c^2*polylog(2,I*(1+I*a*x)^(1/2)/(1-I*a*x)^(1/2))*(a^2*x^2+1)
^(1/2)/a^3/(a^2*c*x^2+c)^(1/2)+1/16*c*(a^2*c*x^2+c)^(1/2)/a^3+1/16*c*x*arctan(a*x)*(a^2*c*x^2+c)^(1/2)/a^2+7/2
4*c*x^3*arctan(a*x)*(a^2*c*x^2+c)^(1/2)+1/6*a^2*c*x^5*arctan(a*x)*(a^2*c*x^2+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.78, antiderivative size = 357, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4950, 4946, 4952, 261, 4890, 4886, 266, 43} \[ -\frac {i c^2 \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{16 a^3 \sqrt {a^2 c x^2+c}}+\frac {i c^2 \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{16 a^3 \sqrt {a^2 c x^2+c}}+\frac {i c^2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a^3 \sqrt {a^2 c x^2+c}}-\frac {\left (a^2 c x^2+c\right )^{5/2}}{30 a^3 c}+\frac {\left (a^2 c x^2+c\right )^{3/2}}{72 a^3}+\frac {c \sqrt {a^2 c x^2+c}}{16 a^3}+\frac {1}{6} a^2 c x^5 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)+\frac {7}{24} c x^3 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)+\frac {c x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{16 a^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[c + a^2*c*x^2])/(16*a^3) + (c + a^2*c*x^2)^(3/2)/(72*a^3) - (c + a^2*c*x^2)^(5/2)/(30*a^3*c) + (c*x*Sq
rt[c + a^2*c*x^2]*ArcTan[a*x])/(16*a^2) + (7*c*x^3*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/24 + (a^2*c*x^5*Sqrt[c + a
^2*c*x^2]*ArcTan[a*x])/6 + ((I/8)*c^2*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*ArcTan[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/(
a^3*Sqrt[c + a^2*c*x^2]) - ((I/16)*c^2*Sqrt[1 + a^2*x^2]*PolyLog[2, ((-I)*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(
a^3*Sqrt[c + a^2*c*x^2]) + ((I/16)*c^2*Sqrt[1 + a^2*x^2]*PolyLog[2, (I*Sqrt[1 + I*a*x])/Sqrt[1 - I*a*x]])/(a^3
*Sqrt[c + a^2*c*x^2])

Rule 43

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 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 4886

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 4890

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 4946

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

Rule 4950

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 4952

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^2 \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x) \, dx &=c \int x^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x) \, dx+\left (a^2 c\right ) \int x^4 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x) \, dx\\ &=\frac {1}{4} c x^3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{6} a^2 c x^5 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{4} c^2 \int \frac {x^2 \tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx-\frac {1}{4} \left (a c^2\right ) \int \frac {x^3}{\sqrt {c+a^2 c x^2}} \, dx+\frac {1}{6} \left (a^2 c^2\right ) \int \frac {x^4 \tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx-\frac {1}{6} \left (a^3 c^2\right ) \int \frac {x^5}{\sqrt {c+a^2 c x^2}} \, dx\\ &=\frac {c x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{8 a^2}+\frac {7}{24} c x^3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{6} a^2 c x^5 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)-\frac {1}{8} c^2 \int \frac {x^2 \tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx-\frac {c^2 \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{8 a^2}-\frac {c^2 \int \frac {x}{\sqrt {c+a^2 c x^2}} \, dx}{8 a}-\frac {1}{24} \left (a c^2\right ) \int \frac {x^3}{\sqrt {c+a^2 c x^2}} \, dx-\frac {1}{8} \left (a c^2\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {c+a^2 c x}} \, dx,x,x^2\right )-\frac {1}{12} \left (a^3 c^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {c+a^2 c x}} \, dx,x,x^2\right )\\ &=-\frac {c \sqrt {c+a^2 c x^2}}{8 a^3}+\frac {c x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{16 a^2}+\frac {7}{24} c x^3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{6} a^2 c x^5 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {c^2 \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{16 a^2}+\frac {c^2 \int \frac {x}{\sqrt {c+a^2 c x^2}} \, dx}{16 a}-\frac {1}{48} \left (a c^2\right ) \operatorname {Subst}\left (\int \frac {x}{\sqrt {c+a^2 c x}} \, dx,x,x^2\right )-\frac {1}{8} \left (a c^2\right ) \operatorname {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 {1}{12} \left (a^3 c^2\right ) \operatorname {Subst}\left (\int \left (\frac {1}{a^4 \sqrt {c+a^2 c x}}-\frac {2 \sqrt {c+a^2 c x}}{a^4 c}+\frac {\left (c+a^2 c x\right )^{3/2}}{a^4 c^2}\right ) \, dx,x,x^2\right )-\frac {\left (c^2 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{8 a^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {c \sqrt {c+a^2 c x^2}}{48 a^3}+\frac {\left (c+a^2 c x^2\right )^{3/2}}{36 a^3}-\frac {\left (c+a^2 c x^2\right )^{5/2}}{30 a^3 c}+\frac {c x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{16 a^2}+\frac {7}{24} c x^3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{6} a^2 c x^5 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {i c^2 \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 )}{4 a^3 \sqrt {c+a^2 c x^2}}-\frac {i c^2 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a^3 \sqrt {c+a^2 c x^2}}+\frac {i c^2 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{8 a^3 \sqrt {c+a^2 c x^2}}-\frac {1}{48} \left (a c^2\right ) \operatorname {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 (c^2 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{\sqrt {1+a^2 x^2}} \, dx}{16 a^2 \sqrt {c+a^2 c x^2}}\\ &=\frac {c \sqrt {c+a^2 c x^2}}{16 a^3}+\frac {\left (c+a^2 c x^2\right )^{3/2}}{72 a^3}-\frac {\left (c+a^2 c x^2\right )^{5/2}}{30 a^3 c}+\frac {c x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{16 a^2}+\frac {7}{24} c x^3 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{6} a^2 c x^5 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {i c^2 \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 )}{8 a^3 \sqrt {c+a^2 c x^2}}-\frac {i c^2 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{16 a^3 \sqrt {c+a^2 c x^2}}+\frac {i c^2 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{16 a^3 \sqrt {c+a^2 c x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 6.34, size = 576, normalized size = 1.61 \[ \frac {c \sqrt {a^2 c x^2+c} \left (\frac {3}{4} \left (a^2 x^2+1\right )^{5/2}+\frac {55}{8} \left (a^2 x^2+1\right )^3 \cos \left (3 \tan ^{-1}(a x)\right )-\frac {45}{8} \left (a^2 x^2+1\right )^3 \cos \left (5 \tan ^{-1}(a x)\right )+\frac {15}{16} \left (a^2 x^2+1\right )^3 \tan ^{-1}(a x) \left (\frac {156 a x}{\sqrt {a^2 x^2+1}}+30 \log \left (1-i e^{i \tan ^{-1}(a x)}\right )-30 \log \left (1+i e^{i \tan ^{-1}(a x)}\right )-94 \sin \left (3 \tan ^{-1}(a x)\right )+6 \sin \left (5 \tan ^{-1}(a x)\right )+3 \log \left (1-i e^{i \tan ^{-1}(a x)}\right ) \cos \left (6 \tan ^{-1}(a x)\right )+45 \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right ) \cos \left (2 \tan ^{-1}(a x)\right )+18 \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right ) \cos \left (4 \tan ^{-1}(a x)\right )-3 \log \left (1+i e^{i \tan ^{-1}(a x)}\right ) \cos \left (6 \tan ^{-1}(a x)\right )\right )-\frac {15}{2} \left (a^2 x^2+1\right )^2 \left (-\frac {2}{\sqrt {a^2 x^2+1}}+3 \tan ^{-1}(a x) \left (-\frac {14 a x}{\sqrt {a^2 x^2+1}}+3 \log \left (1-i e^{i \tan ^{-1}(a x)}\right )-3 \log \left (1+i e^{i \tan ^{-1}(a x)}\right )+2 \sin \left (3 \tan ^{-1}(a x)\right )+4 \left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right ) \cos \left (2 \tan ^{-1}(a x)\right )+\left (\log \left (1-i e^{i \tan ^{-1}(a x)}\right )-\log \left (1+i e^{i \tan ^{-1}(a x)}\right )\right ) \cos \left (4 \tan ^{-1}(a x)\right )\right )-6 \cos \left (3 \tan ^{-1}(a x)\right )\right )-90 i \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )+90 i \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )\right )}{1440 a^3 \sqrt {a^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(c*Sqrt[c + a^2*c*x^2]*((3*(1 + a^2*x^2)^(5/2))/4 + (55*(1 + a^2*x^2)^3*Cos[3*ArcTan[a*x]])/8 - (45*(1 + a^2*x
^2)^3*Cos[5*ArcTan[a*x]])/8 - (90*I)*PolyLog[2, (-I)*E^(I*ArcTan[a*x])] + (90*I)*PolyLog[2, I*E^(I*ArcTan[a*x]
)] - (15*(1 + a^2*x^2)^2*(-2/Sqrt[1 + a^2*x^2] - 6*Cos[3*ArcTan[a*x]] + 3*ArcTan[a*x]*((-14*a*x)/Sqrt[1 + a^2*
x^2] + 3*Log[1 - I*E^(I*ArcTan[a*x])] + 4*Cos[2*ArcTan[a*x]]*(Log[1 - I*E^(I*ArcTan[a*x])] - Log[1 + I*E^(I*Ar
cTan[a*x])]) + Cos[4*ArcTan[a*x]]*(Log[1 - I*E^(I*ArcTan[a*x])] - Log[1 + I*E^(I*ArcTan[a*x])]) - 3*Log[1 + I*
E^(I*ArcTan[a*x])] + 2*Sin[3*ArcTan[a*x]])))/2 + (15*(1 + a^2*x^2)^3*ArcTan[a*x]*((156*a*x)/Sqrt[1 + a^2*x^2]
+ 30*Log[1 - I*E^(I*ArcTan[a*x])] + 3*Cos[6*ArcTan[a*x]]*Log[1 - I*E^(I*ArcTan[a*x])] + 45*Cos[2*ArcTan[a*x]]*
(Log[1 - I*E^(I*ArcTan[a*x])] - Log[1 + I*E^(I*ArcTan[a*x])]) + 18*Cos[4*ArcTan[a*x]]*(Log[1 - I*E^(I*ArcTan[a
*x])] - Log[1 + I*E^(I*ArcTan[a*x])]) - 30*Log[1 + I*E^(I*ArcTan[a*x])] - 3*Cos[6*ArcTan[a*x]]*Log[1 + I*E^(I*
ArcTan[a*x])] - 94*Sin[3*ArcTan[a*x]] + 6*Sin[5*ArcTan[a*x]]))/16))/(1440*a^3*Sqrt[1 + a^2*x^2])

________________________________________________________________________________________

fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a^{2} c x^{4} + c x^{2}\right )} \sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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(x^2*(a^2*c*x^2+c)^(3/2)*arctan(a*x),x, algorithm="giac")

[Out]

sage0*x

________________________________________________________________________________________

maple [A]  time = 1.32, size = 221, normalized size = 0.62 \[ \frac {c \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (120 \arctan \left (a x \right ) x^{5} a^{5}-24 a^{4} x^{4}+210 \arctan \left (a x \right ) x^{3} a^{3}-38 a^{2} x^{2}+45 \arctan \left (a x \right ) x a +31\right )}{720 a^{3}}+\frac {c \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (\arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \dilog \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+i \dilog \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{16 a^{3} \sqrt {a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/720*c/a^3*(c*(a*x-I)*(I+a*x))^(1/2)*(120*arctan(a*x)*x^5*a^5-24*a^4*x^4+210*arctan(a*x)*x^3*a^3-38*a^2*x^2+4
5*arctan(a*x)*x*a+31)+1/16*c*(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*dilog(1-I*(1+I*a*x)/(a^2*
x^2+1)^(1/2)))/a^3/(a^2*x^2+1)^(1/2)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a^2*c*x^2 + c)^(3/2)*x^2*arctan(a*x), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int x^2\,\mathrm {atan}\left (a\,x\right )\,{\left (c\,a^2\,x^2+c\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (c \left (a^{2} x^{2} + 1\right )\right )^{\frac {3}{2}} \operatorname {atan}{\left (a x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________