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

Optimal. Leaf size=329 \[ -\frac {149}{120} c^{5/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )+\frac {i c^3 \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {i c^3 \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {\sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {2 c^3 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {29}{120} a c^2 x \sqrt {a^2 c x^2+c}+c^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)-\frac {1}{20} a c x \left (a^2 c x^2+c\right )^{3/2}+\frac {1}{3} c \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{5} \left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x) \]

[Out]

-1/20*a*c*x*(a^2*c*x^2+c)^(3/2)+1/3*c*(a^2*c*x^2+c)^(3/2)*arctan(a*x)+1/5*(a^2*c*x^2+c)^(5/2)*arctan(a*x)-149/
120*c^(5/2)*arctanh(a*x*c^(1/2)/(a^2*c*x^2+c)^(1/2))-2*c^3*arctan(a*x)*arctanh((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)+I*c^3*polylog(2,-(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)-I*c^3*polylog(2,(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)-29/12
0*a*c^2*x*(a^2*c*x^2+c)^(1/2)+c^2*arctan(a*x)*(a^2*c*x^2+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.55, antiderivative size = 329, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 8, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.364, Rules used = {4950, 4946, 4958, 4954, 217, 206, 4930, 195} \[ \frac {i c^3 \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {i c^3 \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {29}{120} a c^2 x \sqrt {a^2 c x^2+c}+c^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)-\frac {149}{120} c^{5/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )-\frac {2 c^3 \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {a^2 c x^2+c}}-\frac {1}{20} a c x \left (a^2 c x^2+c\right )^{3/2}+\frac {1}{3} c \left (a^2 c x^2+c\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{5} \left (a^2 c x^2+c\right )^{5/2} \tan ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-29*a*c^2*x*Sqrt[c + a^2*c*x^2])/120 - (a*c*x*(c + a^2*c*x^2)^(3/2))/20 + c^2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]
+ (c*(c + a^2*c*x^2)^(3/2)*ArcTan[a*x])/3 + ((c + a^2*c*x^2)^(5/2)*ArcTan[a*x])/5 - (2*c^3*Sqrt[1 + a^2*x^2]*
ArcTan[a*x]*ArcTanh[Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/Sqrt[c + a^2*c*x^2] - (149*c^(5/2)*ArcTanh[(a*Sqrt[c]*x)
/Sqrt[c + a^2*c*x^2]])/120 + (I*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, -(Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x])])/Sqrt[c +
a^2*c*x^2] - (I*c^3*Sqrt[1 + a^2*x^2]*PolyLog[2, Sqrt[1 + I*a*x]/Sqrt[1 - I*a*x]])/Sqrt[c + a^2*c*x^2]

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 4930

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 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 4954

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

Rule 4958

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*Sqrt[(d_) + (e_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + c^2*
x^2]/Sqrt[d + e*x^2], Int[(a + b*ArcTan[c*x])^p/(x*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]

Rubi steps

\begin {align*} \int \frac {\left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)}{x} \, dx &=c \int \frac {\left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)}{x} \, dx+\left (a^2 c\right ) \int x \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x) \, dx\\ &=\frac {1}{5} \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)-\frac {1}{5} (a c) \int \left (c+a^2 c x^2\right )^{3/2} \, dx+c^2 \int \frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{x} \, dx+\left (a^2 c^2\right ) \int x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x) \, dx\\ &=-\frac {1}{20} a c x \left (c+a^2 c x^2\right )^{3/2}+c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{3} c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{5} \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)-\frac {1}{20} \left (3 a c^2\right ) \int \sqrt {c+a^2 c x^2} \, dx-\frac {1}{3} \left (a c^2\right ) \int \sqrt {c+a^2 c x^2} \, dx+c^3 \int \frac {\tan ^{-1}(a x)}{x \sqrt {c+a^2 c x^2}} \, dx-\left (a c^3\right ) \int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {29}{120} a c^2 x \sqrt {c+a^2 c x^2}-\frac {1}{20} a c x \left (c+a^2 c x^2\right )^{3/2}+c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{3} c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{5} \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)-\frac {1}{40} \left (3 a c^3\right ) \int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx-\frac {1}{6} \left (a c^3\right ) \int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx-\left (a c^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )+\frac {\left (c^3 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)}{x \sqrt {1+a^2 x^2}} \, dx}{\sqrt {c+a^2 c x^2}}\\ &=-\frac {29}{120} a c^2 x \sqrt {c+a^2 c x^2}-\frac {1}{20} a c x \left (c+a^2 c x^2\right )^{3/2}+c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{3} c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{5} \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)-\frac {2 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-c^{5/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )+\frac {i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {1}{40} \left (3 a c^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )-\frac {1}{6} \left (a c^3\right ) \operatorname {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )\\ &=-\frac {29}{120} a c^2 x \sqrt {c+a^2 c x^2}-\frac {1}{20} a c x \left (c+a^2 c x^2\right )^{3/2}+c^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)+\frac {1}{3} c \left (c+a^2 c x^2\right )^{3/2} \tan ^{-1}(a x)+\frac {1}{5} \left (c+a^2 c x^2\right )^{5/2} \tan ^{-1}(a x)-\frac {2 c^3 \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \tanh ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {149}{120} c^{5/2} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )+\frac {i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (-\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}-\frac {i c^3 \sqrt {1+a^2 x^2} \text {Li}_2\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{\sqrt {c+a^2 c x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.35, size = 268, normalized size = 0.81 \[ \frac {c^2 \sqrt {a^2 c x^2+c} \left (-35 a x \sqrt {a^2 x^2+1}+88 a^2 x^2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)+184 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)+24 a^4 x^4 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)-6 a^3 x^3 \sqrt {a^2 x^2+1}+120 i \text {Li}_2\left (-e^{i \tan ^{-1}(a x)}\right )-120 i \text {Li}_2\left (e^{i \tan ^{-1}(a x)}\right )-29 \sinh ^{-1}(a x)+120 \tan ^{-1}(a x) \log \left (1-e^{i \tan ^{-1}(a x)}\right )-120 \tan ^{-1}(a x) \log \left (1+e^{i \tan ^{-1}(a x)}\right )+120 \log \left (\cos \left (\frac {1}{2} \tan ^{-1}(a x)\right )-\sin \left (\frac {1}{2} \tan ^{-1}(a x)\right )\right )-120 \log \left (\sin \left (\frac {1}{2} \tan ^{-1}(a x)\right )+\cos \left (\frac {1}{2} \tan ^{-1}(a x)\right )\right )\right )}{120 \sqrt {a^2 x^2+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c^2*Sqrt[c + a^2*c*x^2]*(-35*a*x*Sqrt[1 + a^2*x^2] - 6*a^3*x^3*Sqrt[1 + a^2*x^2] - 29*ArcSinh[a*x] + 184*Sqrt
[1 + a^2*x^2]*ArcTan[a*x] + 88*a^2*x^2*Sqrt[1 + a^2*x^2]*ArcTan[a*x] + 24*a^4*x^4*Sqrt[1 + a^2*x^2]*ArcTan[a*x
] + 120*ArcTan[a*x]*Log[1 - E^(I*ArcTan[a*x])] - 120*ArcTan[a*x]*Log[1 + E^(I*ArcTan[a*x])] + 120*Log[Cos[ArcT
an[a*x]/2] - Sin[ArcTan[a*x]/2]] - 120*Log[Cos[ArcTan[a*x]/2] + Sin[ArcTan[a*x]/2]] + (120*I)*PolyLog[2, -E^(I
*ArcTan[a*x])] - (120*I)*PolyLog[2, E^(I*ArcTan[a*x])]))/(120*Sqrt[1 + a^2*x^2])

________________________________________________________________________________________

fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (a^{4} c^{2} x^{4} + 2 \, a^{2} c^{2} x^{2} + c^{2}\right )} \sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

________________________________________________________________________________________

giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

________________________________________________________________________________________

maple [A]  time = 0.74, size = 198, normalized size = 0.60 \[ \frac {c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (24 \arctan \left (a x \right ) x^{4} a^{4}-6 a^{3} x^{3}+88 \arctan \left (a x \right ) x^{2} a^{2}-35 a x +184 \arctan \left (a x \right )\right )}{120}+\frac {c^{2} \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (149 i \arctan \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+60 i \dilog \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )-60 \arctan \left (a x \right ) \ln \left (1+\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )+60 i \dilog \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{60 \sqrt {a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*c^2*(c*(a*x-I)*(I+a*x))^(1/2)*(24*arctan(a*x)*x^4*a^4-6*a^3*x^3+88*arctan(a*x)*x^2*a^2-35*a*x+184*arctan
(a*x))+1/60*c^2*(c*(a*x-I)*(I+a*x))^(1/2)*(149*I*arctan((1+I*a*x)/(a^2*x^2+1)^(1/2))+60*I*dilog(1+(1+I*a*x)/(a
^2*x^2+1)^(1/2))-60*arctan(a*x)*ln(1+(1+I*a*x)/(a^2*x^2+1)^(1/2))+60*I*dilog((1+I*a*x)/(a^2*x^2+1)^(1/2)))/(a^
2*x^2+1)^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {2}{3} \, {\left (a^{2} c^{2} x^{2} + c^{2}\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {c} \arctan \left (a x\right ) - \frac {1}{3} \, {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} {\left (a c^{2} x \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right ) + 2 \, c^{2} \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, -a^{2} x^{2} + 3\right )\right )\right )} \sqrt {c} - \frac {1}{120} \, {\left ({\left (a {\left (\frac {3 \, {\left (\frac {2 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{a^{2}} - \frac {\sqrt {a^{2} x^{2} + 1} x}{a^{2}} - \frac {\operatorname {arsinh}\left (a x\right )}{a^{3}}\right )}}{a^{2}} - \frac {8 \, {\left (\sqrt {a^{2} x^{2} + 1} x + \frac {\operatorname {arsinh}\left (a x\right )}{a}\right )}}{a^{4}}\right )} - 8 \, {\left (\frac {3 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{a^{2}} - \frac {2 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{4}}\right )} \arctan \left (a x\right )\right )} a^{4} c^{2} - 20 \, c^{2} \arctan \left ({\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right ) + 2, a x + {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right )\right ) - 20 \, c^{2} \arctan \left ({\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \sin \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right ) - 2, -a x + {\left (a^{4} x^{4} + 10 \, a^{2} x^{2} + 9\right )}^{\frac {1}{4}} \cos \left (\frac {1}{2} \, \arctan \left (4 \, a x, a^{2} x^{2} - 3\right )\right )\right ) - 120 \, c^{2} \int \frac {\sqrt {a^{2} x^{2} + 1} \arctan \left (a x\right )}{x}\,{d x}\right )} \sqrt {c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(a^2*c^2*x^2 + c^2)*sqrt(a^2*x^2 + 1)*sqrt(c)*arctan(a*x) - 1/3*(a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*(a*c^2*x*
cos(1/2*arctan2(4*a*x, -a^2*x^2 + 3)) + 2*c^2*sin(1/2*arctan2(4*a*x, -a^2*x^2 + 3)))*sqrt(c) - 1/120*((a*(3*(2
*(a^2*x^2 + 1)^(3/2)*x/a^2 - sqrt(a^2*x^2 + 1)*x/a^2 - arcsinh(a*x)/a^3)/a^2 - 8*(sqrt(a^2*x^2 + 1)*x + arcsin
h(a*x)/a)/a^4) - 8*(3*(a^2*x^2 + 1)^(3/2)*x^2/a^2 - 2*(a^2*x^2 + 1)^(3/2)/a^4)*arctan(a*x))*a^4*c^2 - 20*c^2*a
rctan2((a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*sin(1/2*arctan2(4*a*x, a^2*x^2 - 3)) + 2, a*x + (a^4*x^4 + 10*a^2*x^2
+ 9)^(1/4)*cos(1/2*arctan2(4*a*x, a^2*x^2 - 3))) - 20*c^2*arctan2((a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*sin(1/2*arc
tan2(4*a*x, a^2*x^2 - 3)) - 2, -a*x + (a^4*x^4 + 10*a^2*x^2 + 9)^(1/4)*cos(1/2*arctan2(4*a*x, a^2*x^2 - 3))) -
120*c^2*integrate(sqrt(a^2*x^2 + 1)*arctan(a*x)/x, x))*sqrt(c)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________