3.310 \(\int \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2 \, dx\)

Optimal. Leaf size=340 \[ \frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {c \sqrt {a^2 x^2+1} \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}+\frac {c \sqrt {a^2 x^2+1} \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt {a^2 c x^2+c}}+\frac {1}{2} x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{a}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a} \]

[Out]

arctanh(a*x*c^(1/2)/(a^2*c*x^2+c)^(1/2))*c^(1/2)/a-I*c*arctan((1+I*a*x)/(a^2*x^2+1)^(1/2))*arctan(a*x)^2*(a^2*
x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)+I*c*arctan(a*x)*polylog(2,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)
/a/(a^2*c*x^2+c)^(1/2)-I*c*arctan(a*x)*polylog(2,I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2
+c)^(1/2)-c*polylog(3,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)+c*polylog(3,I*(1
+I*a*x)/(a^2*x^2+1)^(1/2))*(a^2*x^2+1)^(1/2)/a/(a^2*c*x^2+c)^(1/2)-arctan(a*x)*(a^2*c*x^2+c)^(1/2)/a+1/2*x*arc
tan(a*x)^2*(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 340, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {4880, 4890, 4888, 4181, 2531, 2282, 6589, 217, 206} \[ \frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}(a x) \text {PolyLog}\left (2,i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {c \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}+\frac {c \sqrt {a^2 x^2+1} \text {PolyLog}\left (3,i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {a^2 c x^2+c}}-\frac {i c \sqrt {a^2 x^2+1} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt {a^2 c x^2+c}}+\frac {1}{2} x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2-\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{a}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/a) + (x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/2 - (I*c*Sqrt[1 + a^2*x^2]*ArcT
an[E^(I*ArcTan[a*x])]*ArcTan[a*x]^2)/(a*Sqrt[c + a^2*c*x^2]) + (Sqrt[c]*ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x
^2]])/a + (I*c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[2, (-I)*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2]) - (I*
c*Sqrt[1 + a^2*x^2]*ArcTan[a*x]*PolyLog[2, I*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2]) - (c*Sqrt[1 + a^2*x^2
]*PolyLog[3, (-I)*E^(I*ArcTan[a*x])])/(a*Sqrt[c + a^2*c*x^2]) + (c*Sqrt[1 + a^2*x^2]*PolyLog[3, I*E^(I*ArcTan[
a*x])])/(a*Sqrt[c + a^2*c*x^2])

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 2282

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2531

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> -Simp[((
f + g*x)^m*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)])/(b*c*n*Log[F]), x] + Dist[(g*m)/(b*c*n*Log[F]), Int[(f + g*x)
^(m - 1)*PolyLog[2, -(e*(F^(c*(a + b*x)))^n)], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4181

Int[csc[(e_.) + Pi*(k_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E
^(I*k*Pi)*E^(I*(e + f*x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*k*Pi)*E^(I*(e + f*x))],
x], x] + Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 + E^(I*k*Pi)*E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e,
f}, x] && IntegerQ[2*k] && IGtQ[m, 0]

Rule 4880

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

Rule 4888

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Dist[1/(c*Sqrt[d]), Subst
[Int[(a + b*x)^p*Sec[x], x], x, ArcTan[c*x]], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0] &
& 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 6589

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

Rubi steps

\begin {align*} \int \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2 \, dx &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2+\frac {1}{2} c \int \frac {\tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx+c \int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2+c \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 \sqrt {1+a^2 x^2}\right ) \int \frac {\tan ^{-1}(a x)^2}{\sqrt {1+a^2 x^2}} \, dx}{2 \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a}+\frac {\left (c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x^2 \sec (x) \, dx,x,\tan ^{-1}(a x)\right )}{2 a \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a}-\frac {\left (c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1-i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt {c+a^2 c x^2}}+\frac {\left (c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int x \log \left (1+i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a}+\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {\left (i c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (-i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt {c+a^2 c x^2}}+\frac {\left (i c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \text {Li}_2\left (i e^{i x}\right ) \, dx,x,\tan ^{-1}(a x)\right )}{a \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a}+\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {\left (c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(-i x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {\left (c \sqrt {1+a^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\text {Li}_2(i x)}{x} \, dx,x,e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}\\ &=-\frac {\sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{a}+\frac {1}{2} x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2}{a \sqrt {c+a^2 c x^2}}+\frac {\sqrt {c} \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{a}+\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {i c \sqrt {1+a^2 x^2} \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}-\frac {c \sqrt {1+a^2 x^2} \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}+\frac {c \sqrt {1+a^2 x^2} \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )}{a \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.42, size = 201, normalized size = 0.59 \[ \frac {\sqrt {c \left (a^2 x^2+1\right )} \left (a x \sqrt {a^2 x^2+1} \tan ^{-1}(a x)^2-2 \sqrt {a^2 x^2+1} \tan ^{-1}(a x)+2 \tanh ^{-1}\left (\frac {a x}{\sqrt {a^2 x^2+1}}\right )+2 i \tan ^{-1}(a x) \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )-2 i \tan ^{-1}(a x) \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )-2 \text {Li}_3\left (-i e^{i \tan ^{-1}(a x)}\right )+2 \text {Li}_3\left (i e^{i \tan ^{-1}(a x)}\right )-2 i \tan ^{-1}\left (e^{i \tan ^{-1}(a x)}\right ) \tan ^{-1}(a x)^2\right )}{2 a \sqrt {a^2 x^2+1}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[c*(1 + a^2*x^2)]*(-2*Sqrt[1 + a^2*x^2]*ArcTan[a*x] + a*x*Sqrt[1 + a^2*x^2]*ArcTan[a*x]^2 - (2*I)*ArcTan[
E^(I*ArcTan[a*x])]*ArcTan[a*x]^2 + 2*ArcTanh[(a*x)/Sqrt[1 + a^2*x^2]] + (2*I)*ArcTan[a*x]*PolyLog[2, (-I)*E^(I
*ArcTan[a*x])] - (2*I)*ArcTan[a*x]*PolyLog[2, I*E^(I*ArcTan[a*x])] - 2*PolyLog[3, (-I)*E^(I*ArcTan[a*x])] + 2*
PolyLog[3, I*E^(I*ArcTan[a*x])]))/(2*a*Sqrt[1 + a^2*x^2])

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {a^{2} c x^{2} + c} \arctan \left (a x\right )^{2}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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(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,x):;OUTPUT:sym2
poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.84, size = 268, normalized size = 0.79 \[ \frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \arctan \left (a x \right ) \left (\arctan \left (a x \right ) x a -2\right )}{2 a}-\frac {i \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i \arctan \left (a x \right )^{2} \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \arctan \left (a x \right )^{2} \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+2 \arctan \left (a x \right ) \polylog \left (2, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+2 i \polylog \left (3, \frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-2 \arctan \left (a x \right ) \polylog \left (2, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-2 i \polylog \left (3, -\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+4 \arctan \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}\right )\right )}{2 a \sqrt {a^{2} x^{2}+1}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/a*(c*(a*x-I)*(I+a*x))^(1/2)*arctan(a*x)*(arctan(a*x)*x*a-2)-1/2*I*(c*(a*x-I)*(I+a*x))^(1/2)*(I*arctan(a*x)
^2*ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-I*arctan(a*x)^2*ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+2*arctan(a*x)*polyl
og(2,I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+2*I*polylog(3,I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-2*arctan(a*x)*polylog(2,-I*(1
+I*a*x)/(a^2*x^2+1)^(1/2))-2*I*polylog(3,-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+4*arctan((1+I*a*x)/(a^2*x^2+1)^(1/2))
)/a/(a^2*x^2+1)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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