3.331 \(\int \frac {x^3 \tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx\)

Optimal. Leaf size=315 \[ \frac {x^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{3 a^2 c}+\frac {5 i \sqrt {a^2 x^2+1} \text {Li}_2\left (-\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{3 a^4 \sqrt {a^2 c x^2+c}}-\frac {5 i \sqrt {a^2 x^2+1} \text {Li}_2\left (\frac {i \sqrt {i a x+1}}{\sqrt {1-i a x}}\right )}{3 a^4 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 c x^2+c}}{3 a^4 c}-\frac {2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{3 a^4 c}-\frac {10 i \sqrt {a^2 x^2+1} \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right ) \tan ^{-1}(a x)}{3 a^4 \sqrt {a^2 c x^2+c}}-\frac {x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{3 a^3 c} \]

[Out]

-10/3*I*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)+5/3*I*po
lylog(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)-5/3*I*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)+1/3*(a^2*c*x^2+c)^(1/2)/a^4/c-1/3*x*arcta
n(a*x)*(a^2*c*x^2+c)^(1/2)/a^3/c-2/3*arctan(a*x)^2*(a^2*c*x^2+c)^(1/2)/a^4/c+1/3*x^2*arctan(a*x)^2*(a^2*c*x^2+
c)^(1/2)/a^2/c

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Rubi [A]  time = 0.43, antiderivative size = 315, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {4952, 261, 4890, 4886, 4930} \[ \frac {5 i \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,-\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{3 a^4 \sqrt {a^2 c x^2+c}}-\frac {5 i \sqrt {a^2 x^2+1} \text {PolyLog}\left (2,\frac {i \sqrt {1+i a x}}{\sqrt {1-i a x}}\right )}{3 a^4 \sqrt {a^2 c x^2+c}}+\frac {\sqrt {a^2 c x^2+c}}{3 a^4 c}+\frac {x^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{3 a^2 c}-\frac {2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)^2}{3 a^4 c}-\frac {10 i \sqrt {a^2 x^2+1} \tan ^{-1}\left (\frac {\sqrt {1+i a x}}{\sqrt {1-i a x}}\right ) \tan ^{-1}(a x)}{3 a^4 \sqrt {a^2 c x^2+c}}-\frac {x \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{3 a^3 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sqrt[c + a^2*c*x^2]/(3*a^4*c) - (x*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(3*a^3*c) - (2*Sqrt[c + a^2*c*x^2]*ArcTan[
a*x]^2)/(3*a^4*c) + (x^2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x]^2)/(3*a^2*c) - (((10*I)/3)*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]) + (((5*I)/3)*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]) - (((5*I)/3)*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 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 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 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 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 \frac {x^3 \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 c}-\frac {2 \int \frac {x \tan ^{-1}(a x)^2}{\sqrt {c+a^2 c x^2}} \, dx}{3 a^2}-\frac {2 \int \frac {x^2 \tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{3 a}\\ &=-\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 a^3 c}-\frac {2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 a^4 c}+\frac {x^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 a^2 c}+\frac {\int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{3 a^3}+\frac {4 \int \frac {\tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{3 a^3}+\frac {\int \frac {x}{\sqrt {c+a^2 c x^2}} \, dx}{3 a^2}\\ &=\frac {\sqrt {c+a^2 c x^2}}{3 a^4 c}-\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 a^3 c}-\frac {2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 a^4 c}+\frac {x^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 a^2 c}+\frac {\sqrt {1+a^2 x^2} \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 \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}}{3 a^4 c}-\frac {x \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 a^3 c}-\frac {2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 a^4 c}+\frac {x^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)^2}{3 a^2 c}-\frac {10 i \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 \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 \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}}\\ \end {align*}

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Mathematica [A]  time = 0.70, size = 279, normalized size = 0.89 \[ \frac {\left (a^2 x^2+1\right ) \sqrt {c \left (a^2 x^2+1\right )} \left (\frac {20 i \text {Li}_2\left (-i e^{i \tan ^{-1}(a x)}\right )}{\left (a^2 x^2+1\right )^{3/2}}-\frac {20 i \text {Li}_2\left (i e^{i \tan ^{-1}(a x)}\right )}{\left (a^2 x^2+1\right )^{3/2}}+\frac {15 \tan ^{-1}(a x) \log \left (1-i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 x^2+1}}-\frac {15 \tan ^{-1}(a x) \log \left (1+i e^{i \tan ^{-1}(a x)}\right )}{\sqrt {a^2 x^2+1}}-2 \tan ^{-1}(a x)^2-2 \tan ^{-1}(a x) \sin \left (2 \tan ^{-1}(a x)\right )-6 \tan ^{-1}(a x)^2 \cos \left (2 \tan ^{-1}(a x)\right )+2 \cos \left (2 \tan ^{-1}(a x)\right )+5 \tan ^{-1}(a x) \log \left (1-i e^{i \tan ^{-1}(a x)}\right ) \cos \left (3 \tan ^{-1}(a x)\right )-5 \tan ^{-1}(a x) \log \left (1+i e^{i \tan ^{-1}(a x)}\right ) \cos \left (3 \tan ^{-1}(a x)\right )+2\right )}{12 a^4 c} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((1 + a^2*x^2)*Sqrt[c*(1 + a^2*x^2)]*(2 - 2*ArcTan[a*x]^2 + 2*Cos[2*ArcTan[a*x]] - 6*ArcTan[a*x]^2*Cos[2*ArcTa
n[a*x]] + (15*ArcTan[a*x]*Log[1 - I*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] + 5*ArcTan[a*x]*Cos[3*ArcTan[a*x]]*L
og[1 - I*E^(I*ArcTan[a*x])] - (15*ArcTan[a*x]*Log[1 + I*E^(I*ArcTan[a*x])])/Sqrt[1 + a^2*x^2] - 5*ArcTan[a*x]*
Cos[3*ArcTan[a*x]]*Log[1 + I*E^(I*ArcTan[a*x])] + ((20*I)*PolyLog[2, (-I)*E^(I*ArcTan[a*x])])/(1 + a^2*x^2)^(3
/2) - ((20*I)*PolyLog[2, I*E^(I*ArcTan[a*x])])/(1 + a^2*x^2)^(3/2) - 2*ArcTan[a*x]*Sin[2*ArcTan[a*x]]))/(12*a^
4*c)

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fricas [F]  time = 0.71, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {x^{3} \arctan \left (a x\right )^{2}}{\sqrt {a^{2} c x^{2} + c}}, x\right ) \]

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(x^3*arctan(a*x)^2/sqrt(a^2*c*x^2 + c), 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(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,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 = 3.08, size = 206, normalized size = 0.65 \[ \frac {\left (\arctan \left (a x \right )^{2} x^{2} a^{2}-\arctan \left (a x \right ) x a -2 \arctan \left (a x \right )^{2}+1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{3 c \,a^{4}}+\frac {5 i \left (i \arctan \left (a x \right ) \ln \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-i \arctan \left (a x \right ) \ln \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )+\dilog \left (1+\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )-\dilog \left (1-\frac {i \left (i a x +1\right )}{\sqrt {a^{2} x^{2}+1}}\right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{3 \sqrt {a^{2} x^{2}+1}\, a^{4} c} \]

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)

[Out]

1/3*(arctan(a*x)^2*x^2*a^2-arctan(a*x)*x*a-2*arctan(a*x)^2+1)*(c*(a*x-I)*(I+a*x))^(1/2)/c/a^4+5/3*I*(I*arctan(
a*x)*ln(1+I*(1+I*a*x)/(a^2*x^2+1)^(1/2))-I*arctan(a*x)*ln(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2))+dilog(1+I*(1+I*a*x)
/(a^2*x^2+1)^(1/2))-dilog(1-I*(1+I*a*x)/(a^2*x^2+1)^(1/2)))*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)^(1/2)/a^4/c

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

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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