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

Optimal. Leaf size=120 \[ \frac {x^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{3 a^2 c}-\frac {2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{3 a^4 c}+\frac {5 \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{6 a^4 \sqrt {c}}-\frac {x \sqrt {a^2 c x^2+c}}{6 a^3 c} \]

[Out]

5/6*arctanh(a*x*c^(1/2)/(a^2*c*x^2+c)^(1/2))/a^4/c^(1/2)-1/6*x*(a^2*c*x^2+c)^(1/2)/a^3/c-2/3*arctan(a*x)*(a^2*
c*x^2+c)^(1/2)/a^4/c+1/3*x^2*arctan(a*x)*(a^2*c*x^2+c)^(1/2)/a^2/c

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Rubi [A]  time = 0.15, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {4952, 321, 217, 206, 4930} \[ -\frac {x \sqrt {a^2 c x^2+c}}{6 a^3 c}+\frac {x^2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{3 a^2 c}-\frac {2 \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{3 a^4 c}+\frac {5 \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{6 a^4 \sqrt {c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x*Sqrt[c + a^2*c*x^2])/(6*a^3*c) - (2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(3*a^4*c) + (x^2*Sqrt[c + a^2*c*x^2]*
ArcTan[a*x])/(3*a^2*c) + (5*ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]])/(6*a^4*Sqrt[c])

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 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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)}{\sqrt {c+a^2 c x^2}} \, dx &=\frac {x^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 a^2 c}-\frac {2 \int \frac {x \tan ^{-1}(a x)}{\sqrt {c+a^2 c x^2}} \, dx}{3 a^2}-\frac {\int \frac {x^2}{\sqrt {c+a^2 c x^2}} \, dx}{3 a}\\ &=-\frac {x \sqrt {c+a^2 c x^2}}{6 a^3 c}-\frac {2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 a^4 c}+\frac {x^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 a^2 c}+\frac {\int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx}{6 a^3}+\frac {2 \int \frac {1}{\sqrt {c+a^2 c x^2}} \, dx}{3 a^3}\\ &=-\frac {x \sqrt {c+a^2 c x^2}}{6 a^3 c}-\frac {2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 a^4 c}+\frac {x^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 a^2 c}+\frac {\operatorname {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )}{6 a^3}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{1-a^2 c x^2} \, dx,x,\frac {x}{\sqrt {c+a^2 c x^2}}\right )}{3 a^3}\\ &=-\frac {x \sqrt {c+a^2 c x^2}}{6 a^3 c}-\frac {2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 a^4 c}+\frac {x^2 \sqrt {c+a^2 c x^2} \tan ^{-1}(a x)}{3 a^2 c}+\frac {5 \tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {c+a^2 c x^2}}\right )}{6 a^4 \sqrt {c}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 91, normalized size = 0.76 \[ \frac {-a x \sqrt {a^2 c x^2+c}+5 \sqrt {c} \log \left (\sqrt {c} \sqrt {a^2 c x^2+c}+a c x\right )+2 \left (a^2 x^2-2\right ) \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{6 a^4 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*x*Sqrt[c + a^2*c*x^2]) + 2*(-2 + a^2*x^2)*Sqrt[c + a^2*c*x^2]*ArcTan[a*x] + 5*Sqrt[c]*Log[a*c*x + Sqrt[c]
*Sqrt[c + a^2*c*x^2]])/(6*a^4*c)

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fricas [A]  time = 0.66, size = 80, normalized size = 0.67 \[ -\frac {2 \, \sqrt {a^{2} c x^{2} + c} {\left (a x - 2 \, {\left (a^{2} x^{2} - 2\right )} \arctan \left (a x\right )\right )} - 5 \, \sqrt {c} \log \left (-2 \, a^{2} c x^{2} - 2 \, \sqrt {a^{2} c x^{2} + c} a \sqrt {c} x - c\right )}{12 \, a^{4} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(2*sqrt(a^2*c*x^2 + c)*(a*x - 2*(a^2*x^2 - 2)*arctan(a*x)) - 5*sqrt(c)*log(-2*a^2*c*x^2 - 2*sqrt(a^2*c*x
^2 + c)*a*sqrt(c)*x - c))/(a^4*c)

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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)/(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 [C]  time = 3.16, size = 165, normalized size = 1.38 \[ \frac {\left (2 \arctan \left (a x \right ) x^{2} a^{2}-a x -4 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{6 c \,a^{4}}+\frac {5 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{6 \sqrt {a^{2} x^{2}+1}\, a^{4} c}-\frac {5 \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{6 \sqrt {a^{2} x^{2}+1}\, a^{4} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*arctan(a*x)*x^2*a^2-a*x-4*arctan(a*x))*(c*(a*x-I)*(I+a*x))^(1/2)/c/a^4+5/6*ln((1+I*a*x)/(a^2*x^2+1)^(1/
2)+I)*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)^(1/2)/a^4/c-5/6*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-I)*(c*(a*x-I)*(I+a*
x))^(1/2)/(a^2*x^2+1)^(1/2)/a^4/c

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maxima [A]  time = 0.47, size = 89, normalized size = 0.74 \[ -\frac {a {\left (\frac {\frac {\sqrt {a^{2} x^{2} + 1} x}{a^{2}} - \frac {\operatorname {arsinh}\left (a x\right )}{a^{3}}}{a^{2}} - \frac {4 \, \operatorname {arsinh}\left (a x\right )}{a^{5}}\right )} - 2 \, {\left (\frac {\sqrt {a^{2} x^{2} + 1} x^{2}}{a^{2}} - \frac {2 \, \sqrt {a^{2} x^{2} + 1}}{a^{4}}\right )} \arctan \left (a x\right )}{6 \, \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(a*((sqrt(a^2*x^2 + 1)*x/a^2 - arcsinh(a*x)/a^3)/a^2 - 4*arcsinh(a*x)/a^5) - 2*(sqrt(a^2*x^2 + 1)*x^2/a^2
- 2*sqrt(a^2*x^2 + 1)/a^4)*arctan(a*x))/sqrt(c)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\mathrm {atan}\left (a\,x\right )}{\sqrt {c\,a^2\,x^2+c}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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