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

Optimal. Leaf size=107 \[ -\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a^4 c^{3/2}}+\frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{a^4 c^2}+\frac {\tan ^{-1}(a x)}{a^4 c \sqrt {a^2 c x^2+c}}-\frac {x}{a^3 c \sqrt {a^2 c x^2+c}} \]

[Out]

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

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Rubi [A]  time = 0.20, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {4964, 4930, 217, 206, 191} \[ \frac {\sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{a^4 c^2}-\frac {\tanh ^{-1}\left (\frac {a \sqrt {c} x}{\sqrt {a^2 c x^2+c}}\right )}{a^4 c^{3/2}}-\frac {x}{a^3 c \sqrt {a^2 c x^2+c}}+\frac {\tan ^{-1}(a x)}{a^4 c \sqrt {a^2 c x^2+c}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

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 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_)^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Dist[1/e, Int[
x^(m - 2)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x])^p, x], x] - Dist[d/e, Int[x^(m - 2)*(d + e*x^2)^q*(a + b*Arc
Tan[c*x])^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && IntegersQ[p, 2*q] && LtQ[q, -1] && IGtQ[m
, 1] && NeQ[p, -1]

Rubi steps

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

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Mathematica [A]  time = 0.13, size = 107, normalized size = 1.00 \[ \frac {-a x \sqrt {a^2 c x^2+c}-\sqrt {c} \left (a^2 x^2+1\right ) \log \left (\sqrt {c} \sqrt {a^2 c x^2+c}+a c x\right )+\left (a^2 x^2+2\right ) \sqrt {a^2 c x^2+c} \tan ^{-1}(a x)}{a^4 c^2 \left (a^2 x^2+1\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*x*Sqrt[c + a^2*c*x^2]) + (2 + a^2*x^2)*Sqrt[c + a^2*c*x^2]*ArcTan[a*x] - Sqrt[c]*(1 + a^2*x^2)*Log[a*c*x
+ Sqrt[c]*Sqrt[c + a^2*c*x^2]])/(a^4*c^2*(1 + a^2*x^2))

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fricas [A]  time = 0.62, size = 102, normalized size = 0.95 \[ \frac {{\left (a^{2} x^{2} + 1\right )} \sqrt {c} \log \left (-2 \, a^{2} c x^{2} + 2 \, \sqrt {a^{2} c x^{2} + c} a \sqrt {c} x - c\right ) - 2 \, \sqrt {a^{2} c x^{2} + c} {\left (a x - {\left (a^{2} x^{2} + 2\right )} \arctan \left (a x\right )\right )}}{2 \, {\left (a^{6} c^{2} x^{2} + a^{4} c^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^(3/2),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 = 3.02, size = 242, normalized size = 2.26 \[ \frac {\left (i+\arctan \left (a x \right )\right ) \left (i a x +1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{2 \left (a^{2} x^{2}+1\right ) a^{4} c^{2}}-\frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i a x -1\right ) \left (\arctan \left (a x \right )-i\right )}{2 \left (a^{2} x^{2}+1\right ) a^{4} c^{2}}+\frac {\arctan \left (a x \right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{a^{4} c^{2}}+\frac {\ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, a^{4} c^{2}}-\frac {\ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+i\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{\sqrt {a^{2} x^{2}+1}\, a^{4} c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Exception raised: TypeError

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