3.242 \(\int \frac {x^3 \tan ^{-1}(a x)}{(c+a^2 c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=112 \[ -\frac {x^2 \tan ^{-1}(a x)}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}}+\frac {x^3}{9 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac {2 \tan ^{-1}(a x)}{3 a^4 c^2 \sqrt {a^2 c x^2+c}}+\frac {2 x}{3 a^3 c^2 \sqrt {a^2 c x^2+c}} \]

[Out]

1/9*x^3/a/c/(a^2*c*x^2+c)^(3/2)-1/3*x^2*arctan(a*x)/a^2/c/(a^2*c*x^2+c)^(3/2)+2/3*x/a^3/c^2/(a^2*c*x^2+c)^(1/2
)-2/3*arctan(a*x)/a^4/c^2/(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.14, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {4938, 4930, 191} \[ \frac {2 x}{3 a^3 c^2 \sqrt {a^2 c x^2+c}}-\frac {2 \tan ^{-1}(a x)}{3 a^4 c^2 \sqrt {a^2 c x^2+c}}+\frac {x^3}{9 a c \left (a^2 c x^2+c\right )^{3/2}}-\frac {x^2 \tan ^{-1}(a x)}{3 a^2 c \left (a^2 c x^2+c\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*ArcTan[a*x])/(c + a^2*c*x^2)^(5/2),x]

[Out]

x^3/(9*a*c*(c + a^2*c*x^2)^(3/2)) + (2*x)/(3*a^3*c^2*Sqrt[c + a^2*c*x^2]) - (x^2*ArcTan[a*x])/(3*a^2*c*(c + a^
2*c*x^2)^(3/2)) - (2*ArcTan[a*x])/(3*a^4*c^2*Sqrt[c + a^2*c*x^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 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 4938

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(b*(f*x
)^m*(d + e*x^2)^(q + 1))/(c*d*m^2), x] + (Dist[(f^2*(m - 1))/(c^2*d*m), Int[(f*x)^(m - 2)*(d + e*x^2)^(q + 1)*
(a + b*ArcTan[c*x]), x], x] - Simp[(f*(f*x)^(m - 1)*(d + e*x^2)^(q + 1)*(a + b*ArcTan[c*x]))/(c^2*d*m), x]) /;
FreeQ[{a, b, c, d, e, f}, x] && EqQ[e, c^2*d] && EqQ[m + 2*q + 2, 0] && LtQ[q, -1]

Rubi steps

\begin {align*} \int \frac {x^3 \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{5/2}} \, dx &=\frac {x^3}{9 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac {x^2 \tan ^{-1}(a x)}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}+\frac {2 \int \frac {x \tan ^{-1}(a x)}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 a^2 c}\\ &=\frac {x^3}{9 a c \left (c+a^2 c x^2\right )^{3/2}}-\frac {x^2 \tan ^{-1}(a x)}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \tan ^{-1}(a x)}{3 a^4 c^2 \sqrt {c+a^2 c x^2}}+\frac {2 \int \frac {1}{\left (c+a^2 c x^2\right )^{3/2}} \, dx}{3 a^3 c}\\ &=\frac {x^3}{9 a c \left (c+a^2 c x^2\right )^{3/2}}+\frac {2 x}{3 a^3 c^2 \sqrt {c+a^2 c x^2}}-\frac {x^2 \tan ^{-1}(a x)}{3 a^2 c \left (c+a^2 c x^2\right )^{3/2}}-\frac {2 \tan ^{-1}(a x)}{3 a^4 c^2 \sqrt {c+a^2 c x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 65, normalized size = 0.58 \[ \frac {\sqrt {a^2 c x^2+c} \left (a x \left (7 a^2 x^2+6\right )-3 \left (3 a^2 x^2+2\right ) \tan ^{-1}(a x)\right )}{9 a^4 c^3 \left (a^2 x^2+1\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*ArcTan[a*x])/(c + a^2*c*x^2)^(5/2),x]

[Out]

(Sqrt[c + a^2*c*x^2]*(a*x*(6 + 7*a^2*x^2) - 3*(2 + 3*a^2*x^2)*ArcTan[a*x]))/(9*a^4*c^3*(1 + a^2*x^2)^2)

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fricas [A]  time = 0.79, size = 74, normalized size = 0.66 \[ \frac {{\left (7 \, a^{3} x^{3} + 6 \, a x - 3 \, {\left (3 \, a^{2} x^{2} + 2\right )} \arctan \left (a x\right )\right )} \sqrt {a^{2} c x^{2} + c}}{9 \, {\left (a^{8} c^{3} x^{4} + 2 \, a^{6} c^{3} x^{2} + a^{4} c^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(a*x)/(a^2*c*x^2+c)^(5/2),x, algorithm="fricas")

[Out]

1/9*(7*a^3*x^3 + 6*a*x - 3*(3*a^2*x^2 + 2)*arctan(a*x))*sqrt(a^2*c*x^2 + c)/(a^8*c^3*x^4 + 2*a^6*c^3*x^2 + a^4
*c^3)

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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)^(5/2),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 = 3.06, size = 244, normalized size = 2.18 \[ -\frac {\left (i+3 \arctan \left (a x \right )\right ) \left (i x^{3} a^{3}+3 a^{2} x^{2}-3 i a x -1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{72 \left (a^{2} x^{2}+1\right )^{2} c^{3} a^{4}}-\frac {3 \left (i+\arctan \left (a x \right )\right ) \left (i a x +1\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}}{8 a^{4} c^{3} \left (a^{2} x^{2}+1\right )}+\frac {3 \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i a x -1\right ) \left (\arctan \left (a x \right )-i\right )}{8 a^{4} c^{3} \left (a^{2} x^{2}+1\right )}+\frac {\left (-i+3 \arctan \left (a x \right )\right ) \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (i x^{3} a^{3}-3 a^{2} x^{2}-3 i a x +1\right )}{72 \left (a^{4} x^{4}+2 a^{2} x^{2}+1\right ) c^{3} a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctan(a*x)/(a^2*c*x^2+c)^(5/2),x)

[Out]

-1/72*(I+3*arctan(a*x))*(I*x^3*a^3+3*a^2*x^2-3*I*a*x-1)*(c*(a*x-I)*(I+a*x))^(1/2)/(a^2*x^2+1)^2/c^3/a^4-3/8*(I
+arctan(a*x))*(1+I*a*x)*(c*(a*x-I)*(I+a*x))^(1/2)/a^4/c^3/(a^2*x^2+1)+3/8*(c*(a*x-I)*(I+a*x))^(1/2)*(-1+I*a*x)
*(arctan(a*x)-I)/a^4/c^3/(a^2*x^2+1)+1/72*(-I+3*arctan(a*x))*(c*(a*x-I)*(I+a*x))^(1/2)*(I*x^3*a^3-3*a^2*x^2-3*
I*a*x+1)/(a^4*x^4+2*a^2*x^2+1)/c^3/a^4

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maxima [A]  time = 0.46, size = 65, normalized size = 0.58 \[ \frac {7 \, a^{3} x^{3} + 6 \, a x - 3 \, {\left (3 \, a^{2} x^{2} + 2\right )} \arctan \left (a x\right )}{9 \, {\left (a^{6} c^{2} x^{2} + a^{4} c^{2}\right )} \sqrt {a^{2} x^{2} + 1} \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctan(a*x)/(a^2*c*x^2+c)^(5/2),x, algorithm="maxima")

[Out]

1/9*(7*a^3*x^3 + 6*a*x - 3*(3*a^2*x^2 + 2)*arctan(a*x))/((a^6*c^2*x^2 + a^4*c^2)*sqrt(a^2*x^2 + 1)*sqrt(c))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\mathrm {atan}\left (a\,x\right )}{{\left (c\,a^2\,x^2+c\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*atan(a*x))/(c + a^2*c*x^2)^(5/2),x)

[Out]

int((x^3*atan(a*x))/(c + a^2*c*x^2)^(5/2), 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)**(5/2),x)

[Out]

Exception raised: TypeError

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