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

Optimal. Leaf size=160 \[ -\frac{x^3 \sqrt{a^2 c x^2+c}}{20 a}+\frac{x \sqrt{a^2 c x^2+c}}{24 a^3}+\frac{1}{5} x^4 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)+\frac{x^2 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{15 a^2}-\frac{2 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{15 a^4}+\frac{11 \sqrt{c} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{120 a^4} \]

[Out]

(x*Sqrt[c + a^2*c*x^2])/(24*a^3) - (x^3*Sqrt[c + a^2*c*x^2])/(20*a) - (2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(15*
a^4) + (x^2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(15*a^2) + (x^4*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/5 + (11*Sqrt[c]*
ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]])/(120*a^4)

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Rubi [A]  time = 0.271102, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 6, integrand size = 22, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.273, Rules used = {4946, 4952, 321, 217, 206, 4930} \[ -\frac{x^3 \sqrt{a^2 c x^2+c}}{20 a}+\frac{x \sqrt{a^2 c x^2+c}}{24 a^3}+\frac{1}{5} x^4 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)+\frac{x^2 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{15 a^2}-\frac{2 \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{15 a^4}+\frac{11 \sqrt{c} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{a^2 c x^2+c}}\right )}{120 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[c + a^2*c*x^2])/(24*a^3) - (x^3*Sqrt[c + a^2*c*x^2])/(20*a) - (2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(15*
a^4) + (x^2*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/(15*a^2) + (x^4*Sqrt[c + a^2*c*x^2]*ArcTan[a*x])/5 + (11*Sqrt[c]*
ArcTanh[(a*Sqrt[c]*x)/Sqrt[c + a^2*c*x^2]])/(120*a^4)

Rule 4946

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

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]

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 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 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 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]

Rubi steps

\begin{align*} \int x^3 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x) \, dx &=\frac{1}{5} x^4 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)+\frac{1}{5} c \int \frac{x^3 \tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx-\frac{1}{5} (a c) \int \frac{x^4}{\sqrt{c+a^2 c x^2}} \, dx\\ &=-\frac{x^3 \sqrt{c+a^2 c x^2}}{20 a}+\frac{x^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{15 a^2}+\frac{1}{5} x^4 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)-\frac{(2 c) \int \frac{x \tan ^{-1}(a x)}{\sqrt{c+a^2 c x^2}} \, dx}{15 a^2}-\frac{c \int \frac{x^2}{\sqrt{c+a^2 c x^2}} \, dx}{15 a}+\frac{(3 c) \int \frac{x^2}{\sqrt{c+a^2 c x^2}} \, dx}{20 a}\\ &=\frac{x \sqrt{c+a^2 c x^2}}{24 a^3}-\frac{x^3 \sqrt{c+a^2 c x^2}}{20 a}-\frac{2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{15 a^4}+\frac{x^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{15 a^2}+\frac{1}{5} x^4 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)+\frac{c \int \frac{1}{\sqrt{c+a^2 c x^2}} \, dx}{30 a^3}-\frac{(3 c) \int \frac{1}{\sqrt{c+a^2 c x^2}} \, dx}{40 a^3}+\frac{(2 c) \int \frac{1}{\sqrt{c+a^2 c x^2}} \, dx}{15 a^3}\\ &=\frac{x \sqrt{c+a^2 c x^2}}{24 a^3}-\frac{x^3 \sqrt{c+a^2 c x^2}}{20 a}-\frac{2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{15 a^4}+\frac{x^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{15 a^2}+\frac{1}{5} x^4 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)+\frac{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 )}{30 a^3}-\frac{(3 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 )}{40 a^3}+\frac{(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 )}{15 a^3}\\ &=\frac{x \sqrt{c+a^2 c x^2}}{24 a^3}-\frac{x^3 \sqrt{c+a^2 c x^2}}{20 a}-\frac{2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{15 a^4}+\frac{x^2 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)}{15 a^2}+\frac{1}{5} x^4 \sqrt{c+a^2 c x^2} \tan ^{-1}(a x)+\frac{11 \sqrt{c} \tanh ^{-1}\left (\frac{a \sqrt{c} x}{\sqrt{c+a^2 c x^2}}\right )}{120 a^4}\\ \end{align*}

Mathematica [A]  time = 0.123572, size = 105, normalized size = 0.66 \[ \frac{a x \left (5-6 a^2 x^2\right ) \sqrt{a^2 c x^2+c}+11 \sqrt{c} \log \left (\sqrt{c} \sqrt{a^2 c x^2+c}+a c x\right )+8 \left (3 a^4 x^4+a^2 x^2-2\right ) \sqrt{a^2 c x^2+c} \tan ^{-1}(a x)}{120 a^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x*(5 - 6*a^2*x^2)*Sqrt[c + a^2*c*x^2] + 8*Sqrt[c + a^2*c*x^2]*(-2 + a^2*x^2 + 3*a^4*x^4)*ArcTan[a*x] + 11*S
qrt[c]*Log[a*c*x + Sqrt[c]*Sqrt[c + a^2*c*x^2]])/(120*a^4)

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Maple [C]  time = 0.776, size = 176, normalized size = 1.1 \begin{align*}{\frac{24\,\arctan \left ( ax \right ){x}^{4}{a}^{4}-6\,{a}^{3}{x}^{3}+8\,\arctan \left ( ax \right ){a}^{2}{x}^{2}+5\,ax-16\,\arctan \left ( ax \right ) }{120\,{a}^{4}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }}-{\frac{11}{120\,{a}^{4}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }\ln \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}-i \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+{\frac{11}{120\,{a}^{4}}\sqrt{c \left ( ax-i \right ) \left ( ax+i \right ) }\ln \left ({(1+iax){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}}+i \right ){\frac{1}{\sqrt{{a}^{2}{x}^{2}+1}}}} \end{align*}

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]

1/120/a^4*(c*(a*x-I)*(a*x+I))^(1/2)*(24*arctan(a*x)*x^4*a^4-6*a^3*x^3+8*arctan(a*x)*a^2*x^2+5*a*x-16*arctan(a*
x))-11/120/a^4*(c*(a*x-I)*(a*x+I))^(1/2)*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)-I)/(a^2*x^2+1)^(1/2)+11/120/a^4*(c*(a*
x-I)*(a*x+I))^(1/2)*ln((1+I*a*x)/(a^2*x^2+1)^(1/2)+I)/(a^2*x^2+1)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.74631, size = 228, normalized size = 1.42 \begin{align*} -\frac{2 \,{\left (6 \, a^{3} x^{3} - 5 \, a x - 8 \,{\left (3 \, a^{4} x^{4} + a^{2} x^{2} - 2\right )} \arctan \left (a x\right )\right )} \sqrt{a^{2} c x^{2} + c} - 11 \, \sqrt{c} \log \left (-2 \, a^{2} c x^{2} - 2 \, \sqrt{a^{2} c x^{2} + c} a \sqrt{c} x - c\right )}{240 \, a^{4}} \end{align*}

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]

-1/240*(2*(6*a^3*x^3 - 5*a*x - 8*(3*a^4*x^4 + a^2*x^2 - 2)*arctan(a*x))*sqrt(a^2*c*x^2 + c) - 11*sqrt(c)*log(-
2*a^2*c*x^2 - 2*sqrt(a^2*c*x^2 + c)*a*sqrt(c)*x - c))/a^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \sqrt{c \left (a^{2} x^{2} + 1\right )} \operatorname{atan}{\left (a x \right )}\, dx \end{align*}

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

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Giac [A]  time = 1.18046, size = 144, normalized size = 0.9 \begin{align*} -\frac{\sqrt{a^{2} c x^{2} + c}{\left (6 \, a^{2} x^{2} - 5\right )} x + \frac{11 \, \sqrt{c} \log \left ({\left | -\sqrt{a^{2} c} x + \sqrt{a^{2} c x^{2} + c} \right |}\right )}{{\left | a \right |}}}{120 \, a^{3}} + \frac{{\left (3 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{5}{2}} - 5 \,{\left (a^{2} c x^{2} + c\right )}^{\frac{3}{2}} c\right )} \arctan \left (a x\right )}{15 \, a^{4} c^{2}} \end{align*}

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]

-1/120*(sqrt(a^2*c*x^2 + c)*(6*a^2*x^2 - 5)*x + 11*sqrt(c)*log(abs(-sqrt(a^2*c)*x + sqrt(a^2*c*x^2 + c)))/abs(
a))/a^3 + 1/15*(3*(a^2*c*x^2 + c)^(5/2) - 5*(a^2*c*x^2 + c)^(3/2)*c)*arctan(a*x)/(a^4*c^2)