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

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

[Out]

11/120*arctanh(a*x*c^(1/2)/(a^2*c*x^2+c)^(1/2))*c^(1/2)/a^4+1/24*x*(a^2*c*x^2+c)^(1/2)/a^3-1/20*x^3*(a^2*c*x^2
+c)^(1/2)/a-2/15*arctan(a*x)*(a^2*c*x^2+c)^(1/2)/a^4+1/15*x^2*arctan(a*x)*(a^2*c*x^2+c)^(1/2)/a^2+1/5*x^4*arct
an(a*x)*(a^2*c*x^2+c)^(1/2)

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Rubi [A]  time = 0.27, antiderivative size = 160, normalized size of antiderivative = 1.00, 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 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 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]

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*}

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Mathematica [A]  time = 0.13, 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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fricas [A]  time = 0.59, size = 94, normalized size = 0.59 \[ -\frac {2 \, {\left (6 \, a^{3} x^{3} - 5 \, a x - 8 \, {\left (3 \, a^{4} x^{4} + a^{2} x^{2} - 2\right )} \arctan \left (a x\right )\right )} \sqrt {a^{2} c x^{2} + c} - 11 \, \sqrt {c} \log \left (-2 \, a^{2} c x^{2} - 2 \, \sqrt {a^{2} c x^{2} + c} a \sqrt {c} x - c\right )}{240 \, a^{4}} \]

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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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 = 2.72, size = 176, normalized size = 1.10 \[ \frac {\sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \left (24 \arctan \left (a x \right ) x^{4} a^{4}-6 a^{3} x^{3}+8 \arctan \left (a x \right ) x^{2} a^{2}+5 a x -16 \arctan \left (a x \right )\right )}{120 a^{4}}+\frac {11 \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}+i\right )}{120 a^{4} \sqrt {a^{2} x^{2}+1}}-\frac {11 \sqrt {c \left (a x -i\right ) \left (a x +i\right )}\, \ln \left (\frac {i a x +1}{\sqrt {a^{2} x^{2}+1}}-i\right )}{120 a^{4} \sqrt {a^{2} x^{2}+1}} \]

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)*(I+a*x))^(1/2)*(24*arctan(a*x)*x^4*a^4-6*a^3*x^3+8*arctan(a*x)*x^2*a^2+5*a*x-16*arctan(a*
x))+11/120/a^4*(c*(a*x-I)*(I+a*x))^(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)*(I+a*x))^(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 [A]  time = 0.45, size = 127, normalized size = 0.79 \[ -\frac {1}{120} \, {\left (a {\left (\frac {3 \, {\left (\frac {2 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x}{a^{2}} - \frac {\sqrt {a^{2} x^{2} + 1} x}{a^{2}} - \frac {\operatorname {arsinh}\left (a x\right )}{a^{3}}\right )}}{a^{2}} - \frac {8 \, {\left (\sqrt {a^{2} x^{2} + 1} x + \frac {\operatorname {arsinh}\left (a x\right )}{a}\right )}}{a^{4}}\right )} - 8 \, {\left (\frac {3 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}} x^{2}}{a^{2}} - \frac {2 \, {\left (a^{2} x^{2} + 1\right )}^{\frac {3}{2}}}{a^{4}}\right )} \arctan \left (a x\right )\right )} \sqrt {c} \]

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]

-1/120*(a*(3*(2*(a^2*x^2 + 1)^(3/2)*x/a^2 - sqrt(a^2*x^2 + 1)*x/a^2 - arcsinh(a*x)/a^3)/a^2 - 8*(sqrt(a^2*x^2
+ 1)*x + arcsinh(a*x)/a)/a^4) - 8*(3*(a^2*x^2 + 1)^(3/2)*x^2/a^2 - 2*(a^2*x^2 + 1)^(3/2)/a^4)*arctan(a*x))*sqr
t(c)

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

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 \[ \int x^{3} \sqrt {c \left (a^{2} x^{2} + 1\right )} \operatorname {atan}{\left (a x \right )}\, dx \]

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

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