3.13.9 \(\int \frac {x^3 (a+b \text {ArcTan}(c x))}{(d+e x^2)^{3/2}} \, dx\) [1209]

Optimal. Leaf size=137 \[ \frac {d (a+b \text {ArcTan}(c x))}{e^2 \sqrt {d+e x^2}}+\frac {\sqrt {d+e x^2} (a+b \text {ArcTan}(c x))}{e^2}-\frac {b \left (2 c^2 d-e\right ) \text {ArcTan}\left (\frac {\sqrt {c^2 d-e} x}{\sqrt {d+e x^2}}\right )}{c \sqrt {c^2 d-e} e^2}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c e^{3/2}} \]

[Out]

-b*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/c/e^(3/2)-b*(2*c^2*d-e)*arctan(x*(c^2*d-e)^(1/2)/(e*x^2+d)^(1/2))/c/e^2/
(c^2*d-e)^(1/2)+d*(a+b*arctan(c*x))/e^2/(e*x^2+d)^(1/2)+(a+b*arctan(c*x))*(e*x^2+d)^(1/2)/e^2

________________________________________________________________________________________

Rubi [A]
time = 0.14, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {272, 45, 5096, 12, 537, 223, 212, 385, 209} \begin {gather*} \frac {\sqrt {d+e x^2} (a+b \text {ArcTan}(c x))}{e^2}+\frac {d (a+b \text {ArcTan}(c x))}{e^2 \sqrt {d+e x^2}}-\frac {b \left (2 c^2 d-e\right ) \text {ArcTan}\left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{c e^2 \sqrt {c^2 d-e}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(d*(a + b*ArcTan[c*x]))/(e^2*Sqrt[d + e*x^2]) + (Sqrt[d + e*x^2]*(a + b*ArcTan[c*x]))/e^2 - (b*(2*c^2*d - e)*A
rcTan[(Sqrt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/(c*Sqrt[c^2*d - e]*e^2) - (b*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])
/(c*e^(3/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 385

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 537

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 5096

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u =
IntHide[(f*x)^m*(d + e*x^2)^q, x]}, Dist[a + b*ArcTan[c*x], u, x] - Dist[b*c, Int[SimplifyIntegrand[u/(1 + c^
2*x^2), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && ((IGtQ[q, 0] &&  !(ILtQ[(m - 1)/2, 0] && GtQ[m +
2*q + 3, 0])) || (IGtQ[(m + 1)/2, 0] &&  !(ILtQ[q, 0] && GtQ[m + 2*q + 3, 0])) || (ILtQ[(m + 2*q + 1)/2, 0] &&
!ILtQ[(m - 1)/2, 0]))

Rubi steps

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

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 0.45, size = 321, normalized size = 2.34 \begin {gather*} \frac {\frac {2 a \left (2 d+e x^2\right )}{\sqrt {d+e x^2}}+\frac {2 b \left (2 d+e x^2\right ) \text {ArcTan}(c x)}{\sqrt {d+e x^2}}-\frac {i b \left (2 c^2 d-e\right ) \log \left (\frac {4 c^2 e^2 \left (-i c d+e x-i \sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \sqrt {c^2 d-e} \left (2 c^2 d-e\right ) (-i+c x)}\right )}{c \sqrt {c^2 d-e}}+\frac {i b \left (2 c^2 d-e\right ) \log \left (\frac {4 c^2 e^2 \left (i c d+e x+i \sqrt {c^2 d-e} \sqrt {d+e x^2}\right )}{b \sqrt {c^2 d-e} \left (2 c^2 d-e\right ) (i+c x)}\right )}{c \sqrt {c^2 d-e}}-\frac {2 b \sqrt {e} \log \left (e x+\sqrt {e} \sqrt {d+e x^2}\right )}{c}}{2 e^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*(2*d + e*x^2))/Sqrt[d + e*x^2] + (2*b*(2*d + e*x^2)*ArcTan[c*x])/Sqrt[d + e*x^2] - (I*b*(2*c^2*d - e)*Lo
g[(4*c^2*e^2*((-I)*c*d + e*x - I*Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*Sqrt[c^2*d - e]*(2*c^2*d - e)*(-I + c*x)
)])/(c*Sqrt[c^2*d - e]) + (I*b*(2*c^2*d - e)*Log[(4*c^2*e^2*(I*c*d + e*x + I*Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))
/(b*Sqrt[c^2*d - e]*(2*c^2*d - e)*(I + c*x))])/(c*Sqrt[c^2*d - e]) - (2*b*Sqrt[e]*Log[e*x + Sqrt[e]*Sqrt[d + e
*x^2]])/c)/(2*e^2)

________________________________________________________________________________________

Maple [F]
time = 0.14, size = 0, normalized size = 0.00 \[\int \frac {x^{3} \left (a +b \arctan \left (c x \right )\right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(x^2*e^(-1)/sqrt(x^2*e + d) + 2*d*e^(-2)/sqrt(x^2*e + d))*a + 2*b*integrate(1/2*x^3*arctan(c*x)/(x^2*e + d)^(3
/2), x)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 322 vs. \(2 (123) = 246\).
time = 3.28, size = 681, normalized size = 4.97 \begin {gather*} \left [\frac {2 \, {\left (b c^{2} d^{2} - b x^{2} e^{2} + {\left (b c^{2} d x^{2} - b d\right )} e\right )} e^{\frac {1}{2}} \log \left (-2 \, x^{2} e + 2 \, \sqrt {x^{2} e + d} x e^{\frac {1}{2}} - d\right ) + {\left (2 \, b c^{2} d^{2} - b x^{2} e^{2} + {\left (2 \, b c^{2} d x^{2} - b d\right )} e\right )} \sqrt {-c^{2} d + e} \log \left (\frac {c^{4} d^{2} x^{4} - 6 \, c^{2} d^{2} x^{2} + 8 \, x^{4} e^{2} - 4 \, {\left (c^{2} d x^{3} - 2 \, x^{3} e - d x\right )} \sqrt {-c^{2} d + e} \sqrt {x^{2} e + d} + d^{2} - 8 \, {\left (c^{2} d x^{4} - d x^{2}\right )} e}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) + 4 \, {\left (2 \, a c^{3} d^{2} - a c x^{2} e^{2} + {\left (2 \, b c^{3} d^{2} - b c x^{2} e^{2} + {\left (b c^{3} d x^{2} - 2 \, b c d\right )} e\right )} \arctan \left (c x\right ) + {\left (a c^{3} d x^{2} - 2 \, a c d\right )} e\right )} \sqrt {x^{2} e + d}}{4 \, {\left (c^{3} d^{2} e^{2} - c x^{2} e^{4} + {\left (c^{3} d x^{2} - c d\right )} e^{3}\right )}}, \frac {{\left (b c^{2} d^{2} - b x^{2} e^{2} + {\left (b c^{2} d x^{2} - b d\right )} e\right )} e^{\frac {1}{2}} \log \left (-2 \, x^{2} e + 2 \, \sqrt {x^{2} e + d} x e^{\frac {1}{2}} - d\right ) - {\left (2 \, b c^{2} d^{2} - b x^{2} e^{2} + {\left (2 \, b c^{2} d x^{2} - b d\right )} e\right )} \sqrt {c^{2} d - e} \arctan \left (\frac {{\left (c^{2} d x^{2} - 2 \, x^{2} e - d\right )} \sqrt {c^{2} d - e} \sqrt {x^{2} e + d}}{2 \, {\left (c^{2} d^{2} x - x^{3} e^{2} + {\left (c^{2} d x^{3} - d x\right )} e\right )}}\right ) + 2 \, {\left (2 \, a c^{3} d^{2} - a c x^{2} e^{2} + {\left (2 \, b c^{3} d^{2} - b c x^{2} e^{2} + {\left (b c^{3} d x^{2} - 2 \, b c d\right )} e\right )} \arctan \left (c x\right ) + {\left (a c^{3} d x^{2} - 2 \, a c d\right )} e\right )} \sqrt {x^{2} e + d}}{2 \, {\left (c^{3} d^{2} e^{2} - c x^{2} e^{4} + {\left (c^{3} d x^{2} - c d\right )} e^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/4*(2*(b*c^2*d^2 - b*x^2*e^2 + (b*c^2*d*x^2 - b*d)*e)*e^(1/2)*log(-2*x^2*e + 2*sqrt(x^2*e + d)*x*e^(1/2) - d
) + (2*b*c^2*d^2 - b*x^2*e^2 + (2*b*c^2*d*x^2 - b*d)*e)*sqrt(-c^2*d + e)*log((c^4*d^2*x^4 - 6*c^2*d^2*x^2 + 8*
x^4*e^2 - 4*(c^2*d*x^3 - 2*x^3*e - d*x)*sqrt(-c^2*d + e)*sqrt(x^2*e + d) + d^2 - 8*(c^2*d*x^4 - d*x^2)*e)/(c^4
*x^4 + 2*c^2*x^2 + 1)) + 4*(2*a*c^3*d^2 - a*c*x^2*e^2 + (2*b*c^3*d^2 - b*c*x^2*e^2 + (b*c^3*d*x^2 - 2*b*c*d)*e
)*arctan(c*x) + (a*c^3*d*x^2 - 2*a*c*d)*e)*sqrt(x^2*e + d))/(c^3*d^2*e^2 - c*x^2*e^4 + (c^3*d*x^2 - c*d)*e^3),
1/2*((b*c^2*d^2 - b*x^2*e^2 + (b*c^2*d*x^2 - b*d)*e)*e^(1/2)*log(-2*x^2*e + 2*sqrt(x^2*e + d)*x*e^(1/2) - d)
- (2*b*c^2*d^2 - b*x^2*e^2 + (2*b*c^2*d*x^2 - b*d)*e)*sqrt(c^2*d - e)*arctan(1/2*(c^2*d*x^2 - 2*x^2*e - d)*sqr
t(c^2*d - e)*sqrt(x^2*e + d)/(c^2*d^2*x - x^3*e^2 + (c^2*d*x^3 - d*x)*e)) + 2*(2*a*c^3*d^2 - a*c*x^2*e^2 + (2*
b*c^3*d^2 - b*c*x^2*e^2 + (b*c^3*d*x^2 - 2*b*c*d)*e)*arctan(c*x) + (a*c^3*d*x^2 - 2*a*c*d)*e)*sqrt(x^2*e + d))
/(c^3*d^2*e^2 - c*x^2*e^4 + (c^3*d*x^2 - c*d)*e^3)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{3} \left (a + b \operatorname {atan}{\left (c x \right )}\right )}{\left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atan(c*x))/(e*x**2+d)**(3/2),x)

[Out]

Integral(x**3*(a + b*atan(c*x))/(d + e*x**2)**(3/2), x)

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctan(c*x))/(e*x^2+d)^(3/2),x, algorithm="giac")

[Out]

sage0*x

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________