3.1209 \(\int \frac {x^3 (a+b \tan ^{-1}(c x))}{(d+e x^2)^{3/2}} \, dx\)

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

[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 = 0.18, 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 = {266, 43, 4976, 12, 523, 217, 206, 377, 203} \[ \frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{e^2}+\frac {d \left (a+b \tan ^{-1}(c x)\right )}{e^2 \sqrt {d+e x^2}}-\frac {b \left (2 c^2 d-e\right ) \tan ^{-1}\left (\frac {x \sqrt {c^2 d-e}}{\sqrt {d+e x^2}}\right )}{c e^2 \sqrt {c^2 d-e}}-\frac {b \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{c e^{3/2}} \]

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 43

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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 266

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 377

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 523

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 4976

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 \operatorname {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 ) \operatorname {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]  time = 0.66, size = 321, normalized size = 2.34 \[ \frac {\frac {2 a \left (2 d+e x^2\right )}{\sqrt {d+e x^2}}-\frac {i b \left (2 c^2 d-e\right ) \log \left (\frac {4 c^2 e^2 \left (-i \sqrt {c^2 d-e} \sqrt {d+e x^2}-i c d+e x\right )}{b (c x-i) \sqrt {c^2 d-e} \left (2 c^2 d-e\right )}\right )}{c \sqrt {c^2 d-e}}+\frac {i b \left (2 c^2 d-e\right ) \log \left (\frac {4 c^2 e^2 \left (i \sqrt {c^2 d-e} \sqrt {d+e x^2}+i c d+e x\right )}{b (c x+i) \sqrt {c^2 d-e} \left (2 c^2 d-e\right )}\right )}{c \sqrt {c^2 d-e}}-\frac {2 b \sqrt {e} \log \left (\sqrt {e} \sqrt {d+e x^2}+e x\right )}{c}+\frac {2 b \tan ^{-1}(c x) \left (2 d+e x^2\right )}{\sqrt {d+e x^2}}}{2 e^2} \]

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)

________________________________________________________________________________________

fricas [B]  time = 0.89, size = 1291, normalized size = 9.42 \[ \text {result too large to display} \]

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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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

________________________________________________________________________________________

maple [F]  time = 1.12, 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 \[ a {\left (\frac {x^{2}}{\sqrt {e x^{2} + d} e} + \frac {2 \, d}{\sqrt {e x^{2} + d} e^{2}}\right )} + 2 \, b \int \frac {x^{3} \arctan \left (c x\right )}{2 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}}}\,{d x} \]

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]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________