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]
(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))
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Rubi [A] time = 0.184364, antiderivative size = 137, normalized size of antiderivative = 1.,
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 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 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 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]))
Rule 12
Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] && !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]
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 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 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 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])
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 = 0.615544, 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)
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Maple [F] time = 0.605, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3} \left ( a+b\arctan \left ( cx \right ) \right ) \left ( e{x}^{2}+d \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}
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)
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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*(a+b*arctan(c*x))/(e*x^2+d)^(3/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 5.2971, size = 2724, normalized size = 19.88 \begin{align*} \text{result too large to display} \end{align*}
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)]
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}
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)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \arctan \left (c x\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
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]
integrate((b*arctan(c*x) + a)*x^3/(e*x^2 + d)^(3/2), x)