3.1218 \(\int \frac{x^3 (a+b \tan ^{-1}(c x))}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=143 \[ -\frac{a+b \tan ^{-1}(c x)}{e^2 \sqrt{d+e x^2}}+\frac{d \left (a+b \tan ^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}+\frac{b c \left (2 c^2 d-3 e\right ) \tan ^{-1}\left (\frac{x \sqrt{c^2 d-e}}{\sqrt{d+e x^2}}\right )}{3 e^2 \left (c^2 d-e\right )^{3/2}}+\frac{b c x}{3 e \left (c^2 d-e\right ) \sqrt{d+e x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.208724, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {266, 43, 4976, 12, 527, 377, 203} \[ -\frac{a+b \tan ^{-1}(c x)}{e^2 \sqrt{d+e x^2}}+\frac{d \left (a+b \tan ^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}+\frac{b c \left (2 c^2 d-3 e\right ) \tan ^{-1}\left (\frac{x \sqrt{c^2 d-e}}{\sqrt{d+e x^2}}\right )}{3 e^2 \left (c^2 d-e\right )^{3/2}}+\frac{b c x}{3 e \left (c^2 d-e\right ) \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcTan[c*x]))/(d + e*x^2)^(5/2),x]

[Out]

(b*c*x)/(3*(c^2*d - e)*e*Sqrt[d + e*x^2]) + (d*(a + b*ArcTan[c*x]))/(3*e^2*(d + e*x^2)^(3/2)) - (a + b*ArcTan[
c*x])/(e^2*Sqrt[d + e*x^2]) + (b*c*(2*c^2*d - 3*e)*ArcTan[(Sqrt[c^2*d - e]*x)/Sqrt[d + e*x^2]])/(3*(c^2*d - e)
^(3/2)*e^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 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

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 )^{5/2}} \, dx &=\frac{d \left (a+b \tan ^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac{a+b \tan ^{-1}(c x)}{e^2 \sqrt{d+e x^2}}-(b c) \int \frac{-2 d-3 e x^2}{3 e^2 \left (1+c^2 x^2\right ) \left (d+e x^2\right )^{3/2}} \, dx\\ &=\frac{d \left (a+b \tan ^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac{a+b \tan ^{-1}(c x)}{e^2 \sqrt{d+e x^2}}-\frac{(b c) \int \frac{-2 d-3 e x^2}{\left (1+c^2 x^2\right ) \left (d+e x^2\right )^{3/2}} \, dx}{3 e^2}\\ &=\frac{b c x}{3 \left (c^2 d-e\right ) e \sqrt{d+e x^2}}+\frac{d \left (a+b \tan ^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac{a+b \tan ^{-1}(c x)}{e^2 \sqrt{d+e x^2}}+\frac{(b c) \int \frac{d \left (2 c^2 d-3 e\right )}{\left (1+c^2 x^2\right ) \sqrt{d+e x^2}} \, dx}{3 d \left (c^2 d-e\right ) e^2}\\ &=\frac{b c x}{3 \left (c^2 d-e\right ) e \sqrt{d+e x^2}}+\frac{d \left (a+b \tan ^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac{a+b \tan ^{-1}(c x)}{e^2 \sqrt{d+e x^2}}+\frac{\left (b c \left (2 c^2 d-3 e\right )\right ) \int \frac{1}{\left (1+c^2 x^2\right ) \sqrt{d+e x^2}} \, dx}{3 \left (c^2 d-e\right ) e^2}\\ &=\frac{b c x}{3 \left (c^2 d-e\right ) e \sqrt{d+e x^2}}+\frac{d \left (a+b \tan ^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac{a+b \tan ^{-1}(c x)}{e^2 \sqrt{d+e x^2}}+\frac{\left (b c \left (2 c^2 d-3 e\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 )}{3 \left (c^2 d-e\right ) e^2}\\ &=\frac{b c x}{3 \left (c^2 d-e\right ) e \sqrt{d+e x^2}}+\frac{d \left (a+b \tan ^{-1}(c x)\right )}{3 e^2 \left (d+e x^2\right )^{3/2}}-\frac{a+b \tan ^{-1}(c x)}{e^2 \sqrt{d+e x^2}}+\frac{b c \left (2 c^2 d-3 e\right ) \tan ^{-1}\left (\frac{\sqrt{c^2 d-e} x}{\sqrt{d+e x^2}}\right )}{3 \left (c^2 d-e\right )^{3/2} e^2}\\ \end{align*}

Mathematica [C]  time = 0.537823, size = 326, normalized size = 2.28 \[ \frac{2 \sqrt{c^2 d-e} \left (b c e x \left (d+e x^2\right )-a \left (c^2 d-e\right ) \left (2 d+3 e x^2\right )\right )-i b c \left (2 c^2 d-3 e\right ) \left (d+e x^2\right )^{3/2} \log \left (-\frac{12 i e^2 \sqrt{c^2 d-e} \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d-i e x\right )}{b (c x+i) \left (2 c^2 d-3 e\right )}\right )+i b c \left (2 c^2 d-3 e\right ) \left (d+e x^2\right )^{3/2} \log \left (\frac{12 i e^2 \sqrt{c^2 d-e} \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d+i e x\right )}{b (c x-i) \left (2 c^2 d-3 e\right )}\right )-2 b \left (c^2 d-e\right )^{3/2} \tan ^{-1}(c x) \left (2 d+3 e x^2\right )}{6 e^2 \left (c^2 d-e\right )^{3/2} \left (d+e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcTan[c*x]))/(d + e*x^2)^(5/2),x]

[Out]

(2*Sqrt[c^2*d - e]*(b*c*e*x*(d + e*x^2) - a*(c^2*d - e)*(2*d + 3*e*x^2)) - 2*b*(c^2*d - e)^(3/2)*(2*d + 3*e*x^
2)*ArcTan[c*x] - I*b*c*(2*c^2*d - 3*e)*(d + e*x^2)^(3/2)*Log[((-12*I)*Sqrt[c^2*d - e]*e^2*(c*d - I*e*x + Sqrt[
c^2*d - e]*Sqrt[d + e*x^2]))/(b*(2*c^2*d - 3*e)*(I + c*x))] + I*b*c*(2*c^2*d - 3*e)*(d + e*x^2)^(3/2)*Log[((12
*I)*Sqrt[c^2*d - e]*e^2*(c*d + I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(2*c^2*d - 3*e)*(-I + c*x))])/(6*(
c^2*d - e)^(3/2)*e^2*(d + e*x^2)^(3/2))

________________________________________________________________________________________

Maple [F]  time = 0.601, 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{5}{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)^(5/2),x)

[Out]

int(x^3*(a+b*arctan(c*x))/(e*x^2+d)^(5/2),x)

________________________________________________________________________________________

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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 5.67279, size = 1781, normalized size = 12.45 \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)^(5/2),x, algorithm="fricas")

[Out]

[-1/12*((2*b*c^3*d^3 - 3*b*c*d^2*e + (2*b*c^3*d*e^2 - 3*b*c*e^3)*x^4 + 2*(2*b*c^3*d^2*e - 3*b*c*d*e^2)*x^2)*sq
rt(-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^4*d^3 - 4*a*c^2*d^2*e + 2*a*d*
e^2 - (b*c^3*d*e^2 - b*c*e^3)*x^3 + 3*(a*c^4*d^2*e - 2*a*c^2*d*e^2 + a*e^3)*x^2 - (b*c^3*d^2*e - b*c*d*e^2)*x
+ (2*b*c^4*d^3 - 4*b*c^2*d^2*e + 2*b*d*e^2 + 3*(b*c^4*d^2*e - 2*b*c^2*d*e^2 + b*e^3)*x^2)*arctan(c*x))*sqrt(e*
x^2 + d))/(c^4*d^4*e^2 - 2*c^2*d^3*e^3 + d^2*e^4 + (c^4*d^2*e^4 - 2*c^2*d*e^5 + e^6)*x^4 + 2*(c^4*d^3*e^3 - 2*
c^2*d^2*e^4 + d*e^5)*x^2), 1/6*((2*b*c^3*d^3 - 3*b*c*d^2*e + (2*b*c^3*d*e^2 - 3*b*c*e^3)*x^4 + 2*(2*b*c^3*d^2*
e - 3*b*c*d*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*(2*a*c^4*d^3 - 4*a*c^2*d^2*e + 2*a*d*e^2 - (b*c^3*d*e^2 - b*c*e^3)*x
^3 + 3*(a*c^4*d^2*e - 2*a*c^2*d*e^2 + a*e^3)*x^2 - (b*c^3*d^2*e - b*c*d*e^2)*x + (2*b*c^4*d^3 - 4*b*c^2*d^2*e
+ 2*b*d*e^2 + 3*(b*c^4*d^2*e - 2*b*c^2*d*e^2 + b*e^3)*x^2)*arctan(c*x))*sqrt(e*x^2 + d))/(c^4*d^4*e^2 - 2*c^2*
d^3*e^3 + d^2*e^4 + (c^4*d^2*e^4 - 2*c^2*d*e^5 + e^6)*x^4 + 2*(c^4*d^3*e^3 - 2*c^2*d^2*e^4 + d*e^5)*x^2)]

________________________________________________________________________________________

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{5}{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)**(5/2),x)

[Out]

Integral(x**3*(a + b*atan(c*x))/(d + e*x**2)**(5/2), x)

________________________________________________________________________________________

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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate((b*arctan(c*x) + a)*x^3/(e*x^2 + d)^(5/2), x)