3.1221 \(\int \frac{a+b \tan ^{-1}(c x)}{(d+e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=144 \[ \frac{2 x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \sqrt{d+e x^2}}+\frac{x \left (a+b \tan ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac{b \left (3 c^2 d-2 e\right ) \tanh ^{-1}\left (\frac{c \sqrt{d+e x^2}}{\sqrt{c^2 d-e}}\right )}{3 d^2 \left (c^2 d-e\right )^{3/2}}-\frac{b c}{3 d \left (c^2 d-e\right ) \sqrt{d+e x^2}} \]

[Out]

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

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Rubi [A]  time = 0.314336, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 9, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.45, Rules used = {192, 191, 4912, 6688, 12, 571, 78, 63, 208} \[ \frac{2 x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \sqrt{d+e x^2}}+\frac{x \left (a+b \tan ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac{b \left (3 c^2 d-2 e\right ) \tanh ^{-1}\left (\frac{c \sqrt{d+e x^2}}{\sqrt{c^2 d-e}}\right )}{3 d^2 \left (c^2 d-e\right )^{3/2}}-\frac{b c}{3 d \left (c^2 d-e\right ) \sqrt{d+e x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 192

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

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 4912

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

Rule 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 12

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

Rule 571

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x
_Symbol] :> Dist[1/n, Subst[Int[(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^n], x] /; FreeQ[{a, b, c, d, e,
f, m, n, p, q, r}, x] && EqQ[m - n + 1, 0]

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \frac{a+b \tan ^{-1}(c x)}{\left (d+e x^2\right )^{5/2}} \, dx &=\frac{x \left (a+b \tan ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac{2 x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \sqrt{d+e x^2}}-(b c) \int \frac{\frac{x}{3 d \left (d+e x^2\right )^{3/2}}+\frac{2 x}{3 d^2 \sqrt{d+e x^2}}}{1+c^2 x^2} \, dx\\ &=\frac{x \left (a+b \tan ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac{2 x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \sqrt{d+e x^2}}-(b c) \int \frac{x \left (3 d+2 e x^2\right )}{3 d^2 \left (1+c^2 x^2\right ) \left (d+e x^2\right )^{3/2}} \, dx\\ &=\frac{x \left (a+b \tan ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac{2 x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \sqrt{d+e x^2}}-\frac{(b c) \int \frac{x \left (3 d+2 e x^2\right )}{\left (1+c^2 x^2\right ) \left (d+e x^2\right )^{3/2}} \, dx}{3 d^2}\\ &=\frac{x \left (a+b \tan ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac{2 x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \sqrt{d+e x^2}}-\frac{(b c) \operatorname{Subst}\left (\int \frac{3 d+2 e x}{\left (1+c^2 x\right ) (d+e x)^{3/2}} \, dx,x,x^2\right )}{6 d^2}\\ &=-\frac{b c}{3 d \left (c^2 d-e\right ) \sqrt{d+e x^2}}+\frac{x \left (a+b \tan ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac{2 x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \sqrt{d+e x^2}}-\frac{\left (b c \left (3 c^2 d-2 e\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+c^2 x\right ) \sqrt{d+e x}} \, dx,x,x^2\right )}{6 d^2 \left (c^2 d-e\right )}\\ &=-\frac{b c}{3 d \left (c^2 d-e\right ) \sqrt{d+e x^2}}+\frac{x \left (a+b \tan ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac{2 x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \sqrt{d+e x^2}}-\frac{\left (b c \left (3 c^2 d-2 e\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{c^2 d}{e}+\frac{c^2 x^2}{e}} \, dx,x,\sqrt{d+e x^2}\right )}{3 d^2 \left (c^2 d-e\right ) e}\\ &=-\frac{b c}{3 d \left (c^2 d-e\right ) \sqrt{d+e x^2}}+\frac{x \left (a+b \tan ^{-1}(c x)\right )}{3 d \left (d+e x^2\right )^{3/2}}+\frac{2 x \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 \sqrt{d+e x^2}}+\frac{b \left (3 c^2 d-2 e\right ) \tanh ^{-1}\left (\frac{c \sqrt{d+e x^2}}{\sqrt{c^2 d-e}}\right )}{3 d^2 \left (c^2 d-e\right )^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.552874, size = 317, normalized size = 2.2 \[ \frac{2 \sqrt{c^2 d-e} \left (a x \left (c^2 d-e\right ) \left (3 d+2 e x^2\right )-b c d \left (d+e x^2\right )\right )+b \left (3 c^2 d-2 e\right ) \left (d+e x^2\right )^{3/2} \log \left (-\frac{12 c d^2 \sqrt{c^2 d-e} \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d-i e x\right )}{b (c x+i) \left (3 c^2 d-2 e\right )}\right )+b \left (3 c^2 d-2 e\right ) \left (d+e x^2\right )^{3/2} \log \left (-\frac{12 c d^2 \sqrt{c^2 d-e} \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d+i e x\right )}{b (c x-i) \left (3 c^2 d-2 e\right )}\right )+2 b x \left (c^2 d-e\right )^{3/2} \tan ^{-1}(c x) \left (3 d+2 e x^2\right )}{6 d^2 \left (c^2 d-e\right )^{3/2} \left (d+e x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.197, size = 0, normalized size = 0. \begin{align*} \int{(a+b\arctan \left ( cx \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((a+b*arctan(c*x))/(e*x^2+d)^(5/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{3} \, a{\left (\frac{2 \, x}{\sqrt{e x^{2} + d} d^{2}} + \frac{x}{{\left (e x^{2} + d\right )}^{\frac{3}{2}} d}\right )} + 2 \, b \int \frac{\arctan \left (c x\right )}{2 \,{\left (e^{2} x^{4} + 2 \, d e x^{2} + d^{2}\right )} \sqrt{e x^{2} + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/(e*x^2+d)^(5/2),x, algorithm="maxima")

[Out]

1/3*a*(2*x/(sqrt(e*x^2 + d)*d^2) + x/((e*x^2 + d)^(3/2)*d)) + 2*b*integrate(1/2*arctan(c*x)/((e^2*x^4 + 2*d*e*
x^2 + d^2)*sqrt(e*x^2 + d)), x)

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Fricas [B]  time = 5.92089, size = 1778, normalized size = 12.35 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/(e*x^2+d)^(5/2),x, algorithm="fricas")

[Out]

[1/12*((3*b*c^2*d^3 + (3*b*c^2*d*e^2 - 2*b*e^3)*x^4 - 2*b*d^2*e + 2*(3*b*c^2*d^2*e - 2*b*d*e^2)*x^2)*sqrt(c^2*
d - e)*log((c^4*e^2*x^4 + 8*c^4*d^2 - 8*c^2*d*e + 2*(4*c^4*d*e - 3*c^2*e^2)*x^2 + 4*(c^3*e*x^2 + 2*c^3*d - c*e
)*sqrt(c^2*d - e)*sqrt(e*x^2 + d) + e^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) - 4*(b*c^3*d^3 - b*c*d^2*e - 2*(a*c^4*d^2*
e - 2*a*c^2*d*e^2 + a*e^3)*x^3 + (b*c^3*d^2*e - b*c*d*e^2)*x^2 - 3*(a*c^4*d^3 - 2*a*c^2*d^2*e + a*d*e^2)*x - (
2*(b*c^4*d^2*e - 2*b*c^2*d*e^2 + b*e^3)*x^3 + 3*(b*c^4*d^3 - 2*b*c^2*d^2*e + b*d*e^2)*x)*arctan(c*x))*sqrt(e*x
^2 + d))/(c^4*d^6 - 2*c^2*d^5*e + d^4*e^2 + (c^4*d^4*e^2 - 2*c^2*d^3*e^3 + d^2*e^4)*x^4 + 2*(c^4*d^5*e - 2*c^2
*d^4*e^2 + d^3*e^3)*x^2), 1/6*((3*b*c^2*d^3 + (3*b*c^2*d*e^2 - 2*b*e^3)*x^4 - 2*b*d^2*e + 2*(3*b*c^2*d^2*e - 2
*b*d*e^2)*x^2)*sqrt(-c^2*d + e)*arctan(-1/2*(c^2*e*x^2 + 2*c^2*d - e)*sqrt(-c^2*d + e)*sqrt(e*x^2 + d)/(c^3*d^
2 - c*d*e + (c^3*d*e - c*e^2)*x^2)) - 2*(b*c^3*d^3 - b*c*d^2*e - 2*(a*c^4*d^2*e - 2*a*c^2*d*e^2 + a*e^3)*x^3 +
(b*c^3*d^2*e - b*c*d*e^2)*x^2 - 3*(a*c^4*d^3 - 2*a*c^2*d^2*e + a*d*e^2)*x - (2*(b*c^4*d^2*e - 2*b*c^2*d*e^2 +
b*e^3)*x^3 + 3*(b*c^4*d^3 - 2*b*c^2*d^2*e + b*d*e^2)*x)*arctan(c*x))*sqrt(e*x^2 + d))/(c^4*d^6 - 2*c^2*d^5*e
+ d^4*e^2 + (c^4*d^4*e^2 - 2*c^2*d^3*e^3 + d^2*e^4)*x^4 + 2*(c^4*d^5*e - 2*c^2*d^4*e^2 + d^3*e^3)*x^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{b \arctan \left (c x\right ) + a}{{\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((a+b*arctan(c*x))/(e*x^2+d)^(5/2),x, algorithm="giac")

[Out]

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