3.1206 \(\int \frac{a+b \tan ^{-1}(c x)}{x^2 \sqrt{d+e x^2}} \, dx\)

Optimal. Leaf size=100 \[ -\frac{\sqrt{d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{d x}+\frac{b \sqrt{c^2 d-e} \tanh ^{-1}\left (\frac{c \sqrt{d+e x^2}}{\sqrt{c^2 d-e}}\right )}{d}-\frac{b c \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{\sqrt{d}} \]

[Out]

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

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Rubi [A]  time = 0.176807, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {264, 4976, 446, 83, 63, 208} \[ -\frac{\sqrt{d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{d x}+\frac{b \sqrt{c^2 d-e} \tanh ^{-1}\left (\frac{c \sqrt{d+e x^2}}{\sqrt{c^2 d-e}}\right )}{d}-\frac{b c \tanh ^{-1}\left (\frac{\sqrt{d+e x^2}}{\sqrt{d}}\right )}{\sqrt{d}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 264

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

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 446

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

Rule 83

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

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

Mathematica [C]  time = 0.418306, size = 247, normalized size = 2.47 \[ \frac{-2 a \sqrt{d+e x^2}+b x \sqrt{c^2 d-e} \log \left (-\frac{4 c d \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d-i e x\right )}{b (c x+i) \left (c^2 d-e\right )^{3/2}}\right )+b x \sqrt{c^2 d-e} \log \left (-\frac{4 c d \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d+i e x\right )}{b (c x-i) \left (c^2 d-e\right )^{3/2}}\right )-2 b c \sqrt{d} x \log \left (\sqrt{d} \sqrt{d+e x^2}+d\right )-2 b \tan ^{-1}(c x) \sqrt{d+e x^2}+2 b c \sqrt{d} x \log (x)}{2 d x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*Sqrt[d + e*x^2] - 2*b*Sqrt[d + e*x^2]*ArcTan[c*x] + 2*b*c*Sqrt[d]*x*Log[x] - 2*b*c*Sqrt[d]*x*Log[d + Sqr
t[d]*Sqrt[d + e*x^2]] + b*Sqrt[c^2*d - e]*x*Log[(-4*c*d*(c*d - I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))/(b*(c
^2*d - e)^(3/2)*(I + c*x))] + b*Sqrt[c^2*d - e]*x*Log[(-4*c*d*(c*d + I*e*x + Sqrt[c^2*d - e]*Sqrt[d + e*x^2]))
/(b*(c^2*d - e)^(3/2)*(-I + c*x))])/(2*d*x)

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Maple [F]  time = 0.81, size = 0, normalized size = 0. \begin{align*} \int{\frac{a+b\arctan \left ( cx \right ) }{{x}^{2}}{\frac{1}{\sqrt{e{x}^{2}+d}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a+b*arctan(c*x))/x^2/(e*x^2+d)^(1/2),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((a+b*arctan(c*x))/x^2/(e*x^2+d)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.4444, size = 1504, normalized size = 15.04 \begin{align*} \left [\frac{2 \, b c \sqrt{d} x \log \left (-\frac{e x^{2} - 2 \, \sqrt{e x^{2} + d} \sqrt{d} + 2 \, d}{x^{2}}\right ) + \sqrt{c^{2} d - e} b x \log \left (\frac{c^{4} e^{2} x^{4} + 8 \, c^{4} d^{2} - 8 \, c^{2} d e + 2 \,{\left (4 \, c^{4} d e - 3 \, c^{2} e^{2}\right )} x^{2} + 4 \,{\left (c^{3} e x^{2} + 2 \, c^{3} d - c e\right )} \sqrt{c^{2} d - e} \sqrt{e x^{2} + d} + e^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) - 4 \, \sqrt{e x^{2} + d}{\left (b \arctan \left (c x\right ) + a\right )}}{4 \, d x}, \frac{b c \sqrt{d} x \log \left (-\frac{e x^{2} - 2 \, \sqrt{e x^{2} + d} \sqrt{d} + 2 \, d}{x^{2}}\right ) + \sqrt{-c^{2} d + e} b x \arctan \left (-\frac{{\left (c^{2} e x^{2} + 2 \, c^{2} d - e\right )} \sqrt{-c^{2} d + e} \sqrt{e x^{2} + d}}{2 \,{\left (c^{3} d^{2} - c d e +{\left (c^{3} d e - c e^{2}\right )} x^{2}\right )}}\right ) - 2 \, \sqrt{e x^{2} + d}{\left (b \arctan \left (c x\right ) + a\right )}}{2 \, d x}, \frac{4 \, b c \sqrt{-d} x \arctan \left (\frac{\sqrt{-d}}{\sqrt{e x^{2} + d}}\right ) + \sqrt{c^{2} d - e} b x \log \left (\frac{c^{4} e^{2} x^{4} + 8 \, c^{4} d^{2} - 8 \, c^{2} d e + 2 \,{\left (4 \, c^{4} d e - 3 \, c^{2} e^{2}\right )} x^{2} + 4 \,{\left (c^{3} e x^{2} + 2 \, c^{3} d - c e\right )} \sqrt{c^{2} d - e} \sqrt{e x^{2} + d} + e^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) - 4 \, \sqrt{e x^{2} + d}{\left (b \arctan \left (c x\right ) + a\right )}}{4 \, d x}, \frac{2 \, b c \sqrt{-d} x \arctan \left (\frac{\sqrt{-d}}{\sqrt{e x^{2} + d}}\right ) + \sqrt{-c^{2} d + e} b x \arctan \left (-\frac{{\left (c^{2} e x^{2} + 2 \, c^{2} d - e\right )} \sqrt{-c^{2} d + e} \sqrt{e x^{2} + d}}{2 \,{\left (c^{3} d^{2} - c d e +{\left (c^{3} d e - c e^{2}\right )} x^{2}\right )}}\right ) - 2 \, \sqrt{e x^{2} + d}{\left (b \arctan \left (c x\right ) + a\right )}}{2 \, d x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x^2/(e*x^2+d)^(1/2),x, algorithm="fricas")

[Out]

[1/4*(2*b*c*sqrt(d)*x*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) + sqrt(c^2*d - e)*b*x*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)*sqr
t(e*x^2 + d) + e^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) - 4*sqrt(e*x^2 + d)*(b*arctan(c*x) + a))/(d*x), 1/2*(b*c*sqrt(d
)*x*log(-(e*x^2 - 2*sqrt(e*x^2 + d)*sqrt(d) + 2*d)/x^2) + sqrt(-c^2*d + e)*b*x*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*sqrt(e*x^2 + d)*(b*arct
an(c*x) + a))/(d*x), 1/4*(4*b*c*sqrt(-d)*x*arctan(sqrt(-d)/sqrt(e*x^2 + d)) + sqrt(c^2*d - e)*b*x*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)*s
qrt(e*x^2 + d) + e^2)/(c^4*x^4 + 2*c^2*x^2 + 1)) - 4*sqrt(e*x^2 + d)*(b*arctan(c*x) + a))/(d*x), 1/2*(2*b*c*sq
rt(-d)*x*arctan(sqrt(-d)/sqrt(e*x^2 + d)) + sqrt(-c^2*d + e)*b*x*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*sqrt(e*x^2 + d)*(b*arctan(c*x) + a))/
(d*x)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atan(c*x))/x**2/(e*x**2+d)**(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/x^2/(e*x^2+d)^(1/2),x, algorithm="giac")

[Out]

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