3.1212 \(\int \frac{a+b \tan ^{-1}(c x)}{(d+e x^2)^{3/2}} \, dx\)

Optimal. Leaf size=70 \[ \frac{x \left (a+b \tan ^{-1}(c x)\right )}{d \sqrt{d+e x^2}}+\frac{b \tanh ^{-1}\left (\frac{c \sqrt{d+e x^2}}{\sqrt{c^2 d-e}}\right )}{d \sqrt{c^2 d-e}} \]

[Out]

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

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Rubi [A]  time = 0.0769833, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.3, Rules used = {191, 4912, 12, 444, 63, 208} \[ \frac{x \left (a+b \tan ^{-1}(c x)\right )}{d \sqrt{d+e x^2}}+\frac{b \tanh ^{-1}\left (\frac{c \sqrt{d+e x^2}}{\sqrt{c^2 d-e}}\right )}{d \sqrt{c^2 d-e}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 12

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

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(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] && EqQ[m
- n + 1, 0]

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

Mathematica [C]  time = 0.265082, size = 202, normalized size = 2.89 \[ \frac{\frac{2 a x}{\sqrt{d+e x^2}}+\frac{b \log \left (-\frac{4 c d \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d-i e x\right )}{b (c x+i) \sqrt{c^2 d-e}}\right )}{\sqrt{c^2 d-e}}+\frac{b \log \left (-\frac{4 c d \left (\sqrt{c^2 d-e} \sqrt{d+e x^2}+c d+i e x\right )}{b (c x-i) \sqrt{c^2 d-e}}\right )}{\sqrt{c^2 d-e}}+\frac{2 b x \tan ^{-1}(c x)}{\sqrt{d+e x^2}}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.188, size = 0, normalized size = 0. \begin{align*} \int{(a+b\arctan \left ( cx \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((a+b*arctan(c*x))/(e*x^2+d)^(3/2),x)

[Out]

int((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((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 = 2.35641, size = 810, normalized size = 11.57 \begin{align*} \left [\frac{{\left (b e x^{2} + b d\right )} \sqrt{c^{2} d - e} \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 ({\left (b c^{2} d - b e\right )} x \arctan \left (c x\right ) +{\left (a c^{2} d - a e\right )} x\right )}}{4 \,{\left (c^{2} d^{3} - d^{2} e +{\left (c^{2} d^{2} e - d e^{2}\right )} x^{2}\right )}}, \frac{{\left (b e x^{2} + b d\right )} \sqrt{-c^{2} d + e} \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 ({\left (b c^{2} d - b e\right )} x \arctan \left (c x\right ) +{\left (a c^{2} d - a e\right )} x\right )}}{2 \,{\left (c^{2} d^{3} - d^{2} e +{\left (c^{2} d^{2} e - d e^{2}\right )} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctan(c*x))/(e*x^2+d)^(3/2),x, algorithm="fricas")

[Out]

[1/4*((b*e*x^2 + b*d)*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*sqrt(e*
x^2 + d)*((b*c^2*d - b*e)*x*arctan(c*x) + (a*c^2*d - a*e)*x))/(c^2*d^3 - d^2*e + (c^2*d^2*e - d*e^2)*x^2), 1/2
*((b*e*x^2 + b*d)*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*sqrt(e*x^2 + d)*((b*c^2*d - b*e)*x*arctan(c*x) + (a*c^2*d - a*e)*x)
)/(c^2*d^3 - d^2*e + (c^2*d^2*e - d*e^2)*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)**(3/2),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{3}{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)^(3/2),x, algorithm="giac")

[Out]

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