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]

b*arctanh(c*(e*x^2+d)^(1/2)/(c^2*d-e)^(1/2))/d/(c^2*d-e)^(1/2)+x*(a+b*arctan(c*x))/d/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, 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 12

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

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 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 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]

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 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])

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*}

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Mathematica [C]  time = 0.29, 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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fricas [B]  time = 0.54, size = 388, normalized size = 5.54 \[ \left [\frac {{\left (b e x^{2} + b d\right )} \sqrt {c^{2} d - e} \log \left (\frac {c^{4} e^{2} x^{4} + 8 \, c^{4} d^{2} - 8 \, c^{2} d e + 2 \, {\left (4 \, c^{4} d e - 3 \, c^{2} e^{2}\right )} x^{2} + 4 \, {\left (c^{3} e x^{2} + 2 \, c^{3} d - c e\right )} \sqrt {c^{2} d - e} \sqrt {e x^{2} + d} + e^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) + 4 \, \sqrt {e x^{2} + d} {\left ({\left (b c^{2} d - b e\right )} x \arctan \left (c x\right ) + {\left (a c^{2} d - a e\right )} x\right )}}{4 \, {\left (c^{2} d^{3} - d^{2} e + {\left (c^{2} d^{2} e - d e^{2}\right )} x^{2}\right )}}, \frac {{\left (b e x^{2} + b d\right )} \sqrt {-c^{2} d + e} \arctan \left (-\frac {{\left (c^{2} e x^{2} + 2 \, c^{2} d - e\right )} \sqrt {-c^{2} d + e} \sqrt {e x^{2} + d}}{2 \, {\left (c^{3} d^{2} - c d e + {\left (c^{3} d e - c e^{2}\right )} x^{2}\right )}}\right ) + 2 \, \sqrt {e x^{2} + d} {\left ({\left (b c^{2} d - b e\right )} x \arctan \left (c x\right ) + {\left (a c^{2} d - a e\right )} x\right )}}{2 \, {\left (c^{2} d^{3} - d^{2} e + {\left (c^{2} d^{2} e - d e^{2}\right )} x^{2}\right )}}\right ] \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

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]

sage0*x

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maple [F]  time = 2.14, size = 0, normalized size = 0.00 \[ \int \frac {a +b \arctan \left (c x \right )}{\left (e \,x^{2}+d \right )^{\frac {3}{2}}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e-c^2*d>0)', see `assume?` for
more details)Is e-c^2*d positive or negative?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{{\left (e\,x^2+d\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \operatorname {atan}{\left (c x \right )}}{\left (d + e x^{2}\right )^{\frac {3}{2}}}\, dx \]

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]

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

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