3.19.45 \(\int \frac {a+b x}{\sqrt {d+e x} (a^2+2 a b x+b^2 x^2)} \, dx\)

Optimal. Leaf size=47 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} \sqrt {b d-a e}} \]

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Rubi [A]  time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {27, 63, 208} \begin {gather*} -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} \sqrt {b d-a e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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

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Mathematica [A]  time = 0.01, size = 47, normalized size = 1.00 \begin {gather*} -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} \sqrt {b d-a e}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.06, size = 57, normalized size = 1.21 \begin {gather*} -\frac {2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{\sqrt {b} \sqrt {a e-b d}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.47, size = 119, normalized size = 2.53 \begin {gather*} \left [\frac {\log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right )}{\sqrt {b^{2} d - a b e}}, \frac {2 \, \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right )}{b^{2} d - a b e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b*e)*sqrt(e*x + d))/(b*x + a))/sqrt(b^2*d - a*b*e), 2*sqrt(-b^2*d
+ a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d))/(b^2*d - a*b*e)]

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giac [A]  time = 0.23, size = 41, normalized size = 0.87 \begin {gather*} \frac {2 \, \arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/sqrt(-b^2*d + a*b*e)

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maple [A]  time = 0.05, size = 37, normalized size = 0.79 \begin {gather*} \frac {2 \arctan \left (\frac {\sqrt {e x +d}\, b}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(b^2*x^2+2*a*b*x+a^2)/(e*x+d)^(1/2),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(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d positive or negative?

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mupad [B]  time = 2.05, size = 38, normalized size = 0.81 \begin {gather*} \frac {2\,\mathrm {atan}\left (\frac {b\,\sqrt {d+e\,x}}{\sqrt {a\,b\,e-b^2\,d}}\right )}{\sqrt {a\,b\,e-b^2\,d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*atan((b*(d + e*x)^(1/2))/(a*b*e - b^2*d)^(1/2)))/(a*b*e - b^2*d)^(1/2)

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sympy [A]  time = 96.51, size = 44, normalized size = 0.94 \begin {gather*} - \frac {2 \operatorname {atan}{\left (\frac {1}{\sqrt {\frac {b}{a e - b d}} \sqrt {d + e x}} \right )}}{\sqrt {\frac {b}{a e - b d}} \left (a e - b d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*atan(1/(sqrt(b/(a*e - b*d))*sqrt(d + e*x)))/(sqrt(b/(a*e - b*d))*(a*e - b*d))

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