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

Optimal. Leaf size=70 \[ -\frac {e \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2} \sqrt {b d-a e}}-\frac {\sqrt {d+e x}}{b (a+b x)} \]

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Rubi [A]  time = 0.03, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {27, 47, 63, 208} \begin {gather*} -\frac {e \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2} \sqrt {b d-a e}}-\frac {\sqrt {d+e x}}{b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[d + e*x]/(b*(a + b*x))) - (e*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(b^(3/2)*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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, 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 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 {\sqrt {d+e x}}{a^2+2 a b x+b^2 x^2} \, dx &=\int \frac {\sqrt {d+e x}}{(a+b x)^2} \, dx\\ &=-\frac {\sqrt {d+e x}}{b (a+b x)}+\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{2 b}\\ &=-\frac {\sqrt {d+e x}}{b (a+b x)}+\frac {\operatorname {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{b}\\ &=-\frac {\sqrt {d+e x}}{b (a+b x)}-\frac {e \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{b^{3/2} \sqrt {b d-a e}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 69, normalized size = 0.99 \begin {gather*} \frac {e \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {a e-b d}}\right )}{b^{3/2} \sqrt {a e-b d}}-\frac {\sqrt {d+e x}}{b (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.23, size = 91, normalized size = 1.30 \begin {gather*} -\frac {e \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x} \sqrt {a e-b d}}{b d-a e}\right )}{b^{3/2} \sqrt {a e-b d}}-\frac {e \sqrt {d+e x}}{b (a e+b (d+e x)-b d)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 232, normalized size = 3.31 \begin {gather*} \left [\frac {\sqrt {b^{2} d - a b e} {\left (b e x + a e\right )} \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 ) - 2 \, {\left (b^{2} d - a b e\right )} \sqrt {e x + d}}{2 \, {\left (a b^{3} d - a^{2} b^{2} e + {\left (b^{4} d - a b^{3} e\right )} x\right )}}, \frac {\sqrt {-b^{2} d + a b e} {\left (b e x + a e\right )} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) - {\left (b^{2} d - a b e\right )} \sqrt {e x + d}}{a b^{3} d - a^{2} b^{2} e + {\left (b^{4} d - a b^{3} e\right )} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 80, normalized size = 1.14 \begin {gather*} \frac {\arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e}{\sqrt {-b^{2} d + a b e} b} - \frac {\sqrt {x e + d} e}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 64, normalized size = 0.91 \begin {gather*} \frac {e \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}\, b}-\frac {\sqrt {e x +d}\, e}{\left (b e x +a e \right ) b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-e/b*(e*x+d)^(1/2)/(b*e*x+a*e)+e/b/((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((e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^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 = 0.07, size = 61, normalized size = 0.87 \begin {gather*} \frac {e\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{b^{3/2}\,\sqrt {a\,e-b\,d}}-\frac {e\,\sqrt {d+e\,x}}{e\,x\,b^2+a\,e\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(e*atan((b^(1/2)*(d + e*x)^(1/2))/(a*e - b*d)^(1/2)))/(b^(3/2)*(a*e - b*d)^(1/2)) - (e*(d + e*x)^(1/2))/(a*b*e
+ b^2*e*x)

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sympy [B]  time = 39.85, size = 573, normalized size = 8.19 \begin {gather*} - \frac {2 a e^{2} \sqrt {d + e x}}{2 a^{2} b e^{2} - 2 a b^{2} d e + 2 a b^{2} e^{2} x - 2 b^{3} d e x} + \frac {a e^{2} \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} \log {\left (- a^{2} e^{2} \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} + 2 a b d e \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} - b^{2} d^{2} \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} + \sqrt {d + e x} \right )}}{2 b} - \frac {a e^{2} \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} \log {\left (a^{2} e^{2} \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} - 2 a b d e \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} + b^{2} d^{2} \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} + \sqrt {d + e x} \right )}}{2 b} - \frac {d e \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} \log {\left (- a^{2} e^{2} \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} + 2 a b d e \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} - b^{2} d^{2} \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} + \sqrt {d + e x} \right )}}{2} + \frac {d e \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} \log {\left (a^{2} e^{2} \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} - 2 a b d e \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} + b^{2} d^{2} \sqrt {- \frac {1}{b \left (a e - b d\right )^{3}}} + \sqrt {d + e x} \right )}}{2} + \frac {2 d e \sqrt {d + e x}}{2 a^{2} e^{2} - 2 a b d e + 2 a b e^{2} x - 2 b^{2} d e x} + \frac {2 e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e}{b} - d}} \right )}}{b^{2} \sqrt {\frac {a e}{b} - d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*e**2*sqrt(d + e*x)/(2*a**2*b*e**2 - 2*a*b**2*d*e + 2*a*b**2*e**2*x - 2*b**3*d*e*x) + a*e**2*sqrt(-1/(b*(a
*e - b*d)**3))*log(-a**2*e**2*sqrt(-1/(b*(a*e - b*d)**3)) + 2*a*b*d*e*sqrt(-1/(b*(a*e - b*d)**3)) - b**2*d**2*
sqrt(-1/(b*(a*e - b*d)**3)) + sqrt(d + e*x))/(2*b) - a*e**2*sqrt(-1/(b*(a*e - b*d)**3))*log(a**2*e**2*sqrt(-1/
(b*(a*e - b*d)**3)) - 2*a*b*d*e*sqrt(-1/(b*(a*e - b*d)**3)) + b**2*d**2*sqrt(-1/(b*(a*e - b*d)**3)) + sqrt(d +
e*x))/(2*b) - d*e*sqrt(-1/(b*(a*e - b*d)**3))*log(-a**2*e**2*sqrt(-1/(b*(a*e - b*d)**3)) + 2*a*b*d*e*sqrt(-1/
(b*(a*e - b*d)**3)) - b**2*d**2*sqrt(-1/(b*(a*e - b*d)**3)) + sqrt(d + e*x))/2 + d*e*sqrt(-1/(b*(a*e - b*d)**3
))*log(a**2*e**2*sqrt(-1/(b*(a*e - b*d)**3)) - 2*a*b*d*e*sqrt(-1/(b*(a*e - b*d)**3)) + b**2*d**2*sqrt(-1/(b*(a
*e - b*d)**3)) + sqrt(d + e*x))/2 + 2*d*e*sqrt(d + e*x)/(2*a**2*e**2 - 2*a*b*d*e + 2*a*b*e**2*x - 2*b**2*d*e*x
) + 2*e*atan(sqrt(d + e*x)/sqrt(a*e/b - d))/(b**2*sqrt(a*e/b - d))

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