\(\int \frac {1}{\sqrt {d+e x} (a^2+2 a b x+b^2 x^2)} \, dx\) [1650]

Optimal result

Integrand size = 28, antiderivative size = 76 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {\sqrt {d+e x}}{(b d-a e) (a+b x)}+\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {27, 44, 65, 214} \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {e \text {arctanh}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}}-\frac {\sqrt {d+e x}}{(a+b x) (b d-a e)} \]

[In]

Int[1/(Sqrt[d + e*x]*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

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

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 44

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rubi steps \begin{align*} \text {integral}& = \int \frac {1}{(a+b x)^2 \sqrt {d+e x}} \, dx \\ & = -\frac {\sqrt {d+e x}}{(b d-a e) (a+b x)}-\frac {e \int \frac {1}{(a+b x) \sqrt {d+e x}} \, dx}{2 (b d-a e)} \\ & = -\frac {\sqrt {d+e x}}{(b d-a e) (a+b x)}-\frac {\text {Subst}\left (\int \frac {1}{a-\frac {b d}{e}+\frac {b x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{b d-a e} \\ & = -\frac {\sqrt {d+e x}}{(b d-a e) (a+b x)}+\frac {e \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{\sqrt {b} (b d-a e)^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.99 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {\sqrt {d+e x}}{(-b d+a e) (a+b x)}+\frac {e \arctan \left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{\sqrt {b} (-b d+a e)^{3/2}} \]

[In]

Integrate[1/(Sqrt[d + e*x]*(a^2 + 2*a*b*x + b^2*x^2)),x]

[Out]

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

Maple [A] (verified)

Time = 2.27 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.84

method	result	size
pseudoelliptic	\(\frac {\frac {\sqrt {e x +d}}{b x +a}+\frac {e \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{\sqrt {\left (a e -b d \right ) b}}}{a e -b d}\)	\(64\)
derivativedivides	\(2 e \left (\frac {\sqrt {e x +d}}{2 \left (a e -b d \right ) \left (b \left (e x +d \right )+a e -b d \right )}+\frac {\arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}}\right )\)	\(87\)
default	\(2 e \left (\frac {\sqrt {e x +d}}{2 \left (a e -b d \right ) \left (b \left (e x +d \right )+a e -b d \right )}+\frac {\arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {\left (a e -b d \right ) b}}\right )}{2 \left (a e -b d \right ) \sqrt {\left (a e -b d \right ) b}}\right )\)	\(87\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (64) = 128\).

Time = 0.38 (sec) , antiderivative size = 280, normalized size of antiderivative = 3.68 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\left [-\frac {\sqrt {b^{2} d - a b e} {\left (b e x + a e\right )} \log \left (\frac {b e x + 2 \, b d - a e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {e x + d}}{b x + a}\right ) + 2 \, {\left (b^{2} d - a b e\right )} \sqrt {e x + d}}{2 \, {\left (a b^{3} d^{2} - 2 \, a^{2} b^{2} d e + a^{3} b e^{2} + {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} x\right )}}, -\frac {\sqrt {-b^{2} d + a b e} {\left (b e x + a e\right )} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {e x + d}}{b e x + b d}\right ) + {\left (b^{2} d - a b e\right )} \sqrt {e x + d}}{a b^{3} d^{2} - 2 \, a^{2} b^{2} d e + a^{3} b e^{2} + {\left (b^{4} d^{2} - 2 \, a b^{3} d e + a^{2} b^{2} e^{2}\right )} x}\right ] \]

[In]

integrate(1/(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^2 - 2*a^2*b^2*d*e + a^3*b*e^2 + (b^4*d^2 - 2*a*b^3*d*e + a^2
*b^2*e^2)*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^2 - 2*a^2*b^2*d*e + a^3*b*e^2 + (b^4*d^2 - 2*a*b^3*d*e + a^2*b^2*e^2)*
x)]

Sympy [F]

\[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\int \frac {1}{\left (a + b x\right )^{2} \sqrt {d + e x}}\, dx \]

[In]

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

[Out]

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

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(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 detail

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.14 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=-\frac {e \arctan \left (\frac {\sqrt {e x + d} b}{\sqrt {-b^{2} d + a b e}}\right )}{\sqrt {-b^{2} d + a b e} {\left (b d - a e\right )}} - \frac {\sqrt {e x + d} e}{{\left ({\left (e x + d\right )} b - b d + a e\right )} {\left (b d - a e\right )}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.97 \[ \int \frac {1}{\sqrt {d+e x} \left (a^2+2 a b x+b^2 x^2\right )} \, dx=\frac {e\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {d+e\,x}}{\sqrt {a\,e-b\,d}}\right )}{\sqrt {b}\,{\left (a\,e-b\,d\right )}^{3/2}}+\frac {e\,\sqrt {d+e\,x}}{\left (a\,e-b\,d\right )\,\left (a\,e-b\,d+b\,\left (d+e\,x\right )\right )} \]

[In]

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