3.18.15 \(\int \frac {\sqrt {d+e x}}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [1715]

Optimal. Leaf size=168 \[ -\frac {e \sqrt {d+e x}}{4 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {d+e x}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{3/2} (b d-a e)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 168, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {660, 43, 44, 65, 214} \begin {gather*} -\frac {e \sqrt {d+e x}}{4 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {\sqrt {d+e x}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{3/2} \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

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, n
}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, -1] &&  !IntegerQ[n] && GtQ[n, 0]

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]

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int \frac {\sqrt {d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {d+e x}}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {\sqrt {d+e x}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (e \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right )^2 \sqrt {d+e x}} \, dx}{4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {e \sqrt {d+e x}}{4 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {d+e x}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (e^2 \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) \sqrt {d+e x}} \, dx}{8 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {e \sqrt {d+e x}}{4 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {d+e x}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left (e \left (a b+b^2 x\right )\right ) \text {Subst}\left (\int \frac {1}{a b-\frac {b^2 d}{e}+\frac {b^2 x^2}{e}} \, dx,x,\sqrt {d+e x}\right )}{4 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {e \sqrt {d+e x}}{4 b (b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {d+e x}}{2 b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x) \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {b d-a e}}\right )}{4 b^{3/2} (b d-a e)^{3/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]
time = 0.34, size = 121, normalized size = 0.72 \begin {gather*} -\frac {\sqrt {b} \sqrt {-b d+a e} \sqrt {d+e x} (a e-b (2 d+e x))-e^2 (a+b x)^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {d+e x}}{\sqrt {-b d+a e}}\right )}{4 b^{3/2} (-b d+a e)^{3/2} (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.68, size = 200, normalized size = 1.19

method	result	size
default	\(\frac {\left (\arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) b^{2} e^{2} x^{2}+2 \arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a b \,e^{2} x +\left (e x +d \right )^{\frac {3}{2}} \sqrt {b \left (a e -b d \right )}\, b +\arctan \left (\frac {b \sqrt {e x +d}}{\sqrt {b \left (a e -b d \right )}}\right ) a^{2} e^{2}-\sqrt {b \left (a e -b d \right )}\, \sqrt {e x +d}\, a e +\sqrt {e x +d}\, \sqrt {b \left (a e -b d \right )}\, b d \right ) \left (b x +a \right )}{4 \sqrt {b \left (a e -b d \right )}\, \left (a e -b d \right ) b \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(200\)

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)^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 1.30, size = 451, normalized size = 2.68 \begin {gather*} \left [-\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {b^{2} d - a b e} e^{2} \log \left (\frac {2 \, b d + {\left (b x - a\right )} e - 2 \, \sqrt {b^{2} d - a b e} \sqrt {x e + d}}{b x + a}\right ) + 2 \, {\left (2 \, b^{3} d^{2} - {\left (a b^{2} x - a^{2} b\right )} e^{2} + {\left (b^{3} d x - 3 \, a b^{2} d\right )} e\right )} \sqrt {x e + d}}{8 \, {\left (b^{6} d^{2} x^{2} + 2 \, a b^{5} d^{2} x + a^{2} b^{4} d^{2} + {\left (a^{2} b^{4} x^{2} + 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )} e^{2} - 2 \, {\left (a b^{5} d x^{2} + 2 \, a^{2} b^{4} d x + a^{3} b^{3} d\right )} e\right )}}, -\frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )} \sqrt {-b^{2} d + a b e} \arctan \left (\frac {\sqrt {-b^{2} d + a b e} \sqrt {x e + d}}{b x e + b d}\right ) e^{2} + {\left (2 \, b^{3} d^{2} - {\left (a b^{2} x - a^{2} b\right )} e^{2} + {\left (b^{3} d x - 3 \, a b^{2} d\right )} e\right )} \sqrt {x e + d}}{4 \, {\left (b^{6} d^{2} x^{2} + 2 \, a b^{5} d^{2} x + a^{2} b^{4} d^{2} + {\left (a^{2} b^{4} x^{2} + 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )} e^{2} - 2 \, {\left (a b^{5} d x^{2} + 2 \, a^{2} b^{4} d x + a^{3} b^{3} d\right )} e\right )}}\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)^(3/2),x, algorithm="fricas")

[Out]

[-1/8*((b^2*x^2 + 2*a*b*x + a^2)*sqrt(b^2*d - a*b*e)*e^2*log((2*b*d + (b*x - a)*e - 2*sqrt(b^2*d - a*b*e)*sqrt
(x*e + d))/(b*x + a)) + 2*(2*b^3*d^2 - (a*b^2*x - a^2*b)*e^2 + (b^3*d*x - 3*a*b^2*d)*e)*sqrt(x*e + d))/(b^6*d^
2*x^2 + 2*a*b^5*d^2*x + a^2*b^4*d^2 + (a^2*b^4*x^2 + 2*a^3*b^3*x + a^4*b^2)*e^2 - 2*(a*b^5*d*x^2 + 2*a^2*b^4*d
*x + a^3*b^3*d)*e), -1/4*((b^2*x^2 + 2*a*b*x + a^2)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^2*d + a*b*e)*sqrt(x*e
+ d)/(b*x*e + b*d))*e^2 + (2*b^3*d^2 - (a*b^2*x - a^2*b)*e^2 + (b^3*d*x - 3*a*b^2*d)*e)*sqrt(x*e + d))/(b^6*d^
2*x^2 + 2*a*b^5*d^2*x + a^2*b^4*d^2 + (a^2*b^4*x^2 + 2*a^3*b^3*x + a^4*b^2)*e^2 - 2*(a*b^5*d*x^2 + 2*a^2*b^4*d
*x + a^3*b^3*d)*e)]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {d + e x}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \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)**(3/2),x)

[Out]

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

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Giac [A]
time = 0.84, size = 156, normalized size = 0.93 \begin {gather*} -\frac {\arctan \left (\frac {\sqrt {x e + d} b}{\sqrt {-b^{2} d + a b e}}\right ) e^{2}}{4 \, {\left (b^{2} d \mathrm {sgn}\left (b x + a\right ) - a b e \mathrm {sgn}\left (b x + a\right )\right )} \sqrt {-b^{2} d + a b e}} - \frac {{\left (x e + d\right )}^{\frac {3}{2}} b e^{2} + \sqrt {x e + d} b d e^{2} - \sqrt {x e + d} a e^{3}}{4 \, {\left (b^{2} d \mathrm {sgn}\left (b x + a\right ) - a b e \mathrm {sgn}\left (b x + a\right )\right )} {\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \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)^(3/2),x, algorithm="giac")

[Out]

-1/4*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^2*d*sgn(b*x + a) - a*b*e*sgn(b*x + a))*sqrt(-b^2*d +
a*b*e)) - 1/4*((x*e + d)^(3/2)*b*e^2 + sqrt(x*e + d)*b*d*e^2 - sqrt(x*e + d)*a*e^3)/((b^2*d*sgn(b*x + a) - a*
b*e*sgn(b*x + a))*((x*e + d)*b - b*d + a*e)^2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {d+e\,x}}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \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)^(3/2),x)

[Out]

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

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