3.20.13 \(\int \frac {\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}}{(d+e x)^2} \, dx\) [1913]

Optimal. Leaf size=124 \[ -\frac {2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e (d+e x)}+\frac {\sqrt {c} \sqrt {d} \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{e^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.081, Rules used = {676, 635, 212} \begin {gather*} \frac {\sqrt {c} \sqrt {d} \tanh ^{-1}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{e^{3/2}}-\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e (d+e x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 676

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((
a + b*x + c*x^2)^p/(e*(m + p + 1))), x] - Dist[c*(p/(e^2*(m + p + 1))), Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2
)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[
p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rubi steps

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

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Mathematica [A]
time = 0.14, size = 110, normalized size = 0.89 \begin {gather*} \frac {2 \sqrt {(a e+c d x) (d+e x)} \left (-\frac {\sqrt {e}}{d+e x}+\frac {\sqrt {c} \sqrt {d} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {a e+c d x}}{\sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{\sqrt {a e+c d x} \sqrt {d+e x}}\right )}{e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.70, size = 212, normalized size = 1.71

method	result	size
default	\(\frac {-\frac {2 \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{2}}+\frac {2 c d e \left (\sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}+\frac {\left (e^{2} a -c \,d^{2}\right ) \ln \left (\frac {\frac {e^{2} a}{2}-\frac {c \,d^{2}}{2}+c d e \left (x +\frac {d}{e}\right )}{\sqrt {c d e}}+\sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{2 \sqrt {c d e}}\right )}{e^{2} a -c \,d^{2}}}{e^{2}}\)	\(212\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/e^2*(-2/(a*e^2-c*d^2)/(x+d/e)^2*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^(3/2)+2*c*d*e/(a*e^2-c*d^2)*((c*d*e*
(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^(1/2)+1/2*(a*e^2-c*d^2)*ln((1/2*e^2*a-1/2*c*d^2+c*d*e*(x+d/e))/(c*d*e)^(1/2)+
(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(1/2)))

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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((a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/(e*x+d)^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(c*d^2-%e^2*a>0)', see `assume?
` for more d

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Fricas [A]
time = 3.57, size = 321, normalized size = 2.59 \begin {gather*} \left [\frac {\sqrt {c d} {\left (x e + d\right )} e^{\left (-\frac {1}{2}\right )} \log \left (8 \, c^{2} d^{3} x e + c^{2} d^{4} + 8 \, a c d x e^{3} + a^{2} e^{4} + 4 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (2 \, c d x e^{2} + c d^{2} e + a e^{3}\right )} \sqrt {c d} e^{\left (-\frac {1}{2}\right )} + 2 \, {\left (4 \, c^{2} d^{2} x^{2} + 3 \, a c d^{2}\right )} e^{2}\right ) - 4 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}{2 \, {\left (x e^{2} + d e\right )}}, -\frac {\sqrt {-c d e^{\left (-1\right )}} {\left (x e + d\right )} \arctan \left (\frac {\sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} {\left (2 \, c d x e + c d^{2} + a e^{2}\right )} \sqrt {-c d e^{\left (-1\right )}}}{2 \, {\left (c^{2} d^{3} x + a c d x e^{2} + {\left (c^{2} d^{2} x^{2} + a c d^{2}\right )} e\right )}}\right ) + 2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e}}{x e^{2} + d e}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.83, size = 181, normalized size = 1.46 \begin {gather*} -2 \, {\left (\frac {c d \arctan \left (\frac {\sqrt {c d e - \frac {c d^{2} e}{x e + d} + \frac {a e^{3}}{x e + d}}}{\sqrt {-c d e}}\right ) e^{\left (-2\right )} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{\sqrt {-c d e}} + \sqrt {c d e - \frac {c d^{2} e}{x e + d} + \frac {a e^{3}}{x e + d}} e^{\left (-3\right )} \mathrm {sgn}\left (\frac {1}{x e + d}\right ) - \frac {{\left (c d \arctan \left (\frac {\sqrt {c d} e^{\frac {1}{2}}}{\sqrt {-c d e}}\right ) e + \sqrt {-c d e} \sqrt {c d} e^{\frac {1}{2}}\right )} e^{\left (-3\right )} \mathrm {sgn}\left (\frac {1}{x e + d}\right )}{\sqrt {-c d e}}\right )} e \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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