3.2004 \(\int \frac {\sqrt {d+e x}}{a d e+(c d^2+a e^2) x+c d e x^2} \, dx\)

Optimal. Leaf size=65 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c} \sqrt {d} \sqrt {c d^2-a e^2}} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 65, 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 = {626, 63, 208} \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c} \sqrt {d} \sqrt {c d^2-a e^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 626

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

Rubi steps

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

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Mathematica [A]  time = 0.02, size = 65, normalized size = 1.00 \[ -\frac {2 \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c} \sqrt {d} \sqrt {c d^2-a e^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.02, size = 155, normalized size = 2.38 \[ \left [\frac {\log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {c^{2} d^{3} - a c d e^{2}} \sqrt {e x + d}}{c d x + a e}\right )}{\sqrt {c^{2} d^{3} - a c d e^{2}}}, \frac {2 \, \sqrt {-c^{2} d^{3} + a c d e^{2}} \arctan \left (\frac {\sqrt {-c^{2} d^{3} + a c d e^{2}} \sqrt {e x + d}}{c d e x + c d^{2}}\right )}{c^{2} d^{3} - a c d e^{2}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 7.80, size = 1470, normalized size = 22.62 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*c^4*d^6 - 4*a*c^3*d^4*e^2 - sqrt(2)*sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2 - sqr
t(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*c*d)*c^2*d^4 + 2*a^2*c^2*d^2*e^4 - 2*sqrt(2)*sqrt(c^2*d^4 - 2*a*c*d^2*e^2
+ a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2 - sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*c*d)*c^2*d^3 + 2*sqrt(2)*sqrt(
c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2 - sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*c*d)*a*
c*d^2*e^2 - sqrt(2)*sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2 - sqrt(c^2*d^4 - 2*a*c*d
^2*e^2 + a^2*e^4)*c*d)*c^2*d^2 + 2*sqrt(2)*sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2 -
sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*c*d)*a*c*d*e^2 - 2*(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*c^2*d^2 - sqrt
(2)*sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2 - sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4
)*c*d)*a^2*e^4)*arctan(2*sqrt(1/2)*sqrt(x*e + d)/sqrt(-(c*d^2 - a*e^2 + sqrt((c*d^2 - a*e^2)^2))/(c*d)))/((c^4
*d^8 + 2*c^4*d^7 - 4*a*c^3*d^6*e^2 + c^4*d^6 - 6*a*c^3*d^5*e^2 + 6*a^2*c^2*d^4*e^4 - 2*a*c^3*d^4*e^2 + 6*a^2*c
^2*d^3*e^4 - 4*a^3*c*d^2*e^6 + a^2*c^2*d^2*e^4 - 2*a^3*c*d*e^6 + a^4*e^8)*abs(c)*abs(d)) - (2*c^4*d^6 - 4*a*c^
3*d^4*e^2 - sqrt(2)*sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2 + sqrt(c^2*d^4 - 2*a*c*d
^2*e^2 + a^2*e^4)*c*d)*c^2*d^4 + 2*a^2*c^2*d^2*e^4 - 2*sqrt(2)*sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c
^2*d^3 + a*c*d*e^2 + sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*c*d)*c^2*d^3 + 2*sqrt(2)*sqrt(c^2*d^4 - 2*a*c*d^2
*e^2 + a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2 + sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*c*d)*a*c*d^2*e^2 - sqrt(2)
*sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2 + sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*c
*d)*c^2*d^2 + 2*sqrt(2)*sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2 + sqrt(c^2*d^4 - 2*a
*c*d^2*e^2 + a^2*e^4)*c*d)*a*c*d*e^2 - 2*(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*c^2*d^2 - sqrt(2)*sqrt(c^2*d^4 -
2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2 + sqrt(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*c*d)*a^2*e^4)*arc
tan(2*sqrt(1/2)*sqrt(x*e + d)/sqrt(-(c*d^2 - a*e^2 - sqrt((c*d^2 - a*e^2)^2))/(c*d)))/((c^4*d^8 + 2*c^4*d^7 -
4*a*c^3*d^6*e^2 + c^4*d^6 - 6*a*c^3*d^5*e^2 + 6*a^2*c^2*d^4*e^4 - 2*a*c^3*d^4*e^2 + 6*a^2*c^2*d^3*e^4 - 4*a^3*
c*d^2*e^6 + a^2*c^2*d^2*e^4 - 2*a^3*c*d*e^6 + a^4*e^8)*abs(c)*abs(d))

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maple [A]  time = 0.05, size = 48, normalized size = 0.74 \[ \frac {2 \arctan \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}}\right )}{\sqrt {\left (a \,e^{2}-c \,d^{2}\right ) c d}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [B]  time = 0.65, size = 49, normalized size = 0.75 \[ \frac {2\,\mathrm {atan}\left (\frac {c\,d\,\sqrt {d+e\,x}}{\sqrt {a\,c\,d\,e^2-c^2\,d^3}}\right )}{\sqrt {a\,c\,d\,e^2-c^2\,d^3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 4.85, size = 48, normalized size = 0.74 \[ \frac {2 \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e^{2} - c d^{2}}{c d}}} \right )}}{c d \sqrt {\frac {a e^{2} - c d^{2}}{c d}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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