3.21.4 \(\int \frac {\sqrt {d+e x}}{a d e+(c d^2+a e^2) x+c d e x^2} \, dx\) [2004]

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 = 0.03, 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 = {640, 65, 214} \begin {gather*} -\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 {gather*}

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 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 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^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 \text {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 = 0.05, size = 65, normalized size = 1.00 \begin {gather*} \frac {2 \tan ^{-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 {gather*}

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*ArcTan[(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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Maple [A]
time = 0.73, size = 48, normalized size = 0.74

method	result	size
derivativedivides	\(\frac {2 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\)	\(48\)
default	\(\frac {2 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.57, size = 154, normalized size = 2.37 \begin {gather*} \left [\frac {\log \left (\frac {c d x e + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {c^{2} d^{3} - a c d e^{2}} \sqrt {x e + 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 {x e + d}}{c d x e + c d^{2}}\right )}{c^{2} d^{3} - a c d e^{2}}\right ] \end {gather*}

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

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Sympy [A]
time = 1.57, size = 48, normalized size = 0.74 \begin {gather*} \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}}} \end {gather*}

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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Giac [A]
time = 1.09, size = 48, normalized size = 0.74 \begin {gather*} \frac {2 \, \arctan \left (\frac {\sqrt {x e + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{\sqrt {-c^{2} d^{3} + a c d e^{2}}} \end {gather*}

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

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Mupad [B]
time = 0.65, size = 49, normalized size = 0.75 \begin {gather*} \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}} \end {gather*}

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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