3.20.48 \(\int \frac {d+e x}{\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\) [1948]

Optimal. Leaf size=134 \[ \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{c d}+\frac {\left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{2 c^{3/2} d^{3/2} \sqrt {e}} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {654, 635, 212} \begin {gather*} \frac {\left (c d^2-a e^2\right ) \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 )}{2 c^{3/2} d^{3/2} \sqrt {e}}+\frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 654

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

Rubi steps

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

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Mathematica [A]
time = 0.34, size = 127, normalized size = 0.95 \begin {gather*} \frac {e (a e+c d x) (d+e x)+\sqrt {\frac {e}{c d}} \left (-c d^2+a e^2\right ) \sqrt {a e+c d x} \sqrt {d+e x} \log \left (-\sqrt {\frac {e}{c d}} \sqrt {a e+c d x}+\sqrt {d+e x}\right )}{c d e \sqrt {(a e+c d x) (d+e x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.66, size = 186, normalized size = 1.39

method	result	size
default	\(e \left (\frac {\sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}}{c d e}-\frac {\left (e^{2} a +c \,d^{2}\right ) \ln \left (\frac {\frac {1}{2} e^{2} a +\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{2 c d e \sqrt {c d e}}\right )+\frac {d \ln \left (\frac {\frac {1}{2} e^{2} a +\frac {1}{2} c \,d^{2}+c d e x}{\sqrt {c d e}}+\sqrt {a d e +\left (e^{2} a +c \,d^{2}\right ) x +c d e \,x^{2}}\right )}{\sqrt {c d e}}\)	\(186\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

e*(1/c/d/e*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)-1/2*(a*e^2+c*d^2)/c/d/e*ln((1/2*e^2*a+1/2*c*d^2+c*d*e*x)/(c
*d*e)^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2))/(c*d*e)^(1/2))+d*ln((1/2*e^2*a+1/2*c*d^2+c*d*e*x)/(c*d*e)
^(1/2)+(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(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((e*x+d)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),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 = 335, normalized size = 2.50 \begin {gather*} \left [\frac {{\left (4 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} c d e - {\left (c d^{2} - a e^{2}\right )} \sqrt {c d} e^{\frac {1}{2}} \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 + c d^{2} + a e^{2}\right )} \sqrt {c d} e^{\frac {1}{2}} + 2 \, {\left (4 \, c^{2} d^{2} x^{2} + 3 \, a c d^{2}\right )} e^{2}\right )\right )} e^{\left (-1\right )}}{4 \, c^{2} d^{2}}, \frac {{\left (2 \, \sqrt {c d^{2} x + a x e^{2} + {\left (c d x^{2} + a d\right )} e} c d e - {\left (c d^{2} - a e^{2}\right )} \sqrt {-c d e} \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}}{2 \, {\left (c^{2} d^{3} x e + a c d x e^{3} + {\left (c^{2} d^{2} x^{2} + a c d^{2}\right )} e^{2}\right )}}\right )\right )} e^{\left (-1\right )}}{2 \, c^{2} d^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.75, size = 132, normalized size = 0.99 \begin {gather*} -\frac {{\left (c d^{2} - a e^{2}\right )} \sqrt {c d} e^{\left (-\frac {1}{2}\right )} \log \left ({\left | -\sqrt {c d} c d^{2} e^{\frac {1}{2}} - 2 \, {\left (\sqrt {c d} x e^{\frac {1}{2}} - \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}\right )} c d e - \sqrt {c d} a e^{\frac {5}{2}} \right |}\right )}{2 \, c^{2} d^{2}} + \frac {\sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e}}{c d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(c*d^2 - a*e^2)*sqrt(c*d)*e^(-1/2)*log(abs(-sqrt(c*d)*c*d^2*e^(1/2) - 2*(sqrt(c*d)*x*e^(1/2) - sqrt(c*d*x
^2*e + c*d^2*x + a*x*e^2 + a*d*e))*c*d*e - sqrt(c*d)*a*e^(5/2)))/(c^2*d^2) + sqrt(c*d*x^2*e + c*d^2*x + a*x*e^
2 + a*d*e)/(c*d)

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Mupad [B]
time = 1.07, size = 144, normalized size = 1.07 \begin {gather*} \frac {\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{c\,d}-\frac {a\,e^3\,\ln \left (2\,\sqrt {\left (a\,e+c\,d\,x\right )\,\left (d+e\,x\right )}\,\sqrt {c\,d\,e}+a\,e^2+c\,d^2+2\,c\,d\,e\,x\right )}{2\,{\left (c\,d\,e\right )}^{3/2}}+\frac {c\,d^2\,e\,\ln \left (2\,\sqrt {\left (a\,e+c\,d\,x\right )\,\left (d+e\,x\right )}\,\sqrt {c\,d\,e}+a\,e^2+c\,d^2+2\,c\,d\,e\,x\right )}{2\,{\left (c\,d\,e\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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