3.5.40 \(\int \frac {\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}}{d+e x} \, dx\) [440]

Optimal. Leaf size=131 \[ \frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e}-\frac {\left (c d^2-a e^2\right ) \tanh ^{-1}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{2 \sqrt {c} \sqrt {d} e^{3/2}} \]

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

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Rubi [A]
time = 0.04, antiderivative size = 131, 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 = {678, 635, 212} \begin {gather*} \frac {\sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{e}-\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 \sqrt {c} \sqrt {d} e^{3/2}} \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),x]

[Out]

Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]/e - ((c*d^2 - a*e^2)*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])])/(2*Sqrt[c]*Sqrt[d]*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 678

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 + 2*p + 1))), x] - Dist[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1))), Int[(d + e*x)^(m + 1)*
(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] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && NeQ[m + 2*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} \, dx &=\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e}-\frac {\left (2 c d^2 e-e \left (c d^2+a e^2\right )\right ) \int \frac {1}{\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{2 e^2}\\ &=\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e}-\frac {\left (2 c d^2 e-e \left (c d^2+a e^2\right )\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 )}{e^2}\\ &=\frac {\sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{e}-\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 \sqrt {c} \sqrt {d} e^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 107, normalized size = 0.82 \begin {gather*} \frac {\sqrt {(a e+c d x) (d+e x)} \left (e+\frac {\sqrt {\frac {e}{c d}} \left (c d^2-a e^2\right ) \log \left (-\sqrt {\frac {e}{c d}} \sqrt {a e+c d x}+\sqrt {d+e x}\right )}{\sqrt {a e+c d x} \sqrt {d+e x}}\right )}{e^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),x]

[Out]

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

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Maple [A]
time = 0.10, size = 131, normalized size = 1.00

method	result	size
default	\(\frac {\sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}+\frac {\left (a \,e^{2}-c \,d^{2}\right ) \ln \left (\frac {\frac {a \,e^{2}}{2}-\frac {c \,d^{2}}{2}+c d e \left (x +\frac {d}{e}\right )}{\sqrt {c d e}}+\sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{2 \sqrt {c d e}}}{e}\)	\(131\)

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),x,method=_RETURNVERBOSE)

[Out]

1/e*((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*a*e^2-1/2*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),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 = 1.39, size = 334, normalized size = 2.55 \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 (-2\right )}}{4 \, c d}, \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 (-2\right )}}{2 \, c d}\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),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^(-2)/(c*d), 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^(-2)/(c*d)
]

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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 )}}{d + e x}\, 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),x)

[Out]

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

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Giac [A]
time = 2.28, size = 128, normalized size = 0.98 \begin {gather*} \sqrt {c d x^{2} e + c d^{2} x + a x e^{2} + a d e} e^{\left (-1\right )} + \frac {{\left (c d^{2} - a e^{2}\right )} \sqrt {c d} e^{\left (-\frac {3}{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 d} \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),x, algorithm="giac")

[Out]

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

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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}}{d+e\,x} \,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),x)

[Out]

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

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