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

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

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Rubi [A]  time = 0.05, 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 = {662, 621, 206} \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 206

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

Rule 621

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 662

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) \operatorname {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 = 0.70, size = 164, normalized size = 1.32 \begin {gather*} \frac {2 \sqrt {(d+e x) (a e+c d x)} \left (\frac {\sqrt {c} \sqrt {d} \sqrt {c d} \sinh ^{-1}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a e+c d x}}{\sqrt {c d} \sqrt {c d^2-a e^2}}\right )}{\sqrt {c d^2-a e^2} \sqrt {a e+c d x} \sqrt {\frac {c d (d+e x)}{c d^2-a e^2}}}-\frac {\sqrt {e}}{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]*Sqrt[c*d]*ArcSinh[(Sqrt[c]*Sqrt[d]*S
qrt[e]*Sqrt[a*e + c*d*x])/(Sqrt[c*d]*Sqrt[c*d^2 - a*e^2])])/(Sqrt[c*d^2 - a*e^2]*Sqrt[a*e + c*d*x]*Sqrt[(c*d*(
d + e*x))/(c*d^2 - a*e^2)])))/e^(3/2)

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IntegrateAlgebraic [B]  time = 0.79, size = 275, normalized size = 2.22 \begin {gather*} -\frac {\sqrt {c d e} \log \left (a^2 e^4+8 c d e x \sqrt {c d e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}-2 a c d^2 e^2-4 a c d e^3 x+c^2 d^4-4 c^2 d^3 e x-8 c^2 d^2 e^2 x^2\right )}{2 e^2}-\frac {2 \sqrt {a d e+a e^2 x+c d^2 x+c d e x^2}}{e (d+e x)}-\frac {\sqrt {c} \sqrt {d} \tanh ^{-1}\left (\frac {2 \sqrt {c} \sqrt {d} \sqrt {e} x \sqrt {c d e}}{a e^2+c d^2}-\frac {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}}{a e^2+c d^2}\right )}{e^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[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*x + a*e^2*x + c*d*e*x^2])/(e*(d + e*x)) - (Sqrt[c]*Sqrt[d]*ArcTanh[(2*Sqrt[c]*Sqrt[d]*S
qrt[e]*Sqrt[c*d*e]*x)/(c*d^2 + a*e^2) - (2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]
)/(c*d^2 + a*e^2)])/e^(3/2) - (Sqrt[c*d*e]*Log[c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4 - 4*c^2*d^3*e*x - 4*a*c*d*e^3
*x - 8*c^2*d^2*e^2*x^2 + 8*c*d*e*Sqrt[c*d*e]*x*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]])/(2*e^2)

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fricas [A]  time = 0.49, size = 326, normalized size = 2.63 \begin {gather*} \left [\frac {{\left (e x + d\right )} \sqrt {\frac {c d}{e}} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} + 4 \, {\left (2 \, c d e^{2} x + c d^{2} e + a e^{3}\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} \sqrt {\frac {c d}{e}} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) - 4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{2 \, {\left (e^{2} x + d e\right )}}, -\frac {{\left (e x + d\right )} \sqrt {-\frac {c d}{e}} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-\frac {c d}{e}}}{2 \, {\left (c^{2} d^{2} e x^{2} + a c d^{2} e + {\left (c^{2} d^{3} + a c d e^{2}\right )} x\right )}}\right ) + 2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{e^{2} x + 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*((e*x + d)*sqrt(c*d/e)*log(8*c^2*d^2*e^2*x^2 + c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4 + 4*(2*c*d*e^2*x + c*d^2
*e + a*e^3)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(c*d/e) + 8*(c^2*d^3*e + a*c*d*e^3)*x) - 4*sqrt(c*
d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^2*x + d*e), -((e*x + d)*sqrt(-c*d/e)*arctan(1/2*sqrt(c*d*e*x^2 + a*d*
e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(-c*d/e)/(c^2*d^2*e*x^2 + a*c*d^2*e + (c^2*d^3 + a*c*d*
e^2)*x)) + 2*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(e^2*x + d*e)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Evaluation time: 0.48Error: Bad Argument Type

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maple [B]  time = 0.05, size = 317, normalized size = 2.56 \begin {gather*} \frac {a c d e \ln \left (\frac {\frac {a \,e^{2}}{2}-\frac {c \,d^{2}}{2}+\left (x +\frac {d}{e}\right ) c d e}{\sqrt {c d e}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{\left (a \,e^{2}-c \,d^{2}\right ) \sqrt {c d e}}-\frac {c^{2} d^{3} \ln \left (\frac {\frac {a \,e^{2}}{2}-\frac {c \,d^{2}}{2}+\left (x +\frac {d}{e}\right ) c d e}{\sqrt {c d e}}+\sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{\left (a \,e^{2}-c \,d^{2}\right ) \sqrt {c d e}\, e}+\frac {2 \sqrt {\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\, c d}{\left (a \,e^{2}-c \,d^{2}\right ) e}-\frac {2 \left (\left (x +\frac {d}{e}\right )^{2} c d e +\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{\left (a \,e^{2}-c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{2} e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/e^2/(a*e^2-c*d^2)/(x+d/e)^2*((x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(3/2)+2/e*c*d/(a*e^2-c*d^2)*((x+d/e)^2*
c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2)+e*c*d/(a*e^2-c*d^2)*ln((1/2*a*e^2-1/2*c*d^2+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((
x+d/e)^2*c*d*e+(a*e^2-c*d^2)*(x+d/e))^(1/2))/(c*d*e)^(1/2)*a-1/e*c^2*d^3/(a*e^2-c*d^2)*ln((1/2*a*e^2-1/2*c*d^2
+(x+d/e)*c*d*e)/(c*d*e)^(1/2)+((x+d/e)^2*c*d*e+(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(a*e^2-c*d^2>0)', see `assume?`
for more details)Is a*e^2-c*d^2 zero or nonzero?

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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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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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