3.16.100 \(\int \frac {\sqrt {a d e+(c d^2+a e^2) x+c d e x^2}}{(d+e x)^3} \, dx\)

Optimal. Leaf size=54 \[ \frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 (d+e x)^3 \left (c d^2-a e^2\right )} \]

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Rubi [A]  time = 0.02, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.027, Rules used = {650} \begin {gather*} \frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 (d+e x)^3 \left (c d^2-a e^2\right )} \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)^3,x]

[Out]

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

Rule 650

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^m*(a +
b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)), x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] &&
EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + 2*p + 2, 0]

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

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Mathematica [A]  time = 0.02, size = 43, normalized size = 0.80 \begin {gather*} \frac {2 ((d+e x) (a e+c d x))^{3/2}}{3 (d+e x)^3 \left (c d^2-a e^2\right )} \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)^3,x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.81, size = 62, normalized size = 1.15 \begin {gather*} \frac {2 (a e+c d x) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 (d+e x)^2 \left (c d^2-a e^2\right )} \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)^3,x]

[Out]

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

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fricas [A]  time = 0.53, size = 90, normalized size = 1.67 \begin {gather*} \frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (c d x + a e\right )}}{3 \, {\left (c d^{4} - a d^{2} e^{2} + {\left (c d^{2} e^{2} - a e^{4}\right )} x^{2} + 2 \, {\left (c d^{3} e - a d e^{3}\right )} x\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)^3,x, algorithm="fricas")

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \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)^3,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 2*((-exp(1)^2*(sqrt(a*d*exp(1)+a*x*exp(2
)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)^3*a^2*exp(2)^2+6*c*d^2*exp(1)^2*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^
2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)^3*a*exp(2)-4*c*d^2*exp(1)^4*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^
2*exp(1))-sqrt(c*d*exp(1))*x)^3*a-c^2*d^4*exp(1)^2*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*
d*exp(1))*x)^3-5*d*exp(1)*sqrt(c*d*exp(1))*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1)
)*x)^2*a^2*exp(2)^2+8*d*exp(1)^3*sqrt(c*d*exp(1))*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d
*exp(1))*x)^2*a^2*exp(2)-10*c*d^3*exp(1)*sqrt(c*d*exp(1))*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-
sqrt(c*d*exp(1))*x)^2*a*exp(2)+4*c*d^3*exp(1)^3*sqrt(c*d*exp(1))*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*e
xp(1))-sqrt(c*d*exp(1))*x)^2*a+3*c^2*d^5*exp(1)*sqrt(c*d*exp(1))*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*e
xp(1))-sqrt(c*d*exp(1))*x)^2+d*exp(1)*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)*
a^3*exp(2)^3-d*exp(1)^3*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)*a^3*exp(2)^2+7
*c*d^3*exp(1)*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)*a^2*exp(2)^2-6*c*d^3*exp
(1)^3*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)*a^2*exp(2)-4*c*d^3*exp(1)^5*(sqr
t(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)*a^2+3*c^2*d^5*exp(1)*(sqrt(a*d*exp(1)+a*x*
exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)*a*exp(2)+3*c^2*d^5*exp(1)^3*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d
^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)*a-3*c^3*d^7*exp(1)*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1)
)-sqrt(c*d*exp(1))*x)-d^2*sqrt(c*d*exp(1))*a^3*exp(2)^3+d^2*exp(1)^2*sqrt(c*d*exp(1))*a^3*exp(2)^2-c*d^4*sqrt(
c*d*exp(1))*a^2*exp(2)^2-2*c*d^4*exp(1)^2*sqrt(c*d*exp(1))*a^2*exp(2)+4*c*d^4*exp(1)^4*sqrt(c*d*exp(1))*a^2+c^
2*d^6*sqrt(c*d*exp(1))*a*exp(2)-3*c^2*d^6*exp(1)^2*sqrt(c*d*exp(1))*a+c^3*d^8*sqrt(c*d*exp(1)))/(-8*d*exp(1)^2
*a*exp(2)+8*d*exp(1)^4*a)/(-exp(1)*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)^2+2
*d*sqrt(c*d*exp(1))*(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2*exp(1))-sqrt(c*d*exp(1))*x)-d*a*exp(2)+d*exp(1
)^2*a-c*d^3)^2+(a^2*exp(2)^2+2*c*d^2*a*exp(2)-4*c*d^2*exp(1)^2*a+c^2*d^4)/2/(4*d*exp(1)*a*exp(2)-4*d*exp(1)^3*
a)/sqrt(-a*d*exp(1)^3+a*d*exp(1)*exp(2))*atan((-d*sqrt(c*d*exp(1))+(sqrt(a*d*exp(1)+a*x*exp(2)+c*d^2*x+c*d*x^2
*exp(1))-sqrt(c*d*exp(1))*x)*exp(1))/sqrt(-a*d*exp(1)^3+a*d*exp(1)*exp(2))))

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maple [A]  time = 0.05, size = 58, normalized size = 1.07 \begin {gather*} -\frac {2 \left (c d x +a e \right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}}{3 \left (e x +d \right )^{2} \left (a \,e^{2}-c \,d^{2}\right )} \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)^3,x)

[Out]

-2/3*(c*d*x+a*e)/(e*x+d)^2/(a*e^2-c*d^2)*(c*d*e*x^2+a*e^2*x+c*d^2*x+a*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)^3,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 [B]  time = 1.08, size = 58, normalized size = 1.07 \begin {gather*} -\frac {2\,\left (a\,e+c\,d\,x\right )\,\sqrt {c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e}}{3\,\left (a\,e^2-c\,d^2\right )\,{\left (d+e\,x\right )}^2} \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)^3,x)

[Out]

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

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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 )^{3}}\, 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)**3,x)

[Out]

Integral(sqrt((d + e*x)*(a*e + c*d*x))/(d + e*x)**3, x)

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