3.17.37 \(\int \frac {1}{(d+e x)^2 \sqrt {a d e+(c d^2+a e^2) x+c d e x^2}} \, dx\)

Optimal. Leaf size=111 \[ \frac {4 c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 (d+e x) \left (c d^2-a e^2\right )^2}+\frac {2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 (d+e x)^2 \left (c d^2-a e^2\right )} \]

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Rubi [A]  time = 0.05, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.054, Rules used = {658, 650} \begin {gather*} \frac {4 c d \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 (d+e x) \left (c d^2-a e^2\right )^2}+\frac {2 \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]

Int[1/((d + e*x)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]),x]

[Out]

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

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]

Rule 658

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))/((m + p + 1)*(2*c*d - b*e)), x] + Dist[(c*Simplify[m + 2*p + 2])/((m + p + 1)*(2*c*d -
b*e)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], 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] && ILtQ[Simplify[m + 2*p + 2], 0]

Rubi steps

\begin {align*} \int \frac {1}{(d+e x)^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c 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}}{3 \left (c d^2-a e^2\right ) (d+e x)^2}+\frac {(2 c d) \int \frac {1}{(d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{3 \left (c d^2-a e^2\right )}\\ &=\frac {2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 \left (c d^2-a e^2\right ) (d+e x)^2}+\frac {4 c d \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 \left (c d^2-a e^2\right )^2 (d+e x)}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 61, normalized size = 0.55 \begin {gather*} \frac {2 \sqrt {(d+e x) (a e+c d x)} \left (c d (3 d+2 e x)-a e^2\right )}{3 (d+e x)^2 \left (c d^2-a e^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.73, size = 73, normalized size = 0.66 \begin {gather*} \frac {2 \left (-a e^2+3 c d^2+2 c d e x\right ) \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 )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((d + e*x)^2*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]),x]

[Out]

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

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fricas [A]  time = 0.75, size = 140, normalized size = 1.26 \begin {gather*} \frac {2 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + 3 \, c d^{2} - a e^{2}\right )}}{3 \, {\left (c^{2} d^{6} - 2 \, a c d^{4} e^{2} + a^{2} d^{2} e^{4} + {\left (c^{2} d^{4} e^{2} - 2 \, a c d^{2} e^{4} + a^{2} e^{6}\right )} x^{2} + 2 \, {\left (c^{2} d^{5} e - 2 \, a c d^{3} e^{3} + a^{2} d e^{5}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + 3*c*d^2 - a*e^2)/(c^2*d^6 - 2*a*c*d^4*e^2 + a^2*d
^2*e^4 + (c^2*d^4*e^2 - 2*a*c*d^2*e^4 + a^2*e^6)*x^2 + 2*(c^2*d^5*e - 2*a*c*d^3*e^3 + a^2*d*e^5)*x)

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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(1/(e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2),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: 0.7Error: Bad Argument Type

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maple [A]  time = 0.05, size = 89, normalized size = 0.80 \begin {gather*} -\frac {2 \left (c d x +a e \right ) \left (-2 c d e x +a \,e^{2}-3 c \,d^{2}\right )}{3 \left (e x +d \right ) \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \sqrt {c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(c*d*x+a*e)*(-2*c*d*e*x+a*e^2-3*c*d^2)/(e*x+d)/(a^2*e^4-2*a*c*d^2*e^2+c^2*d^4)/(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(1/(e*x+d)^2/(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(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 = 0.72, size = 69, normalized size = 0.62 \begin {gather*} \frac {2\,\left (3\,c\,d^2+2\,c\,x\,d\,e-a\,e^2\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 )}^2\,{\left (d+e\,x\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\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(1/(e*x+d)**2/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(1/2),x)

[Out]

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

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