3.17.54 \(\int \frac {(d+e x)^2}{(a d e+(c d^2+a e^2) x+c d e x^2)^{5/2}} \, dx\)

Optimal. Leaf size=116 \[ \frac {2 e \left (a e^2+c d^2+2 c d e x\right )}{3 c d \left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 (d+e x)}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \]

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Rubi [A]  time = 0.04, antiderivative size = 116, 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 = {652, 613} \begin {gather*} \frac {2 e \left (a e^2+c d^2+2 c d e x\right )}{3 c d \left (c d^2-a e^2\right )^2 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {2 (d+e x)}{3 c d \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^2/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2),x]

[Out]

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

Rule 613

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[(-2*(b + 2*c*x))/((b^2 - 4*a*c)*Sqrt[a + b*x
+ c*x^2]), x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 652

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

Rubi steps

\begin {align*} \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}} \, dx &=-\frac {2 (d+e x)}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}-\frac {e \int \frac {1}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{3 c d}\\ &=-\frac {2 (d+e x)}{3 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}+\frac {2 e \left (c d^2+a e^2+2 c d e x\right )}{3 c d \left (c d^2-a e^2\right )^2 \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 59, normalized size = 0.51 \begin {gather*} -\frac {2 (d+e x)^2 \left (c d (d-2 e x)-3 a e^2\right )}{3 \left (c d^2-a e^2\right )^2 ((d+e x) (a e+c d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^2/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.81, size = 76, normalized size = 0.66 \begin {gather*} \frac {2 \left (3 a e^2-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 \left (a e^2-c d^2\right )^2 (a e+c d x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^2/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2),x]

[Out]

(2*(-(c*d^2) + 3*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*(a*e
+ c*d*x)^2)

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fricas [A]  time = 1.10, size = 156, normalized size = 1.34 \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 - c d^{2} + 3 \, a e^{2}\right )}}{3 \, {\left (a^{2} c^{2} d^{4} e^{2} - 2 \, a^{3} c d^{2} e^{4} + a^{4} e^{6} + {\left (c^{4} d^{6} - 2 \, a c^{3} d^{4} e^{2} + a^{2} c^{2} d^{2} e^{4}\right )} x^{2} + 2 \, {\left (a c^{3} d^{5} e - 2 \, a^{2} c^{2} d^{3} e^{3} + a^{3} c d e^{5}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [B]  time = 0.65, size = 386, normalized size = 3.33 \begin {gather*} \frac {2 \, {\left ({\left ({\left (\frac {2 \, {\left (c^{3} d^{5} e^{3} - 2 \, a c^{2} d^{3} e^{5} + a^{2} c d e^{7}\right )} x}{c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}} + \frac {3 \, {\left (c^{3} d^{6} e^{2} - a c^{2} d^{4} e^{4} - a^{2} c d^{2} e^{6} + a^{3} e^{8}\right )}}{c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}}\right )} x + \frac {6 \, {\left (a c^{2} d^{5} e^{3} - 2 \, a^{2} c d^{3} e^{5} + a^{3} d e^{7}\right )}}{c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}}\right )} x - \frac {c^{3} d^{8} - 5 \, a c^{2} d^{6} e^{2} + 7 \, a^{2} c d^{4} e^{4} - 3 \, a^{3} d^{2} e^{6}}{c^{4} d^{8} - 4 \, a c^{3} d^{6} e^{2} + 6 \, a^{2} c^{2} d^{4} e^{4} - 4 \, a^{3} c d^{2} e^{6} + a^{4} e^{8}}\right )}}{3 \, {\left (c d x^{2} e + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),x, algorithm="giac")

[Out]

2/3*(((2*(c^3*d^5*e^3 - 2*a*c^2*d^3*e^5 + a^2*c*d*e^7)*x/(c^4*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^
3*c*d^2*e^6 + a^4*e^8) + 3*(c^3*d^6*e^2 - a*c^2*d^4*e^4 - a^2*c*d^2*e^6 + a^3*e^8)/(c^4*d^8 - 4*a*c^3*d^6*e^2
+ 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8))*x + 6*(a*c^2*d^5*e^3 - 2*a^2*c*d^3*e^5 + a^3*d*e^7)/(c^4*d^8
- 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8))*x - (c^3*d^8 - 5*a*c^2*d^6*e^2 + 7*a^2*c*
d^4*e^4 - 3*a^3*d^2*e^6)/(c^4*d^8 - 4*a*c^3*d^6*e^2 + 6*a^2*c^2*d^4*e^4 - 4*a^3*c*d^2*e^6 + a^4*e^8))/(c*d*x^2
*e + a*d*e + (c*d^2 + a*e^2)*x)^(3/2)

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maple [A]  time = 0.05, size = 90, normalized size = 0.78 \begin {gather*} \frac {2 \left (c d x +a e \right ) \left (e x +d \right )^{3} \left (2 c d e x +3 a \,e^{2}-c \,d^{2}\right )}{3 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/(c*d*e*x^2+a*d*e+(a*e^2+c*d^2)*x)^(5/2),x)

[Out]

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

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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)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/2),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 = 0.95, size = 72, normalized size = 0.62 \begin {gather*} \frac {2\,\left (-c\,d^2+2\,c\,x\,d\,e+3\,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+c\,d\,x\right )}^2\,{\left (a\,e^2-c\,d^2\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^2/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{2}}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/2),x)

[Out]

Integral((d + e*x)**2/((d + e*x)*(a*e + c*d*x))**(5/2), x)

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