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

Optimal. Leaf size=126 \[ \frac {2 x^2}{3 (d+e x) \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {8 a e \left (x \left (a e^2+c d^2\right )+2 a d e\right )}{3 \left (c d^2-a e^2\right )^3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \]

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Rubi [A]  time = 0.11, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {854, 12, 636} \begin {gather*} \frac {2 x^2}{3 (d+e x) \left (c d^2-a e^2\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}-\frac {8 a e \left (x \left (a e^2+c d^2\right )+2 a d e\right )}{3 \left (c d^2-a e^2\right )^3 \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/((d + e*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

(2*x^2)/(3*(c*d^2 - a*e^2)*(d + e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2]) - (8*a*e*(2*a*d*e + (c*d^2 +
a*e^2)*x))/(3*(c*d^2 - a*e^2)^3*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 636

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

Rule 854

Int[(((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*(x_)), x_Symbol] :> -Si
mp[((2*c*d - b*e)*(f + g*x)^n*(a + b*x + c*x^2)^(p + 1))/(e*p*(b^2 - 4*a*c)*(d + e*x)), x] - Dist[1/(d*e*p*(b^
2 - 4*a*c)), Int[(f + g*x)^(n - 1)*(a + b*x + c*x^2)^p*Simp[b*(a*e*g*n - c*d*f*(2*p + 1)) - 2*a*c*(d*g*n - e*f
*(2*p + 1)) - c*g*(b*d - 2*a*e)*(n + 2*p + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*
g, 0] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] && IGtQ[n, 0] && ILtQ[n + 2*p,
0]

Rubi steps

\begin {align*} \int \frac {x^2}{(d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx &=\frac {2 x^2}{3 \left (c d^2-a e^2\right ) (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}+\frac {2 \int -\frac {2 a d e^2 \left (c d^2-a e^2\right ) x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{3 d e \left (c d^2-a e^2\right )^2}\\ &=\frac {2 x^2}{3 \left (c d^2-a e^2\right ) (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {(4 a e) \int \frac {x}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}} \, dx}{3 \left (c d^2-a e^2\right )}\\ &=\frac {2 x^2}{3 \left (c d^2-a e^2\right ) (d+e x) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}-\frac {8 a e \left (2 a d e+\left (c d^2+a e^2\right ) x\right )}{3 \left (c d^2-a e^2\right )^3 \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 = 0.07, size = 99, normalized size = 0.79 \begin {gather*} \frac {-2 a^2 e^2 \left (8 d^2+12 d e x+3 e^2 x^2\right )-4 a c d^2 e x (2 d+3 e x)+2 c^2 d^4 x^2}{3 (d+e x) \left (c d^2-a e^2\right )^3 \sqrt {(d+e x) (a e+c d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/((d + e*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

(2*c^2*d^4*x^2 - 4*a*c*d^2*e*x*(2*d + 3*e*x) - 2*a^2*e^2*(8*d^2 + 12*d*e*x + 3*e^2*x^2))/(3*(c*d^2 - a*e^2)^3*
(d + e*x)*Sqrt[(a*e + c*d*x)*(d + e*x)])

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IntegrateAlgebraic [B]  time = 152.02, size = 20752, normalized size = 164.70 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^2/((d + e*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)),x]

[Out]

Result too large to show

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fricas [B]  time = 2.83, size = 308, normalized size = 2.44 \begin {gather*} -\frac {2 \, {\left (8 \, a^{2} d^{2} e^{2} - {\left (c^{2} d^{4} - 6 \, a c d^{2} e^{2} - 3 \, a^{2} e^{4}\right )} x^{2} + 4 \, {\left (a c d^{3} e + 3 \, a^{2} d e^{3}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{3 \, {\left (a c^{3} d^{8} e - 3 \, a^{2} c^{2} d^{6} e^{3} + 3 \, a^{3} c d^{4} e^{5} - a^{4} d^{2} e^{7} + {\left (c^{4} d^{7} e^{2} - 3 \, a c^{3} d^{5} e^{4} + 3 \, a^{2} c^{2} d^{3} e^{6} - a^{3} c d e^{8}\right )} x^{3} + {\left (2 \, c^{4} d^{8} e - 5 \, a c^{3} d^{6} e^{3} + 3 \, a^{2} c^{2} d^{4} e^{5} + a^{3} c d^{2} e^{7} - a^{4} e^{9}\right )} x^{2} + {\left (c^{4} d^{9} - a c^{3} d^{7} e^{2} - 3 \, a^{2} c^{2} d^{5} e^{4} + 5 \, a^{3} c d^{3} e^{6} - 2 \, a^{4} d e^{8}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(e*x+d)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="fricas")

[Out]

-2/3*(8*a^2*d^2*e^2 - (c^2*d^4 - 6*a*c*d^2*e^2 - 3*a^2*e^4)*x^2 + 4*(a*c*d^3*e + 3*a^2*d*e^3)*x)*sqrt(c*d*e*x^
2 + a*d*e + (c*d^2 + a*e^2)*x)/(a*c^3*d^8*e - 3*a^2*c^2*d^6*e^3 + 3*a^3*c*d^4*e^5 - a^4*d^2*e^7 + (c^4*d^7*e^2
- 3*a*c^3*d^5*e^4 + 3*a^2*c^2*d^3*e^6 - a^3*c*d*e^8)*x^3 + (2*c^4*d^8*e - 5*a*c^3*d^6*e^3 + 3*a^2*c^2*d^4*e^5
+ a^3*c*d^2*e^7 - a^4*e^9)*x^2 + (c^4*d^9 - a*c^3*d^7*e^2 - 3*a^2*c^2*d^5*e^4 + 5*a^3*c*d^3*e^6 - 2*a^4*d*e^8
)*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(x^2/(e*x+d)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to transpose Error: Bad Argument Value

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maple [A]  time = 0.01, size = 145, normalized size = 1.15 \begin {gather*} \frac {2 \left (c d x +a e \right ) \left (3 a^{2} e^{4} x^{2}+6 a c \,d^{2} e^{2} x^{2}-c^{2} d^{4} x^{2}+12 a^{2} d \,e^{3} x +4 a c \,d^{3} e x +8 a^{2} d^{2} e^{2}\right )}{3 \left (a^{3} e^{6}-3 a^{2} c \,d^{2} e^{4}+3 a \,c^{2} d^{4} e^{2}-c^{3} d^{6}\right ) \left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(c*d*x+a*e)*(3*a^2*e^4*x^2+6*a*c*d^2*e^2*x^2-c^2*d^4*x^2+12*a^2*d*e^3*x+4*a*c*d^3*e*x+8*a^2*d^2*e^2)/(a^3*
e^6-3*a^2*c*d^2*e^4+3*a*c^2*d^4*e^2-c^3*d^6)/(c*d*e*x^2+a*e^2*x+c*d^2*x+a*d*e)^(3/2)

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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(x^2/(e*x+d)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2),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 = 3.60, size = 1071, normalized size = 8.50 \begin {gather*} \frac {4\,c\,d^3\,\sqrt {c\,d^2\,x+c\,d\,e\,x^2+a\,d\,e+a\,e^2\,x}}{3\,\left (a^3\,d\,e^7+x\,a^3\,e^8-3\,a^2\,c\,d^3\,e^5-3\,x\,a^2\,c\,d^2\,e^6+3\,a\,c^2\,d^5\,e^3+3\,x\,a\,c^2\,d^4\,e^4-c^3\,d^7\,e-x\,c^3\,d^6\,e^2\right )}-\frac {2\,d^2\,\sqrt {c\,d^2\,x+c\,d\,e\,x^2+a\,d\,e+a\,e^2\,x}}{3\,a^2\,d^2\,e^5+6\,a^2\,d\,e^6\,x+3\,a^2\,e^7\,x^2-6\,a\,c\,d^4\,e^3-12\,a\,c\,d^3\,e^4\,x-6\,a\,c\,d^2\,e^5\,x^2+3\,c^2\,d^6\,e+6\,c^2\,d^5\,e^2\,x+3\,c^2\,d^4\,e^3\,x^2}-\frac {4\,a\,d\,e^2\,\sqrt {c\,d^2\,x+c\,d\,e\,x^2+a\,d\,e+a\,e^2\,x}}{3\,\left (a^3\,d\,e^7+x\,a^3\,e^8-3\,a^2\,c\,d^3\,e^5-3\,x\,a^2\,c\,d^2\,e^6+3\,a\,c^2\,d^5\,e^3+3\,x\,a\,c^2\,d^4\,e^4-c^3\,d^7\,e-x\,c^3\,d^6\,e^2\right )}+\frac {2\,c^4\,d^7\,x}{\sqrt {c\,d^2\,x+c\,d\,e\,x^2+a\,d\,e+a\,e^2\,x}\,\left (a^4\,c\,d\,e^9-4\,a^3\,c^2\,d^3\,e^7+6\,a^2\,c^3\,d^5\,e^5-4\,a\,c^4\,d^7\,e^3+c^5\,d^9\,e\right )}+\frac {22\,a^3\,c\,d^2\,e^5}{3\,\sqrt {c\,d^2\,x+c\,d\,e\,x^2+a\,d\,e+a\,e^2\,x}\,\left (a^4\,c\,d\,e^9-4\,a^3\,c^2\,d^3\,e^7+6\,a^2\,c^3\,d^5\,e^5-4\,a\,c^4\,d^7\,e^3+c^5\,d^9\,e\right )}-\frac {28\,a^2\,c^2\,d^4\,e^3}{3\,\sqrt {c\,d^2\,x+c\,d\,e\,x^2+a\,d\,e+a\,e^2\,x}\,\left (a^4\,c\,d\,e^9-4\,a^3\,c^2\,d^3\,e^7+6\,a^2\,c^3\,d^5\,e^5-4\,a\,c^4\,d^7\,e^3+c^5\,d^9\,e\right )}+\frac {2\,a\,c^3\,d^6\,e}{\sqrt {c\,d^2\,x+c\,d\,e\,x^2+a\,d\,e+a\,e^2\,x}\,\left (a^4\,c\,d\,e^9-4\,a^3\,c^2\,d^3\,e^7+6\,a^2\,c^3\,d^5\,e^5-4\,a\,c^4\,d^7\,e^3+c^5\,d^9\,e\right )}+\frac {10\,a^2\,c^2\,d^3\,e^4\,x}{3\,\sqrt {c\,d^2\,x+c\,d\,e\,x^2+a\,d\,e+a\,e^2\,x}\,\left (a^4\,c\,d\,e^9-4\,a^3\,c^2\,d^3\,e^7+6\,a^2\,c^3\,d^5\,e^5-4\,a\,c^4\,d^7\,e^3+c^5\,d^9\,e\right )}+\frac {2\,a^3\,c\,d\,e^6\,x}{\sqrt {c\,d^2\,x+c\,d\,e\,x^2+a\,d\,e+a\,e^2\,x}\,\left (a^4\,c\,d\,e^9-4\,a^3\,c^2\,d^3\,e^7+6\,a^2\,c^3\,d^5\,e^5-4\,a\,c^4\,d^7\,e^3+c^5\,d^9\,e\right )}-\frac {22\,a\,c^3\,d^5\,e^2\,x}{3\,\sqrt {c\,d^2\,x+c\,d\,e\,x^2+a\,d\,e+a\,e^2\,x}\,\left (a^4\,c\,d\,e^9-4\,a^3\,c^2\,d^3\,e^7+6\,a^2\,c^3\,d^5\,e^5-4\,a\,c^4\,d^7\,e^3+c^5\,d^9\,e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(4*c*d^3*(a*d*e + a*e^2*x + c*d^2*x + c*d*e*x^2)^(1/2))/(3*(a^3*d*e^7 - c^3*d^7*e + a^3*e^8*x + 3*a*c^2*d^5*e^
3 - 3*a^2*c*d^3*e^5 - c^3*d^6*e^2*x + 3*a*c^2*d^4*e^4*x - 3*a^2*c*d^2*e^6*x)) - (2*d^2*(a*d*e + a*e^2*x + c*d^
2*x + c*d*e*x^2)^(1/2))/(3*c^2*d^6*e + 3*a^2*d^2*e^5 + 3*a^2*e^7*x^2 + 6*c^2*d^5*e^2*x + 3*c^2*d^4*e^3*x^2 - 6
*a*c*d^4*e^3 + 6*a^2*d*e^6*x - 12*a*c*d^3*e^4*x - 6*a*c*d^2*e^5*x^2) - (4*a*d*e^2*(a*d*e + a*e^2*x + c*d^2*x +
c*d*e*x^2)^(1/2))/(3*(a^3*d*e^7 - c^3*d^7*e + a^3*e^8*x + 3*a*c^2*d^5*e^3 - 3*a^2*c*d^3*e^5 - c^3*d^6*e^2*x +
3*a*c^2*d^4*e^4*x - 3*a^2*c*d^2*e^6*x)) + (2*c^4*d^7*x)/((a*d*e + a*e^2*x + c*d^2*x + c*d*e*x^2)^(1/2)*(c^5*d
^9*e - 4*a*c^4*d^7*e^3 + 6*a^2*c^3*d^5*e^5 - 4*a^3*c^2*d^3*e^7 + a^4*c*d*e^9)) + (22*a^3*c*d^2*e^5)/(3*(a*d*e
+ a*e^2*x + c*d^2*x + c*d*e*x^2)^(1/2)*(c^5*d^9*e - 4*a*c^4*d^7*e^3 + 6*a^2*c^3*d^5*e^5 - 4*a^3*c^2*d^3*e^7 +
a^4*c*d*e^9)) - (28*a^2*c^2*d^4*e^3)/(3*(a*d*e + a*e^2*x + c*d^2*x + c*d*e*x^2)^(1/2)*(c^5*d^9*e - 4*a*c^4*d^7
*e^3 + 6*a^2*c^3*d^5*e^5 - 4*a^3*c^2*d^3*e^7 + a^4*c*d*e^9)) + (2*a*c^3*d^6*e)/((a*d*e + a*e^2*x + c*d^2*x + c
*d*e*x^2)^(1/2)*(c^5*d^9*e - 4*a*c^4*d^7*e^3 + 6*a^2*c^3*d^5*e^5 - 4*a^3*c^2*d^3*e^7 + a^4*c*d*e^9)) + (10*a^2
*c^2*d^3*e^4*x)/(3*(a*d*e + a*e^2*x + c*d^2*x + c*d*e*x^2)^(1/2)*(c^5*d^9*e - 4*a*c^4*d^7*e^3 + 6*a^2*c^3*d^5*
e^5 - 4*a^3*c^2*d^3*e^7 + a^4*c*d*e^9)) + (2*a^3*c*d*e^6*x)/((a*d*e + a*e^2*x + c*d^2*x + c*d*e*x^2)^(1/2)*(c^
5*d^9*e - 4*a*c^4*d^7*e^3 + 6*a^2*c^3*d^5*e^5 - 4*a^3*c^2*d^3*e^7 + a^4*c*d*e^9)) - (22*a*c^3*d^5*e^2*x)/(3*(a
*d*e + a*e^2*x + c*d^2*x + c*d*e*x^2)^(1/2)*(c^5*d^9*e - 4*a*c^4*d^7*e^3 + 6*a^2*c^3*d^5*e^5 - 4*a^3*c^2*d^3*e
^7 + a^4*c*d*e^9))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {x^{2}}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {3}{2}} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(e*x+d)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/2),x)

[Out]

Integral(x**2/(((d + e*x)*(a*e + c*d*x))**(3/2)*(d + e*x)), x)

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