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

Optimal. Leaf size=67 \[ -\frac {4 b \sqrt {d+e x} (b d-a e)}{e^3}-\frac {2 (b d-a e)^2}{e^3 \sqrt {d+e x}}+\frac {2 b^2 (d+e x)^{3/2}}{3 e^3} \]

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Rubi [A]  time = 0.02, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {27, 43} \begin {gather*} -\frac {4 b \sqrt {d+e x} (b d-a e)}{e^3}-\frac {2 (b d-a e)^2}{e^3 \sqrt {d+e x}}+\frac {2 b^2 (d+e x)^{3/2}}{3 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 59, normalized size = 0.88 \begin {gather*} \frac {2 \left (-3 a^2 e^2+6 a b e (2 d+e x)+b^2 \left (-8 d^2-4 d e x+e^2 x^2\right )\right )}{3 e^3 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-3*a^2*e^2 + 6*a*b*e*(2*d + e*x) + b^2*(-8*d^2 - 4*d*e*x + e^2*x^2)))/(3*e^3*Sqrt[d + e*x])

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IntegrateAlgebraic [A]  time = 0.05, size = 71, normalized size = 1.06 \begin {gather*} \frac {2 \left (-3 a^2 e^2+6 a b e (d+e x)+6 a b d e-3 b^2 d^2+b^2 (d+e x)^2-6 b^2 d (d+e x)\right )}{3 e^3 \sqrt {d+e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-3*b^2*d^2 + 6*a*b*d*e - 3*a^2*e^2 - 6*b^2*d*(d + e*x) + 6*a*b*e*(d + e*x) + b^2*(d + e*x)^2))/(3*e^3*Sqrt
[d + e*x])

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fricas [A]  time = 0.39, size = 73, normalized size = 1.09 \begin {gather*} \frac {2 \, {\left (b^{2} e^{2} x^{2} - 8 \, b^{2} d^{2} + 12 \, a b d e - 3 \, a^{2} e^{2} - 2 \, {\left (2 \, b^{2} d e - 3 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{4} x + d e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.19, size = 83, normalized size = 1.24 \begin {gather*} \frac {2}{3} \, {\left ({\left (x e + d\right )}^{\frac {3}{2}} b^{2} e^{6} - 6 \, \sqrt {x e + d} b^{2} d e^{6} + 6 \, \sqrt {x e + d} a b e^{7}\right )} e^{\left (-9\right )} - \frac {2 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} e^{\left (-3\right )}}{\sqrt {x e + d}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*((x*e + d)^(3/2)*b^2*e^6 - 6*sqrt(x*e + d)*b^2*d*e^6 + 6*sqrt(x*e + d)*a*b*e^7)*e^(-9) - 2*(b^2*d^2 - 2*a*
b*d*e + a^2*e^2)*e^(-3)/sqrt(x*e + d)

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maple [A]  time = 0.05, size = 63, normalized size = 0.94 \begin {gather*} -\frac {2 \left (-b^{2} e^{2} x^{2}-6 a b \,e^{2} x +4 b^{2} d e x +3 a^{2} e^{2}-12 a b d e +8 b^{2} d^{2}\right )}{3 \sqrt {e x +d}\, e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(-b^2*e^2*x^2-6*a*b*e^2*x+4*b^2*d*e*x+3*a^2*e^2-12*a*b*d*e+8*b^2*d^2)/(e*x+d)^(1/2)/e^3

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maxima [A]  time = 1.16, size = 75, normalized size = 1.12 \begin {gather*} \frac {2 \, {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}} b^{2} - 6 \, {\left (b^{2} d - a b e\right )} \sqrt {e x + d}}{e^{2}} - \frac {3 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )}}{\sqrt {e x + d} e^{2}}\right )}}{3 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*(((e*x + d)^(3/2)*b^2 - 6*(b^2*d - a*b*e)*sqrt(e*x + d))/e^2 - 3*(b^2*d^2 - 2*a*b*d*e + a^2*e^2)/(sqrt(e*x
+ d)*e^2))/e

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mupad [B]  time = 0.56, size = 67, normalized size = 1.00 \begin {gather*} \frac {\frac {2\,b^2\,{\left (d+e\,x\right )}^2}{3}-2\,a^2\,e^2-2\,b^2\,d^2-4\,b^2\,d\,\left (d+e\,x\right )+4\,a\,b\,e\,\left (d+e\,x\right )+4\,a\,b\,d\,e}{e^3\,\sqrt {d+e\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*b^2*(d + e*x)^2)/3 - 2*a^2*e^2 - 2*b^2*d^2 - 4*b^2*d*(d + e*x) + 4*a*b*e*(d + e*x) + 4*a*b*d*e)/(e^3*(d +
e*x)^(1/2))

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sympy [A]  time = 13.71, size = 65, normalized size = 0.97 \begin {gather*} \frac {2 b^{2} \left (d + e x\right )^{\frac {3}{2}}}{3 e^{3}} + \frac {\sqrt {d + e x} \left (4 a b e - 4 b^{2} d\right )}{e^{3}} - \frac {2 \left (a e - b d\right )^{2}}{e^{3} \sqrt {d + e x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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