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

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

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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 (b d-a e)}{e^3 \sqrt {d+e x}}-\frac {2 (b d-a e)^2}{3 e^3 (d+e x)^{3/2}}+\frac {2 b^2 \sqrt {d+e x}}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*d - a*e)^2)/(3*e^3*(d + e*x)^(3/2)) + (4*b*(b*d - a*e))/(e^3*Sqrt[d + e*x]) + (2*b^2*Sqrt[d + e*x])/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)^{5/2}} \, dx &=\int \frac {(a+b x)^2}{(d+e x)^{5/2}} \, dx\\ &=\int \left (\frac {(-b d+a e)^2}{e^2 (d+e x)^{5/2}}-\frac {2 b (b d-a e)}{e^2 (d+e x)^{3/2}}+\frac {b^2}{e^2 \sqrt {d+e x}}\right ) \, dx\\ &=-\frac {2 (b d-a e)^2}{3 e^3 (d+e x)^{3/2}}+\frac {4 b (b d-a e)}{e^3 \sqrt {d+e x}}+\frac {2 b^2 \sqrt {d+e x}}{e^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 62, normalized size = 0.93 \begin {gather*} \frac {-2 a^2 e^2-4 a b e (2 d+3 e x)+2 b^2 \left (8 d^2+12 d e x+3 e^2 x^2\right )}{3 e^3 (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.06, size = 72, normalized size = 1.07 \begin {gather*} \frac {2 \left (-a^2 e^2-6 a b e (d+e x)+2 a b d e+b^2 \left (-d^2\right )+3 b^2 (d+e x)^2+6 b^2 d (d+e x)\right )}{3 e^3 (d+e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.39, size = 85, normalized size = 1.27 \begin {gather*} \frac {2 \, {\left (3 \, b^{2} e^{2} x^{2} + 8 \, b^{2} d^{2} - 4 \, a b d e - a^{2} e^{2} + 6 \, {\left (2 \, b^{2} d e - a b e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} 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)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.18, size = 75, normalized size = 1.12 \begin {gather*} 2 \, \sqrt {x e + d} b^{2} e^{\left (-3\right )} + \frac {2 \, {\left (6 \, {\left (x e + d\right )} b^{2} d - b^{2} d^{2} - 6 \, {\left (x e + d\right )} a b e + 2 \, a b d e - a^{2} e^{2}\right )} e^{\left (-3\right )}}{3 \, {\left (x e + d\right )}^{\frac {3}{2}}} \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)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 62, normalized size = 0.93 \begin {gather*} -\frac {2 \left (-3 b^{2} e^{2} x^{2}+6 a b \,e^{2} x -12 b^{2} d e x +a^{2} e^{2}+4 a b d e -8 b^{2} d^{2}\right )}{3 \left (e x +d \right )^{\frac {3}{2}} 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)^(5/2),x)

[Out]

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

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maxima [A]  time = 1.16, size = 72, normalized size = 1.07 \begin {gather*} \frac {2 \, {\left (\frac {3 \, \sqrt {e x + d} b^{2}}{e^{2}} - \frac {b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2} - 6 \, {\left (b^{2} d - a b e\right )} {\left (e x + d\right )}}{{\left (e x + d\right )}^{\frac {3}{2}} 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)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.05, size = 68, normalized size = 1.01 \begin {gather*} \frac {6\,b^2\,{\left (d+e\,x\right )}^2-2\,a^2\,e^2-2\,b^2\,d^2+12\,b^2\,d\,\left (d+e\,x\right )-12\,a\,b\,e\,\left (d+e\,x\right )+4\,a\,b\,d\,e}{3\,e^3\,{\left (d+e\,x\right )}^{3/2}} \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)^(5/2),x)

[Out]

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

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sympy [A]  time = 1.32, size = 265, normalized size = 3.96 \begin {gather*} \begin {cases} - \frac {2 a^{2} e^{2}}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} - \frac {8 a b d e}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} - \frac {12 a b e^{2} x}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} + \frac {16 b^{2} d^{2}}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} + \frac {24 b^{2} d e x}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} + \frac {6 b^{2} e^{2} x^{2}}{3 d e^{3} \sqrt {d + e x} + 3 e^{4} x \sqrt {d + e x}} & \text {for}\: e \neq 0 \\\frac {a^{2} x + a b x^{2} + \frac {b^{2} x^{3}}{3}}{d^{\frac {5}{2}}} & \text {otherwise} \end {cases} \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)**(5/2),x)

[Out]

Piecewise((-2*a**2*e**2/(3*d*e**3*sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x)) - 8*a*b*d*e/(3*d*e**3*sqrt(d + e*x)
+ 3*e**4*x*sqrt(d + e*x)) - 12*a*b*e**2*x/(3*d*e**3*sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x)) + 16*b**2*d**2/(3*
d*e**3*sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x)) + 24*b**2*d*e*x/(3*d*e**3*sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x
)) + 6*b**2*e**2*x**2/(3*d*e**3*sqrt(d + e*x) + 3*e**4*x*sqrt(d + e*x)), Ne(e, 0)), ((a**2*x + a*b*x**2 + b**2
*x**3/3)/d**(5/2), True))

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