\(\int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^{3/2}} \, dx\) [1626]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 26, antiderivative size = 67 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^{3/2}} \, dx=-\frac {2 (b d-a e)^2}{e^3 \sqrt {d+e x}}-\frac {4 b (b d-a e) \sqrt {d+e x}}{e^3}+\frac {2 b^2 (d+e x)^{3/2}}{3 e^3} \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {27, 45} \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^{3/2}} \, dx=-\frac {4 b \sqrt {d+e x} (b d-a e)}{e^3}-\frac {2 (b d-a e)^2}{e^3 \sqrt {d+e x}}+\frac {2 b^2 (d+e x)^{3/2}}{3 e^3} \]

[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 45

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*} \text {integral}& = \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*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^{3/2}} \, dx=\frac {2 \left (-3 a^2 e^2+6 a b e (2 d+e x)+b^2 \left (-8 d^2-4 d e x+e^2 x^2\right )\right )}{3 e^3 \sqrt {d+e x}} \]

[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])

Maple [A] (verified)

Time = 2.33 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.82

method result size
pseudoelliptic \(\frac {\frac {2 \left (x^{2} e^{2}-4 d e x -8 d^{2}\right ) b^{2}}{3}+8 \left (\frac {e x}{2}+d \right ) e a b -2 a^{2} e^{2}}{\sqrt {e x +d}\, e^{3}}\) \(55\)
risch \(\frac {2 b \left (b e x +6 a e -5 b d \right ) \sqrt {e x +d}}{3 e^{3}}-\frac {2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}{e^{3} \sqrt {e x +d}}\) \(61\)
gosper \(-\frac {2 \left (-x^{2} b^{2} e^{2}-6 x a b \,e^{2}+4 b^{2} d e x +3 a^{2} e^{2}-12 a b d e +8 b^{2} d^{2}\right )}{3 \sqrt {e x +d}\, e^{3}}\) \(63\)
trager \(-\frac {2 \left (-x^{2} b^{2} e^{2}-6 x a b \,e^{2}+4 b^{2} d e x +3 a^{2} e^{2}-12 a b d e +8 b^{2} d^{2}\right )}{3 \sqrt {e x +d}\, e^{3}}\) \(63\)
derivativedivides \(\frac {\frac {2 \left (e x +d \right )^{\frac {3}{2}} b^{2}}{3}+4 a b e \sqrt {e x +d}-4 b^{2} d \sqrt {e x +d}-\frac {2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}{\sqrt {e x +d}}}{e^{3}}\) \(74\)
default \(\frac {\frac {2 \left (e x +d \right )^{\frac {3}{2}} b^{2}}{3}+4 a b e \sqrt {e x +d}-4 b^{2} d \sqrt {e x +d}-\frac {2 \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}{\sqrt {e x +d}}}{e^{3}}\) \(74\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.09 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (b^{2} e^{2} x^{2} - 8 \, b^{2} d^{2} + 12 \, a b d e - 3 \, a^{2} e^{2} - 2 \, {\left (2 \, b^{2} d e - 3 \, a b e^{2}\right )} x\right )} \sqrt {e x + d}}{3 \, {\left (e^{4} x + d e^{3}\right )}} \]

[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)

Sympy [A] (verification not implemented)

Time = 1.35 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.40 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^{3/2}} \, dx=\begin {cases} \frac {2 \left (\frac {b^{2} \left (d + e x\right )^{\frac {3}{2}}}{3 e^{2}} + \frac {\sqrt {d + e x} \left (2 a b e - 2 b^{2} d\right )}{e^{2}} - \frac {\left (a e - b d\right )^{2}}{e^{2} \sqrt {d + e x}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {a^{2} x + a b x^{2} + \frac {b^{2} x^{3}}{3}}{d^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(b**2*(d + e*x)**(3/2)/(3*e**2) + sqrt(d + e*x)*(2*a*b*e - 2*b**2*d)/e**2 - (a*e - b*d)**2/(e**2*
sqrt(d + e*x)))/e, Ne(e, 0)), ((a**2*x + a*b*x**2 + b**2*x**3/3)/d**(3/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.12 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {{\left (e x + d\right )}^{\frac {3}{2}} b^{2} - 6 \, {\left (b^{2} d - a b e\right )} \sqrt {e x + d}}{e^{2}} - \frac {3 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )}}{\sqrt {e x + d} e^{2}}\right )}}{3 \, e} \]

[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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.25 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^{3/2}} \, dx=-\frac {2 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )}}{\sqrt {e x + d} e^{3}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} b^{2} e^{6} - 6 \, \sqrt {e x + d} b^{2} d e^{6} + 6 \, \sqrt {e x + d} a b e^{7}\right )}}{3 \, e^{9}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.55 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00 \[ \int \frac {a^2+2 a b x+b^2 x^2}{(d+e x)^{3/2}} \, dx=\frac {\frac {2\,b^2\,{\left (d+e\,x\right )}^2}{3}-2\,a^2\,e^2-2\,b^2\,d^2-4\,b^2\,d\,\left (d+e\,x\right )+4\,a\,b\,e\,\left (d+e\,x\right )+4\,a\,b\,d\,e}{e^3\,\sqrt {d+e\,x}} \]

[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))