\(\int \frac {(d+e x)^2}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [1608]

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

Optimal result

Integrand size = 28, antiderivative size = 125 \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {(b d-a e)^2}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 e (b d-a e)}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {660, 45} \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {2 e (b d-a e)}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^2}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

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

[Out]

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

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])

Rule 660

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
 c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^2}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac {(b d-a e)^2}{b^7 (a+b x)^5}+\frac {2 e (b d-a e)}{b^7 (a+b x)^4}+\frac {e^2}{b^7 (a+b x)^3}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {(b d-a e)^2}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 e (b d-a e)}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(417\) vs. \(2(125)=250\).

Time = 1.30 (sec) , antiderivative size = 417, normalized size of antiderivative = 3.34 \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {x \left (6 a^2 b^6 d e x^7-3 a^4 \sqrt {a^2} b^2 x^2 \sqrt {(a+b x)^2} \left (2 d^2+2 d e x+e^2 x^2\right )+4 a^8 \left (3 d^2+3 d e x+e^2 x^2\right )+a^7 b x \left (18 d^2+8 d e x+e^2 x^2\right )-3 \sqrt {a^2} b^4 x^4 \sqrt {(a+b x)^2} \left (b^2 d^2 x^2+a^2 (d+e x)^2\right )+3 a^3 b^3 x^3 \left (-b^2 e^2 x^4+\sqrt {a^2} \sqrt {(a+b x)^2} (d+e x)^2\right )+3 a b^5 d x^5 \left (-b^2 d x^2+\sqrt {a^2} \sqrt {(a+b x)^2} (d+2 e x)\right )-2 a^6 \left (-b^2 d x^2 (6 d+e x)+2 \sqrt {a^2} \sqrt {(a+b x)^2} \left (3 d^2+3 d e x+e^2 x^2\right )\right )+a^5 \left (3 b^3 d^2 x^3+\sqrt {a^2} b x \sqrt {(a+b x)^2} \left (-6 d^2+4 d e x+3 e^2 x^2\right )\right )\right )}{12 a^8 (a+b x)^3 \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )} \]

[In]

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

[Out]

-1/12*(x*(6*a^2*b^6*d*e*x^7 - 3*a^4*Sqrt[a^2]*b^2*x^2*Sqrt[(a + b*x)^2]*(2*d^2 + 2*d*e*x + e^2*x^2) + 4*a^8*(3
*d^2 + 3*d*e*x + e^2*x^2) + a^7*b*x*(18*d^2 + 8*d*e*x + e^2*x^2) - 3*Sqrt[a^2]*b^4*x^4*Sqrt[(a + b*x)^2]*(b^2*
d^2*x^2 + a^2*(d + e*x)^2) + 3*a^3*b^3*x^3*(-(b^2*e^2*x^4) + Sqrt[a^2]*Sqrt[(a + b*x)^2]*(d + e*x)^2) + 3*a*b^
5*d*x^5*(-(b^2*d*x^2) + Sqrt[a^2]*Sqrt[(a + b*x)^2]*(d + 2*e*x)) - 2*a^6*(-(b^2*d*x^2*(6*d + e*x)) + 2*Sqrt[a^
2]*Sqrt[(a + b*x)^2]*(3*d^2 + 3*d*e*x + e^2*x^2)) + a^5*(3*b^3*d^2*x^3 + Sqrt[a^2]*b*x*Sqrt[(a + b*x)^2]*(-6*d
^2 + 4*d*e*x + 3*e^2*x^2))))/(a^8*(a + b*x)^3*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[(a + b*x)^2])))

Maple [A] (verified)

Time = 2.46 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.55

method result size
gosper \(-\frac {\left (b x +a \right ) \left (6 x^{2} b^{2} e^{2}+4 x a b \,e^{2}+8 b^{2} d e x +a^{2} e^{2}+2 a b d e +3 b^{2} d^{2}\right )}{12 b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) \(69\)
default \(-\frac {\left (b x +a \right ) \left (6 x^{2} b^{2} e^{2}+4 x a b \,e^{2}+8 b^{2} d e x +a^{2} e^{2}+2 a b d e +3 b^{2} d^{2}\right )}{12 b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\) \(69\)
risch \(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {e^{2} x^{2}}{2 b}-\frac {e \left (a e +2 b d \right ) x}{3 b^{2}}-\frac {a^{2} e^{2}+2 a b d e +3 b^{2} d^{2}}{12 b^{3}}\right )}{\left (b x +a \right )^{5}}\) \(72\)

[In]

int((e*x+d)^2/(b^2*x^2+2*a*b*x+a^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/12*(b*x+a)/b^3*(6*b^2*e^2*x^2+4*a*b*e^2*x+8*b^2*d*e*x+a^2*e^2+2*a*b*d*e+3*b^2*d^2)/((b*x+a)^2)^(5/2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.78 \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {6 \, b^{2} e^{2} x^{2} + 3 \, b^{2} d^{2} + 2 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b^{2} d e + a b e^{2}\right )} x}{12 \, {\left (b^{7} x^{4} + 4 \, a b^{6} x^{3} + 6 \, a^{2} b^{5} x^{2} + 4 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} \]

[In]

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

[Out]

-1/12*(6*b^2*e^2*x^2 + 3*b^2*d^2 + 2*a*b*d*e + a^2*e^2 + 4*(2*b^2*d*e + a*b*e^2)*x)/(b^7*x^4 + 4*a*b^6*x^3 + 6
*a^2*b^5*x^2 + 4*a^3*b^4*x + a^4*b^3)

Sympy [F]

\[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (d + e x\right )^{2}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 115, normalized size of antiderivative = 0.92 \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, d e}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {e^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, a e^{2}}{3 \, b^{6} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {d^{2}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {a d e}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {a^{2} e^{2}}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} \]

[In]

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

[Out]

-2/3*d*e/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 1/2*e^2/(b^5*(x + a/b)^2) + 2/3*a*e^2/(b^6*(x + a/b)^3) - 1/4
*d^2/(b^5*(x + a/b)^4) + 1/2*a*d*e/(b^6*(x + a/b)^4) - 1/4*a^2*e^2/(b^7*(x + a/b)^4)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.55 \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {6 \, b^{2} e^{2} x^{2} + 8 \, b^{2} d e x + 4 \, a b e^{2} x + 3 \, b^{2} d^{2} + 2 \, a b d e + a^{2} e^{2}}{12 \, {\left (b x + a\right )}^{4} b^{3} \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 9.63 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.63 \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^2\,e^2+2\,a\,b\,d\,e+4\,a\,b\,e^2\,x+3\,b^2\,d^2+8\,b^2\,d\,e\,x+6\,b^2\,e^2\,x^2\right )}{12\,b^3\,{\left (a+b\,x\right )}^5} \]

[In]

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

[Out]

-((a^2 + b^2*x^2 + 2*a*b*x)^(1/2)*(a^2*e^2 + 3*b^2*d^2 + 6*b^2*e^2*x^2 + 4*a*b*e^2*x + 8*b^2*d*e*x + 2*a*b*d*e
))/(12*b^3*(a + b*x)^5)