\(\int \frac {(d+e x)^2}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [1598]

Optimal result

Integrand size = 28, antiderivative size = 117 \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {2 e (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 117, 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} \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {2 e (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

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

[Out]

(-2*e*(b*d - a*e))/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b*d - a*e)^2/(2*b^3*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b
^2*x^2]) + (e^2*(a + b*x)*Log[a + b*x])/(b^3*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^2 \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^2}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac {(b d-a e)^2}{b^5 (a+b x)^3}+\frac {2 e (b d-a e)}{b^5 (a+b x)^2}+\frac {e^2}{b^5 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {2 e (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x) \log (a+b x)}{b^3 \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. \(241\) vs. \(2(117)=234\).

Time = 1.18 (sec) , antiderivative size = 241, normalized size of antiderivative = 2.06 \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {\frac {b (-b d+a e) x \left (2 a^5 e+a^2 b^3 e x^3+a^4 b (2 d+3 e x)+a^3 \left (b^2 d x-2 \sqrt {a^2} e \sqrt {(a+b x)^2}\right )+a \left (-b^4 d x^3+\sqrt {a^2} b^2 x \sqrt {(a+b x)^2} (d+e x)\right )-\sqrt {a^2} b \sqrt {(a+b x)^2} \left (b^2 d x^2+a^2 (2 d+e x)\right )\right )}{a^4 (a+b x) \left (\sqrt {a^2} b x+a \left (\sqrt {a^2}-\sqrt {(a+b x)^2}\right )\right )}-4 e^2 \text {arctanh}\left (\frac {b x}{\sqrt {a^2}-\sqrt {(a+b x)^2}}\right )}{2 b^3} \]

[In]

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

[Out]

((b*(-(b*d) + a*e)*x*(2*a^5*e + a^2*b^3*e*x^3 + a^4*b*(2*d + 3*e*x) + a^3*(b^2*d*x - 2*Sqrt[a^2]*e*Sqrt[(a + b
*x)^2]) + a*(-(b^4*d*x^3) + Sqrt[a^2]*b^2*x*Sqrt[(a + b*x)^2]*(d + e*x)) - Sqrt[a^2]*b*Sqrt[(a + b*x)^2]*(b^2*
d*x^2 + a^2*(2*d + e*x))))/(a^4*(a + b*x)*(Sqrt[a^2]*b*x + a*(Sqrt[a^2] - Sqrt[(a + b*x)^2]))) - 4*e^2*ArcTanh
[(b*x)/(Sqrt[a^2] - Sqrt[(a + b*x)^2])])/(2*b^3)

Maple [A] (verified)

Time = 2.34 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (\frac {2 e \left (a e -b d \right ) x}{b^{2}}+\frac {3 a^{2} e^{2}-2 a b d e -b^{2} d^{2}}{2 b^{3}}\right )}{\left (b x +a \right )^{3}}+\frac {\sqrt {\left (b x +a \right )^{2}}\, e^{2} \ln \left (b x +a \right )}{\left (b x +a \right ) b^{3}}\)	\(92\)
default	\(\frac {\left (2 \ln \left (b x +a \right ) b^{2} e^{2} x^{2}+4 \ln \left (b x +a \right ) x a b \,e^{2}+2 \ln \left (b x +a \right ) a^{2} e^{2}+4 x a b \,e^{2}-4 b^{2} d e x +3 a^{2} e^{2}-2 a b d e -b^{2} d^{2}\right ) \left (b x +a \right )}{2 b^{3} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(104\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.85 \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {b^{2} d^{2} + 2 \, a b d e - 3 \, a^{2} e^{2} + 4 \, {\left (b^{2} d e - a b e^{2}\right )} x - 2 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.97 \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {2 \, d e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {2 \, a e^{2} x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {d^{2}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {a d e}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, a^{2} e^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.72 \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {e^{2} \log \left ({\left | b x + a \right |}\right )}{b^{3} \mathrm {sgn}\left (b x + a\right )} - \frac {4 \, {\left (b d e - a e^{2}\right )} x + \frac {b^{2} d^{2} + 2 \, a b d e - 3 \, a^{2} e^{2}}{b}}{2 \, {\left (b x + a\right )}^{2} b^{2} \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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