3.1598 \(\int \frac {(d+e x)^2}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=117 \[ -\frac {2 e (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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)

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Rubi [A]  time = 0.07, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {646, 43} \[ -\frac {2 e (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^2 (a+b x) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[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 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])

Rule 646

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

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Mathematica [A]  time = 0.03, size = 67, normalized size = 0.57 \[ \frac {2 e^2 (a+b x)^2 \log (a+b x)-(b d-a e) (3 a e+b (d+4 e x))}{2 b^3 (a+b x) \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.91, size = 99, normalized size = 0.85 \[ -\frac {b^{2} d^{2} + 2 \, a b d e - 3 \, a^{2} e^{2} + 4 \, {\left (b^{2} d e - a b e^{2}\right )} x - 2 \, {\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.06, size = 104, normalized size = 0.89 \[ \frac {\left (2 b^{2} e^{2} x^{2} \ln \left (b x +a \right )+4 a b \,e^{2} x \ln \left (b x +a \right )+2 a^{2} e^{2} \ln \left (b x +a \right )+4 a b \,e^{2} x -4 b^{2} d e x +3 a^{2} e^{2}-2 a b d e -b^{2} d^{2}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.08, size = 113, normalized size = 0.97 \[ \frac {e^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {2 \, d e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {2 \, a e^{2} x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {d^{2}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {a d e}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, a^{2} e^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (d+e\,x\right )}^2}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{2}}{\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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