3.1608 \(\int \frac {(d+e x)^2}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=125 \[ -\frac {2 e (b d-a e)}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^2}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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)

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Rubi [A]  time = 0.07, antiderivative size = 125, 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} \[ -\frac {2 e (b d-a e)}{3 b^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(b d-a e)^2}{4 b^3 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^2}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*d - a*e)^2/(4*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 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 )^{5/2}} \, dx &=\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*}

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Mathematica [A]  time = 0.03, size = 69, normalized size = 0.55 \[ \frac {-a^2 e^2-2 a b e (d+2 e x)-\left (b^2 \left (3 d^2+8 d e x+6 e^2 x^2\right )\right )}{12 b^3 (a+b x)^3 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.02, size = 98, normalized size = 0.78 \[ -\frac {6 \, b^{2} e^{2} x^{2} + 3 \, b^{2} d^{2} + 2 \, a b d e + a^{2} e^{2} + 4 \, {\left (2 \, b^{2} d e + a b e^{2}\right )} x}{12 \, {\left (b^{7} x^{4} + 4 \, a b^{6} x^{3} + 6 \, a^{2} b^{5} x^{2} + 4 \, a^{3} b^{4} x + a^{4} b^{3}\right )}} \]

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)^(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)

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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)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [A]  time = 0.05, size = 69, normalized size = 0.55 \[ -\frac {\left (b x +a \right ) \left (6 b^{2} e^{2} x^{2}+4 a b \,e^{2} x +8 b^{2} d e x +a^{2} e^{2}+2 a b d e +3 b^{2} d^{2}\right )}{12 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{3}} \]

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)^(5/2),x)

[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)

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maxima [A]  time = 1.06, size = 115, normalized size = 0.92 \[ -\frac {2 \, d e}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {e^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {2 \, a e^{2}}{3 \, b^{6} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {d^{2}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {a d e}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {a^{2} e^{2}}{4 \, b^{7} {\left (x + \frac {a}{b}\right )}^{4}} \]

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)^(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)

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mupad [B]  time = 0.71, size = 79, normalized size = 0.63 \[ -\frac {\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}\,\left (a^2\,e^2+2\,a\,b\,d\,e+4\,a\,b\,e^2\,x+3\,b^2\,d^2+8\,b^2\,d\,e\,x+6\,b^2\,e^2\,x^2\right )}{12\,b^3\,{\left (a+b\,x\right )}^5} \]

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)^(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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{2}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

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)**(5/2),x)

[Out]

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

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