### 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]

-(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])

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Rubi [A]  time = 0.0697736, antiderivative size = 125, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [A]  time = 0.0281796, size = 69, normalized size = 0.55 $\frac{-a^2 e^2-2 a b e (d+2 e x)+b^2 \left (-\left (3 d^2+8 d e x+6 e^2 x^2\right )\right )}{12 b^3 (a+b x)^3 \sqrt{(a+b x)^2}}$

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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Maple [A]  time = 0.154, size = 69, normalized size = 0.6 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( 6\,{e}^{2}{x}^{2}{b}^{2}+4\,xab{e}^{2}+8\,x{b}^{2}de+{a}^{2}{e}^{2}+2\,abde+3\,{b}^{2}{d}^{2} \right ) }{12\,{b}^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

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 = 1.14905, size = 178, normalized size = 1.42 \begin{align*} -\frac{2 \, d e}{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}} b^{2}} - \frac{a^{2} b^{2} e^{2}}{4 \,{\left (b^{2}\right )}^{\frac{9}{2}}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{2 \, a b e^{2}}{3 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{3}} - \frac{e^{2}}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{d^{2}}{4 \,{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{4}} + \frac{a d e}{2 \,{\left (b^{2}\right )}^{\frac{5}{2}} b{\left (x + \frac{a}{b}\right )}^{4}} \end{align*}

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) - 1/4*a^2*b^2*e^2/((b^2)^(9/2)*(x + a/b)^4) + 2/3*a*b*e^2/((b^2
)^(7/2)*(x + a/b)^3) - 1/2*e^2/((b^2)^(5/2)*(x + a/b)^2) - 1/4*d^2/((b^2)^(5/2)*(x + a/b)^4) + 1/2*a*d*e/((b^2
)^(5/2)*b*(x + a/b)^4)

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Fricas [A]  time = 1.51971, size = 201, normalized size = 1.61 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{2}}{\left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

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