### 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*(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])

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Rubi [A]  time = 0.0688655, antiderivative size = 117, 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} $-\frac{2 e (b d-a e)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(b d-a e)^2}{2 b^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{e^2 (a+b x) \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}$

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 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 )^{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*}

Mathematica [A]  time = 0.0324117, 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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Maple [A]  time = 0.201, size = 104, normalized size = 0.9 \begin{align*}{\frac{ \left ( 2\,\ln \left ( bx+a \right ){x}^{2}{b}^{2}{e}^{2}+4\,\ln \left ( bx+a \right ) xab{e}^{2}+2\,\ln \left ( bx+a \right ){a}^{2}{e}^{2}+4\,xab{e}^{2}-4\,x{b}^{2}de+3\,{a}^{2}{e}^{2}-2\,abde-{b}^{2}{d}^{2} \right ) \left ( bx+a \right ) }{2\,{b}^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{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)^(3/2),x)

[Out]

1/2*(2*ln(b*x+a)*x^2*b^2*e^2+4*ln(b*x+a)*x*a*b*e^2+2*ln(b*x+a)*a^2*e^2+4*x*a*b*e^2-4*x*b^2*d*e+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 = 1.089, size = 176, normalized size = 1.5 \begin{align*} \frac{e^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{3 \, a^{2} b^{2} e^{2}}{2 \,{\left (b^{2}\right )}^{\frac{7}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{2 \, a b e^{2} x}{{\left (b^{2}\right )}^{\frac{5}{2}}{\left (x + \frac{a}{b}\right )}^{2}} - \frac{2 \, d e}{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac{d^{2}}{2 \,{\left (b^{2}\right )}^{\frac{3}{2}}{\left (x + \frac{a}{b}\right )}^{2}} + \frac{a d e}{{\left (b^{2}\right )}^{\frac{3}{2}} b{\left (x + \frac{a}{b}\right )}^{2}} \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)^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.53551, size = 207, normalized size = 1.77 \begin{align*} -\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 )}} \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)^(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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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{3}{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)**(3/2),x)

[Out]

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

[Out]

sage0*x