3.1589 \(\int \frac{(d+e x)^2}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=124 \[ \frac{e x (a+b x) (b d-a e)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (d+e x)^2}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (b d-a e)^2 \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(e*(b*d - a*e)*x*(a + b*x))/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((a + b*x)*(d + e*x)^2)/(2*b*Sqrt[a^2 + 2*a*
b*x + b^2*x^2]) + ((b*d - a*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 = 0.0458448, antiderivative size = 124, 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} \[ \frac{e x (a+b x) (b d-a e)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (d+e x)^2}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (b d-a e)^2 \log (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*(b*d - a*e)*x*(a + b*x))/(b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((a + b*x)*(d + e*x)^2)/(2*b*Sqrt[a^2 + 2*a*
b*x + b^2*x^2]) + ((b*d - a*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}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{(d+e x)^2}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \left (\frac{e (b d-a e)}{b^3}+\frac{(b d-a e)^2}{b^2 \left (a b+b^2 x\right )}+\frac{e (d+e x)}{b^2}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{e (b d-a e) x (a+b x)}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (d+e x)^2}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(b d-a 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 = 0.0337432, size = 59, normalized size = 0.48 \[ \frac{(a+b x) \left (b e x (-2 a e+4 b d+b e x)+2 (b d-a e)^2 \log (a+b x)\right )}{2 b^3 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.154, size = 87, normalized size = 0.7 \begin{align*}{\frac{ \left ( bx+a \right ) \left ({x}^{2}{b}^{2}{e}^{2}+2\,\ln \left ( bx+a \right ){a}^{2}{e}^{2}-4\,\ln \left ( bx+a \right ) abde+2\,\ln \left ( bx+a \right ){b}^{2}{d}^{2}-2\,xab{e}^{2}+4\,x{b}^{2}de \right ) }{2\,{b}^{3}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^2/((b*x+a)^2)^(1/2),x)

[Out]

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

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Maxima [A]  time = 1.06228, size = 153, normalized size = 1.23 \begin{align*} \frac{a^{2} b^{2} e^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{a b e^{2} x}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{e^{2} x^{2}}{2 \, \sqrt{b^{2}}} + \sqrt{\frac{1}{b^{2}}} d^{2} \log \left (x + \frac{a}{b}\right ) - \frac{2 \, a \sqrt{\frac{1}{b^{2}}} d e \log \left (x + \frac{a}{b}\right )}{b} + \frac{2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} d e}{b^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.56328, size = 135, normalized size = 1.09 \begin{align*} \frac{b^{2} e^{2} x^{2} + 2 \,{\left (2 \, b^{2} d e - a b e^{2}\right )} x + 2 \,{\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left (b x + a\right )}{2 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.452613, size = 44, normalized size = 0.35 \begin{align*} \frac{e^{2} x^{2}}{2 b} - \frac{x \left (a e^{2} - 2 b d e\right )}{b^{2}} + \frac{\left (a e - b d\right )^{2} \log{\left (a + b x \right )}}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**2/((b*x+a)**2)**(1/2),x)

[Out]

e**2*x**2/(2*b) - x*(a*e**2 - 2*b*d*e)/b**2 + (a*e - b*d)**2*log(a + b*x)/b**3

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Giac [A]  time = 1.18255, size = 128, normalized size = 1.03 \begin{align*} \frac{b x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 4 \, b d x e \mathrm{sgn}\left (b x + a\right ) - 2 \, a x e^{2} \mathrm{sgn}\left (b x + a\right )}{2 \, b^{2}} + \frac{{\left (b^{2} d^{2} \mathrm{sgn}\left (b x + a\right ) - 2 \, a b d e \mathrm{sgn}\left (b x + a\right ) + a^{2} e^{2} \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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