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

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Rubi [A]  time = 0.05, antiderivative size = 124, 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} \[ \frac {e x (a+b x) (b d-a e)}{b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (d+e x)^2}{2 b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (b d-a e)^2 \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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 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}{\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*}

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Mathematica [A]  time = 0.03, 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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fricas [A]  time = 0.75, size = 63, normalized size = 0.51 \[ \frac {b^{2} e^{2} x^{2} + 2 \, {\left (2 \, b^{2} d e - a b e^{2}\right )} x + 2 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left (b x + a\right )}{2 \, b^{3}} \]

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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giac [A]  time = 0.17, size = 95, normalized size = 0.77 \[ \frac {b x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 4 \, b d x e \mathrm {sgn}\left (b x + a\right ) - 2 \, a x e^{2} \mathrm {sgn}\left (b x + a\right )}{2 \, b^{2}} + \frac {{\left (b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b d e \mathrm {sgn}\left (b x + a\right ) + a^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3}} \]

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

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maple [A]  time = 0.05, size = 87, normalized size = 0.70 \[ \frac {\left (b x +a \right ) \left (b^{2} e^{2} x^{2}+2 a^{2} e^{2} \ln \left (b x +a \right )-4 a b d e \ln \left (b x +a \right )-2 a b \,e^{2} x +2 b^{2} d^{2} \ln \left (b x +a \right )+4 b^{2} d e x \right )}{2 \sqrt {\left (b x +a \right )^{2}}\, b^{3}} \]

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 = 1.07, size = 96, normalized size = 0.77 \[ \frac {e^{2} x^{2}}{2 \, b} - \frac {a e^{2} x}{b^{2}} + \frac {d^{2} \log \left (x + \frac {a}{b}\right )}{b} - \frac {2 \, a d e \log \left (x + \frac {a}{b}\right )}{b^{2}} + \frac {a^{2} e^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} + \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} d e}{b^{2}} \]

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]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.25, size = 44, normalized size = 0.35 \[ x \left (- \frac {a e^{2}}{b^{2}} + \frac {2 d e}{b}\right ) + \frac {e^{2} x^{2}}{2 b} + \frac {\left (a e - b d\right )^{2} \log {\left (a + b x \right )}}{b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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