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

Optimal. Leaf size=69 \[ \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{2 b^2}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 b^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {640, 609} \begin {gather*} \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{2 b^2}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 609

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((b + 2*c*x)*(a + b*x + c*x^2)^p)/(2*c*(2*p + 1
)), x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && NeQ[p, -2^(-1)]

Rule 640

Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(a + b*x + c*x^2)^(p +
1))/(2*c*(p + 1)), x] + Dist[(2*c*d - b*e)/(2*c), Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}
, x] && NeQ[2*c*d - b*e, 0] && NeQ[p, -1]

Rubi steps

\begin {align*} \int (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx &=\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 b^2}+\frac {\left (2 b^2 d-2 a b e\right ) \int \sqrt {a^2+2 a b x+b^2 x^2} \, dx}{2 b^2}\\ &=\frac {(b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 b^2}+\frac {e \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 b^2}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 45, normalized size = 0.65 \begin {gather*} \frac {x \sqrt {(a+b x)^2} (3 a (2 d+e x)+b x (3 d+2 e x))}{6 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 0.54, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

Defer[IntegrateAlgebraic][(d + e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2], x]

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fricas [A]  time = 0.40, size = 24, normalized size = 0.35 \begin {gather*} \frac {1}{3} \, b e x^{3} + a d x + \frac {1}{2} \, {\left (b d + a e\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 52, normalized size = 0.75 \begin {gather*} \frac {1}{3} \, b x^{3} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, b d x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, a x^{2} e \mathrm {sgn}\left (b x + a\right ) + a d x \mathrm {sgn}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 42, normalized size = 0.61 \begin {gather*} \frac {\left (2 b e \,x^{2}+3 a e x +3 x b d +6 a d \right ) \sqrt {\left (b x +a \right )^{2}}\, x}{6 b x +6 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*x*(2*b*e*x^2+3*a*e*x+3*b*d*x+6*a*d)*((b*x+a)^2)^(1/2)/(b*x+a)

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maxima [B]  time = 1.04, size = 125, normalized size = 1.81 \begin {gather*} \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} d x - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a e x}{2 \, b} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a d}{2 \, b} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} e}{2 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} e}{3 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.72, size = 77, normalized size = 1.12 \begin {gather*} \frac {e\,\left (8\,b^2\,\left (a^2+b^2\,x^2\right )-12\,a^2\,b^2+4\,a\,b^3\,x\right )\,\sqrt {a^2+2\,a\,b\,x+b^2\,x^2}}{24\,b^4}+\frac {d\,\sqrt {{\left (a+b\,x\right )}^2}\,\left (a+b\,x\right )}{2\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^2)^(1/2)*(d + e*x),x)

[Out]

(e*(8*b^2*(a^2 + b^2*x^2) - 12*a^2*b^2 + 4*a*b^3*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(1/2))/(24*b^4) + (d*((a + b*x)^
2)^(1/2)*(a + b*x))/(2*b)

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sympy [A]  time = 0.09, size = 26, normalized size = 0.38 \begin {gather*} a d x + \frac {b e x^{3}}{3} + x^{2} \left (\frac {a e}{2} + \frac {b d}{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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