3.15.92 \(\int (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx\)

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

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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} (A b-a B)}{2 b^2}+\frac {B \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[(A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

((A*b - a*B)*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b^2) + (B*(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 (A+B x) \sqrt {a^2+2 a b x+b^2 x^2} \, dx &=\frac {B \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 b^2}+\frac {\left (2 A b^2-2 a b B\right ) \int \sqrt {a^2+2 a b x+b^2 x^2} \, dx}{2 b^2}\\ &=\frac {(A b-a B) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 b^2}+\frac {B \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 = 0.01, size = 45, normalized size = 0.65 \begin {gather*} \frac {x \sqrt {(a+b x)^2} (3 a (2 A+B x)+b x (3 A+2 B x))}{6 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

(x*Sqrt[(a + b*x)^2]*(3*a*(2*A + B*x) + b*x*(3*A + 2*B*x)))/(6*(a + b*x))

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

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2],x]

[Out]

Defer[IntegrateAlgebraic][(A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2], x]

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fricas [A]  time = 0.42, size = 24, normalized size = 0.35 \begin {gather*} \frac {1}{3} \, B b x^{3} + A a x + \frac {1}{2} \, {\left (B a + A b\right )} x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*B*b*x^3 + A*a*x + 1/2*(B*a + A*b)*x^2

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giac [A]  time = 0.21, size = 74, normalized size = 1.07 \begin {gather*} \frac {1}{3} \, B b x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, B a x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{2} \, A b x^{2} \mathrm {sgn}\left (b x + a\right ) + A a x \mathrm {sgn}\left (b x + a\right ) - \frac {{\left (B a^{3} - 3 \, A a^{2} b\right )} \mathrm {sgn}\left (b x + a\right )}{6 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*B*b*x^3*sgn(b*x + a) + 1/2*B*a*x^2*sgn(b*x + a) + 1/2*A*b*x^2*sgn(b*x + a) + A*a*x*sgn(b*x + a) - 1/6*(B*a
^3 - 3*A*a^2*b)*sgn(b*x + a)/b^2

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maple [A]  time = 0.04, size = 42, normalized size = 0.61 \begin {gather*} \frac {\left (2 B b \,x^{2}+3 A b x +3 B a x +6 A a \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((B*x+A)*((b*x+a)^2)^(1/2),x)

[Out]

1/6*x*(2*B*b*x^2+3*A*b*x+3*B*a*x+6*A*a)*((b*x+a)^2)^(1/2)/(b*x+a)

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maxima [B]  time = 0.56, size = 125, normalized size = 1.81 \begin {gather*} \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A x - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a x}{2 \, b} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B a^{2}}{2 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A a}{2 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} B}{3 \, b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*x - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*a*x/b - 1/2*sqrt(b^2*x^2 + 2*a*b*x
+ a^2)*B*a^2/b^2 + 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*a/b + 1/3*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*B/b^2

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mupad [B]  time = 2.42, size = 77, normalized size = 1.12 \begin {gather*} \frac {A\,\sqrt {{\left (a+b\,x\right )}^2}\,\left (a+b\,x\right )}{2\,b}+\frac {B\,\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*x)^2)^(1/2)*(A + B*x),x)

[Out]

(A*((a + b*x)^2)^(1/2)*(a + b*x))/(2*b) + (B*(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)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2),x)

[Out]

A*a*x + B*b*x**3/3 + x**2*(A*b/2 + B*a/2)

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