3.1958 \(\int (a+b x) (d+e x) \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=78 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^2 (b d-a e)}{3 b^2}+\frac{e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3}{4 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0578095, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {770, 21, 43} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^2 (b d-a e)}{3 b^2}+\frac{e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3}{4 b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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 (a+b x) (d+e x) \sqrt{a^2+2 a b x+b^2 x^2} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int (a+b x) \left (a b+b^2 x\right ) (d+e x) \, dx}{a b+b^2 x}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int (a+b x)^2 (d+e x) \, dx}{a b+b^2 x}\\ &=\frac{\left (b \sqrt{a^2+2 a b x+b^2 x^2}\right ) \int \left (\frac{(b d-a e) (a+b x)^2}{b}+\frac{e (a+b x)^3}{b}\right ) \, dx}{a b+b^2 x}\\ &=\frac{(b d-a e) (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}{3 b^2}+\frac{e (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{4 b^2}\\ \end{align*}

Mathematica [A]  time = 0.0242511, size = 64, normalized size = 0.82 \[ \frac{x \sqrt{(a+b x)^2} \left (6 a^2 (2 d+e x)+4 a b x (3 d+2 e x)+b^2 x^2 (4 d+3 e x)\right )}{12 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(6*a^2*(2*d + e*x) + 4*a*b*x*(3*d + 2*e*x) + b^2*x^2*(4*d + 3*e*x)))/(12*(a + b*x))

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Maple [A]  time = 0.003, size = 66, normalized size = 0.9 \begin{align*}{\frac{x \left ( 3\,{b}^{2}e{x}^{3}+8\,{x}^{2}abe+4\,{x}^{2}{b}^{2}d+6\,xe{a}^{2}+12\,abdx+12\,d{a}^{2} \right ) }{12\,bx+12\,a}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*x*(3*b^2*e*x^3+8*a*b*e*x^2+4*b^2*d*x^2+6*a^2*e*x+12*a*b*d*x+12*a^2*d)*((b*x+a)^2)^(1/2)/(b*x+a)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66913, size = 109, normalized size = 1.4 \begin{align*} \frac{1}{4} \, b^{2} e x^{4} + a^{2} d x + \frac{1}{3} \,{\left (b^{2} d + 2 \, a b e\right )} x^{3} + \frac{1}{2} \,{\left (2 \, a b d + a^{2} e\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 0.096458, size = 49, normalized size = 0.63 \begin{align*} a^{2} d x + \frac{b^{2} e x^{4}}{4} + x^{3} \left (\frac{2 a b e}{3} + \frac{b^{2} d}{3}\right ) + x^{2} \left (\frac{a^{2} e}{2} + a b d\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*d*x + b**2*e*x**4/4 + x**3*(2*a*b*e/3 + b**2*d/3) + x**2*(a**2*e/2 + a*b*d)

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Giac [A]  time = 1.11986, size = 119, normalized size = 1.53 \begin{align*} \frac{1}{4} \, b^{2} x^{4} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, b^{2} d x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{2}{3} \, a b x^{3} e \mathrm{sgn}\left (b x + a\right ) + a b d x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, a^{2} x^{2} e \mathrm{sgn}\left (b x + a\right ) + a^{2} d x \mathrm{sgn}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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