3.1957 \(\int (a+b x) (d+e x)^2 \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=125 \[ \frac{e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (b d-a e)}{2 b^3}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^2 (b d-a e)^2}{3 b^3}+\frac{e^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^4}{5 b^3} \]

[Out]

((b*d - a*e)^2*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*b^3) + (e*(b*d - a*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*
b*x + b^2*x^2])/(2*b^3) + (e^2*(a + b*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*b^3)

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Rubi [A]  time = 0.0969148, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {770, 21, 43} \[ \frac{e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (b d-a e)}{2 b^3}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^2 (b d-a e)^2}{3 b^3}+\frac{e^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^4}{5 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - a*e)^2*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*b^3) + (e*(b*d - a*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*
b*x + b^2*x^2])/(2*b^3) + (e^2*(a + b*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*b^3)

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)^2 \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)^2 \, 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)^2 \, 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)^2 (a+b x)^2}{b^2}+\frac{2 e (b d-a e) (a+b x)^3}{b^2}+\frac{e^2 (a+b x)^4}{b^2}\right ) \, dx}{a b+b^2 x}\\ &=\frac{(b d-a e)^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}{3 b^3}+\frac{e (b d-a e) (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{2 b^3}+\frac{e^2 (a+b x)^4 \sqrt{a^2+2 a b x+b^2 x^2}}{5 b^3}\\ \end{align*}

Mathematica [A]  time = 0.0349204, size = 97, normalized size = 0.78 \[ \frac{x \sqrt{(a+b x)^2} \left (10 a^2 \left (3 d^2+3 d e x+e^2 x^2\right )+5 a b x \left (6 d^2+8 d e x+3 e^2 x^2\right )+b^2 x^2 \left (10 d^2+15 d e x+6 e^2 x^2\right )\right )}{30 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(10*a^2*(3*d^2 + 3*d*e*x + e^2*x^2) + 5*a*b*x*(6*d^2 + 8*d*e*x + 3*e^2*x^2) + b^2*x^2*(10
*d^2 + 15*d*e*x + 6*e^2*x^2)))/(30*(a + b*x))

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Maple [A]  time = 0.005, size = 107, normalized size = 0.9 \begin{align*}{\frac{x \left ( 6\,{b}^{2}{e}^{2}{x}^{4}+15\,{x}^{3}ab{e}^{2}+15\,{x}^{3}{b}^{2}de+10\,{x}^{2}{a}^{2}{e}^{2}+40\,{x}^{2}abde+10\,{x}^{2}{b}^{2}{d}^{2}+30\,{a}^{2}dex+30\,ab{d}^{2}x+30\,{a}^{2}{d}^{2} \right ) }{30\,bx+30\,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)^2*((b*x+a)^2)^(1/2),x)

[Out]

1/30*x*(6*b^2*e^2*x^4+15*a*b*e^2*x^3+15*b^2*d*e*x^3+10*a^2*e^2*x^2+40*a*b*d*e*x^2+10*b^2*d^2*x^2+30*a^2*d*e*x+
30*a*b*d^2*x+30*a^2*d^2)*((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)^2*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52618, size = 171, normalized size = 1.37 \begin{align*} \frac{1}{5} \, b^{2} e^{2} x^{5} + a^{2} d^{2} x + \frac{1}{2} \,{\left (b^{2} d e + a b e^{2}\right )} x^{4} + \frac{1}{3} \,{\left (b^{2} d^{2} + 4 \, a b d e + a^{2} e^{2}\right )} x^{3} +{\left (a b d^{2} + a^{2} d e\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*b^2*e^2*x^5 + a^2*d^2*x + 1/2*(b^2*d*e + a*b*e^2)*x^4 + 1/3*(b^2*d^2 + 4*a*b*d*e + a^2*e^2)*x^3 + (a*b*d^2
+ a^2*d*e)*x^2

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Sympy [A]  time = 0.10872, size = 87, normalized size = 0.7 \begin{align*} a^{2} d^{2} x + \frac{b^{2} e^{2} x^{5}}{5} + x^{4} \left (\frac{a b e^{2}}{2} + \frac{b^{2} d e}{2}\right ) + x^{3} \left (\frac{a^{2} e^{2}}{3} + \frac{4 a b d e}{3} + \frac{b^{2} d^{2}}{3}\right ) + x^{2} \left (a^{2} d e + a b d^{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*d**2*x + b**2*e**2*x**5/5 + x**4*(a*b*e**2/2 + b**2*d*e/2) + x**3*(a**2*e**2/3 + 4*a*b*d*e/3 + b**2*d**2/
3) + x**2*(a**2*d*e + a*b*d**2)

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Giac [A]  time = 1.11559, size = 193, normalized size = 1.54 \begin{align*} \frac{1}{5} \, b^{2} x^{5} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, b^{2} d x^{4} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, b^{2} d^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, a b x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{4}{3} \, a b d x^{3} e \mathrm{sgn}\left (b x + a\right ) + a b d^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, a^{2} x^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) + a^{2} d x^{2} e \mathrm{sgn}\left (b x + a\right ) + a^{2} d^{2} 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)^2*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/5*b^2*x^5*e^2*sgn(b*x + a) + 1/2*b^2*d*x^4*e*sgn(b*x + a) + 1/3*b^2*d^2*x^3*sgn(b*x + a) + 1/2*a*b*x^4*e^2*s
gn(b*x + a) + 4/3*a*b*d*x^3*e*sgn(b*x + a) + a*b*d^2*x^2*sgn(b*x + a) + 1/3*a^2*x^3*e^2*sgn(b*x + a) + a^2*d*x
^2*e*sgn(b*x + a) + a^2*d^2*x*sgn(b*x + a)