3.1545 \(\int (d+e x)^2 \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=92 \[ \frac{b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^4}{4 e^2 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^3 (b d-a e)}{3 e^2 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0489773, antiderivative size = 92, normalized size of antiderivative = 1., 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} \[ \frac{b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^4}{4 e^2 (a+b x)}-\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^3 (b d-a e)}{3 e^2 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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

Mathematica [A]  time = 0.0219973, size = 67, normalized size = 0.73 \[ \frac{x \sqrt{(a+b x)^2} \left (4 a \left (3 d^2+3 d e x+e^2 x^2\right )+b x \left (6 d^2+8 d e x+3 e^2 x^2\right )\right )}{12 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.041, size = 66, normalized size = 0.7 \begin{align*}{\frac{x \left ( 3\,b{e}^{2}{x}^{3}+4\,{x}^{2}a{e}^{2}+8\,{x}^{2}bde+12\,adex+6\,b{d}^{2}x+12\,a{d}^{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((e*x+d)^2*((b*x+a)^2)^(1/2),x)

[Out]

1/12*x*(3*b*e^2*x^3+4*a*e^2*x^2+8*b*d*e*x^2+12*a*d*e*x+6*b*d^2*x+12*a*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((e*x+d)^2*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.64003, size = 109, normalized size = 1.18 \begin{align*} \frac{1}{4} \, b e^{2} x^{4} + a d^{2} x + \frac{1}{3} \,{\left (2 \, b d e + a e^{2}\right )} x^{3} + \frac{1}{2} \,{\left (b d^{2} + 2 \, a d e\right )} x^{2} \end{align*}

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]

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

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Sympy [A]  time = 0.111704, size = 49, normalized size = 0.53 \begin{align*} a d^{2} x + \frac{b e^{2} x^{4}}{4} + x^{3} \left (\frac{a e^{2}}{3} + \frac{2 b d e}{3}\right ) + x^{2} \left (a d e + \frac{b d^{2}}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.16699, size = 115, normalized size = 1.25 \begin{align*} \frac{1}{4} \, b x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{2}{3} \, b d x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, b d^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, a x^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) + a d x^{2} e \mathrm{sgn}\left (b x + a\right ) + a d^{2} x \mathrm{sgn}\left (b x + a\right ) \end{align*}

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]

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