3.1558 \(\int (d+e x) (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=69 \[ \frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (b d-a e)}{4 b^2}+\frac{e \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2} \]

[Out]

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

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Rubi [A]  time = 0.0211858, antiderivative size = 69, normalized size of antiderivative = 1., 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} \[ \frac{(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (b d-a e)}{4 b^2}+\frac{e \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{5 b^2} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

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

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]

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)]

Rubi steps

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

Mathematica [A]  time = 0.032854, size = 83, normalized size = 1.2 \[ \frac{x \sqrt{(a+b x)^2} \left (10 a^2 b x (3 d+2 e x)+10 a^3 (2 d+e x)+5 a b^2 x^2 (4 d+3 e x)+b^3 x^3 (5 d+4 e x)\right )}{20 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(x*Sqrt[(a + b*x)^2]*(10*a^3*(2*d + e*x) + 10*a^2*b*x*(3*d + 2*e*x) + 5*a*b^2*x^2*(4*d + 3*e*x) + b^3*x^3*(5*d
+ 4*e*x)))/(20*(a + b*x))

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Maple [A]  time = 0.178, size = 90, normalized size = 1.3 \begin{align*}{\frac{x \left ( 4\,e{b}^{3}{x}^{4}+15\,{x}^{3}e{b}^{2}a+5\,{x}^{3}d{b}^{3}+20\,{a}^{2}be{x}^{2}+20\,a{b}^{2}d{x}^{2}+10\,{a}^{3}ex+30\,xdb{a}^{2}+20\,d{a}^{3} \right ) }{20\, \left ( bx+a \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20*x*(4*b^3*e*x^4+15*a*b^2*e*x^3+5*b^3*d*x^3+20*a^2*b*e*x^2+20*a*b^2*d*x^2+10*a^3*e*x+30*a^2*b*d*x+20*a^3*d)
*((b*x+a)^2)^(3/2)/(b*x+a)^3

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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)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.52932, size = 150, normalized size = 2.17 \begin{align*} \frac{1}{5} \, b^{3} e x^{5} + a^{3} d x + \frac{1}{4} \,{\left (b^{3} d + 3 \, a b^{2} e\right )} x^{4} +{\left (a b^{2} d + a^{2} b e\right )} x^{3} + \frac{1}{2} \,{\left (3 \, a^{2} b d + a^{3} e\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d + e x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.13498, size = 167, normalized size = 2.42 \begin{align*} \frac{1}{5} \, b^{3} x^{5} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, b^{3} d x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{4} \, a b^{2} x^{4} e \mathrm{sgn}\left (b x + a\right ) + a b^{2} d x^{3} \mathrm{sgn}\left (b x + a\right ) + a^{2} b x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, a^{2} b d x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, a^{3} x^{2} e \mathrm{sgn}\left (b x + a\right ) + a^{3} d 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)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

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