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

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

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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) \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} \end {gather*}

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 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 (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*}

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Mathematica [A]  time = 0.03, size = 83, normalized size = 1.20 \begin {gather*} \frac {x \sqrt {(a+b x)^2} \left (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)\right )}{20 (a+b x)} \end {gather*}

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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IntegrateAlgebraic [F]  time = 0.95, size = 0, normalized size = 0.00 \begin {gather*} \int (d+e x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.40, size = 69, normalized size = 1.00 \begin {gather*} \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 {gather*}

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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giac [B]  time = 0.16, size = 124, normalized size = 1.80 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.04, size = 90, normalized size = 1.30 \begin {gather*} \frac {\left (4 e \,b^{3} x^{4}+15 x^{3} e a \,b^{2}+5 x^{3} b^{3} d +20 a^{2} b e \,x^{2}+20 a \,b^{2} d \,x^{2}+10 a^{3} e x +30 x d \,a^{2} b +20 a^{3} d \right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x}{20 \left (b x +a \right )^{3}} \end {gather*}

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 [B]  time = 1.01, size = 125, normalized size = 1.81 \begin {gather*} \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d x - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a e x}{4 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d}{4 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} e}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} e}{5 \, b^{2}} \end {gather*}

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]

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

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mupad [B]  time = 0.67, size = 42, normalized size = 0.61 \begin {gather*} \frac {\left (a+b\,x\right )\,\left (5\,b\,d-a\,e+4\,b\,e\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}}{20\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a + b*x)*(5*b*d - a*e + 4*b*e*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(20*b^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (d + e x\right ) \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}\, dx \end {gather*}

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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