3.1557 \(\int (d+e x)^2 (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=114 \[ \frac {2 e \left (a^2+2 a b x+b^2 x^2\right )^{5/2} (b d-a e)}{5 b^3}+\frac {(a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{3/2} (b d-a e)^2}{4 b^3}+\frac {e^2 (a+b x) \left (a^2+2 a b x+b^2 x^2\right )^{5/2}}{6 b^3} \]

[Out]

1/4*(-a*e+b*d)^2*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(3/2)/b^3+2/5*e*(-a*e+b*d)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/b^3+1/6*
e^2*(b*x+a)*(b^2*x^2+2*a*b*x+a^2)^(5/2)/b^3

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Rubi [A]  time = 0.05, antiderivative size = 125, normalized size of antiderivative = 1.10, number of steps used = 2, number of rules used = 1, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.036, Rules used = {645} \[ \frac {2 e \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e)}{5 b^3}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^3 (b d-a e)^2}{4 b^3}+\frac {e^2 \sqrt {a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - a*e)^2*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*b^3) + (2*e*(b*d - a*e)*(a + b*x)^4*Sqrt[a^2 + 2*
a*b*x + b^2*x^2])/(5*b^3) + (e^2*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*b^3)

Rule 645

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[ExpandLinearProduct[(b/2 + c*x)^(2*p), (d + e*x)^m, b
/2, c, x], 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] && IGtQ[m, 0] && EqQ[m - 2*p + 1, 0]

Rubi steps

\begin {align*} \int (d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {(b d-a e)^2 \left (a b+b^2 x\right )^3}{b^2}+\frac {2 e (b d-a e) \left (a b+b^2 x\right )^4}{b^3}+\frac {e^2 \left (a b+b^2 x\right )^5}{b^4}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {(b d-a e)^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 b^3}+\frac {2 e (b d-a e) (a+b x)^4 \sqrt {a^2+2 a b x+b^2 x^2}}{5 b^3}+\frac {e^2 (a+b x)^5 \sqrt {a^2+2 a b x+b^2 x^2}}{6 b^3}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 127, normalized size = 1.11 \[ \frac {x \sqrt {(a+b x)^2} \left (20 a^3 \left (3 d^2+3 d e x+e^2 x^2\right )+15 a^2 b x \left (6 d^2+8 d e x+3 e^2 x^2\right )+6 a b^2 x^2 \left (10 d^2+15 d e x+6 e^2 x^2\right )+b^3 x^3 \left (15 d^2+24 d e x+10 e^2 x^2\right )\right )}{60 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(20*a^3*(3*d^2 + 3*d*e*x + e^2*x^2) + 15*a^2*b*x*(6*d^2 + 8*d*e*x + 3*e^2*x^2) + 6*a*b^2*
x^2*(10*d^2 + 15*d*e*x + 6*e^2*x^2) + b^3*x^3*(15*d^2 + 24*d*e*x + 10*e^2*x^2)))/(60*(a + b*x))

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fricas [A]  time = 0.96, size = 124, normalized size = 1.09 \[ \frac {1}{6} \, b^{3} e^{2} x^{6} + a^{3} d^{2} x + \frac {1}{5} \, {\left (2 \, b^{3} d e + 3 \, a b^{2} e^{2}\right )} x^{5} + \frac {1}{4} \, {\left (b^{3} d^{2} + 6 \, a b^{2} d e + 3 \, a^{2} b e^{2}\right )} x^{4} + \frac {1}{3} \, {\left (3 \, a b^{2} d^{2} + 6 \, a^{2} b d e + a^{3} e^{2}\right )} x^{3} + \frac {1}{2} \, {\left (3 \, a^{2} b d^{2} + 2 \, a^{3} d e\right )} x^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^3*e^2*x^6 + a^3*d^2*x + 1/5*(2*b^3*d*e + 3*a*b^2*e^2)*x^5 + 1/4*(b^3*d^2 + 6*a*b^2*d*e + 3*a^2*b*e^2)*x^
4 + 1/3*(3*a*b^2*d^2 + 6*a^2*b*d*e + a^3*e^2)*x^3 + 1/2*(3*a^2*b*d^2 + 2*a^3*d*e)*x^2

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giac [A]  time = 0.16, size = 202, normalized size = 1.77 \[ \frac {1}{6} \, b^{3} x^{6} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {2}{5} \, b^{3} d x^{5} e \mathrm {sgn}\left (b x + a\right ) + \frac {1}{4} \, b^{3} d^{2} x^{4} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{5} \, a b^{2} x^{5} e^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, a b^{2} d x^{4} e \mathrm {sgn}\left (b x + a\right ) + a b^{2} d^{2} x^{3} \mathrm {sgn}\left (b x + a\right ) + \frac {3}{4} \, a^{2} b x^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + 2 \, a^{2} b d x^{3} e \mathrm {sgn}\left (b x + a\right ) + \frac {3}{2} \, a^{2} b d^{2} x^{2} \mathrm {sgn}\left (b x + a\right ) + \frac {1}{3} \, a^{3} x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + a^{3} d x^{2} e \mathrm {sgn}\left (b x + a\right ) + a^{3} d^{2} x \mathrm {sgn}\left (b x + a\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*b^3*x^6*e^2*sgn(b*x + a) + 2/5*b^3*d*x^5*e*sgn(b*x + a) + 1/4*b^3*d^2*x^4*sgn(b*x + a) + 3/5*a*b^2*x^5*e^2
*sgn(b*x + a) + 3/2*a*b^2*d*x^4*e*sgn(b*x + a) + a*b^2*d^2*x^3*sgn(b*x + a) + 3/4*a^2*b*x^4*e^2*sgn(b*x + a) +
2*a^2*b*d*x^3*e*sgn(b*x + a) + 3/2*a^2*b*d^2*x^2*sgn(b*x + a) + 1/3*a^3*x^3*e^2*sgn(b*x + a) + a^3*d*x^2*e*sg
n(b*x + a) + a^3*d^2*x*sgn(b*x + a)

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maple [A]  time = 0.05, size = 148, normalized size = 1.30 \[ \frac {\left (10 b^{3} e^{2} x^{5}+36 x^{4} e^{2} a \,b^{2}+24 x^{4} b^{3} d e +45 x^{3} e^{2} a^{2} b +90 x^{3} d e a \,b^{2}+15 x^{3} d^{2} b^{3}+20 x^{2} a^{3} e^{2}+120 x^{2} d e \,a^{2} b +60 x^{2} a \,b^{2} d^{2}+60 x d e \,a^{3}+90 x \,d^{2} a^{2} b +60 a^{3} d^{2}\right ) \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} x}{60 \left (b x +a \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*x*(10*b^3*e^2*x^5+36*a*b^2*e^2*x^4+24*b^3*d*e*x^4+45*a^2*b*e^2*x^3+90*a*b^2*d*e*x^3+15*b^3*d^2*x^3+20*a^3
*e^2*x^2+120*a^2*b*d*e*x^2+60*a*b^2*d^2*x^2+60*a^3*d*e*x+90*a^2*b*d^2*x+60*a^3*d^2)*((b*x+a)^2)^(3/2)/(b*x+a)^
3

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maxima [B]  time = 1.13, size = 245, normalized size = 2.15 \[ \frac {1}{4} \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} d^{2} x - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d e x}{2 \, b} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} e^{2} x}{4 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a d^{2}}{4 \, b} - \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} d e}{2 \, b^{2}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3} e^{2}}{4 \, b^{3}} + \frac {{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} e^{2} x}{6 \, b^{2}} + \frac {2 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} d e}{5 \, b^{2}} - \frac {7 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} a e^{2}}{30 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^2*(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^2*x - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*d*e*x/b + 1/4*(b^2*x^2 + 2*a
*b*x + a^2)^(3/2)*a^2*e^2*x/b^2 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a*d^2/b - 1/2*(b^2*x^2 + 2*a*b*x + a^2)^
(3/2)*a^2*d*e/b^2 + 1/4*(b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^3*e^2/b^3 + 1/6*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*e^2*
x/b^2 + 2/5*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*d*e/b^2 - 7/30*(b^2*x^2 + 2*a*b*x + a^2)^(5/2)*a*e^2/b^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d+e\,x\right )}^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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