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]

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

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Rubi [A]  time = 0.0499471, antiderivative size = 125, normalized size of antiderivative = 1.1, 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*}

Mathematica [A]  time = 0.0449353, size = 127, normalized size = 1.11 \[ \frac{x \sqrt{(a+b x)^2} \left (15 a^2 b x \left (6 d^2+8 d e x+3 e^2 x^2\right )+20 a^3 \left (3 d^2+3 d e x+e^2 x^2\right )+6 a b^2 x^2 \left (10 d^2+15 d e x+6 e^2 x^2\right )+b^3 x^3 \left (15 d^2+24 d e x+10 e^2 x^2\right )\right )}{60 (a+b x)} \]

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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Maple [A]  time = 0.154, size = 148, normalized size = 1.3 \begin{align*}{\frac{x \left ( 10\,{b}^{3}{e}^{2}{x}^{5}+36\,{x}^{4}{e}^{2}{b}^{2}a+24\,{x}^{4}de{b}^{3}+45\,{x}^{3}{e}^{2}b{a}^{2}+90\,{x}^{3}de{b}^{2}a+15\,{x}^{3}{d}^{2}{b}^{3}+20\,{x}^{2}{e}^{2}{a}^{3}+120\,{x}^{2}deb{a}^{2}+60\,{x}^{2}a{b}^{2}{d}^{2}+60\,xde{a}^{3}+90\,x{d}^{2}b{a}^{2}+60\,{a}^{3}{d}^{2} \right ) }{60\, \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)^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 [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^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 = 1.44664, size = 266, normalized size = 2.33 \begin{align*} \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} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d + e x\right )^{2} \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)**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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Giac [A]  time = 1.173, size = 273, normalized size = 2.39 \begin{align*} \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 ) \end{align*}

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)