3.1956 \(\int (a+b x) (d+e x)^3 \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=146 \[ \frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^6}{6 e^3 (a+b x)}-\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^5 (b d-a e)}{5 e^3 (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^4 (b d-a e)^2}{4 e^3 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.108517, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.091, Rules used = {770, 21, 43} \[ \frac{b^2 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^6}{6 e^3 (a+b x)}-\frac{2 b \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^5 (b d-a e)}{5 e^3 (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^4 (b d-a e)^2}{4 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

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

Mathematica [A]  time = 0.0465692, size = 130, normalized size = 0.89 \[ \frac{x \sqrt{(a+b x)^2} \left (15 a^2 \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+6 a b x \left (20 d^2 e x+10 d^3+15 d e^2 x^2+4 e^3 x^3\right )+b^2 x^2 \left (45 d^2 e x+20 d^3+36 d e^2 x^2+10 e^3 x^3\right )\right )}{60 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(15*a^2*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + 6*a*b*x*(10*d^3 + 20*d^2*e*x + 15*d
*e^2*x^2 + 4*e^3*x^3) + b^2*x^2*(20*d^3 + 45*d^2*e*x + 36*d*e^2*x^2 + 10*e^3*x^3)))/(60*(a + b*x))

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Maple [A]  time = 0.006, size = 148, normalized size = 1. \begin{align*}{\frac{x \left ( 10\,{b}^{2}{e}^{3}{x}^{5}+24\,{x}^{4}ab{e}^{3}+36\,{x}^{4}{b}^{2}d{e}^{2}+15\,{x}^{3}{a}^{2}{e}^{3}+90\,{x}^{3}abd{e}^{2}+45\,{x}^{3}{b}^{2}{d}^{2}e+60\,{x}^{2}{a}^{2}d{e}^{2}+120\,{x}^{2}ab{d}^{2}e+20\,{x}^{2}{b}^{2}{d}^{3}+90\,x{a}^{2}{d}^{2}e+60\,xab{d}^{3}+60\,{a}^{2}{d}^{3} \right ) }{60\,bx+60\,a}\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.51623, size = 266, normalized size = 1.82 \begin{align*} \frac{1}{6} \, b^{2} e^{3} x^{6} + a^{2} d^{3} x + \frac{1}{5} \,{\left (3 \, b^{2} d e^{2} + 2 \, a b e^{3}\right )} x^{5} + \frac{1}{4} \,{\left (3 \, b^{2} d^{2} e + 6 \, a b d e^{2} + a^{2} e^{3}\right )} x^{4} + \frac{1}{3} \,{\left (b^{2} d^{3} + 6 \, a b d^{2} e + 3 \, a^{2} d e^{2}\right )} x^{3} + \frac{1}{2} \,{\left (2 \, a b d^{3} + 3 \, a^{2} d^{2} e\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^3*((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.122157, size = 133, normalized size = 0.91 \begin{align*} a^{2} d^{3} x + \frac{b^{2} e^{3} x^{6}}{6} + x^{5} \left (\frac{2 a b e^{3}}{5} + \frac{3 b^{2} d e^{2}}{5}\right ) + x^{4} \left (\frac{a^{2} e^{3}}{4} + \frac{3 a b d e^{2}}{2} + \frac{3 b^{2} d^{2} e}{4}\right ) + x^{3} \left (a^{2} d e^{2} + 2 a b d^{2} e + \frac{b^{2} d^{3}}{3}\right ) + x^{2} \left (\frac{3 a^{2} d^{2} e}{2} + a b d^{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)**3*((b*x+a)**2)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.19355, size = 269, normalized size = 1.84 \begin{align*} \frac{1}{6} \, b^{2} x^{6} e^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{5} \, b^{2} d x^{5} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{4} \, b^{2} d^{2} x^{4} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, b^{2} d^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{2}{5} \, a b x^{5} e^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, a b d x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + 2 \, a b d^{2} x^{3} e \mathrm{sgn}\left (b x + a\right ) + a b d^{3} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, a^{2} x^{4} e^{3} \mathrm{sgn}\left (b x + a\right ) + a^{2} d x^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, a^{2} d^{2} x^{2} e \mathrm{sgn}\left (b x + a\right ) + a^{2} d^{3} x \mathrm{sgn}\left (b x + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*(e*x+d)^3*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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