3.1709 \(\int (A+B x) (d+e x)^3 \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

Optimal. Leaf size=158 \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^5 (-a B e-A b e+2 b B d)}{5 e^3 (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^4 (b d-a e) (B d-A e)}{4 e^3 (a+b x)}+\frac{b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^6}{6 e^3 (a+b x)} \]

[Out]

((b*d - a*e)*(B*d - A*e)*(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*b*e - a*
B*e)*(d + e*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)) + (b*B*(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 = 0.144537, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.061, Rules used = {770, 77} \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^5 (-a B e-A b e+2 b B d)}{5 e^3 (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^4 (b d-a e) (B d-A e)}{4 e^3 (a+b x)}+\frac{b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^6}{6 e^3 (a+b x)} \]

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)*(B*d - A*e)*(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*b*e - a*
B*e)*(d + e*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*e^3*(a + b*x)) + (b*B*(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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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 \left (a b+b^2 x\right ) (A+B x) (d+e x)^3 \, dx}{a b+b^2 x}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (-\frac{b (b d-a e) (-B d+A e) (d+e x)^3}{e^2}+\frac{b (-2 b B d+A b e+a B e) (d+e x)^4}{e^2}+\frac{b^2 B (d+e x)^5}{e^2}\right ) \, dx}{a b+b^2 x}\\ &=\frac{(b d-a e) (B d-A e) (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 b e-a B e) (d+e x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)}+\frac{b B (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 = 0.0695418, size = 163, normalized size = 1.03 \[ \frac{x \sqrt{(a+b x)^2} \left (3 a \left (5 A \left (6 d^2 e x+4 d^3+4 d e^2 x^2+e^3 x^3\right )+B x \left (20 d^2 e x+10 d^3+15 d e^2 x^2+4 e^3 x^3\right )\right )+b x \left (3 A \left (20 d^2 e x+10 d^3+15 d e^2 x^2+4 e^3 x^3\right )+B x \left (45 d^2 e x+20 d^3+36 d e^2 x^2+10 e^3 x^3\right )\right )\right )}{60 (a+b x)} \]

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]*(3*a*(5*A*(4*d^3 + 6*d^2*e*x + 4*d*e^2*x^2 + e^3*x^3) + B*x*(10*d^3 + 20*d^2*e*x + 15*d*e
^2*x^2 + 4*e^3*x^3)) + b*x*(3*A*(10*d^3 + 20*d^2*e*x + 15*d*e^2*x^2 + 4*e^3*x^3) + B*x*(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 = 180, normalized size = 1.1 \begin{align*}{\frac{x \left ( 10\,bB{e}^{3}{x}^{5}+12\,{x}^{4}Ab{e}^{3}+12\,{x}^{4}Ba{e}^{3}+36\,{x}^{4}bBd{e}^{2}+15\,{x}^{3}aA{e}^{3}+45\,{x}^{3}Abd{e}^{2}+45\,{x}^{3}aBd{e}^{2}+45\,{x}^{3}bB{d}^{2}e+60\,{x}^{2}Aad{e}^{2}+60\,{x}^{2}Ab{d}^{2}e+60\,{x}^{2}Ba{d}^{2}e+20\,{x}^{2}bB{d}^{3}+90\,xaA{d}^{2}e+30\,xAb{d}^{3}+30\,xBa{d}^{3}+60\,aA{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*b*e^3*x^5+12*A*b*e^3*x^4+12*B*a*e^3*x^4+36*B*b*d*e^2*x^4+15*A*a*e^3*x^3+45*A*b*d*e^2*x^3+45*B*a*d
*e^2*x^3+45*B*b*d^2*e*x^3+60*A*a*d*e^2*x^2+60*A*b*d^2*e*x^2+60*B*a*d^2*e*x^2+20*B*b*d^3*x^2+90*A*a*d^2*e*x+30*
A*b*d^3*x+30*B*a*d^3*x+60*A*a*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 = 1.54744, size = 304, normalized size = 1.92 \begin{align*} \frac{1}{6} \, B b e^{3} x^{6} + A a d^{3} x + \frac{1}{5} \,{\left (3 \, B b d e^{2} +{\left (B a + A b\right )} e^{3}\right )} x^{5} + \frac{1}{4} \,{\left (3 \, B b d^{2} e + A a e^{3} + 3 \,{\left (B a + A b\right )} d e^{2}\right )} x^{4} + \frac{1}{3} \,{\left (B b d^{3} + 3 \, A a d e^{2} + 3 \,{\left (B a + A b\right )} d^{2} e\right )} x^{3} + \frac{1}{2} \,{\left (3 \, A a d^{2} e +{\left (B a + A b\right )} d^{3}\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*b*e^3*x^6 + A*a*d^3*x + 1/5*(3*B*b*d*e^2 + (B*a + A*b)*e^3)*x^5 + 1/4*(3*B*b*d^2*e + A*a*e^3 + 3*(B*a +
A*b)*d*e^2)*x^4 + 1/3*(B*b*d^3 + 3*A*a*d*e^2 + 3*(B*a + A*b)*d^2*e)*x^3 + 1/2*(3*A*a*d^2*e + (B*a + A*b)*d^3)*
x^2

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Sympy [A]  time = 0.131362, size = 168, normalized size = 1.06 \begin{align*} A a d^{3} x + \frac{B b e^{3} x^{6}}{6} + x^{5} \left (\frac{A b e^{3}}{5} + \frac{B a e^{3}}{5} + \frac{3 B b d e^{2}}{5}\right ) + x^{4} \left (\frac{A a e^{3}}{4} + \frac{3 A b d e^{2}}{4} + \frac{3 B a d e^{2}}{4} + \frac{3 B b d^{2} e}{4}\right ) + x^{3} \left (A a d e^{2} + A b d^{2} e + B a d^{2} e + \frac{B b d^{3}}{3}\right ) + x^{2} \left (\frac{3 A a d^{2} e}{2} + \frac{A b d^{3}}{2} + \frac{B a d^{3}}{2}\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*a*d**3*x + B*b*e**3*x**6/6 + x**5*(A*b*e**3/5 + B*a*e**3/5 + 3*B*b*d*e**2/5) + x**4*(A*a*e**3/4 + 3*A*b*d*e*
*2/4 + 3*B*a*d*e**2/4 + 3*B*b*d**2*e/4) + x**3*(A*a*d*e**2 + A*b*d**2*e + B*a*d**2*e + B*b*d**3/3) + x**2*(3*A
*a*d**2*e/2 + A*b*d**3/2 + B*a*d**3/2)

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Giac [B]  time = 1.18925, size = 344, normalized size = 2.18 \begin{align*} \frac{1}{6} \, B b x^{6} e^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{5} \, B b d x^{5} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{4} \, B b d^{2} x^{4} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, B b d^{3} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{5} \, B a x^{5} e^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{5} \, A b x^{5} e^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{4} \, B a d x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{4} \, A b d x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + B a d^{2} x^{3} e \mathrm{sgn}\left (b x + a\right ) + A b d^{2} x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, B a d^{3} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, A b d^{3} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, A a x^{4} e^{3} \mathrm{sgn}\left (b x + a\right ) + A a d x^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, A a d^{2} x^{2} e \mathrm{sgn}\left (b x + a\right ) + A a 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*b*x^6*e^3*sgn(b*x + a) + 3/5*B*b*d*x^5*e^2*sgn(b*x + a) + 3/4*B*b*d^2*x^4*e*sgn(b*x + a) + 1/3*B*b*d^3*x
^3*sgn(b*x + a) + 1/5*B*a*x^5*e^3*sgn(b*x + a) + 1/5*A*b*x^5*e^3*sgn(b*x + a) + 3/4*B*a*d*x^4*e^2*sgn(b*x + a)
+ 3/4*A*b*d*x^4*e^2*sgn(b*x + a) + B*a*d^2*x^3*e*sgn(b*x + a) + A*b*d^2*x^3*e*sgn(b*x + a) + 1/2*B*a*d^3*x^2*
sgn(b*x + a) + 1/2*A*b*d^3*x^2*sgn(b*x + a) + 1/4*A*a*x^4*e^3*sgn(b*x + a) + A*a*d*x^3*e^2*sgn(b*x + a) + 3/2*
A*a*d^2*x^2*e*sgn(b*x + a) + A*a*d^3*x*sgn(b*x + a)