3.1710 \(\int (A+B x) (d+e x)^2 \sqrt{a^2+2 a b x+b^2 x^2} \, dx\)

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.118247, 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} \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^4 (-a B e-A b e+2 b B d)}{4 e^3 (a+b x)}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^3 (b d-a e) (B d-A e)}{3 e^3 (a+b x)}+\frac{b B \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^5}{5 e^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*d - a*e)*(B*d - A*e)*(d + e*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^3*(a + b*x)) - ((2*b*B*d - A*b*e - a*
B*e)*(d + e*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*e^3*(a + b*x)) + (b*B*(d + e*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*
x^2])/(5*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)^2 \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)^2 \, 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)^2}{e^2}+\frac{b (-2 b B d+A b e+a B e) (d+e x)^3}{e^2}+\frac{b^2 B (d+e x)^4}{e^2}\right ) \, dx}{a b+b^2 x}\\ &=\frac{(b d-a e) (B d-A e) (d+e x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{3 e^3 (a+b x)}-\frac{(2 b B d-A b e-a B e) (d+e x)^4 \sqrt{a^2+2 a b x+b^2 x^2}}{4 e^3 (a+b x)}+\frac{b B (d+e x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{5 e^3 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.0532615, size = 120, normalized size = 0.76 \[ \frac{x \sqrt{(a+b x)^2} \left (5 a \left (4 A \left (3 d^2+3 d e x+e^2 x^2\right )+B x \left (6 d^2+8 d e x+3 e^2 x^2\right )\right )+b x \left (5 A \left (6 d^2+8 d e x+3 e^2 x^2\right )+2 B x \left (10 d^2+15 d e x+6 e^2 x^2\right )\right )\right )}{60 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(5*a*(4*A*(3*d^2 + 3*d*e*x + e^2*x^2) + B*x*(6*d^2 + 8*d*e*x + 3*e^2*x^2)) + b*x*(5*A*(6*
d^2 + 8*d*e*x + 3*e^2*x^2) + 2*B*x*(10*d^2 + 15*d*e*x + 6*e^2*x^2))))/(60*(a + b*x))

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 128, normalized size = 0.8 \begin{align*}{\frac{x \left ( 12\,bB{e}^{2}{x}^{4}+15\,{x}^{3}Ab{e}^{2}+15\,{x}^{3}aB{e}^{2}+30\,{x}^{3}bBde+20\,{x}^{2}aA{e}^{2}+40\,{x}^{2}Abde+40\,{x}^{2}aBde+20\,{x}^{2}bB{d}^{2}+60\,xaAde+30\,xAb{d}^{2}+30\,xBa{d}^{2}+60\,aA{d}^{2} \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)^2*((b*x+a)^2)^(1/2),x)

[Out]

1/60*x*(12*B*b*e^2*x^4+15*A*b*e^2*x^3+15*B*a*e^2*x^3+30*B*b*d*e*x^3+20*A*a*e^2*x^2+40*A*b*d*e*x^2+40*B*a*d*e*x
^2+20*B*b*d^2*x^2+60*A*a*d*e*x+30*A*b*d^2*x+30*B*a*d^2*x+60*A*a*d^2)*((b*x+a)^2)^(1/2)/(b*x+a)

________________________________________________________________________________________

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)^2*((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.53035, size = 215, normalized size = 1.36 \begin{align*} \frac{1}{5} \, B b e^{2} x^{5} + A a d^{2} x + \frac{1}{4} \,{\left (2 \, B b d e +{\left (B a + A b\right )} e^{2}\right )} x^{4} + \frac{1}{3} \,{\left (B b d^{2} + A a e^{2} + 2 \,{\left (B a + A b\right )} d e\right )} x^{3} + \frac{1}{2} \,{\left (2 \, A a d e +{\left (B a + A b\right )} d^{2}\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 0.115422, size = 116, normalized size = 0.73 \begin{align*} A a d^{2} x + \frac{B b e^{2} x^{5}}{5} + x^{4} \left (\frac{A b e^{2}}{4} + \frac{B a e^{2}}{4} + \frac{B b d e}{2}\right ) + x^{3} \left (\frac{A a e^{2}}{3} + \frac{2 A b d e}{3} + \frac{2 B a d e}{3} + \frac{B b d^{2}}{3}\right ) + x^{2} \left (A a d e + \frac{A b d^{2}}{2} + \frac{B a d^{2}}{2}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.12298, size = 250, normalized size = 1.58 \begin{align*} \frac{1}{5} \, B b x^{5} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, B b d x^{4} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, B b d^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, B a x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, A b x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{2}{3} \, B a d x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{2}{3} \, A b d x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, B a d^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, A b d^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, A a x^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) + A a d x^{2} e \mathrm{sgn}\left (b x + a\right ) + A a d^{2} 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)^2*((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/5*B*b*x^5*e^2*sgn(b*x + a) + 1/2*B*b*d*x^4*e*sgn(b*x + a) + 1/3*B*b*d^2*x^3*sgn(b*x + a) + 1/4*B*a*x^4*e^2*s
gn(b*x + a) + 1/4*A*b*x^4*e^2*sgn(b*x + a) + 2/3*B*a*d*x^3*e*sgn(b*x + a) + 2/3*A*b*d*x^3*e*sgn(b*x + a) + 1/2
*B*a*d^2*x^2*sgn(b*x + a) + 1/2*A*b*d^2*x^2*sgn(b*x + a) + 1/3*A*a*x^3*e^2*sgn(b*x + a) + A*a*d*x^2*e*sgn(b*x
+ a) + A*a*d^2*x*sgn(b*x + a)