3.1724 \(\int (A+B x) (d+e x) (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=135 \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^4 (-2 a B e+A b e+b B d)}{5 b^3}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B) (b d-a e)}{4 b^3}+\frac{B e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.143975, antiderivative size = 135, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {770, 77} \[ \frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^4 (-2 a B e+A b e+b B d)}{5 b^3}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B) (b d-a e)}{4 b^3}+\frac{B e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5}{6 b^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0702317, size = 144, normalized size = 1.07 \[ \frac{x \sqrt{(a+b x)^2} \left (15 a^2 b x (A (6 d+4 e x)+B x (4 d+3 e x))+10 a^3 (3 A (2 d+e x)+B x (3 d+2 e x))+3 a b^2 x^2 (5 A (4 d+3 e x)+3 B x (5 d+4 e x))+b^3 x^3 (3 A (5 d+4 e x)+2 B x (6 d+5 e x))\right )}{60 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.004, size = 180, normalized size = 1.3 \begin{align*}{\frac{x \left ( 10\,{b}^{3}Be{x}^{5}+12\,{x}^{4}A{b}^{3}e+36\,{x}^{4}Be{b}^{2}a+12\,{x}^{4}B{b}^{3}d+45\,{x}^{3}Aa{b}^{2}e+15\,{x}^{3}Ad{b}^{3}+45\,{x}^{3}Be{a}^{2}b+45\,{x}^{3}Ba{b}^{2}d+60\,{x}^{2}A{a}^{2}be+60\,{x}^{2}Ad{b}^{2}a+20\,{x}^{2}Be{a}^{3}+60\,{x}^{2}B{a}^{2}bd+30\,xA{a}^{3}e+90\,xAd{a}^{2}b+30\,xB{a}^{3}d+60\,Ad{a}^{3} \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((B*x+A)*(e*x+d)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/60*x*(10*B*b^3*e*x^5+12*A*b^3*e*x^4+36*B*a*b^2*e*x^4+12*B*b^3*d*x^4+45*A*a*b^2*e*x^3+15*A*b^3*d*x^3+45*B*a^2
*b*e*x^3+45*B*a*b^2*d*x^3+60*A*a^2*b*e*x^2+60*A*a*b^2*d*x^2+20*B*a^3*e*x^2+60*B*a^2*b*d*x^2+30*A*a^3*e*x+90*A*
a^2*b*d*x+30*B*a^3*d*x+60*A*a^3*d)*((b*x+a)^2)^(3/2)/(b*x+a)^3

________________________________________________________________________________________

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)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.51386, size = 320, normalized size = 2.37 \begin{align*} \frac{1}{6} \, B b^{3} e x^{6} + A a^{3} d x + \frac{1}{5} \,{\left (B b^{3} d +{\left (3 \, B a b^{2} + A b^{3}\right )} e\right )} x^{5} + \frac{1}{4} \,{\left ({\left (3 \, B a b^{2} + A b^{3}\right )} d + 3 \,{\left (B a^{2} b + A a b^{2}\right )} e\right )} x^{4} + \frac{1}{3} \,{\left (3 \,{\left (B a^{2} b + A a b^{2}\right )} d +{\left (B a^{3} + 3 \, A a^{2} b\right )} e\right )} x^{3} + \frac{1}{2} \,{\left (A a^{3} e +{\left (B a^{3} + 3 \, A a^{2} b\right )} d\right )} x^{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B x\right ) \left (d + e x\right ) \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((B*x+A)*(e*x+d)*(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.1185, size = 360, normalized size = 2.67 \begin{align*} \frac{1}{6} \, B b^{3} x^{6} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{5} \, B b^{3} d x^{5} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{5} \, B a b^{2} x^{5} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{5} \, A b^{3} x^{5} e \mathrm{sgn}\left (b x + a\right ) + \frac{3}{4} \, B a b^{2} d x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, A b^{3} d x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{4} \, B a^{2} b x^{4} e \mathrm{sgn}\left (b x + a\right ) + \frac{3}{4} \, A a b^{2} x^{4} e \mathrm{sgn}\left (b x + a\right ) + B a^{2} b d x^{3} \mathrm{sgn}\left (b x + a\right ) + A a b^{2} d x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, B a^{3} x^{3} e \mathrm{sgn}\left (b x + a\right ) + A a^{2} b x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, B a^{3} d x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, A a^{2} b d x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, A a^{3} x^{2} e \mathrm{sgn}\left (b x + a\right ) + A a^{3} d 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)*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

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