3.1723 \(\int (A+B x) (d+e x)^2 (a^2+2 a b x+b^2 x^2)^{3/2} \, dx\)

Optimal. Leaf size=198 \[ \frac{e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (-3 a B e+A b e+2 b B d)}{6 b^4}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e) (-3 a B e+2 A b e+b B d)}{5 b^4}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B) (b d-a e)^2}{4 b^4}+\frac{B e^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^4} \]

[Out]

((A*b - a*B)*(b*d - a*e)^2*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*b^4) + ((b*d - a*e)*(b*B*d + 2*A*b*e
- 3*a*B*e)*(a + b*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*b^4) + (e*(2*b*B*d + A*b*e - 3*a*B*e)*(a + b*x)^5*Sqr
t[a^2 + 2*a*b*x + b^2*x^2])/(6*b^4) + (B*e^2*(a + b*x)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*b^4)

________________________________________________________________________________________

Rubi [A]  time = 0.235755, antiderivative size = 198, 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} \[ \frac{e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (-3 a B e+A b e+2 b B d)}{6 b^4}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^4 (b d-a e) (-3 a B e+2 A b e+b B d)}{5 b^4}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^3 (A b-a B) (b d-a e)^2}{4 b^4}+\frac{B e^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6}{7 b^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*(b*d - a*e)^2*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*b^4) + ((b*d - a*e)*(b*B*d + 2*A*b*e
- 3*a*B*e)*(a + b*x)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(5*b^4) + (e*(2*b*B*d + A*b*e - 3*a*B*e)*(a + b*x)^5*Sqr
t[a^2 + 2*a*b*x + b^2*x^2])/(6*b^4) + (B*e^2*(a + b*x)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*b^4)

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 \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)^2 \, 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)^2 \left (a b+b^2 x\right )^3}{b^3}+\frac{(b d-a e) (b B d+2 A b e-3 a B e) \left (a b+b^2 x\right )^4}{b^4}+\frac{e (2 b B d+A b e-3 a B e) \left (a b+b^2 x\right )^5}{b^5}+\frac{B e^2 \left (a b+b^2 x\right )^6}{b^6}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac{(A b-a B) (b d-a e)^2 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}{4 b^4}+\frac{(b d-a e) (b B d+2 A b e-3 a B e) (a+b x)^4 \sqrt{a^2+2 a b x+b^2 x^2}}{5 b^4}+\frac{e (2 b B d+A b e-3 a B e) (a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{6 b^4}+\frac{B e^2 (a+b x)^6 \sqrt{a^2+2 a b x+b^2 x^2}}{7 b^4}\\ \end{align*}

Mathematica [A]  time = 0.108587, size = 233, normalized size = 1.18 \[ \frac{x \sqrt{(a+b x)^2} \left (21 a^2 b x \left (5 A \left (6 d^2+8 d e x+3 e^2 x^2\right )+2 B x \left (10 d^2+15 d e x+6 e^2 x^2\right )\right )+35 a^3 \left (4 A \left (3 d^2+3 d e x+e^2 x^2\right )+B x \left (6 d^2+8 d e x+3 e^2 x^2\right )\right )+21 a b^2 x^2 \left (2 A \left (10 d^2+15 d e x+6 e^2 x^2\right )+B x \left (15 d^2+24 d e x+10 e^2 x^2\right )\right )+b^3 x^3 \left (7 A \left (15 d^2+24 d e x+10 e^2 x^2\right )+4 B x \left (21 d^2+35 d e x+15 e^2 x^2\right )\right )\right )}{420 (a+b x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x*Sqrt[(a + b*x)^2]*(35*a^3*(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)) + 21*a^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)) + 21*a*b^2*x^2*(2*A*(10*d^2 + 15
*d*e*x + 6*e^2*x^2) + B*x*(15*d^2 + 24*d*e*x + 10*e^2*x^2)) + b^3*x^3*(7*A*(15*d^2 + 24*d*e*x + 10*e^2*x^2) +
4*B*x*(21*d^2 + 35*d*e*x + 15*e^2*x^2))))/(420*(a + b*x))

________________________________________________________________________________________

Maple [B]  time = 0.006, size = 304, normalized size = 1.5 \begin{align*}{\frac{x \left ( 60\,{b}^{3}B{e}^{2}{x}^{6}+70\,{x}^{5}A{b}^{3}{e}^{2}+210\,{x}^{5}B{e}^{2}{b}^{2}a+140\,{x}^{5}B{b}^{3}de+252\,{x}^{4}Aa{b}^{2}{e}^{2}+168\,{x}^{4}A{b}^{3}de+252\,{x}^{4}B{e}^{2}{a}^{2}b+504\,{x}^{4}Ba{b}^{2}de+84\,{x}^{4}B{b}^{3}{d}^{2}+315\,{x}^{3}A{a}^{2}b{e}^{2}+630\,{x}^{3}Aa{b}^{2}de+105\,{x}^{3}A{d}^{2}{b}^{3}+105\,{x}^{3}B{e}^{2}{a}^{3}+630\,{x}^{3}B{a}^{2}bde+315\,{x}^{3}Ba{b}^{2}{d}^{2}+140\,{x}^{2}A{a}^{3}{e}^{2}+840\,{x}^{2}A{a}^{2}bde+420\,{x}^{2}A{d}^{2}{b}^{2}a+280\,{x}^{2}B{a}^{3}de+420\,{x}^{2}B{a}^{2}b{d}^{2}+420\,xA{a}^{3}de+630\,xA{d}^{2}{a}^{2}b+210\,xB{a}^{3}{d}^{2}+420\,A{d}^{2}{a}^{3} \right ) }{420\, \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)^2*(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/420*x*(60*B*b^3*e^2*x^6+70*A*b^3*e^2*x^5+210*B*a*b^2*e^2*x^5+140*B*b^3*d*e*x^5+252*A*a*b^2*e^2*x^4+168*A*b^3
*d*e*x^4+252*B*a^2*b*e^2*x^4+504*B*a*b^2*d*e*x^4+84*B*b^3*d^2*x^4+315*A*a^2*b*e^2*x^3+630*A*a*b^2*d*e*x^3+105*
A*b^3*d^2*x^3+105*B*a^3*e^2*x^3+630*B*a^2*b*d*e*x^3+315*B*a*b^2*d^2*x^3+140*A*a^3*e^2*x^2+840*A*a^2*b*d*e*x^2+
420*A*a*b^2*d^2*x^2+280*B*a^3*d*e*x^2+420*B*a^2*b*d^2*x^2+420*A*a^3*d*e*x+630*A*a^2*b*d^2*x+210*B*a^3*d^2*x+42
0*A*a^3*d^2)*((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)^2*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.50392, size = 512, normalized size = 2.59 \begin{align*} \frac{1}{7} \, B b^{3} e^{2} x^{7} + A a^{3} d^{2} x + \frac{1}{6} \,{\left (2 \, B b^{3} d e +{\left (3 \, B a b^{2} + A b^{3}\right )} e^{2}\right )} x^{6} + \frac{1}{5} \,{\left (B b^{3} d^{2} + 2 \,{\left (3 \, B a b^{2} + A b^{3}\right )} d e + 3 \,{\left (B a^{2} b + A a b^{2}\right )} e^{2}\right )} x^{5} + \frac{1}{4} \,{\left ({\left (3 \, B a b^{2} + A b^{3}\right )} d^{2} + 6 \,{\left (B a^{2} b + A a b^{2}\right )} d e +{\left (B a^{3} + 3 \, A a^{2} b\right )} e^{2}\right )} x^{4} + \frac{1}{3} \,{\left (A a^{3} e^{2} + 3 \,{\left (B a^{2} b + A a b^{2}\right )} d^{2} + 2 \,{\left (B a^{3} + 3 \, A a^{2} b\right )} d e\right )} x^{3} + \frac{1}{2} \,{\left (2 \, A a^{3} d e +{\left (B a^{3} + 3 \, A a^{2} 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^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

1/7*B*b^3*e^2*x^7 + A*a^3*d^2*x + 1/6*(2*B*b^3*d*e + (3*B*a*b^2 + A*b^3)*e^2)*x^6 + 1/5*(B*b^3*d^2 + 2*(3*B*a*
b^2 + A*b^3)*d*e + 3*(B*a^2*b + A*a*b^2)*e^2)*x^5 + 1/4*((3*B*a*b^2 + A*b^3)*d^2 + 6*(B*a^2*b + A*a*b^2)*d*e +
(B*a^3 + 3*A*a^2*b)*e^2)*x^4 + 1/3*(A*a^3*e^2 + 3*(B*a^2*b + A*a*b^2)*d^2 + 2*(B*a^3 + 3*A*a^2*b)*d*e)*x^3 +
1/2*(2*A*a^3*d*e + (B*a^3 + 3*A*a^2*b)*d^2)*x^2

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.1225, size = 582, normalized size = 2.94 \begin{align*} \frac{1}{7} \, B b^{3} x^{7} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, B b^{3} d x^{6} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{5} \, B b^{3} d^{2} x^{5} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, B a b^{2} x^{6} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{6} \, A b^{3} x^{6} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{6}{5} \, B a b^{2} d x^{5} e \mathrm{sgn}\left (b x + a\right ) + \frac{2}{5} \, A b^{3} d x^{5} e \mathrm{sgn}\left (b x + a\right ) + \frac{3}{4} \, B a b^{2} d^{2} x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, A b^{3} d^{2} x^{4} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{5} \, B a^{2} b x^{5} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{5} \, A a b^{2} x^{5} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, B a^{2} b d x^{4} e \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, A a b^{2} d x^{4} e \mathrm{sgn}\left (b x + a\right ) + B a^{2} b d^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + A a b^{2} d^{2} x^{3} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{4} \, B a^{3} x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{4} \, A a^{2} b x^{4} e^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{2}{3} \, B a^{3} d x^{3} e \mathrm{sgn}\left (b x + a\right ) + 2 \, A a^{2} b d x^{3} e \mathrm{sgn}\left (b x + a\right ) + \frac{1}{2} \, B a^{3} d^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{3}{2} \, A a^{2} b d^{2} x^{2} \mathrm{sgn}\left (b x + a\right ) + \frac{1}{3} \, A a^{3} x^{3} e^{2} \mathrm{sgn}\left (b x + a\right ) + A a^{3} d x^{2} e \mathrm{sgn}\left (b x + a\right ) + A 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((B*x+A)*(e*x+d)^2*(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

1/7*B*b^3*x^7*e^2*sgn(b*x + a) + 1/3*B*b^3*d*x^6*e*sgn(b*x + a) + 1/5*B*b^3*d^2*x^5*sgn(b*x + a) + 1/2*B*a*b^2
*x^6*e^2*sgn(b*x + a) + 1/6*A*b^3*x^6*e^2*sgn(b*x + a) + 6/5*B*a*b^2*d*x^5*e*sgn(b*x + a) + 2/5*A*b^3*d*x^5*e*
sgn(b*x + a) + 3/4*B*a*b^2*d^2*x^4*sgn(b*x + a) + 1/4*A*b^3*d^2*x^4*sgn(b*x + a) + 3/5*B*a^2*b*x^5*e^2*sgn(b*x
+ a) + 3/5*A*a*b^2*x^5*e^2*sgn(b*x + a) + 3/2*B*a^2*b*d*x^4*e*sgn(b*x + a) + 3/2*A*a*b^2*d*x^4*e*sgn(b*x + a)
+ B*a^2*b*d^2*x^3*sgn(b*x + a) + A*a*b^2*d^2*x^3*sgn(b*x + a) + 1/4*B*a^3*x^4*e^2*sgn(b*x + a) + 3/4*A*a^2*b*
x^4*e^2*sgn(b*x + a) + 2/3*B*a^3*d*x^3*e*sgn(b*x + a) + 2*A*a^2*b*d*x^3*e*sgn(b*x + a) + 1/2*B*a^3*d^2*x^2*sgn
(b*x + a) + 3/2*A*a^2*b*d^2*x^2*sgn(b*x + a) + 1/3*A*a^3*x^3*e^2*sgn(b*x + a) + A*a^3*d*x^2*e*sgn(b*x + a) + A
*a^3*d^2*x*sgn(b*x + a)