3.1742 \(\int (A+B x) (d+e x)^2 (a^2+2 a b x+b^2 x^2)^{5/2} \, dx\)

Optimal. Leaf size=198 \[ \frac{e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^7 (-3 a B e+A b e+2 b B d)}{8 b^4}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6 (b d-a e) (-3 a B e+2 A b e+b B d)}{7 b^4}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-a B) (b d-a e)^2}{6 b^4}+\frac{B e^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^8}{9 b^4} \]

[Out]

((A*b - a*B)*(b*d - a*e)^2*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*b^4) + ((b*d - a*e)*(b*B*d + 2*A*b*e
- 3*a*B*e)*(a + b*x)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*b^4) + (e*(2*b*B*d + A*b*e - 3*a*B*e)*(a + b*x)^7*Sqr
t[a^2 + 2*a*b*x + b^2*x^2])/(8*b^4) + (B*e^2*(a + b*x)^8*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*b^4)

________________________________________________________________________________________

Rubi [A]  time = 0.339145, 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} \[ \frac{e \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^7 (-3 a B e+A b e+2 b B d)}{8 b^4}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^6 (b d-a e) (-3 a B e+2 A b e+b B d)}{7 b^4}+\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^5 (A b-a B) (b d-a e)^2}{6 b^4}+\frac{B e^2 \sqrt{a^2+2 a b x+b^2 x^2} (a+b x)^8}{9 b^4} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)*(d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

((A*b - a*B)*(b*d - a*e)^2*(a + b*x)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(6*b^4) + ((b*d - a*e)*(b*B*d + 2*A*b*e
- 3*a*B*e)*(a + b*x)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(7*b^4) + (e*(2*b*B*d + A*b*e - 3*a*B*e)*(a + b*x)^7*Sqr
t[a^2 + 2*a*b*x + b^2*x^2])/(8*b^4) + (B*e^2*(a + b*x)^8*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(9*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 )^{5/2} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (a b+b^2 x\right )^5 (A+B x) (d+e x)^2 \, dx}{b^4 \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 )^5}{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 )^6}{b^4}+\frac{e (2 b B d+A b e-3 a B e) \left (a b+b^2 x\right )^7}{b^5}+\frac{B e^2 \left (a b+b^2 x\right )^8}{b^6}\right ) \, dx}{b^4 \left (a b+b^2 x\right )}\\ &=\frac{(A b-a B) (b d-a e)^2 (a+b x)^5 \sqrt{a^2+2 a b x+b^2 x^2}}{6 b^4}+\frac{(b d-a e) (b B d+2 A b e-3 a B e) (a+b x)^6 \sqrt{a^2+2 a b x+b^2 x^2}}{7 b^4}+\frac{e (2 b B d+A b e-3 a B e) (a+b x)^7 \sqrt{a^2+2 a b x+b^2 x^2}}{8 b^4}+\frac{B e^2 (a+b x)^8 \sqrt{a^2+2 a b x+b^2 x^2}}{9 b^4}\\ \end{align*}

Mathematica [A]  time = 0.161725, size = 347, normalized size = 1.75 \[ \frac{x \sqrt{(a+b x)^2} \left (84 a^3 b^2 x^2 \left (2 A \left (10 d^2+15 d e x+6 e^2 x^2\right )+B x \left (15 d^2+24 d e x+10 e^2 x^2\right )\right )+12 a^2 b^3 x^3 \left (7 A \left (15 d^2+24 d e x+10 e^2 x^2\right )+4 B x \left (21 d^2+35 d e x+15 e^2 x^2\right )\right )+42 a^4 b x \left (5 A \left (6 d^2+8 d e x+3 e^2 x^2\right )+2 B x \left (10 d^2+15 d e x+6 e^2 x^2\right )\right )+42 a^5 \left (4 A \left (3 d^2+3 d e x+e^2 x^2\right )+B x \left (6 d^2+8 d e x+3 e^2 x^2\right )\right )+3 a b^4 x^4 \left (8 A \left (21 d^2+35 d e x+15 e^2 x^2\right )+5 B x \left (28 d^2+48 d e x+21 e^2 x^2\right )\right )+b^5 x^5 \left (3 A \left (28 d^2+48 d e x+21 e^2 x^2\right )+2 B x \left (36 d^2+63 d e x+28 e^2 x^2\right )\right )\right )}{504 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)*(d + e*x)^2*(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]

[Out]

(x*Sqrt[(a + b*x)^2]*(42*a^5*(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)) + 42*a^4*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)) + 84*a^3*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)) + 12*a^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)) + 3*a*b^4*x^4*(8*A*(21*d^2 + 35*d*e*x + 15*e^2*x^2) + 5*B*x*(
28*d^2 + 48*d*e*x + 21*e^2*x^2)) + b^5*x^5*(3*A*(28*d^2 + 48*d*e*x + 21*e^2*x^2) + 2*B*x*(36*d^2 + 63*d*e*x +
28*e^2*x^2))))/(504*(a + b*x))

________________________________________________________________________________________

Maple [B]  time = 0.007, size = 480, normalized size = 2.4 \begin{align*}{\frac{x \left ( 56\,B{e}^{2}{b}^{5}{x}^{8}+63\,{x}^{7}A{b}^{5}{e}^{2}+315\,{x}^{7}B{e}^{2}a{b}^{4}+126\,{x}^{7}B{b}^{5}de+360\,{x}^{6}Aa{b}^{4}{e}^{2}+144\,{x}^{6}A{b}^{5}de+720\,{x}^{6}B{e}^{2}{a}^{2}{b}^{3}+720\,{x}^{6}Ba{b}^{4}de+72\,{x}^{6}B{b}^{5}{d}^{2}+840\,{x}^{5}A{a}^{2}{b}^{3}{e}^{2}+840\,{x}^{5}Aa{b}^{4}de+84\,{x}^{5}A{d}^{2}{b}^{5}+840\,{x}^{5}B{e}^{2}{a}^{3}{b}^{2}+1680\,{x}^{5}B{a}^{2}{b}^{3}de+420\,{x}^{5}Ba{b}^{4}{d}^{2}+1008\,A{a}^{3}{b}^{2}{e}^{2}{x}^{4}+2016\,A{a}^{2}{b}^{3}de{x}^{4}+504\,Aa{b}^{4}{d}^{2}{x}^{4}+504\,B{a}^{4}b{e}^{2}{x}^{4}+2016\,B{a}^{3}{b}^{2}de{x}^{4}+1008\,B{a}^{2}{b}^{3}{d}^{2}{x}^{4}+630\,{x}^{3}A{a}^{4}b{e}^{2}+2520\,{x}^{3}A{a}^{3}{b}^{2}de+1260\,{x}^{3}A{d}^{2}{a}^{2}{b}^{3}+126\,{x}^{3}B{e}^{2}{a}^{5}+1260\,{x}^{3}B{a}^{4}bde+1260\,{x}^{3}B{a}^{3}{b}^{2}{d}^{2}+168\,{x}^{2}A{a}^{5}{e}^{2}+1680\,{x}^{2}A{a}^{4}bde+1680\,{x}^{2}A{d}^{2}{a}^{3}{b}^{2}+336\,{x}^{2}B{a}^{5}de+840\,{x}^{2}B{a}^{4}b{d}^{2}+504\,xA{a}^{5}de+1260\,xA{d}^{2}{a}^{4}b+252\,xB{a}^{5}{d}^{2}+504\,A{d}^{2}{a}^{5} \right ) }{504\, \left ( bx+a \right ) ^{5}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{{\frac{5}{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)^(5/2),x)

[Out]

1/504*x*(56*B*b^5*e^2*x^8+63*A*b^5*e^2*x^7+315*B*a*b^4*e^2*x^7+126*B*b^5*d*e*x^7+360*A*a*b^4*e^2*x^6+144*A*b^5
*d*e*x^6+720*B*a^2*b^3*e^2*x^6+720*B*a*b^4*d*e*x^6+72*B*b^5*d^2*x^6+840*A*a^2*b^3*e^2*x^5+840*A*a*b^4*d*e*x^5+
84*A*b^5*d^2*x^5+840*B*a^3*b^2*e^2*x^5+1680*B*a^2*b^3*d*e*x^5+420*B*a*b^4*d^2*x^5+1008*A*a^3*b^2*e^2*x^4+2016*
A*a^2*b^3*d*e*x^4+504*A*a*b^4*d^2*x^4+504*B*a^4*b*e^2*x^4+2016*B*a^3*b^2*d*e*x^4+1008*B*a^2*b^3*d^2*x^4+630*A*
a^4*b*e^2*x^3+2520*A*a^3*b^2*d*e*x^3+1260*A*a^2*b^3*d^2*x^3+126*B*a^5*e^2*x^3+1260*B*a^4*b*d*e*x^3+1260*B*a^3*
b^2*d^2*x^3+168*A*a^5*e^2*x^2+1680*A*a^4*b*d*e*x^2+1680*A*a^3*b^2*d^2*x^2+336*B*a^5*d*e*x^2+840*B*a^4*b*d^2*x^
2+504*A*a^5*d*e*x+1260*A*a^4*b*d^2*x+252*B*a^5*d^2*x+504*A*a^5*d^2)*((b*x+a)^2)^(5/2)/(b*x+a)^5

________________________________________________________________________________________

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)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 1.55379, size = 809, normalized size = 4.09 \begin{align*} \frac{1}{9} \, B b^{5} e^{2} x^{9} + A a^{5} d^{2} x + \frac{1}{8} \,{\left (2 \, B b^{5} d e +{\left (5 \, B a b^{4} + A b^{5}\right )} e^{2}\right )} x^{8} + \frac{1}{7} \,{\left (B b^{5} d^{2} + 2 \,{\left (5 \, B a b^{4} + A b^{5}\right )} d e + 5 \,{\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} e^{2}\right )} x^{7} + \frac{1}{6} \,{\left ({\left (5 \, B a b^{4} + A b^{5}\right )} d^{2} + 10 \,{\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} d e + 10 \,{\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} e^{2}\right )} x^{6} +{\left ({\left (2 \, B a^{2} b^{3} + A a b^{4}\right )} d^{2} + 4 \,{\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} d e +{\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} e^{2}\right )} x^{5} + \frac{1}{4} \,{\left (10 \,{\left (B a^{3} b^{2} + A a^{2} b^{3}\right )} d^{2} + 10 \,{\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} d e +{\left (B a^{5} + 5 \, A a^{4} b\right )} e^{2}\right )} x^{4} + \frac{1}{3} \,{\left (A a^{5} e^{2} + 5 \,{\left (B a^{4} b + 2 \, A a^{3} b^{2}\right )} d^{2} + 2 \,{\left (B a^{5} + 5 \, A a^{4} b\right )} d e\right )} x^{3} + \frac{1}{2} \,{\left (2 \, A a^{5} d e +{\left (B a^{5} + 5 \, A a^{4} 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)^(5/2),x, algorithm="fricas")

[Out]

1/9*B*b^5*e^2*x^9 + A*a^5*d^2*x + 1/8*(2*B*b^5*d*e + (5*B*a*b^4 + A*b^5)*e^2)*x^8 + 1/7*(B*b^5*d^2 + 2*(5*B*a*
b^4 + A*b^5)*d*e + 5*(2*B*a^2*b^3 + A*a*b^4)*e^2)*x^7 + 1/6*((5*B*a*b^4 + A*b^5)*d^2 + 10*(2*B*a^2*b^3 + A*a*b
^4)*d*e + 10*(B*a^3*b^2 + A*a^2*b^3)*e^2)*x^6 + ((2*B*a^2*b^3 + A*a*b^4)*d^2 + 4*(B*a^3*b^2 + A*a^2*b^3)*d*e +
(B*a^4*b + 2*A*a^3*b^2)*e^2)*x^5 + 1/4*(10*(B*a^3*b^2 + A*a^2*b^3)*d^2 + 10*(B*a^4*b + 2*A*a^3*b^2)*d*e + (B*
a^5 + 5*A*a^4*b)*e^2)*x^4 + 1/3*(A*a^5*e^2 + 5*(B*a^4*b + 2*A*a^3*b^2)*d^2 + 2*(B*a^5 + 5*A*a^4*b)*d*e)*x^3 +
1/2*(2*A*a^5*d*e + (B*a^5 + 5*A*a^4*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{5}{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)**(5/2),x)

[Out]

Integral((A + B*x)*(d + e*x)**2*((a + b*x)**2)**(5/2), x)

________________________________________________________________________________________

Giac [B]  time = 1.15753, size = 917, normalized size = 4.63 \begin{align*} \text{result too large to display} \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)^(5/2),x, algorithm="giac")

[Out]

1/9*B*b^5*x^9*e^2*sgn(b*x + a) + 1/4*B*b^5*d*x^8*e*sgn(b*x + a) + 1/7*B*b^5*d^2*x^7*sgn(b*x + a) + 5/8*B*a*b^4
*x^8*e^2*sgn(b*x + a) + 1/8*A*b^5*x^8*e^2*sgn(b*x + a) + 10/7*B*a*b^4*d*x^7*e*sgn(b*x + a) + 2/7*A*b^5*d*x^7*e
*sgn(b*x + a) + 5/6*B*a*b^4*d^2*x^6*sgn(b*x + a) + 1/6*A*b^5*d^2*x^6*sgn(b*x + a) + 10/7*B*a^2*b^3*x^7*e^2*sgn
(b*x + a) + 5/7*A*a*b^4*x^7*e^2*sgn(b*x + a) + 10/3*B*a^2*b^3*d*x^6*e*sgn(b*x + a) + 5/3*A*a*b^4*d*x^6*e*sgn(b
*x + a) + 2*B*a^2*b^3*d^2*x^5*sgn(b*x + a) + A*a*b^4*d^2*x^5*sgn(b*x + a) + 5/3*B*a^3*b^2*x^6*e^2*sgn(b*x + a)
+ 5/3*A*a^2*b^3*x^6*e^2*sgn(b*x + a) + 4*B*a^3*b^2*d*x^5*e*sgn(b*x + a) + 4*A*a^2*b^3*d*x^5*e*sgn(b*x + a) +
5/2*B*a^3*b^2*d^2*x^4*sgn(b*x + a) + 5/2*A*a^2*b^3*d^2*x^4*sgn(b*x + a) + B*a^4*b*x^5*e^2*sgn(b*x + a) + 2*A*a
^3*b^2*x^5*e^2*sgn(b*x + a) + 5/2*B*a^4*b*d*x^4*e*sgn(b*x + a) + 5*A*a^3*b^2*d*x^4*e*sgn(b*x + a) + 5/3*B*a^4*
b*d^2*x^3*sgn(b*x + a) + 10/3*A*a^3*b^2*d^2*x^3*sgn(b*x + a) + 1/4*B*a^5*x^4*e^2*sgn(b*x + a) + 5/4*A*a^4*b*x^
4*e^2*sgn(b*x + a) + 2/3*B*a^5*d*x^3*e*sgn(b*x + a) + 10/3*A*a^4*b*d*x^3*e*sgn(b*x + a) + 1/2*B*a^5*d^2*x^2*sg
n(b*x + a) + 5/2*A*a^4*b*d^2*x^2*sgn(b*x + a) + 1/3*A*a^5*x^3*e^2*sgn(b*x + a) + A*a^5*d*x^2*e*sgn(b*x + a) +
A*a^5*d^2*x*sgn(b*x + a)