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

Optimal. Leaf size=111 \[ \frac {2}{3} a^4 A x^{3/2}+\frac {2}{5} a^3 x^{5/2} (a B+4 A b)+\frac {4}{7} a^2 b x^{7/2} (2 a B+3 A b)+\frac {2}{11} b^3 x^{11/2} (4 a B+A b)+\frac {4}{9} a b^2 x^{9/2} (3 a B+2 A b)+\frac {2}{13} b^4 B x^{13/2} \]

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Rubi [A]  time = 0.05, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {27, 76} \begin {gather*} \frac {4}{7} a^2 b x^{7/2} (2 a B+3 A b)+\frac {2}{5} a^3 x^{5/2} (a B+4 A b)+\frac {2}{3} a^4 A x^{3/2}+\frac {2}{11} b^3 x^{11/2} (4 a B+A b)+\frac {4}{9} a b^2 x^{9/2} (3 a B+2 A b)+\frac {2}{13} b^4 B x^{13/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^4*A*x^(3/2))/3 + (2*a^3*(4*A*b + a*B)*x^(5/2))/5 + (4*a^2*b*(3*A*b + 2*a*B)*x^(7/2))/7 + (4*a*b^2*(2*A*b
+ 3*a*B)*x^(9/2))/9 + (2*b^3*(A*b + 4*a*B)*x^(11/2))/11 + (2*b^4*B*x^(13/2))/13

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

\begin {align*} \int \sqrt {x} (A+B x) \left (a^2+2 a b x+b^2 x^2\right )^2 \, dx &=\int \sqrt {x} (a+b x)^4 (A+B x) \, dx\\ &=\int \left (a^4 A \sqrt {x}+a^3 (4 A b+a B) x^{3/2}+2 a^2 b (3 A b+2 a B) x^{5/2}+2 a b^2 (2 A b+3 a B) x^{7/2}+b^3 (A b+4 a B) x^{9/2}+b^4 B x^{11/2}\right ) \, dx\\ &=\frac {2}{3} a^4 A x^{3/2}+\frac {2}{5} a^3 (4 A b+a B) x^{5/2}+\frac {4}{7} a^2 b (3 A b+2 a B) x^{7/2}+\frac {4}{9} a b^2 (2 A b+3 a B) x^{9/2}+\frac {2}{11} b^3 (A b+4 a B) x^{11/2}+\frac {2}{13} b^4 B x^{13/2}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 81, normalized size = 0.73 \begin {gather*} \frac {2 \left (\frac {x^{3/2} \left (1155 a^4+2772 a^3 b x+2970 a^2 b^2 x^2+1540 a b^3 x^3+315 b^4 x^4\right ) (13 A b-3 a B)}{3465}+B x^{3/2} (a+b x)^5\right )}{13 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(B*x^(3/2)*(a + b*x)^5 + ((13*A*b - 3*a*B)*x^(3/2)*(1155*a^4 + 2772*a^3*b*x + 2970*a^2*b^2*x^2 + 1540*a*b^3
*x^3 + 315*b^4*x^4))/3465))/(13*b)

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IntegrateAlgebraic [A]  time = 0.05, size = 125, normalized size = 1.13 \begin {gather*} \frac {2 \left (15015 a^4 A x^{3/2}+9009 a^4 B x^{5/2}+36036 a^3 A b x^{5/2}+25740 a^3 b B x^{7/2}+38610 a^2 A b^2 x^{7/2}+30030 a^2 b^2 B x^{9/2}+20020 a A b^3 x^{9/2}+16380 a b^3 B x^{11/2}+4095 A b^4 x^{11/2}+3465 b^4 B x^{13/2}\right )}{45045} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[x]*(A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^2,x]

[Out]

(2*(15015*a^4*A*x^(3/2) + 36036*a^3*A*b*x^(5/2) + 9009*a^4*B*x^(5/2) + 38610*a^2*A*b^2*x^(7/2) + 25740*a^3*b*B
*x^(7/2) + 20020*a*A*b^3*x^(9/2) + 30030*a^2*b^2*B*x^(9/2) + 4095*A*b^4*x^(11/2) + 16380*a*b^3*B*x^(11/2) + 34
65*b^4*B*x^(13/2)))/45045

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fricas [A]  time = 0.42, size = 102, normalized size = 0.92 \begin {gather*} \frac {2}{45045} \, {\left (3465 \, B b^{4} x^{6} + 15015 \, A a^{4} x + 4095 \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{5} + 10010 \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{4} + 12870 \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{3} + 9009 \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{2}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45045*(3465*B*b^4*x^6 + 15015*A*a^4*x + 4095*(4*B*a*b^3 + A*b^4)*x^5 + 10010*(3*B*a^2*b^2 + 2*A*a*b^3)*x^4 +
12870*(2*B*a^3*b + 3*A*a^2*b^2)*x^3 + 9009*(B*a^4 + 4*A*a^3*b)*x^2)*sqrt(x)

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giac [A]  time = 0.15, size = 101, normalized size = 0.91 \begin {gather*} \frac {2}{13} \, B b^{4} x^{\frac {13}{2}} + \frac {8}{11} \, B a b^{3} x^{\frac {11}{2}} + \frac {2}{11} \, A b^{4} x^{\frac {11}{2}} + \frac {4}{3} \, B a^{2} b^{2} x^{\frac {9}{2}} + \frac {8}{9} \, A a b^{3} x^{\frac {9}{2}} + \frac {8}{7} \, B a^{3} b x^{\frac {7}{2}} + \frac {12}{7} \, A a^{2} b^{2} x^{\frac {7}{2}} + \frac {2}{5} \, B a^{4} x^{\frac {5}{2}} + \frac {8}{5} \, A a^{3} b x^{\frac {5}{2}} + \frac {2}{3} \, A a^{4} x^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/13*B*b^4*x^(13/2) + 8/11*B*a*b^3*x^(11/2) + 2/11*A*b^4*x^(11/2) + 4/3*B*a^2*b^2*x^(9/2) + 8/9*A*a*b^3*x^(9/2
) + 8/7*B*a^3*b*x^(7/2) + 12/7*A*a^2*b^2*x^(7/2) + 2/5*B*a^4*x^(5/2) + 8/5*A*a^3*b*x^(5/2) + 2/3*A*a^4*x^(3/2)

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maple [A]  time = 0.06, size = 100, normalized size = 0.90 \begin {gather*} \frac {2 \left (3465 b^{4} B \,x^{5}+4095 A \,b^{4} x^{4}+16380 x^{4} B a \,b^{3}+20020 A a \,b^{3} x^{3}+30030 B \,a^{2} b^{2} x^{3}+38610 A \,a^{2} b^{2} x^{2}+25740 B \,a^{3} b \,x^{2}+36036 A \,a^{3} b x +9009 B \,a^{4} x +15015 A \,a^{4}\right ) x^{\frac {3}{2}}}{45045} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(b^2*x^2+2*a*b*x+a^2)^2*x^(1/2),x)

[Out]

2/45045*x^(3/2)*(3465*B*b^4*x^5+4095*A*b^4*x^4+16380*B*a*b^3*x^4+20020*A*a*b^3*x^3+30030*B*a^2*b^2*x^3+38610*A
*a^2*b^2*x^2+25740*B*a^3*b*x^2+36036*A*a^3*b*x+9009*B*a^4*x+15015*A*a^4)

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maxima [A]  time = 0.56, size = 99, normalized size = 0.89 \begin {gather*} \frac {2}{13} \, B b^{4} x^{\frac {13}{2}} + \frac {2}{3} \, A a^{4} x^{\frac {3}{2}} + \frac {2}{11} \, {\left (4 \, B a b^{3} + A b^{4}\right )} x^{\frac {11}{2}} + \frac {4}{9} \, {\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{\frac {9}{2}} + \frac {4}{7} \, {\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{\frac {7}{2}} + \frac {2}{5} \, {\left (B a^{4} + 4 \, A a^{3} b\right )} x^{\frac {5}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/13*B*b^4*x^(13/2) + 2/3*A*a^4*x^(3/2) + 2/11*(4*B*a*b^3 + A*b^4)*x^(11/2) + 4/9*(3*B*a^2*b^2 + 2*A*a*b^3)*x^
(9/2) + 4/7*(2*B*a^3*b + 3*A*a^2*b^2)*x^(7/2) + 2/5*(B*a^4 + 4*A*a^3*b)*x^(5/2)

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mupad [B]  time = 0.04, size = 91, normalized size = 0.82 \begin {gather*} x^{5/2}\,\left (\frac {2\,B\,a^4}{5}+\frac {8\,A\,b\,a^3}{5}\right )+x^{11/2}\,\left (\frac {2\,A\,b^4}{11}+\frac {8\,B\,a\,b^3}{11}\right )+\frac {2\,A\,a^4\,x^{3/2}}{3}+\frac {2\,B\,b^4\,x^{13/2}}{13}+\frac {4\,a^2\,b\,x^{7/2}\,\left (3\,A\,b+2\,B\,a\right )}{7}+\frac {4\,a\,b^2\,x^{9/2}\,\left (2\,A\,b+3\,B\,a\right )}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^2,x)

[Out]

x^(5/2)*((2*B*a^4)/5 + (8*A*a^3*b)/5) + x^(11/2)*((2*A*b^4)/11 + (8*B*a*b^3)/11) + (2*A*a^4*x^(3/2))/3 + (2*B*
b^4*x^(13/2))/13 + (4*a^2*b*x^(7/2)*(3*A*b + 2*B*a))/7 + (4*a*b^2*x^(9/2)*(2*A*b + 3*B*a))/9

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sympy [A]  time = 4.38, size = 124, normalized size = 1.12 \begin {gather*} \frac {2 A a^{4} x^{\frac {3}{2}}}{3} + \frac {2 B b^{4} x^{\frac {13}{2}}}{13} + \frac {2 x^{\frac {11}{2}} \left (A b^{4} + 4 B a b^{3}\right )}{11} + \frac {2 x^{\frac {9}{2}} \left (4 A a b^{3} + 6 B a^{2} b^{2}\right )}{9} + \frac {2 x^{\frac {7}{2}} \left (6 A a^{2} b^{2} + 4 B a^{3} b\right )}{7} + \frac {2 x^{\frac {5}{2}} \left (4 A a^{3} b + B a^{4}\right )}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*a**4*x**(3/2)/3 + 2*B*b**4*x**(13/2)/13 + 2*x**(11/2)*(A*b**4 + 4*B*a*b**3)/11 + 2*x**(9/2)*(4*A*a*b**3 +
6*B*a**2*b**2)/9 + 2*x**(7/2)*(6*A*a**2*b**2 + 4*B*a**3*b)/7 + 2*x**(5/2)*(4*A*a**3*b + B*a**4)/5

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