3.7.78 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^3}{\sqrt {x}} \, dx\)

Optimal. Leaf size=157 \[ 2 a^6 A \sqrt {x}+\frac {2}{3} a^5 x^{3/2} (a B+6 A b)+\frac {6}{5} a^4 b x^{5/2} (2 a B+5 A b)+\frac {10}{7} a^3 b^2 x^{7/2} (3 a B+4 A b)+\frac {10}{9} a^2 b^3 x^{9/2} (4 a B+3 A b)+\frac {2}{13} b^5 x^{13/2} (6 a B+A b)+\frac {6}{11} a b^4 x^{11/2} (5 a B+2 A b)+\frac {2}{15} b^6 B x^{15/2} \]

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Rubi [A]  time = 0.08, antiderivative size = 157, 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 {10}{9} a^2 b^3 x^{9/2} (4 a B+3 A b)+\frac {10}{7} a^3 b^2 x^{7/2} (3 a B+4 A b)+\frac {6}{5} a^4 b x^{5/2} (2 a B+5 A b)+\frac {2}{3} a^5 x^{3/2} (a B+6 A b)+2 a^6 A \sqrt {x}+\frac {2}{13} b^5 x^{13/2} (6 a B+A b)+\frac {6}{11} a b^4 x^{11/2} (5 a B+2 A b)+\frac {2}{15} b^6 B x^{15/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

2*a^6*A*Sqrt[x] + (2*a^5*(6*A*b + a*B)*x^(3/2))/3 + (6*a^4*b*(5*A*b + 2*a*B)*x^(5/2))/5 + (10*a^3*b^2*(4*A*b +
 3*a*B)*x^(7/2))/7 + (10*a^2*b^3*(3*A*b + 4*a*B)*x^(9/2))/9 + (6*a*b^4*(2*A*b + 5*a*B)*x^(11/2))/11 + (2*b^5*(
A*b + 6*a*B)*x^(13/2))/13 + (2*b^6*B*x^(15/2))/15

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 \frac {(A+B x) \left (a^2+2 a b x+b^2 x^2\right )^3}{\sqrt {x}} \, dx &=\int \frac {(a+b x)^6 (A+B x)}{\sqrt {x}} \, dx\\ &=\int \left (\frac {a^6 A}{\sqrt {x}}+a^5 (6 A b+a B) \sqrt {x}+3 a^4 b (5 A b+2 a B) x^{3/2}+5 a^3 b^2 (4 A b+3 a B) x^{5/2}+5 a^2 b^3 (3 A b+4 a B) x^{7/2}+3 a b^4 (2 A b+5 a B) x^{9/2}+b^5 (A b+6 a B) x^{11/2}+b^6 B x^{13/2}\right ) \, dx\\ &=2 a^6 A \sqrt {x}+\frac {2}{3} a^5 (6 A b+a B) x^{3/2}+\frac {6}{5} a^4 b (5 A b+2 a B) x^{5/2}+\frac {10}{7} a^3 b^2 (4 A b+3 a B) x^{7/2}+\frac {10}{9} a^2 b^3 (3 A b+4 a B) x^{9/2}+\frac {6}{11} a b^4 (2 A b+5 a B) x^{11/2}+\frac {2}{13} b^5 (A b+6 a B) x^{13/2}+\frac {2}{15} b^6 B x^{15/2}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 103, normalized size = 0.66 \begin {gather*} \frac {2 \left (\frac {\sqrt {x} \left (3003 a^6+6006 a^5 b x+9009 a^4 b^2 x^2+8580 a^3 b^3 x^3+5005 a^2 b^4 x^4+1638 a b^5 x^5+231 b^6 x^6\right ) (15 A b-a B)}{3003}+B \sqrt {x} (a+b x)^7\right )}{15 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(B*Sqrt[x]*(a + b*x)^7 + ((15*A*b - a*B)*Sqrt[x]*(3003*a^6 + 6006*a^5*b*x + 9009*a^4*b^2*x^2 + 8580*a^3*b^3
*x^3 + 5005*a^2*b^4*x^4 + 1638*a*b^5*x^5 + 231*b^6*x^6))/3003))/(15*b)

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IntegrateAlgebraic [A]  time = 0.07, size = 181, normalized size = 1.15 \begin {gather*} \frac {2 \left (45045 a^6 A \sqrt {x}+15015 a^6 B x^{3/2}+90090 a^5 A b x^{3/2}+54054 a^5 b B x^{5/2}+135135 a^4 A b^2 x^{5/2}+96525 a^4 b^2 B x^{7/2}+128700 a^3 A b^3 x^{7/2}+100100 a^3 b^3 B x^{9/2}+75075 a^2 A b^4 x^{9/2}+61425 a^2 b^4 B x^{11/2}+24570 a A b^5 x^{11/2}+20790 a b^5 B x^{13/2}+3465 A b^6 x^{13/2}+3003 b^6 B x^{15/2}\right )}{45045} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3)/Sqrt[x],x]

[Out]

(2*(45045*a^6*A*Sqrt[x] + 90090*a^5*A*b*x^(3/2) + 15015*a^6*B*x^(3/2) + 135135*a^4*A*b^2*x^(5/2) + 54054*a^5*b
*B*x^(5/2) + 128700*a^3*A*b^3*x^(7/2) + 96525*a^4*b^2*B*x^(7/2) + 75075*a^2*A*b^4*x^(9/2) + 100100*a^3*b^3*B*x
^(9/2) + 24570*a*A*b^5*x^(11/2) + 61425*a^2*b^4*B*x^(11/2) + 3465*A*b^6*x^(13/2) + 20790*a*b^5*B*x^(13/2) + 30
03*b^6*B*x^(15/2)))/45045

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fricas [A]  time = 0.40, size = 147, normalized size = 0.94 \begin {gather*} \frac {2}{45045} \, {\left (3003 \, B b^{6} x^{7} + 45045 \, A a^{6} + 3465 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{6} + 12285 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} + 25025 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} + 32175 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} + 27027 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} + 15015 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x\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)^3/x^(1/2),x, algorithm="fricas")

[Out]

2/45045*(3003*B*b^6*x^7 + 45045*A*a^6 + 3465*(6*B*a*b^5 + A*b^6)*x^6 + 12285*(5*B*a^2*b^4 + 2*A*a*b^5)*x^5 + 2
5025*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^4 + 32175*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^3 + 27027*(2*B*a^5*b + 5*A*a^4*b^2)
*x^2 + 15015*(B*a^6 + 6*A*a^5*b)*x)*sqrt(x)

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giac [A]  time = 0.16, size = 149, normalized size = 0.95 \begin {gather*} \frac {2}{15} \, B b^{6} x^{\frac {15}{2}} + \frac {12}{13} \, B a b^{5} x^{\frac {13}{2}} + \frac {2}{13} \, A b^{6} x^{\frac {13}{2}} + \frac {30}{11} \, B a^{2} b^{4} x^{\frac {11}{2}} + \frac {12}{11} \, A a b^{5} x^{\frac {11}{2}} + \frac {40}{9} \, B a^{3} b^{3} x^{\frac {9}{2}} + \frac {10}{3} \, A a^{2} b^{4} x^{\frac {9}{2}} + \frac {30}{7} \, B a^{4} b^{2} x^{\frac {7}{2}} + \frac {40}{7} \, A a^{3} b^{3} x^{\frac {7}{2}} + \frac {12}{5} \, B a^{5} b x^{\frac {5}{2}} + 6 \, A a^{4} b^{2} x^{\frac {5}{2}} + \frac {2}{3} \, B a^{6} x^{\frac {3}{2}} + 4 \, A a^{5} b x^{\frac {3}{2}} + 2 \, A a^{6} \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)^3/x^(1/2),x, algorithm="giac")

[Out]

2/15*B*b^6*x^(15/2) + 12/13*B*a*b^5*x^(13/2) + 2/13*A*b^6*x^(13/2) + 30/11*B*a^2*b^4*x^(11/2) + 12/11*A*a*b^5*
x^(11/2) + 40/9*B*a^3*b^3*x^(9/2) + 10/3*A*a^2*b^4*x^(9/2) + 30/7*B*a^4*b^2*x^(7/2) + 40/7*A*a^3*b^3*x^(7/2) +
 12/5*B*a^5*b*x^(5/2) + 6*A*a^4*b^2*x^(5/2) + 2/3*B*a^6*x^(3/2) + 4*A*a^5*b*x^(3/2) + 2*A*a^6*sqrt(x)

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maple [A]  time = 0.05, size = 148, normalized size = 0.94 \begin {gather*} \frac {2 \left (3003 B \,b^{6} x^{7}+3465 A \,b^{6} x^{6}+20790 x^{6} B a \,b^{5}+24570 A a \,b^{5} x^{5}+61425 x^{5} B \,a^{2} b^{4}+75075 A \,a^{2} b^{4} x^{4}+100100 x^{4} B \,a^{3} b^{3}+128700 A \,a^{3} b^{3} x^{3}+96525 B \,a^{4} b^{2} x^{3}+135135 A \,a^{4} b^{2} x^{2}+54054 x^{2} B \,a^{5} b +90090 A \,a^{5} b x +15015 x B \,a^{6}+45045 A \,a^{6}\right ) \sqrt {x}}{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)^3/x^(1/2),x)

[Out]

2/45045*x^(1/2)*(3003*B*b^6*x^7+3465*A*b^6*x^6+20790*B*a*b^5*x^6+24570*A*a*b^5*x^5+61425*B*a^2*b^4*x^5+75075*A
*a^2*b^4*x^4+100100*B*a^3*b^3*x^4+128700*A*a^3*b^3*x^3+96525*B*a^4*b^2*x^3+135135*A*a^4*b^2*x^2+54054*B*a^5*b*
x^2+90090*A*a^5*b*x+15015*B*a^6*x+45045*A*a^6)

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maxima [A]  time = 0.60, size = 147, normalized size = 0.94 \begin {gather*} \frac {2}{15} \, B b^{6} x^{\frac {15}{2}} + 2 \, A a^{6} \sqrt {x} + \frac {2}{13} \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{\frac {13}{2}} + \frac {6}{11} \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{\frac {11}{2}} + \frac {10}{9} \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{\frac {9}{2}} + \frac {10}{7} \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{\frac {7}{2}} + \frac {6}{5} \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{\frac {5}{2}} + \frac {2}{3} \, {\left (B a^{6} + 6 \, A a^{5} b\right )} 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)^3/x^(1/2),x, algorithm="maxima")

[Out]

2/15*B*b^6*x^(15/2) + 2*A*a^6*sqrt(x) + 2/13*(6*B*a*b^5 + A*b^6)*x^(13/2) + 6/11*(5*B*a^2*b^4 + 2*A*a*b^5)*x^(
11/2) + 10/9*(4*B*a^3*b^3 + 3*A*a^2*b^4)*x^(9/2) + 10/7*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^(7/2) + 6/5*(2*B*a^5*b +
 5*A*a^4*b^2)*x^(5/2) + 2/3*(B*a^6 + 6*A*a^5*b)*x^(3/2)

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mupad [B]  time = 0.05, size = 131, normalized size = 0.83 \begin {gather*} x^{3/2}\,\left (\frac {2\,B\,a^6}{3}+4\,A\,b\,a^5\right )+x^{13/2}\,\left (\frac {2\,A\,b^6}{13}+\frac {12\,B\,a\,b^5}{13}\right )+2\,A\,a^6\,\sqrt {x}+\frac {2\,B\,b^6\,x^{15/2}}{15}+\frac {10\,a^3\,b^2\,x^{7/2}\,\left (4\,A\,b+3\,B\,a\right )}{7}+\frac {10\,a^2\,b^3\,x^{9/2}\,\left (3\,A\,b+4\,B\,a\right )}{9}+\frac {6\,a^4\,b\,x^{5/2}\,\left (5\,A\,b+2\,B\,a\right )}{5}+\frac {6\,a\,b^4\,x^{11/2}\,\left (2\,A\,b+5\,B\,a\right )}{11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(a^2 + b^2*x^2 + 2*a*b*x)^3)/x^(1/2),x)

[Out]

x^(3/2)*((2*B*a^6)/3 + 4*A*a^5*b) + x^(13/2)*((2*A*b^6)/13 + (12*B*a*b^5)/13) + 2*A*a^6*x^(1/2) + (2*B*b^6*x^(
15/2))/15 + (10*a^3*b^2*x^(7/2)*(4*A*b + 3*B*a))/7 + (10*a^2*b^3*x^(9/2)*(3*A*b + 4*B*a))/9 + (6*a^4*b*x^(5/2)
*(5*A*b + 2*B*a))/5 + (6*a*b^4*x^(11/2)*(2*A*b + 5*B*a))/11

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sympy [A]  time = 4.11, size = 211, normalized size = 1.34 \begin {gather*} 2 A a^{6} \sqrt {x} + 4 A a^{5} b x^{\frac {3}{2}} + 6 A a^{4} b^{2} x^{\frac {5}{2}} + \frac {40 A a^{3} b^{3} x^{\frac {7}{2}}}{7} + \frac {10 A a^{2} b^{4} x^{\frac {9}{2}}}{3} + \frac {12 A a b^{5} x^{\frac {11}{2}}}{11} + \frac {2 A b^{6} x^{\frac {13}{2}}}{13} + \frac {2 B a^{6} x^{\frac {3}{2}}}{3} + \frac {12 B a^{5} b x^{\frac {5}{2}}}{5} + \frac {30 B a^{4} b^{2} x^{\frac {7}{2}}}{7} + \frac {40 B a^{3} b^{3} x^{\frac {9}{2}}}{9} + \frac {30 B a^{2} b^{4} x^{\frac {11}{2}}}{11} + \frac {12 B a b^{5} x^{\frac {13}{2}}}{13} + \frac {2 B b^{6} x^{\frac {15}{2}}}{15} \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)**3/x**(1/2),x)

[Out]

2*A*a**6*sqrt(x) + 4*A*a**5*b*x**(3/2) + 6*A*a**4*b**2*x**(5/2) + 40*A*a**3*b**3*x**(7/2)/7 + 10*A*a**2*b**4*x
**(9/2)/3 + 12*A*a*b**5*x**(11/2)/11 + 2*A*b**6*x**(13/2)/13 + 2*B*a**6*x**(3/2)/3 + 12*B*a**5*b*x**(5/2)/5 +
30*B*a**4*b**2*x**(7/2)/7 + 40*B*a**3*b**3*x**(9/2)/9 + 30*B*a**2*b**4*x**(11/2)/11 + 12*B*a*b**5*x**(13/2)/13
 + 2*B*b**6*x**(15/2)/15

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