3.754 \(\int \frac {(A+B x) (a^2+2 a b x+b^2 x^2)^3}{x^{7/2}} \, dx\)

Optimal. Leaf size=155 \[ -\frac {2 a^6 A}{5 x^{5/2}}-\frac {2 a^5 (a B+6 A b)}{3 x^{3/2}}-\frac {6 a^4 b (2 a B+5 A b)}{\sqrt {x}}+10 a^3 b^2 \sqrt {x} (3 a B+4 A b)+\frac {10}{3} a^2 b^3 x^{3/2} (4 a B+3 A b)+\frac {2}{7} b^5 x^{7/2} (6 a B+A b)+\frac {6}{5} a b^4 x^{5/2} (5 a B+2 A b)+\frac {2}{9} b^6 B x^{9/2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 155, 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} \[ \frac {10}{3} a^2 b^3 x^{3/2} (4 a B+3 A b)+10 a^3 b^2 \sqrt {x} (3 a B+4 A b)-\frac {2 a^5 (a B+6 A b)}{3 x^{3/2}}-\frac {6 a^4 b (2 a B+5 A b)}{\sqrt {x}}-\frac {2 a^6 A}{5 x^{5/2}}+\frac {6}{5} a b^4 x^{5/2} (5 a B+2 A b)+\frac {2}{7} b^5 x^{7/2} (6 a B+A b)+\frac {2}{9} b^6 B x^{9/2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3)/x^(7/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.04, size = 124, normalized size = 0.80 \[ \frac {2 \left (-21 a^6 (3 A+5 B x)-630 a^5 b x (A+3 B x)+4725 a^4 b^2 x^2 (B x-A)+2100 a^3 b^3 x^3 (3 A+B x)+315 a^2 b^4 x^4 (5 A+3 B x)+54 a b^5 x^5 (7 A+5 B x)+5 b^6 x^6 (9 A+7 B x)\right )}{315 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3)/x^(7/2),x]

[Out]

(2*(4725*a^4*b^2*x^2*(-A + B*x) + 2100*a^3*b^3*x^3*(3*A + B*x) - 630*a^5*b*x*(A + 3*B*x) + 315*a^2*b^4*x^4*(5*
A + 3*B*x) - 21*a^6*(3*A + 5*B*x) + 54*a*b^5*x^5*(7*A + 5*B*x) + 5*b^6*x^6*(9*A + 7*B*x)))/(315*x^(5/2))

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fricas [A]  time = 0.78, size = 147, normalized size = 0.95 \[ \frac {2 \, {\left (35 \, B b^{6} x^{7} - 63 \, A a^{6} + 45 \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{6} + 189 \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} + 525 \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} + 1575 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} - 945 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} - 105 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x\right )}}{315 \, x^{\frac {5}{2}}} \]

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^(7/2),x, algorithm="fricas")

[Out]

2/315*(35*B*b^6*x^7 - 63*A*a^6 + 45*(6*B*a*b^5 + A*b^6)*x^6 + 189*(5*B*a^2*b^4 + 2*A*a*b^5)*x^5 + 525*(4*B*a^3
*b^3 + 3*A*a^2*b^4)*x^4 + 1575*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^3 - 945*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 - 105*(B*a^
6 + 6*A*a^5*b)*x)/x^(5/2)

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giac [A]  time = 0.20, size = 148, normalized size = 0.95 \[ \frac {2}{9} \, B b^{6} x^{\frac {9}{2}} + \frac {12}{7} \, B a b^{5} x^{\frac {7}{2}} + \frac {2}{7} \, A b^{6} x^{\frac {7}{2}} + 6 \, B a^{2} b^{4} x^{\frac {5}{2}} + \frac {12}{5} \, A a b^{5} x^{\frac {5}{2}} + \frac {40}{3} \, B a^{3} b^{3} x^{\frac {3}{2}} + 10 \, A a^{2} b^{4} x^{\frac {3}{2}} + 30 \, B a^{4} b^{2} \sqrt {x} + 40 \, A a^{3} b^{3} \sqrt {x} - \frac {2 \, {\left (90 \, B a^{5} b x^{2} + 225 \, A a^{4} b^{2} x^{2} + 5 \, B a^{6} x + 30 \, A a^{5} b x + 3 \, A a^{6}\right )}}{15 \, x^{\frac {5}{2}}} \]

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^(7/2),x, algorithm="giac")

[Out]

2/9*B*b^6*x^(9/2) + 12/7*B*a*b^5*x^(7/2) + 2/7*A*b^6*x^(7/2) + 6*B*a^2*b^4*x^(5/2) + 12/5*A*a*b^5*x^(5/2) + 40
/3*B*a^3*b^3*x^(3/2) + 10*A*a^2*b^4*x^(3/2) + 30*B*a^4*b^2*sqrt(x) + 40*A*a^3*b^3*sqrt(x) - 2/15*(90*B*a^5*b*x
^2 + 225*A*a^4*b^2*x^2 + 5*B*a^6*x + 30*A*a^5*b*x + 3*A*a^6)/x^(5/2)

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maple [A]  time = 0.05, size = 148, normalized size = 0.95 \[ -\frac {2 \left (-35 B \,b^{6} x^{7}-45 A \,b^{6} x^{6}-270 x^{6} B a \,b^{5}-378 A a \,b^{5} x^{5}-945 x^{5} B \,a^{2} b^{4}-1575 A \,a^{2} b^{4} x^{4}-2100 x^{4} B \,a^{3} b^{3}-6300 A \,a^{3} b^{3} x^{3}-4725 B \,a^{4} b^{2} x^{3}+4725 A \,a^{4} b^{2} x^{2}+1890 x^{2} B \,a^{5} b +630 A \,a^{5} b x +105 x B \,a^{6}+63 A \,a^{6}\right )}{315 x^{\frac {5}{2}}} \]

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^(7/2),x)

[Out]

-2/315*(-35*B*b^6*x^7-45*A*b^6*x^6-270*B*a*b^5*x^6-378*A*a*b^5*x^5-945*B*a^2*b^4*x^5-1575*A*a^2*b^4*x^4-2100*B
*a^3*b^3*x^4-6300*A*a^3*b^3*x^3-4725*B*a^4*b^2*x^3+4725*A*a^4*b^2*x^2+1890*B*a^5*b*x^2+630*A*a^5*b*x+105*B*a^6
*x+63*A*a^6)/x^(5/2)

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maxima [A]  time = 0.54, size = 148, normalized size = 0.95 \[ \frac {2}{9} \, B b^{6} x^{\frac {9}{2}} + \frac {2}{7} \, {\left (6 \, B a b^{5} + A b^{6}\right )} x^{\frac {7}{2}} + \frac {6}{5} \, {\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{\frac {5}{2}} + \frac {10}{3} \, {\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{\frac {3}{2}} + 10 \, {\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} \sqrt {x} - \frac {2 \, {\left (3 \, A a^{6} + 45 \, {\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} + 5 \, {\left (B a^{6} + 6 \, A a^{5} b\right )} x\right )}}{15 \, x^{\frac {5}{2}}} \]

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^(7/2),x, algorithm="maxima")

[Out]

2/9*B*b^6*x^(9/2) + 2/7*(6*B*a*b^5 + A*b^6)*x^(7/2) + 6/5*(5*B*a^2*b^4 + 2*A*a*b^5)*x^(5/2) + 10/3*(4*B*a^3*b^
3 + 3*A*a^2*b^4)*x^(3/2) + 10*(3*B*a^4*b^2 + 4*A*a^3*b^3)*sqrt(x) - 2/15*(3*A*a^6 + 45*(2*B*a^5*b + 5*A*a^4*b^
2)*x^2 + 5*(B*a^6 + 6*A*a^5*b)*x)/x^(5/2)

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mupad [B]  time = 0.05, size = 135, normalized size = 0.87 \[ x^{7/2}\,\left (\frac {2\,A\,b^6}{7}+\frac {12\,B\,a\,b^5}{7}\right )-\frac {x\,\left (\frac {2\,B\,a^6}{3}+4\,A\,b\,a^5\right )+\frac {2\,A\,a^6}{5}+x^2\,\left (12\,B\,a^5\,b+30\,A\,a^4\,b^2\right )}{x^{5/2}}+\frac {2\,B\,b^6\,x^{9/2}}{9}+10\,a^3\,b^2\,\sqrt {x}\,\left (4\,A\,b+3\,B\,a\right )+\frac {10\,a^2\,b^3\,x^{3/2}\,\left (3\,A\,b+4\,B\,a\right )}{3}+\frac {6\,a\,b^4\,x^{5/2}\,\left (2\,A\,b+5\,B\,a\right )}{5} \]

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^(7/2),x)

[Out]

x^(7/2)*((2*A*b^6)/7 + (12*B*a*b^5)/7) - (x*((2*B*a^6)/3 + 4*A*a^5*b) + (2*A*a^6)/5 + x^2*(30*A*a^4*b^2 + 12*B
*a^5*b))/x^(5/2) + (2*B*b^6*x^(9/2))/9 + 10*a^3*b^2*x^(1/2)*(4*A*b + 3*B*a) + (10*a^2*b^3*x^(3/2)*(3*A*b + 4*B
*a))/3 + (6*a*b^4*x^(5/2)*(2*A*b + 5*B*a))/5

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sympy [A]  time = 7.32, size = 204, normalized size = 1.32 \[ - \frac {2 A a^{6}}{5 x^{\frac {5}{2}}} - \frac {4 A a^{5} b}{x^{\frac {3}{2}}} - \frac {30 A a^{4} b^{2}}{\sqrt {x}} + 40 A a^{3} b^{3} \sqrt {x} + 10 A a^{2} b^{4} x^{\frac {3}{2}} + \frac {12 A a b^{5} x^{\frac {5}{2}}}{5} + \frac {2 A b^{6} x^{\frac {7}{2}}}{7} - \frac {2 B a^{6}}{3 x^{\frac {3}{2}}} - \frac {12 B a^{5} b}{\sqrt {x}} + 30 B a^{4} b^{2} \sqrt {x} + \frac {40 B a^{3} b^{3} x^{\frac {3}{2}}}{3} + 6 B a^{2} b^{4} x^{\frac {5}{2}} + \frac {12 B a b^{5} x^{\frac {7}{2}}}{7} + \frac {2 B b^{6} x^{\frac {9}{2}}}{9} \]

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**(7/2),x)

[Out]

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

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