3.753 \(\int \frac{(A+B x) (a^2+2 a b x+b^2 x^2)^3}{x^{5/2}} \, dx\)

Optimal. Leaf size=153 \[ \frac{10}{3} a^3 b^2 x^{3/2} (3 a B+4 A b)+2 a^2 b^3 x^{5/2} (4 a B+3 A b)-\frac{2 a^5 (a B+6 A b)}{\sqrt{x}}+6 a^4 b \sqrt{x} (2 a B+5 A b)-\frac{2 a^6 A}{3 x^{3/2}}+\frac{6}{7} a b^4 x^{7/2} (5 a B+2 A b)+\frac{2}{9} b^5 x^{9/2} (6 a B+A b)+\frac{2}{11} b^6 B x^{11/2} \]

[Out]

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

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Rubi [A]  time = 0.078745, antiderivative size = 153, normalized size of antiderivative = 1., 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} \[ \frac{10}{3} a^3 b^2 x^{3/2} (3 a B+4 A b)+2 a^2 b^3 x^{5/2} (4 a B+3 A b)-\frac{2 a^5 (a B+6 A b)}{\sqrt{x}}+6 a^4 b \sqrt{x} (2 a B+5 A b)-\frac{2 a^6 A}{3 x^{3/2}}+\frac{6}{7} a b^4 x^{7/2} (5 a B+2 A b)+\frac{2}{9} b^5 x^{9/2} (6 a B+A b)+\frac{2}{11} b^6 B x^{11/2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3)/x^(5/2),x]

[Out]

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

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

Mathematica [A]  time = 0.0716734, size = 97, normalized size = 0.63 \[ \frac{2 \left (x \left (1155 a^4 b^2 x^2+924 a^3 b^3 x^3+495 a^2 b^4 x^4+1386 a^5 b x-231 a^6+154 a b^5 x^5+21 b^6 x^6\right ) (3 a B+11 A b)-231 A (a+b x)^7\right )}{693 a x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a^2 + 2*a*b*x + b^2*x^2)^3)/x^(5/2),x]

[Out]

(2*(-231*A*(a + b*x)^7 + (11*A*b + 3*a*B)*x*(-231*a^6 + 1386*a^5*b*x + 1155*a^4*b^2*x^2 + 924*a^3*b^3*x^3 + 49
5*a^2*b^4*x^4 + 154*a*b^5*x^5 + 21*b^6*x^6)))/(693*a*x^(3/2))

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Maple [A]  time = 0.006, size = 148, normalized size = 1. \begin{align*} -{\frac{-126\,B{b}^{6}{x}^{7}-154\,A{b}^{6}{x}^{6}-924\,B{x}^{6}a{b}^{5}-1188\,aA{b}^{5}{x}^{5}-2970\,B{x}^{5}{a}^{2}{b}^{4}-4158\,{a}^{2}A{b}^{4}{x}^{4}-5544\,B{x}^{4}{a}^{3}{b}^{3}-9240\,{a}^{3}A{b}^{3}{x}^{3}-6930\,B{x}^{3}{a}^{4}{b}^{2}-20790\,{a}^{4}A{b}^{2}{x}^{2}-8316\,B{x}^{2}{a}^{5}b+8316\,{a}^{5}Abx+1386\,B{a}^{6}x+462\,A{a}^{6}}{693}{x}^{-{\frac{3}{2}}}} \end{align*}

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^(5/2),x)

[Out]

-2/693*(-63*B*b^6*x^7-77*A*b^6*x^6-462*B*a*b^5*x^6-594*A*a*b^5*x^5-1485*B*a^2*b^4*x^5-2079*A*a^2*b^4*x^4-2772*
B*a^3*b^3*x^4-4620*A*a^3*b^3*x^3-3465*B*a^4*b^2*x^3-10395*A*a^4*b^2*x^2-4158*B*a^5*b*x^2+4158*A*a^5*b*x+693*B*
a^6*x+231*A*a^6)/x^(3/2)

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Maxima [A]  time = 1.04088, size = 198, normalized size = 1.29 \begin{align*} \frac{2}{11} \, B b^{6} x^{\frac{11}{2}} + \frac{2}{9} \,{\left (6 \, B a b^{5} + A b^{6}\right )} x^{\frac{9}{2}} + \frac{6}{7} \,{\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{\frac{7}{2}} + 2 \,{\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{\frac{5}{2}} + \frac{10}{3} \,{\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{\frac{3}{2}} + 6 \,{\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} \sqrt{x} - \frac{2 \,{\left (A a^{6} + 3 \,{\left (B a^{6} + 6 \, A a^{5} b\right )} x\right )}}{3 \, x^{\frac{3}{2}}} \end{align*}

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^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.57064, size = 336, normalized size = 2.2 \begin{align*} \frac{2 \,{\left (63 \, B b^{6} x^{7} - 231 \, A a^{6} + 77 \,{\left (6 \, B a b^{5} + A b^{6}\right )} x^{6} + 297 \,{\left (5 \, B a^{2} b^{4} + 2 \, A a b^{5}\right )} x^{5} + 693 \,{\left (4 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{4} + 1155 \,{\left (3 \, B a^{4} b^{2} + 4 \, A a^{3} b^{3}\right )} x^{3} + 2079 \,{\left (2 \, B a^{5} b + 5 \, A a^{4} b^{2}\right )} x^{2} - 693 \,{\left (B a^{6} + 6 \, A a^{5} b\right )} x\right )}}{693 \, x^{\frac{3}{2}}} \end{align*}

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^(5/2),x, algorithm="fricas")

[Out]

2/693*(63*B*b^6*x^7 - 231*A*a^6 + 77*(6*B*a*b^5 + A*b^6)*x^6 + 297*(5*B*a^2*b^4 + 2*A*a*b^5)*x^5 + 693*(4*B*a^
3*b^3 + 3*A*a^2*b^4)*x^4 + 1155*(3*B*a^4*b^2 + 4*A*a^3*b^3)*x^3 + 2079*(2*B*a^5*b + 5*A*a^4*b^2)*x^2 - 693*(B*
a^6 + 6*A*a^5*b)*x)/x^(3/2)

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Sympy [A]  time = 5.49756, size = 204, normalized size = 1.33 \begin{align*} - \frac{2 A a^{6}}{3 x^{\frac{3}{2}}} - \frac{12 A a^{5} b}{\sqrt{x}} + 30 A a^{4} b^{2} \sqrt{x} + \frac{40 A a^{3} b^{3} x^{\frac{3}{2}}}{3} + 6 A a^{2} b^{4} x^{\frac{5}{2}} + \frac{12 A a b^{5} x^{\frac{7}{2}}}{7} + \frac{2 A b^{6} x^{\frac{9}{2}}}{9} - \frac{2 B a^{6}}{\sqrt{x}} + 12 B a^{5} b \sqrt{x} + 10 B a^{4} b^{2} x^{\frac{3}{2}} + 8 B a^{3} b^{3} x^{\frac{5}{2}} + \frac{30 B a^{2} b^{4} x^{\frac{7}{2}}}{7} + \frac{4 B a b^{5} x^{\frac{9}{2}}}{3} + \frac{2 B b^{6} x^{\frac{11}{2}}}{11} \end{align*}

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**(5/2),x)

[Out]

-2*A*a**6/(3*x**(3/2)) - 12*A*a**5*b/sqrt(x) + 30*A*a**4*b**2*sqrt(x) + 40*A*a**3*b**3*x**(3/2)/3 + 6*A*a**2*b
**4*x**(5/2) + 12*A*a*b**5*x**(7/2)/7 + 2*A*b**6*x**(9/2)/9 - 2*B*a**6/sqrt(x) + 12*B*a**5*b*sqrt(x) + 10*B*a*
*4*b**2*x**(3/2) + 8*B*a**3*b**3*x**(5/2) + 30*B*a**2*b**4*x**(7/2)/7 + 4*B*a*b**5*x**(9/2)/3 + 2*B*b**6*x**(1
1/2)/11

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Giac [A]  time = 1.16092, size = 198, normalized size = 1.29 \begin{align*} \frac{2}{11} \, B b^{6} x^{\frac{11}{2}} + \frac{4}{3} \, B a b^{5} x^{\frac{9}{2}} + \frac{2}{9} \, A b^{6} x^{\frac{9}{2}} + \frac{30}{7} \, B a^{2} b^{4} x^{\frac{7}{2}} + \frac{12}{7} \, A a b^{5} x^{\frac{7}{2}} + 8 \, B a^{3} b^{3} x^{\frac{5}{2}} + 6 \, A a^{2} b^{4} x^{\frac{5}{2}} + 10 \, B a^{4} b^{2} x^{\frac{3}{2}} + \frac{40}{3} \, A a^{3} b^{3} x^{\frac{3}{2}} + 12 \, B a^{5} b \sqrt{x} + 30 \, A a^{4} b^{2} \sqrt{x} - \frac{2 \,{\left (3 \, B a^{6} x + 18 \, A a^{5} b x + A a^{6}\right )}}{3 \, x^{\frac{3}{2}}} \end{align*}

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^(5/2),x, algorithm="giac")

[Out]

2/11*B*b^6*x^(11/2) + 4/3*B*a*b^5*x^(9/2) + 2/9*A*b^6*x^(9/2) + 30/7*B*a^2*b^4*x^(7/2) + 12/7*A*a*b^5*x^(7/2)
+ 8*B*a^3*b^3*x^(5/2) + 6*A*a^2*b^4*x^(5/2) + 10*B*a^4*b^2*x^(3/2) + 40/3*A*a^3*b^3*x^(3/2) + 12*B*a^5*b*sqrt(
x) + 30*A*a^4*b^2*sqrt(x) - 2/3*(3*B*a^6*x + 18*A*a^5*b*x + A*a^6)/x^(3/2)