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

Optimal. Leaf size=107 \[ -\frac{2 a^3 (a B+4 A b)}{3 x^{3/2}}-\frac{4 a^2 b (2 a B+3 A b)}{\sqrt{x}}-\frac{2 a^4 A}{5 x^{5/2}}+\frac{2}{3} b^3 x^{3/2} (4 a B+A b)+4 a b^2 \sqrt{x} (3 a B+2 A b)+\frac{2}{5} b^4 B x^{5/2} \]

[Out]

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

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Rubi [A]  time = 0.0524969, antiderivative size = 107, 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} \[ -\frac{2 a^3 (a B+4 A b)}{3 x^{3/2}}-\frac{4 a^2 b (2 a B+3 A b)}{\sqrt{x}}-\frac{2 a^4 A}{5 x^{5/2}}+\frac{2}{3} b^3 x^{3/2} (4 a B+A b)+4 a b^2 \sqrt{x} (3 a B+2 A b)+\frac{2}{5} b^4 B x^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0266585, size = 85, normalized size = 0.79 \[ \frac{2 \left (90 a^2 b^2 x^2 (B x-A)-20 a^3 b x (A+3 B x)+a^4 (-(3 A+5 B x))+20 a b^3 x^3 (3 A+B x)+b^4 x^4 (5 A+3 B x)\right )}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(90*a^2*b^2*x^2*(-A + B*x) + 20*a*b^3*x^3*(3*A + B*x) - 20*a^3*b*x*(A + 3*B*x) + b^4*x^4*(5*A + 3*B*x) - a^
4*(3*A + 5*B*x)))/(15*x^(5/2))

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Maple [A]  time = 0.009, size = 100, normalized size = 0.9 \begin{align*} -{\frac{-6\,{b}^{4}B{x}^{5}-10\,A{b}^{4}{x}^{4}-40\,B{x}^{4}a{b}^{3}-120\,aA{b}^{3}{x}^{3}-180\,B{x}^{3}{a}^{2}{b}^{2}+180\,{a}^{2}A{b}^{2}{x}^{2}+120\,B{x}^{2}{a}^{3}b+40\,{a}^{3}Abx+10\,{a}^{4}Bx+6\,A{a}^{4}}{15}{x}^{-{\frac{5}{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)^2/x^(7/2),x)

[Out]

-2/15*(-3*B*b^4*x^5-5*A*b^4*x^4-20*B*a*b^3*x^4-60*A*a*b^3*x^3-90*B*a^2*b^2*x^3+90*A*a^2*b^2*x^2+60*B*a^3*b*x^2
+20*A*a^3*b*x+5*B*a^4*x+3*A*a^4)/x^(5/2)

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Maxima [A]  time = 1.01181, size = 135, normalized size = 1.26 \begin{align*} \frac{2}{5} \, B b^{4} x^{\frac{5}{2}} + \frac{2}{3} \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{\frac{3}{2}} + 4 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} \sqrt{x} - \frac{2 \,{\left (3 \, A a^{4} + 30 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} + 5 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} x\right )}}{15 \, x^{\frac{5}{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)^2/x^(7/2),x, algorithm="maxima")

[Out]

2/5*B*b^4*x^(5/2) + 2/3*(4*B*a*b^3 + A*b^4)*x^(3/2) + 4*(3*B*a^2*b^2 + 2*A*a*b^3)*sqrt(x) - 2/15*(3*A*a^4 + 30
*(2*B*a^3*b + 3*A*a^2*b^2)*x^2 + 5*(B*a^4 + 4*A*a^3*b)*x)/x^(5/2)

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Fricas [A]  time = 1.52594, size = 219, normalized size = 2.05 \begin{align*} \frac{2 \,{\left (3 \, B b^{4} x^{5} - 3 \, A a^{4} + 5 \,{\left (4 \, B a b^{3} + A b^{4}\right )} x^{4} + 30 \,{\left (3 \, B a^{2} b^{2} + 2 \, A a b^{3}\right )} x^{3} - 30 \,{\left (2 \, B a^{3} b + 3 \, A a^{2} b^{2}\right )} x^{2} - 5 \,{\left (B a^{4} + 4 \, A a^{3} b\right )} x\right )}}{15 \, x^{\frac{5}{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)^2/x^(7/2),x, algorithm="fricas")

[Out]

2/15*(3*B*b^4*x^5 - 3*A*a^4 + 5*(4*B*a*b^3 + A*b^4)*x^4 + 30*(3*B*a^2*b^2 + 2*A*a*b^3)*x^3 - 30*(2*B*a^3*b + 3
*A*a^2*b^2)*x^2 - 5*(B*a^4 + 4*A*a^3*b)*x)/x^(5/2)

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Sympy [A]  time = 3.54769, size = 141, normalized size = 1.32 \begin{align*} - \frac{2 A a^{4}}{5 x^{\frac{5}{2}}} - \frac{8 A a^{3} b}{3 x^{\frac{3}{2}}} - \frac{12 A a^{2} b^{2}}{\sqrt{x}} + 8 A a b^{3} \sqrt{x} + \frac{2 A b^{4} x^{\frac{3}{2}}}{3} - \frac{2 B a^{4}}{3 x^{\frac{3}{2}}} - \frac{8 B a^{3} b}{\sqrt{x}} + 12 B a^{2} b^{2} \sqrt{x} + \frac{8 B a b^{3} x^{\frac{3}{2}}}{3} + \frac{2 B b^{4} x^{\frac{5}{2}}}{5} \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)**2/x**(7/2),x)

[Out]

-2*A*a**4/(5*x**(5/2)) - 8*A*a**3*b/(3*x**(3/2)) - 12*A*a**2*b**2/sqrt(x) + 8*A*a*b**3*sqrt(x) + 2*A*b**4*x**(
3/2)/3 - 2*B*a**4/(3*x**(3/2)) - 8*B*a**3*b/sqrt(x) + 12*B*a**2*b**2*sqrt(x) + 8*B*a*b**3*x**(3/2)/3 + 2*B*b**
4*x**(5/2)/5

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Giac [A]  time = 1.10517, size = 135, normalized size = 1.26 \begin{align*} \frac{2}{5} \, B b^{4} x^{\frac{5}{2}} + \frac{8}{3} \, B a b^{3} x^{\frac{3}{2}} + \frac{2}{3} \, A b^{4} x^{\frac{3}{2}} + 12 \, B a^{2} b^{2} \sqrt{x} + 8 \, A a b^{3} \sqrt{x} - \frac{2 \,{\left (60 \, B a^{3} b x^{2} + 90 \, A a^{2} b^{2} x^{2} + 5 \, B a^{4} x + 20 \, A a^{3} b x + 3 \, A a^{4}\right )}}{15 \, x^{\frac{5}{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)^2/x^(7/2),x, algorithm="giac")

[Out]

2/5*B*b^4*x^(5/2) + 8/3*B*a*b^3*x^(3/2) + 2/3*A*b^4*x^(3/2) + 12*B*a^2*b^2*sqrt(x) + 8*A*a*b^3*sqrt(x) - 2/15*
(60*B*a^3*b*x^2 + 90*A*a^2*b^2*x^2 + 5*B*a^4*x + 20*A*a^3*b*x + 3*A*a^4)/x^(5/2)