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

Optimal. Leaf size=59 \[ -\frac {2 a^2 A}{\sqrt {x}}+\frac {2}{3} b x^{3/2} (2 a B+A b)+2 a \sqrt {x} (a B+2 A b)+\frac {2}{5} b^2 B x^{5/2} \]

[Out]

2/3*b*(A*b+2*B*a)*x^(3/2)+2/5*b^2*B*x^(5/2)-2*a^2*A/x^(1/2)+2*a*(2*A*b+B*a)*x^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {27, 76} \[ -\frac {2 a^2 A}{\sqrt {x}}+\frac {2}{3} b x^{3/2} (2 a B+A b)+2 a \sqrt {x} (a B+2 A b)+\frac {2}{5} b^2 B x^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 49, normalized size = 0.83 \[ \frac {-30 a^2 (A-B x)+20 a b x (3 A+B x)+2 b^2 x^2 (5 A+3 B x)}{15 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-30*a^2*(A - B*x) + 20*a*b*x*(3*A + B*x) + 2*b^2*x^2*(5*A + 3*B*x))/(15*Sqrt[x])

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fricas [A]  time = 0.85, size = 51, normalized size = 0.86 \[ \frac {2 \, {\left (3 \, B b^{2} x^{3} - 15 \, A a^{2} + 5 \, {\left (2 \, B a b + A b^{2}\right )} x^{2} + 15 \, {\left (B a^{2} + 2 \, A a b\right )} x\right )}}{15 \, \sqrt {x}} \]

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)/x^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 53, normalized size = 0.90 \[ \frac {2}{5} \, B b^{2} x^{\frac {5}{2}} + \frac {4}{3} \, B a b x^{\frac {3}{2}} + \frac {2}{3} \, A b^{2} x^{\frac {3}{2}} + 2 \, B a^{2} \sqrt {x} + 4 \, A a b \sqrt {x} - \frac {2 \, A a^{2}}{\sqrt {x}} \]

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)/x^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 52, normalized size = 0.88 \[ -\frac {2 \left (-3 B \,b^{2} x^{3}-5 A \,b^{2} x^{2}-10 B a b \,x^{2}-30 A a b x -15 B \,a^{2} x +15 A \,a^{2}\right )}{15 \sqrt {x}} \]

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)/x^(3/2),x)

[Out]

-2/15*(-3*B*b^2*x^3-5*A*b^2*x^2-10*B*a*b*x^2-30*A*a*b*x-15*B*a^2*x+15*A*a^2)/x^(1/2)

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maxima [A]  time = 0.52, size = 51, normalized size = 0.86 \[ \frac {2}{5} \, B b^{2} x^{\frac {5}{2}} - \frac {2 \, A a^{2}}{\sqrt {x}} + \frac {2}{3} \, {\left (2 \, B a b + A b^{2}\right )} x^{\frac {3}{2}} + 2 \, {\left (B a^{2} + 2 \, A a b\right )} \sqrt {x} \]

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)/x^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.05, size = 51, normalized size = 0.86 \[ \sqrt {x}\,\left (2\,B\,a^2+4\,A\,b\,a\right )+x^{3/2}\,\left (\frac {2\,A\,b^2}{3}+\frac {4\,B\,a\,b}{3}\right )-\frac {2\,A\,a^2}{\sqrt {x}}+\frac {2\,B\,b^2\,x^{5/2}}{5} \]

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))/x^(3/2),x)

[Out]

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

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sympy [A]  time = 0.66, size = 75, normalized size = 1.27 \[ - \frac {2 A a^{2}}{\sqrt {x}} + 4 A a b \sqrt {x} + \frac {2 A b^{2} x^{\frac {3}{2}}}{3} + 2 B a^{2} \sqrt {x} + \frac {4 B a b x^{\frac {3}{2}}}{3} + \frac {2 B b^{2} x^{\frac {5}{2}}}{5} \]

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)/x**(3/2),x)

[Out]

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

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