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

Optimal. Leaf size=37 \[ \frac{2}{3} x^{3/2} (A c+b B)+2 A b \sqrt{x}+\frac{2}{5} B c x^{5/2} \]

[Out]

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

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Rubi [A]  time = 0.0149419, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.05, Rules used = {765} \[ \frac{2}{3} x^{3/2} (A c+b B)+2 A b \sqrt{x}+\frac{2}{5} B c x^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 765

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[Expand
Integrand[(e*x)^m*(f + g*x)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, e, f, g, m}, x] && IntegerQ[p] && (
GtQ[p, 0] || (EqQ[a, 0] && IntegerQ[m]))

Rubi steps

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

Mathematica [A]  time = 0.010796, size = 31, normalized size = 0.84 \[ \frac{2}{15} \sqrt{x} (5 A (3 b+c x)+B x (5 b+3 c x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[x]*(5*A*(3*b + c*x) + B*x*(5*b + 3*c*x)))/15

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Maple [A]  time = 0.003, size = 28, normalized size = 0.8 \begin{align*}{\frac{6\,Bc{x}^{2}+10\,Acx+10\,bBx+30\,Ab}{15}\sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*x^(1/2)*(3*B*c*x^2+5*A*c*x+5*B*b*x+15*A*b)

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Maxima [A]  time = 1.02592, size = 36, normalized size = 0.97 \begin{align*} \frac{2}{5} \, B c x^{\frac{5}{2}} + 2 \, A b \sqrt{x} + \frac{2}{3} \,{\left (B b + A c\right )} x^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)/x^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.83261, size = 72, normalized size = 1.95 \begin{align*} \frac{2}{15} \,{\left (3 \, B c x^{2} + 15 \, A b + 5 \,{\left (B b + A c\right )} x\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)/x^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.637288, size = 44, normalized size = 1.19 \begin{align*} 2 A b \sqrt{x} + \frac{2 A c x^{\frac{3}{2}}}{3} + \frac{2 B b x^{\frac{3}{2}}}{3} + \frac{2 B c x^{\frac{5}{2}}}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)/x**(3/2),x)

[Out]

2*A*b*sqrt(x) + 2*A*c*x**(3/2)/3 + 2*B*b*x**(3/2)/3 + 2*B*c*x**(5/2)/5

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Giac [A]  time = 1.12281, size = 39, normalized size = 1.05 \begin{align*} \frac{2}{5} \, B c x^{\frac{5}{2}} + \frac{2}{3} \, B b x^{\frac{3}{2}} + \frac{2}{3} \, A c x^{\frac{3}{2}} + 2 \, A b \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)/x^(3/2),x, algorithm="giac")

[Out]

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