3.144 \(\int \frac{(A+B x) (b x+c x^2)}{\sqrt{x}} \, dx\)

Optimal. Leaf size=39 \[ \frac{2}{5} x^{5/2} (A c+b B)+\frac{2}{3} A b x^{3/2}+\frac{2}{7} B c x^{7/2} \]

[Out]

(2*A*b*x^(3/2))/3 + (2*(b*B + A*c)*x^(5/2))/5 + (2*B*c*x^(7/2))/7

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Rubi [A]  time = 0.0156641, antiderivative size = 39, 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} \[ \frac{2}{5} x^{5/2} (A c+b B)+\frac{2}{3} A b x^{3/2}+\frac{2}{7} B c x^{7/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*b*x^(3/2))/3 + (2*(b*B + A*c)*x^(5/2))/5 + (2*B*c*x^(7/2))/7

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

Mathematica [A]  time = 0.0106106, size = 33, normalized size = 0.85 \[ \frac{2}{105} x^{3/2} (7 A (5 b+3 c x)+3 B x (7 b+5 c x)) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(3/2)*(7*A*(5*b + 3*c*x) + 3*B*x*(7*b + 5*c*x)))/105

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Maple [A]  time = 0.003, size = 28, normalized size = 0.7 \begin{align*}{\frac{30\,Bc{x}^{2}+42\,Acx+42\,bBx+70\,Ab}{105}{x}^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)/x^(1/2),x)

[Out]

2/105*x^(3/2)*(15*B*c*x^2+21*A*c*x+21*B*b*x+35*A*b)

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Maxima [A]  time = 0.996198, size = 36, normalized size = 0.92 \begin{align*} \frac{2}{7} \, B c x^{\frac{7}{2}} + \frac{2}{3} \, A b x^{\frac{3}{2}} + \frac{2}{5} \,{\left (B b + A c\right )} x^{\frac{5}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x)/x^(1/2),x, algorithm="maxima")

[Out]

2/7*B*c*x^(7/2) + 2/3*A*b*x^(3/2) + 2/5*(B*b + A*c)*x^(5/2)

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Fricas [A]  time = 1.86162, size = 81, normalized size = 2.08 \begin{align*} \frac{2}{105} \,{\left (15 \, B c x^{3} + 35 \, A b x + 21 \,{\left (B b + A c\right )} x^{2}\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^(1/2),x, algorithm="fricas")

[Out]

2/105*(15*B*c*x^3 + 35*A*b*x + 21*(B*b + A*c)*x^2)*sqrt(x)

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Sympy [A]  time = 0.466973, size = 46, normalized size = 1.18 \begin{align*} \frac{2 A b x^{\frac{3}{2}}}{3} + \frac{2 A c x^{\frac{5}{2}}}{5} + \frac{2 B b x^{\frac{5}{2}}}{5} + \frac{2 B c x^{\frac{7}{2}}}{7} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+b*x)/x**(1/2),x)

[Out]

2*A*b*x**(3/2)/3 + 2*A*c*x**(5/2)/5 + 2*B*b*x**(5/2)/5 + 2*B*c*x**(7/2)/7

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Giac [A]  time = 1.0977, size = 39, normalized size = 1. \begin{align*} \frac{2}{7} \, B c x^{\frac{7}{2}} + \frac{2}{5} \, B b x^{\frac{5}{2}} + \frac{2}{5} \, A c x^{\frac{5}{2}} + \frac{2}{3} \, A b 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^(1/2),x, algorithm="giac")

[Out]

2/7*B*c*x^(7/2) + 2/5*B*b*x^(5/2) + 2/5*A*c*x^(5/2) + 2/3*A*b*x^(3/2)