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

Optimal. Leaf size=51 \[ 2 \sqrt{x} (a B+A b)-\frac{2 a A}{\sqrt{x}}+\frac{2}{3} x^{3/2} (A c+b B)+\frac{2}{5} B c x^{5/2} \]

[Out]

(-2*a*A)/Sqrt[x] + 2*(A*b + 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 = 0.024403, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {765} \[ 2 \sqrt{x} (a B+A b)-\frac{2 a A}{\sqrt{x}}+\frac{2}{3} x^{3/2} (A c+b B)+\frac{2}{5} B c x^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*A)/Sqrt[x] + 2*(A*b + 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 (a+b x+c x^2\right )}{x^{3/2}} \, dx &=\int \left (\frac{a A}{x^{3/2}}+\frac{A b+a B}{\sqrt{x}}+(b B+A c) \sqrt{x}+B c x^{3/2}\right ) \, dx\\ &=-\frac{2 a A}{\sqrt{x}}+2 (A b+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 = 0.0422304, size = 44, normalized size = 0.86 \[ \frac{2 x (5 A (3 b+c x)+B x (5 b+3 c x))-30 a (A-B x)}{15 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.005, size = 42, normalized size = 0.8 \begin{align*} -{\frac{-6\,Bc{x}^{3}-10\,Ac{x}^{2}-10\,Bb{x}^{2}-30\,Abx-30\,aBx+30\,aA}{15}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.18409, size = 53, normalized size = 1.04 \begin{align*} \frac{2}{5} \, B c x^{\frac{5}{2}} + \frac{2}{3} \,{\left (B b + A c\right )} x^{\frac{3}{2}} - \frac{2 \, A a}{\sqrt{x}} + 2 \,{\left (B a + A b\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+a)/x^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.19802, size = 100, normalized size = 1.96 \begin{align*} \frac{2 \,{\left (3 \, B c x^{3} + 5 \,{\left (B b + A c\right )} x^{2} - 15 \, A a + 15 \,{\left (B a + A b\right )} x\right )}}{15 \, \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 1.3948, size = 65, normalized size = 1.27 \begin{align*} - \frac{2 A a}{\sqrt{x}} + 2 A b \sqrt{x} + \frac{2 A c x^{\frac{3}{2}}}{3} + 2 B a \sqrt{x} + \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+a)/x**(3/2),x)

[Out]

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

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Giac [A]  time = 1.19673, size = 58, normalized size = 1.14 \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 \, B a \sqrt{x} + 2 \, A b \sqrt{x} - \frac{2 \, A a}{\sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+b*x+a)/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*B*a*sqrt(x) + 2*A*b*sqrt(x) - 2*A*a/sqrt(x)