3.148 \(\int \frac{(A+B x) (b x+c x^2)}{x^{9/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0169217, 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} \[ -\frac{2 (A c+b B)}{3 x^{3/2}}-\frac{2 A b}{5 x^{5/2}}-\frac{2 B c}{\sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(b*x + c*x^2))/x^(9/2),x]

[Out]

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

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^{9/2}} \, dx &=\int \left (\frac{A b}{x^{7/2}}+\frac{b B+A c}{x^{5/2}}+\frac{B c}{x^{3/2}}\right ) \, dx\\ &=-\frac{2 A b}{5 x^{5/2}}-\frac{2 (b B+A c)}{3 x^{3/2}}-\frac{2 B c}{\sqrt{x}}\\ \end{align*}

Mathematica [A]  time = 0.0099976, size = 30, normalized size = 0.81 \[ -\frac{2 (A (3 b+5 c x)+5 B x (b+3 c x))}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(b*x + c*x^2))/x^(9/2),x]

[Out]

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

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Maple [A]  time = 0.003, size = 28, normalized size = 0.8 \begin{align*} -{\frac{30\,Bc{x}^{2}+10\,Acx+10\,bBx+6\,Ab}{15}{x}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+b*x)/x^(9/2),x)

[Out]

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

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Maxima [A]  time = 1.06149, size = 36, normalized size = 0.97 \begin{align*} -\frac{2 \,{\left (15 \, B c x^{2} + 3 \, A b + 5 \,{\left (B b + A c\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)*(c*x^2+b*x)/x^(9/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.78538, size = 73, normalized size = 1.97 \begin{align*} -\frac{2 \,{\left (15 \, B c x^{2} + 3 \, A b + 5 \,{\left (B b + A c\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)*(c*x^2+b*x)/x^(9/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.37419, size = 46, normalized size = 1.24 \begin{align*} - \frac{2 A b}{5 x^{\frac{5}{2}}} - \frac{2 A c}{3 x^{\frac{3}{2}}} - \frac{2 B b}{3 x^{\frac{3}{2}}} - \frac{2 B c}{\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**(9/2),x)

[Out]

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

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Giac [A]  time = 1.12486, size = 36, normalized size = 0.97 \begin{align*} -\frac{2 \,{\left (15 \, B c x^{2} + 5 \, B b x + 5 \, A c x + 3 \, A b\right )}}{15 \, 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^(9/2),x, algorithm="giac")

[Out]

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