3.391 \(\int \frac{(A+B x) (a+c x^2)}{x^{7/2}} \, dx\)

Optimal. Leaf size=41 \[ -\frac{2 a A}{5 x^{5/2}}-\frac{2 a B}{3 x^{3/2}}-\frac{2 A c}{\sqrt{x}}+2 B c \sqrt{x} \]

[Out]

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

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Rubi [A]  time = 0.0124574, antiderivative size = 41, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 1, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {766} \[ -\frac{2 a A}{5 x^{5/2}}-\frac{2 a B}{3 x^{3/2}}-\frac{2 A c}{\sqrt{x}}+2 B c \sqrt{x} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2))/x^(7/2),x]

[Out]

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

Rule 766

Int[((e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(e*x
)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.0121654, size = 32, normalized size = 0.78 \[ -\frac{2 \left (a (3 A+5 B x)+15 c x^2 (A-B x)\right )}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2))/x^(7/2),x]

[Out]

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

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Maple [A]  time = 0.004, size = 30, normalized size = 0.7 \begin{align*} -{\frac{-30\,Bc{x}^{3}+30\,Ac{x}^{2}+10\,aBx+6\,aA}{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+a)/x^(7/2),x)

[Out]

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

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Maxima [A]  time = 0.991885, size = 41, normalized size = 1. \begin{align*} 2 \, B c \sqrt{x} - \frac{2 \,{\left (15 \, A c x^{2} + 5 \, B a x + 3 \, A a\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+a)/x^(7/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.35741, size = 78, normalized size = 1.9 \begin{align*} \frac{2 \,{\left (15 \, B c x^{3} - 15 \, A c x^{2} - 5 \, B a x - 3 \, A a\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+a)/x^(7/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 1.92602, size = 42, normalized size = 1.02 \begin{align*} - \frac{2 A a}{5 x^{\frac{5}{2}}} - \frac{2 A c}{\sqrt{x}} - \frac{2 B a}{3 x^{\frac{3}{2}}} + 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+a)/x**(7/2),x)

[Out]

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

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Giac [A]  time = 1.26297, size = 41, normalized size = 1. \begin{align*} 2 \, B c \sqrt{x} - \frac{2 \,{\left (15 \, A c x^{2} + 5 \, B a x + 3 \, A a\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+a)/x^(7/2),x, algorithm="giac")

[Out]

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