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

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

[Out]

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

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Rubi [A]  time = 0.01, antiderivative size = 41, normalized size of antiderivative = 1.00, 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} \[ -\frac {2 a A}{\sqrt {x}}+2 a B \sqrt {x}+\frac {2}{3} A c x^{3/2}+\frac {2}{5} B c x^{5/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 32, normalized size = 0.78 \[ \frac {2 \left (c x^2 (5 A+3 B x)-15 a (A-B x)\right )}{15 \sqrt {x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.26, size = 29, normalized size = 0.71 \[ \frac {2 \, {\left (3 \, B c x^{3} + 5 \, A c x^{2} + 15 \, B a x - 15 \, A a\right )}}{15 \, \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 29, normalized size = 0.71 \[ \frac {2}{5} \, B c x^{\frac {5}{2}} + \frac {2}{3} \, A c x^{\frac {3}{2}} + 2 \, B a \sqrt {x} - \frac {2 \, A a}{\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 30, normalized size = 0.73 \[ -\frac {2 \left (-3 B c \,x^{3}-5 A c \,x^{2}-15 B a x +15 a A \right )}{15 \sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 29, normalized size = 0.71 \[ \frac {2}{5} \, B c x^{\frac {5}{2}} + \frac {2}{3} \, A c x^{\frac {3}{2}} + 2 \, B a \sqrt {x} - \frac {2 \, A a}{\sqrt {x}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 29, normalized size = 0.71 \[ 2\,B\,a\,\sqrt {x}-\frac {2\,A\,a}{\sqrt {x}}+\frac {2\,A\,c\,x^{3/2}}{3}+\frac {2\,B\,c\,x^{5/2}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.62, size = 42, normalized size = 1.02 \[ - \frac {2 A a}{\sqrt {x}} + \frac {2 A c x^{\frac {3}{2}}}{3} + 2 B a \sqrt {x} + \frac {2 B c x^{\frac {5}{2}}}{5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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