3.4.90 \(\int \frac {(A+B x) (a+c x^2)}{x^{5/2}} \, dx\)

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

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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} \begin {gather*} -\frac {2 a A}{3 x^{3/2}}-\frac {2 a B}{\sqrt {x}}+2 A c \sqrt {x}+\frac {2}{3} B c x^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2))/x^(5/2),x]

[Out]

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

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

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Mathematica [A]  time = 0.01, size = 32, normalized size = 0.78 \begin {gather*} \frac {2 c x^2 (3 A+B x)-2 a (A+3 B x)}{3 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2))/x^(5/2),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 32, normalized size = 0.78 \begin {gather*} \frac {2 \left (-a A-3 a B x+3 A c x^2+B c x^3\right )}{3 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((A + B*x)*(a + c*x^2))/x^(5/2),x]

[Out]

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

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fricas [A]  time = 0.40, size = 28, normalized size = 0.68 \begin {gather*} \frac {2 \, {\left (B c x^{3} + 3 \, A c x^{2} - 3 \, B a x - A a\right )}}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)/x^(5/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 29, normalized size = 0.71 \begin {gather*} \frac {2}{3} \, B c x^{\frac {3}{2}} + 2 \, A c \sqrt {x} - \frac {2 \, {\left (3 \, B a x + A a\right )}}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)/x^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 29, normalized size = 0.71 \begin {gather*} -\frac {2 \left (-B c \,x^{3}-3 A c \,x^{2}+3 B a x +a A \right )}{3 x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)/x^(5/2),x)

[Out]

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

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maxima [A]  time = 0.71, size = 29, normalized size = 0.71 \begin {gather*} \frac {2}{3} \, B c x^{\frac {3}{2}} + 2 \, A c \sqrt {x} - \frac {2 \, {\left (3 \, B a x + A a\right )}}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)/x^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.05, size = 29, normalized size = 0.71 \begin {gather*} -\frac {-2\,B\,c\,x^3-6\,A\,c\,x^2+6\,B\,a\,x+2\,A\,a}{3\,x^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)*(A + B*x))/x^(5/2),x)

[Out]

-(2*A*a + 6*B*a*x - 6*A*c*x^2 - 2*B*c*x^3)/(3*x^(3/2))

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sympy [A]  time = 0.85, size = 42, normalized size = 1.02 \begin {gather*} - \frac {2 A a}{3 x^{\frac {3}{2}}} + 2 A c \sqrt {x} - \frac {2 B a}{\sqrt {x}} + \frac {2 B c x^{\frac {3}{2}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)/x**(5/2),x)

[Out]

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

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