3.2.33 \(\int \frac {(a+b x^3) (A+B x^3)}{x^{5/2}} \, dx\)

Optimal. Leaf size=39 \[ \frac {2}{3} x^{3/2} (a B+A b)-\frac {2 a A}{3 x^{3/2}}+\frac {2}{9} b B x^{9/2} \]

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Rubi [A]  time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {448} \begin {gather*} \frac {2}{3} x^{3/2} (a B+A b)-\frac {2 a A}{3 x^{3/2}}+\frac {2}{9} b B x^{9/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)*(A + B*x^3))/x^(5/2),x]

[Out]

(-2*a*A)/(3*x^(3/2)) + (2*(A*b + a*B)*x^(3/2))/3 + (2*b*B*x^(9/2))/9

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 34, normalized size = 0.87 \begin {gather*} \frac {2 \left (-3 a A+3 a B x^3+3 A b x^3+b B x^6\right )}{9 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)*(A + B*x^3))/x^(5/2),x]

[Out]

(2*(-3*a*A + 3*A*b*x^3 + 3*a*B*x^3 + b*B*x^6))/(9*x^(3/2))

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IntegrateAlgebraic [A]  time = 0.02, size = 34, normalized size = 0.87 \begin {gather*} \frac {2 \left (-3 a A+3 a B x^3+3 A b x^3+b B x^6\right )}{9 x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[((a + b*x^3)*(A + B*x^3))/x^(5/2),x]

[Out]

(2*(-3*a*A + 3*A*b*x^3 + 3*a*B*x^3 + b*B*x^6))/(9*x^(3/2))

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fricas [A]  time = 0.89, size = 28, normalized size = 0.72 \begin {gather*} \frac {2 \, {\left (B b x^{6} + 3 \, {\left (B a + A b\right )} x^{3} - 3 \, A a\right )}}{9 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*(B*x^3+A)/x^(5/2),x, algorithm="fricas")

[Out]

2/9*(B*b*x^6 + 3*(B*a + A*b)*x^3 - 3*A*a)/x^(3/2)

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giac [A]  time = 0.16, size = 29, normalized size = 0.74 \begin {gather*} \frac {2}{9} \, B b x^{\frac {9}{2}} + \frac {2}{3} \, B a x^{\frac {3}{2}} + \frac {2}{3} \, A b x^{\frac {3}{2}} - \frac {2 \, A a}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*(B*x^3+A)/x^(5/2),x, algorithm="giac")

[Out]

2/9*B*b*x^(9/2) + 2/3*B*a*x^(3/2) + 2/3*A*b*x^(3/2) - 2/3*A*a/x^(3/2)

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maple [A]  time = 0.04, size = 32, normalized size = 0.82 \begin {gather*} -\frac {2 \left (-B b \,x^{6}-3 A b \,x^{3}-3 B a \,x^{3}+3 A a \right )}{9 x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)*(B*x^3+A)/x^(5/2),x)

[Out]

-2/9*(-B*b*x^6-3*A*b*x^3-3*B*a*x^3+3*A*a)/x^(3/2)

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maxima [A]  time = 0.48, size = 27, normalized size = 0.69 \begin {gather*} \frac {2}{9} \, B b x^{\frac {9}{2}} + \frac {2}{3} \, {\left (B a + A b\right )} x^{\frac {3}{2}} - \frac {2 \, A a}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*(B*x^3+A)/x^(5/2),x, algorithm="maxima")

[Out]

2/9*B*b*x^(9/2) + 2/3*(B*a + A*b)*x^(3/2) - 2/3*A*a/x^(3/2)

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mupad [B]  time = 0.04, size = 31, normalized size = 0.79 \begin {gather*} \frac {6\,A\,b\,x^3-6\,A\,a+6\,B\,a\,x^3+2\,B\,b\,x^6}{9\,x^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^3)*(a + b*x^3))/x^(5/2),x)

[Out]

(6*A*b*x^3 - 6*A*a + 6*B*a*x^3 + 2*B*b*x^6)/(9*x^(3/2))

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sympy [A]  time = 3.74, size = 46, normalized size = 1.18 \begin {gather*} - \frac {2 A a}{3 x^{\frac {3}{2}}} + \frac {2 A b x^{\frac {3}{2}}}{3} + \frac {2 B a x^{\frac {3}{2}}}{3} + \frac {2 B b x^{\frac {9}{2}}}{9} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)*(B*x**3+A)/x**(5/2),x)

[Out]

-2*A*a/(3*x**(3/2)) + 2*A*b*x**(3/2)/3 + 2*B*a*x**(3/2)/3 + 2*B*b*x**(9/2)/9

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