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

Optimal. Leaf size=37 \[ \frac{2}{5} x^{5/2} (a B+A b)-\frac{2 a A}{\sqrt{x}}+\frac{2}{11} b B x^{11/2} \]

[Out]

(-2*a*A)/Sqrt[x] + (2*(A*b + a*B)*x^(5/2))/5 + (2*b*B*x^(11/2))/11

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Rubi [A]  time = 0.0148993, 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 = {448} \[ \frac{2}{5} x^{5/2} (a B+A b)-\frac{2 a A}{\sqrt{x}}+\frac{2}{11} b B x^{11/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*A)/Sqrt[x] + (2*(A*b + a*B)*x^(5/2))/5 + (2*b*B*x^(11/2))/11

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^{3/2}} \, dx &=\int \left (\frac{a A}{x^{3/2}}+(A b+a B) x^{3/2}+b B x^{9/2}\right ) \, dx\\ &=-\frac{2 a A}{\sqrt{x}}+\frac{2}{5} (A b+a B) x^{5/2}+\frac{2}{11} b B x^{11/2}\\ \end{align*}

Mathematica [A]  time = 0.0099154, size = 35, normalized size = 0.95 \[ \frac{2 \left (-55 a A+11 a B x^3+11 A b x^3+5 b B x^6\right )}{55 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(-55*a*A + 11*A*b*x^3 + 11*a*B*x^3 + 5*b*B*x^6))/(55*Sqrt[x])

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Maple [A]  time = 0.004, size = 32, normalized size = 0.9 \begin{align*} -{\frac{-10\,bB{x}^{6}-22\,A{x}^{3}b-22\,B{x}^{3}a+110\,Aa}{55}{\frac{1}{\sqrt{x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/55*(-5*B*b*x^6-11*A*b*x^3-11*B*a*x^3+55*A*a)/x^(1/2)

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Maxima [A]  time = 0.947561, size = 36, normalized size = 0.97 \begin{align*} \frac{2}{11} \, B b x^{\frac{11}{2}} + \frac{2}{5} \,{\left (B a + A b\right )} x^{\frac{5}{2}} - \frac{2 \, A a}{\sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*B*b*x^(11/2) + 2/5*(B*a + A*b)*x^(5/2) - 2*A*a/sqrt(x)

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Fricas [A]  time = 1.70209, size = 76, normalized size = 2.05 \begin{align*} \frac{2 \,{\left (5 \, B b x^{6} + 11 \,{\left (B a + A b\right )} x^{3} - 55 \, A a\right )}}{55 \, \sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/55*(5*B*b*x^6 + 11*(B*a + A*b)*x^3 - 55*A*a)/sqrt(x)

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Sympy [A]  time = 2.52094, size = 44, normalized size = 1.19 \begin{align*} - \frac{2 A a}{\sqrt{x}} + \frac{2 A b x^{\frac{5}{2}}}{5} + \frac{2 B a x^{\frac{5}{2}}}{5} + \frac{2 B b x^{\frac{11}{2}}}{11} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*A*a/sqrt(x) + 2*A*b*x**(5/2)/5 + 2*B*a*x**(5/2)/5 + 2*B*b*x**(11/2)/11

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Giac [A]  time = 1.09866, size = 39, normalized size = 1.05 \begin{align*} \frac{2}{11} \, B b x^{\frac{11}{2}} + \frac{2}{5} \, B a x^{\frac{5}{2}} + \frac{2}{5} \, A b x^{\frac{5}{2}} - \frac{2 \, A a}{\sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/11*B*b*x^(11/2) + 2/5*B*a*x^(5/2) + 2/5*A*b*x^(5/2) - 2*A*a/sqrt(x)