3.138 \(\int \frac{(a+b x^3) (A+B x^3)}{x^{7/2}} \, dx\)

Optimal. Leaf size=37 \[ 2 \sqrt{x} (a B+A b)-\frac{2 a A}{5 x^{5/2}}+\frac{2}{7} b B x^{7/2} \]

[Out]

(-2*a*A)/(5*x^(5/2)) + 2*(A*b + a*B)*Sqrt[x] + (2*b*B*x^(7/2))/7

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Rubi [A]  time = 0.0161729, 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} \[ 2 \sqrt{x} (a B+A b)-\frac{2 a A}{5 x^{5/2}}+\frac{2}{7} b B x^{7/2} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^3)*(A + B*x^3))/x^(7/2),x]

[Out]

(-2*a*A)/(5*x^(5/2)) + 2*(A*b + a*B)*Sqrt[x] + (2*b*B*x^(7/2))/7

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

Mathematica [A]  time = 0.0110783, size = 36, normalized size = 0.97 \[ \frac{2 \left (5 b x^3 \left (7 A+B x^3\right )-7 a \left (A-5 B x^3\right )\right )}{35 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^3)*(A + B*x^3))/x^(7/2),x]

[Out]

(2*(-7*a*(A - 5*B*x^3) + 5*b*x^3*(7*A + B*x^3)))/(35*x^(5/2))

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Maple [A]  time = 0.004, size = 32, normalized size = 0.9 \begin{align*} -{\frac{-10\,bB{x}^{6}-70\,A{x}^{3}b-70\,B{x}^{3}a+14\,Aa}{35}{x}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^3+a)*(B*x^3+A)/x^(7/2),x)

[Out]

-2/35*(-5*B*b*x^6-35*A*b*x^3-35*B*a*x^3+7*A*a)/x^(5/2)

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Maxima [A]  time = 0.983294, size = 36, normalized size = 0.97 \begin{align*} \frac{2}{7} \, B b x^{\frac{7}{2}} + 2 \,{\left (B a + A b\right )} \sqrt{x} - \frac{2 \, A a}{5 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*(B*x^3+A)/x^(7/2),x, algorithm="maxima")

[Out]

2/7*B*b*x^(7/2) + 2*(B*a + A*b)*sqrt(x) - 2/5*A*a/x^(5/2)

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Fricas [A]  time = 1.74647, size = 74, normalized size = 2. \begin{align*} \frac{2 \,{\left (5 \, B b x^{6} + 35 \,{\left (B a + A b\right )} x^{3} - 7 \, A a\right )}}{35 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*(B*x^3+A)/x^(7/2),x, algorithm="fricas")

[Out]

2/35*(5*B*b*x^6 + 35*(B*a + A*b)*x^3 - 7*A*a)/x^(5/2)

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Sympy [A]  time = 4.23594, size = 42, normalized size = 1.14 \begin{align*} - \frac{2 A a}{5 x^{\frac{5}{2}}} + 2 A b \sqrt{x} + 2 B a \sqrt{x} + \frac{2 B b x^{\frac{7}{2}}}{7} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**3+a)*(B*x**3+A)/x**(7/2),x)

[Out]

-2*A*a/(5*x**(5/2)) + 2*A*b*sqrt(x) + 2*B*a*sqrt(x) + 2*B*b*x**(7/2)/7

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Giac [A]  time = 1.10656, size = 39, normalized size = 1.05 \begin{align*} \frac{2}{7} \, B b x^{\frac{7}{2}} + 2 \, B a \sqrt{x} + 2 \, A b \sqrt{x} - \frac{2 \, A a}{5 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^3+a)*(B*x^3+A)/x^(7/2),x, algorithm="giac")

[Out]

2/7*B*b*x^(7/2) + 2*B*a*sqrt(x) + 2*A*b*sqrt(x) - 2/5*A*a/x^(5/2)