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

Optimal. Leaf size=37 \[ -\frac{2 (a B+A b)}{\sqrt{x}}-\frac{2 a A}{5 x^{5/2}}+\frac{2}{3} b B x^{3/2} \]

[Out]

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

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Rubi [A]  time = 0.0156172, 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} \[ -\frac{2 (a B+A b)}{\sqrt{x}}-\frac{2 a A}{5 x^{5/2}}+\frac{2}{3} b B x^{3/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0126628, size = 36, normalized size = 0.97 \[ \frac{10 b x^2 \left (B x^2-3 A\right )-6 a \left (A+5 B x^2\right )}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*b*x^2*(-3*A + B*x^2) - 6*a*(A + 5*B*x^2))/(15*x^(5/2))

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Maple [A]  time = 0.003, size = 32, normalized size = 0.9 \begin{align*} -{\frac{-10\,bB{x}^{4}+30\,A{x}^{2}b+30\,B{x}^{2}a+6\,Aa}{15}{x}^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15*(-5*B*b*x^4+15*A*b*x^2+15*B*a*x^2+3*A*a)/x^(5/2)

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Maxima [A]  time = 1.03911, size = 39, normalized size = 1.05 \begin{align*} \frac{2}{3} \, B b x^{\frac{3}{2}} - \frac{2 \,{\left (5 \,{\left (B a + A b\right )} x^{2} + A a\right )}}{5 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 0.858949, size = 74, normalized size = 2. \begin{align*} \frac{2 \,{\left (5 \, B b x^{4} - 15 \,{\left (B a + A b\right )} x^{2} - 3 \, A a\right )}}{15 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*(5*B*b*x^4 - 15*(B*a + A*b)*x^2 - 3*A*a)/x^(5/2)

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Sympy [A]  time = 1.86614, size = 42, normalized size = 1.14 \begin{align*} - \frac{2 A a}{5 x^{\frac{5}{2}}} - \frac{2 A b}{\sqrt{x}} - \frac{2 B a}{\sqrt{x}} + \frac{2 B b x^{\frac{3}{2}}}{3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Giac [A]  time = 1.13884, size = 42, normalized size = 1.14 \begin{align*} \frac{2}{3} \, B b x^{\frac{3}{2}} - \frac{2 \,{\left (5 \, B a x^{2} + 5 \, A b x^{2} + A a\right )}}{5 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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