3.165 \(\int \frac{(A+B x^2) (b x^2+c x^4)}{x^{5/2}} \, dx\)

Optimal. Leaf size=37 \[ \frac{2}{5} x^{5/2} (A c+b B)+2 A b \sqrt{x}+\frac{2}{9} B c x^{9/2} \]

[Out]

2*A*b*Sqrt[x] + (2*(b*B + A*c)*x^(5/2))/5 + (2*B*c*x^(9/2))/9

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Rubi [A]  time = 0.0212557, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.083, Rules used = {1584, 448} \[ \frac{2}{5} x^{5/2} (A c+b B)+2 A b \sqrt{x}+\frac{2}{9} B c x^{9/2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4))/x^(5/2),x]

[Out]

2*A*b*Sqrt[x] + (2*(b*B + A*c)*x^(5/2))/5 + (2*B*c*x^(9/2))/9

Rule 1584

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(m + n*p)*(a + b*x^(q -
p))^n, x] /; FreeQ[{a, b, m, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 (b x^2+c x^4\right )}{x^{5/2}} \, dx &=\int \frac{\left (A+B x^2\right ) \left (b+c x^2\right )}{\sqrt{x}} \, dx\\ &=\int \left (\frac{A b}{\sqrt{x}}+(b B+A c) x^{3/2}+B c x^{7/2}\right ) \, dx\\ &=2 A b \sqrt{x}+\frac{2}{5} (b B+A c) x^{5/2}+\frac{2}{9} B c x^{9/2}\\ \end{align*}

Mathematica [A]  time = 0.0141017, size = 33, normalized size = 0.89 \[ \frac{2}{45} \sqrt{x} \left (9 x^2 (A c+b B)+45 A b+5 B c x^4\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4))/x^(5/2),x]

[Out]

(2*Sqrt[x]*(45*A*b + 9*(b*B + A*c)*x^2 + 5*B*c*x^4))/45

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Maple [A]  time = 0.003, size = 32, normalized size = 0.9 \begin{align*}{\frac{10\,Bc{x}^{4}+18\,A{x}^{2}c+18\,B{x}^{2}b+90\,Ab}{45}\sqrt{x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)*(c*x^4+b*x^2)/x^(5/2),x)

[Out]

2/45*x^(1/2)*(5*B*c*x^4+9*A*c*x^2+9*B*b*x^2+45*A*b)

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Maxima [A]  time = 1.12695, size = 36, normalized size = 0.97 \begin{align*} \frac{2}{9} \, B c x^{\frac{9}{2}} + \frac{2}{5} \,{\left (B b + A c\right )} x^{\frac{5}{2}} + 2 \, A b \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^(5/2),x, algorithm="maxima")

[Out]

2/9*B*c*x^(9/2) + 2/5*(B*b + A*c)*x^(5/2) + 2*A*b*sqrt(x)

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Fricas [A]  time = 1.5373, size = 74, normalized size = 2. \begin{align*} \frac{2}{45} \,{\left (5 \, B c x^{4} + 9 \,{\left (B b + A c\right )} x^{2} + 45 \, A b\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^(5/2),x, algorithm="fricas")

[Out]

2/45*(5*B*c*x^4 + 9*(B*b + A*c)*x^2 + 45*A*b)*sqrt(x)

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Sympy [A]  time = 4.54852, size = 44, normalized size = 1.19 \begin{align*} 2 A b \sqrt{x} + \frac{2 A c x^{\frac{5}{2}}}{5} + \frac{2 B b x^{\frac{5}{2}}}{5} + \frac{2 B c x^{\frac{9}{2}}}{9} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)/x**(5/2),x)

[Out]

2*A*b*sqrt(x) + 2*A*c*x**(5/2)/5 + 2*B*b*x**(5/2)/5 + 2*B*c*x**(9/2)/9

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Giac [A]  time = 1.13098, size = 39, normalized size = 1.05 \begin{align*} \frac{2}{9} \, B c x^{\frac{9}{2}} + \frac{2}{5} \, B b x^{\frac{5}{2}} + \frac{2}{5} \, A c x^{\frac{5}{2}} + 2 \, A b \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)*(c*x^4+b*x^2)/x^(5/2),x, algorithm="giac")

[Out]

2/9*B*c*x^(9/2) + 2/5*B*b*x^(5/2) + 2/5*A*c*x^(5/2) + 2*A*b*sqrt(x)