3.163 \(\int \frac{(A+B x^2) (b x^2+c x^4)}{\sqrt{x}} \, dx\)

Optimal. Leaf size=39 \[ \frac{2}{9} x^{9/2} (A c+b B)+\frac{2}{5} A b x^{5/2}+\frac{2}{13} B c x^{13/2} \]

[Out]

(2*A*b*x^(5/2))/5 + (2*(b*B + A*c)*x^(9/2))/9 + (2*B*c*x^(13/2))/13

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Rubi [A]  time = 0.0210477, antiderivative size = 39, 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} \[ \frac{2}{9} x^{9/2} (A c+b B)+\frac{2}{5} A b x^{5/2}+\frac{2}{13} B c x^{13/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*b*x^(5/2))/5 + (2*(b*B + A*c)*x^(9/2))/9 + (2*B*c*x^(13/2))/13

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 )}{\sqrt{x}} \, dx &=\int x^{3/2} \left (A+B x^2\right ) \left (b+c x^2\right ) \, dx\\ &=\int \left (A b x^{3/2}+(b B+A c) x^{7/2}+B c x^{11/2}\right ) \, dx\\ &=\frac{2}{5} A b x^{5/2}+\frac{2}{9} (b B+A c) x^{9/2}+\frac{2}{13} B c x^{13/2}\\ \end{align*}

Mathematica [A]  time = 0.0143855, size = 33, normalized size = 0.85 \[ \frac{2}{585} x^{5/2} \left (65 x^2 (A c+b B)+117 A b+45 B c x^4\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(5/2)*(117*A*b + 65*(b*B + A*c)*x^2 + 45*B*c*x^4))/585

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Maple [A]  time = 0.003, size = 32, normalized size = 0.8 \begin{align*}{\frac{90\,Bc{x}^{4}+130\,A{x}^{2}c+130\,B{x}^{2}b+234\,Ab}{585}{x}^{{\frac{5}{2}}}} \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^(1/2),x)

[Out]

2/585*x^(5/2)*(45*B*c*x^4+65*A*c*x^2+65*B*b*x^2+117*A*b)

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Maxima [A]  time = 1.02385, size = 36, normalized size = 0.92 \begin{align*} \frac{2}{13} \, B c x^{\frac{13}{2}} + \frac{2}{9} \,{\left (B b + A c\right )} x^{\frac{9}{2}} + \frac{2}{5} \, A b x^{\frac{5}{2}} \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^(1/2),x, algorithm="maxima")

[Out]

2/13*B*c*x^(13/2) + 2/9*(B*b + A*c)*x^(9/2) + 2/5*A*b*x^(5/2)

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Fricas [A]  time = 1.52941, size = 85, normalized size = 2.18 \begin{align*} \frac{2}{585} \,{\left (45 \, B c x^{6} + 65 \,{\left (B b + A c\right )} x^{4} + 117 \, A b x^{2}\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^(1/2),x, algorithm="fricas")

[Out]

2/585*(45*B*c*x^6 + 65*(B*b + A*c)*x^4 + 117*A*b*x^2)*sqrt(x)

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Sympy [A]  time = 3.31652, size = 46, normalized size = 1.18 \begin{align*} \frac{2 A b x^{\frac{5}{2}}}{5} + \frac{2 A c x^{\frac{9}{2}}}{9} + \frac{2 B b x^{\frac{9}{2}}}{9} + \frac{2 B c x^{\frac{13}{2}}}{13} \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**(1/2),x)

[Out]

2*A*b*x**(5/2)/5 + 2*A*c*x**(9/2)/9 + 2*B*b*x**(9/2)/9 + 2*B*c*x**(13/2)/13

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Giac [A]  time = 1.14759, size = 39, normalized size = 1. \begin{align*} \frac{2}{13} \, B c x^{\frac{13}{2}} + \frac{2}{9} \, B b x^{\frac{9}{2}} + \frac{2}{9} \, A c x^{\frac{9}{2}} + \frac{2}{5} \, A b x^{\frac{5}{2}} \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^(1/2),x, algorithm="giac")

[Out]

2/13*B*c*x^(13/2) + 2/9*B*b*x^(9/2) + 2/9*A*c*x^(9/2) + 2/5*A*b*x^(5/2)