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

Optimal. Leaf size=39 \[ \frac{2}{11} x^{11/2} (A c+b B)+\frac{2}{7} A b x^{7/2}+\frac{2}{15} B c x^{15/2} \]

[Out]

(2*A*b*x^(7/2))/7 + (2*(b*B + A*c)*x^(11/2))/11 + (2*B*c*x^(15/2))/15

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Rubi [A]  time = 0.0213687, 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} \[ \frac{2}{11} x^{11/2} (A c+b B)+\frac{2}{7} A b x^{7/2}+\frac{2}{15} B c x^{15/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*b*x^(7/2))/7 + (2*(b*B + A*c)*x^(11/2))/11 + (2*B*c*x^(15/2))/15

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 \sqrt{x} \left (A+B x^2\right ) \left (b x^2+c x^4\right ) \, dx &=\int x^{5/2} \left (A+B x^2\right ) \left (b+c x^2\right ) \, dx\\ &=\int \left (A b x^{5/2}+(b B+A c) x^{9/2}+B c x^{13/2}\right ) \, dx\\ &=\frac{2}{7} A b x^{7/2}+\frac{2}{11} (b B+A c) x^{11/2}+\frac{2}{15} B c x^{15/2}\\ \end{align*}

Mathematica [A]  time = 0.0138779, size = 33, normalized size = 0.85 \[ \frac{2 x^{7/2} \left (105 x^2 (A c+b B)+165 A b+77 B c x^4\right )}{1155} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(7/2)*(165*A*b + 105*(b*B + A*c)*x^2 + 77*B*c*x^4))/1155

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Maple [A]  time = 0.003, size = 32, normalized size = 0.8 \begin{align*}{\frac{154\,Bc{x}^{4}+210\,A{x}^{2}c+210\,B{x}^{2}b+330\,Ab}{1155}{x}^{{\frac{7}{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]

2/1155*x^(7/2)*(77*B*c*x^4+105*A*c*x^2+105*B*b*x^2+165*A*b)

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Maxima [A]  time = 1.18663, size = 36, normalized size = 0.92 \begin{align*} \frac{2}{15} \, B c x^{\frac{15}{2}} + \frac{2}{11} \,{\left (B b + A c\right )} x^{\frac{11}{2}} + \frac{2}{7} \, A b x^{\frac{7}{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]

2/15*B*c*x^(15/2) + 2/11*(B*b + A*c)*x^(11/2) + 2/7*A*b*x^(7/2)

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Fricas [A]  time = 1.63969, size = 88, normalized size = 2.26 \begin{align*} \frac{2}{1155} \,{\left (77 \, B c x^{7} + 105 \,{\left (B b + A c\right )} x^{5} + 165 \, A b x^{3}\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]

2/1155*(77*B*c*x^7 + 105*(B*b + A*c)*x^5 + 165*A*b*x^3)*sqrt(x)

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Sympy [A]  time = 3.65863, size = 37, normalized size = 0.95 \begin{align*} \frac{2 A b x^{\frac{7}{2}}}{7} + \frac{2 B c x^{\frac{15}{2}}}{15} + \frac{2 x^{\frac{11}{2}} \left (A c + B b\right )}{11} \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**(7/2)/7 + 2*B*c*x**(15/2)/15 + 2*x**(11/2)*(A*c + B*b)/11

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Giac [A]  time = 1.14034, size = 39, normalized size = 1. \begin{align*} \frac{2}{15} \, B c x^{\frac{15}{2}} + \frac{2}{11} \, B b x^{\frac{11}{2}} + \frac{2}{11} \, A c x^{\frac{11}{2}} + \frac{2}{7} \, A b x^{\frac{7}{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]

2/15*B*c*x^(15/2) + 2/11*B*b*x^(11/2) + 2/11*A*c*x^(11/2) + 2/7*A*b*x^(7/2)