3.180 \(\int \frac{(A+B x^2) (b x^2+c x^4)^3}{x^{3/2}} \, dx\)

Optimal. Leaf size=85 \[ \frac{2}{15} b^2 x^{15/2} (3 A c+b B)+\frac{2}{11} A b^3 x^{11/2}+\frac{2}{23} c^2 x^{23/2} (A c+3 b B)+\frac{6}{19} b c x^{19/2} (A c+b B)+\frac{2}{27} B c^3 x^{27/2} \]

[Out]

(2*A*b^3*x^(11/2))/11 + (2*b^2*(b*B + 3*A*c)*x^(15/2))/15 + (6*b*c*(b*B + A*c)*x^(19/2))/19 + (2*c^2*(3*b*B +
A*c)*x^(23/2))/23 + (2*B*c^3*x^(27/2))/27

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Rubi [A]  time = 0.0496143, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {1584, 448} \[ \frac{2}{15} b^2 x^{15/2} (3 A c+b B)+\frac{2}{11} A b^3 x^{11/2}+\frac{2}{23} c^2 x^{23/2} (A c+3 b B)+\frac{6}{19} b c x^{19/2} (A c+b B)+\frac{2}{27} B c^3 x^{27/2} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^(3/2),x]

[Out]

(2*A*b^3*x^(11/2))/11 + (2*b^2*(b*B + 3*A*c)*x^(15/2))/15 + (6*b*c*(b*B + A*c)*x^(19/2))/19 + (2*c^2*(3*b*B +
A*c)*x^(23/2))/23 + (2*B*c^3*x^(27/2))/27

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 )^3}{x^{3/2}} \, dx &=\int x^{9/2} \left (A+B x^2\right ) \left (b+c x^2\right )^3 \, dx\\ &=\int \left (A b^3 x^{9/2}+b^2 (b B+3 A c) x^{13/2}+3 b c (b B+A c) x^{17/2}+c^2 (3 b B+A c) x^{21/2}+B c^3 x^{25/2}\right ) \, dx\\ &=\frac{2}{11} A b^3 x^{11/2}+\frac{2}{15} b^2 (b B+3 A c) x^{15/2}+\frac{6}{19} b c (b B+A c) x^{19/2}+\frac{2}{23} c^2 (3 b B+A c) x^{23/2}+\frac{2}{27} B c^3 x^{27/2}\\ \end{align*}

Mathematica [A]  time = 0.0367638, size = 85, normalized size = 1. \[ \frac{2}{15} b^2 x^{15/2} (3 A c+b B)+\frac{2}{11} A b^3 x^{11/2}+\frac{2}{23} c^2 x^{23/2} (A c+3 b B)+\frac{6}{19} b c x^{19/2} (A c+b B)+\frac{2}{27} B c^3 x^{27/2} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x^2)*(b*x^2 + c*x^4)^3)/x^(3/2),x]

[Out]

(2*A*b^3*x^(11/2))/11 + (2*b^2*(b*B + 3*A*c)*x^(15/2))/15 + (6*b*c*(b*B + A*c)*x^(19/2))/19 + (2*c^2*(3*b*B +
A*c)*x^(23/2))/23 + (2*B*c^3*x^(27/2))/27

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Maple [A]  time = 0.005, size = 80, normalized size = 0.9 \begin{align*}{\frac{48070\,B{c}^{3}{x}^{8}+56430\,A{c}^{3}{x}^{6}+169290\,B{x}^{6}b{c}^{2}+204930\,Ab{c}^{2}{x}^{4}+204930\,B{x}^{4}{b}^{2}c+259578\,A{b}^{2}c{x}^{2}+86526\,B{x}^{2}{b}^{3}+117990\,A{b}^{3}}{648945}{x}^{{\frac{11}{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)^3/x^(3/2),x)

[Out]

2/648945*x^(11/2)*(24035*B*c^3*x^8+28215*A*c^3*x^6+84645*B*b*c^2*x^6+102465*A*b*c^2*x^4+102465*B*b^2*c*x^4+129
789*A*b^2*c*x^2+43263*B*b^3*x^2+58995*A*b^3)

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Maxima [A]  time = 1.10736, size = 99, normalized size = 1.16 \begin{align*} \frac{2}{27} \, B c^{3} x^{\frac{27}{2}} + \frac{2}{23} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac{23}{2}} + \frac{6}{19} \,{\left (B b^{2} c + A b c^{2}\right )} x^{\frac{19}{2}} + \frac{2}{11} \, A b^{3} x^{\frac{11}{2}} + \frac{2}{15} \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{\frac{15}{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)^3/x^(3/2),x, algorithm="maxima")

[Out]

2/27*B*c^3*x^(27/2) + 2/23*(3*B*b*c^2 + A*c^3)*x^(23/2) + 6/19*(B*b^2*c + A*b*c^2)*x^(19/2) + 2/11*A*b^3*x^(11
/2) + 2/15*(B*b^3 + 3*A*b^2*c)*x^(15/2)

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Fricas [A]  time = 1.85796, size = 207, normalized size = 2.44 \begin{align*} \frac{2}{648945} \,{\left (24035 \, B c^{3} x^{13} + 28215 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{11} + 102465 \,{\left (B b^{2} c + A b c^{2}\right )} x^{9} + 58995 \, A b^{3} x^{5} + 43263 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{7}\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)^3/x^(3/2),x, algorithm="fricas")

[Out]

2/648945*(24035*B*c^3*x^13 + 28215*(3*B*b*c^2 + A*c^3)*x^11 + 102465*(B*b^2*c + A*b*c^2)*x^9 + 58995*A*b^3*x^5
 + 43263*(B*b^3 + 3*A*b^2*c)*x^7)*sqrt(x)

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Sympy [A]  time = 59.9208, size = 114, normalized size = 1.34 \begin{align*} \frac{2 A b^{3} x^{\frac{11}{2}}}{11} + \frac{2 A b^{2} c x^{\frac{15}{2}}}{5} + \frac{6 A b c^{2} x^{\frac{19}{2}}}{19} + \frac{2 A c^{3} x^{\frac{23}{2}}}{23} + \frac{2 B b^{3} x^{\frac{15}{2}}}{15} + \frac{6 B b^{2} c x^{\frac{19}{2}}}{19} + \frac{6 B b c^{2} x^{\frac{23}{2}}}{23} + \frac{2 B c^{3} x^{\frac{27}{2}}}{27} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)*(c*x**4+b*x**2)**3/x**(3/2),x)

[Out]

2*A*b**3*x**(11/2)/11 + 2*A*b**2*c*x**(15/2)/5 + 6*A*b*c**2*x**(19/2)/19 + 2*A*c**3*x**(23/2)/23 + 2*B*b**3*x*
*(15/2)/15 + 6*B*b**2*c*x**(19/2)/19 + 6*B*b*c**2*x**(23/2)/23 + 2*B*c**3*x**(27/2)/27

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Giac [A]  time = 1.13949, size = 104, normalized size = 1.22 \begin{align*} \frac{2}{27} \, B c^{3} x^{\frac{27}{2}} + \frac{6}{23} \, B b c^{2} x^{\frac{23}{2}} + \frac{2}{23} \, A c^{3} x^{\frac{23}{2}} + \frac{6}{19} \, B b^{2} c x^{\frac{19}{2}} + \frac{6}{19} \, A b c^{2} x^{\frac{19}{2}} + \frac{2}{15} \, B b^{3} x^{\frac{15}{2}} + \frac{2}{5} \, A b^{2} c x^{\frac{15}{2}} + \frac{2}{11} \, A b^{3} x^{\frac{11}{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)^3/x^(3/2),x, algorithm="giac")

[Out]

2/27*B*c^3*x^(27/2) + 6/23*B*b*c^2*x^(23/2) + 2/23*A*c^3*x^(23/2) + 6/19*B*b^2*c*x^(19/2) + 6/19*A*b*c^2*x^(19
/2) + 2/15*B*b^3*x^(15/2) + 2/5*A*b^2*c*x^(15/2) + 2/11*A*b^3*x^(11/2)