3.176 \(\int x^{5/2} (A+B x^2) (b x^2+c x^4)^3 \, dx\)

Optimal. Leaf size=85 \[ \frac{2}{23} b^2 x^{23/2} (3 A c+b B)+\frac{2}{19} A b^3 x^{19/2}+\frac{2}{31} c^2 x^{31/2} (A c+3 b B)+\frac{2}{9} b c x^{27/2} (A c+b B)+\frac{2}{35} B c^3 x^{35/2} \]

[Out]

(2*A*b^3*x^(19/2))/19 + (2*b^2*(b*B + 3*A*c)*x^(23/2))/23 + (2*b*c*(b*B + A*c)*x^(27/2))/9 + (2*c^2*(3*b*B + A
*c)*x^(31/2))/31 + (2*B*c^3*x^(35/2))/35

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Rubi [A]  time = 0.0497119, 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} \[ \frac{2}{23} b^2 x^{23/2} (3 A c+b B)+\frac{2}{19} A b^3 x^{19/2}+\frac{2}{31} c^2 x^{31/2} (A c+3 b B)+\frac{2}{9} b c x^{27/2} (A c+b B)+\frac{2}{35} B c^3 x^{35/2} \]

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(A + B*x^2)*(b*x^2 + c*x^4)^3,x]

[Out]

(2*A*b^3*x^(19/2))/19 + (2*b^2*(b*B + 3*A*c)*x^(23/2))/23 + (2*b*c*(b*B + A*c)*x^(27/2))/9 + (2*c^2*(3*b*B + A
*c)*x^(31/2))/31 + (2*B*c^3*x^(35/2))/35

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 x^{5/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx &=\int x^{17/2} \left (A+B x^2\right ) \left (b+c x^2\right )^3 \, dx\\ &=\int \left (A b^3 x^{17/2}+b^2 (b B+3 A c) x^{21/2}+3 b c (b B+A c) x^{25/2}+c^2 (3 b B+A c) x^{29/2}+B c^3 x^{33/2}\right ) \, dx\\ &=\frac{2}{19} A b^3 x^{19/2}+\frac{2}{23} b^2 (b B+3 A c) x^{23/2}+\frac{2}{9} b c (b B+A c) x^{27/2}+\frac{2}{31} c^2 (3 b B+A c) x^{31/2}+\frac{2}{35} B c^3 x^{35/2}\\ \end{align*}

Mathematica [A]  time = 0.0431332, size = 85, normalized size = 1. \[ \frac{2}{23} b^2 x^{23/2} (3 A c+b B)+\frac{2}{19} A b^3 x^{19/2}+\frac{2}{31} c^2 x^{31/2} (A c+3 b B)+\frac{2}{9} b c x^{27/2} (A c+b B)+\frac{2}{35} B c^3 x^{35/2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(A + B*x^2)*(b*x^2 + c*x^4)^3,x]

[Out]

(2*A*b^3*x^(19/2))/19 + (2*b^2*(b*B + 3*A*c)*x^(23/2))/23 + (2*b*c*(b*B + A*c)*x^(27/2))/9 + (2*c^2*(3*b*B + A
*c)*x^(31/2))/31 + (2*B*c^3*x^(35/2))/35

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Maple [A]  time = 0.007, size = 80, normalized size = 0.9 \begin{align*}{\frac{243846\,B{c}^{3}{x}^{8}+275310\,A{c}^{3}{x}^{6}+825930\,B{x}^{6}b{c}^{2}+948290\,Ab{c}^{2}{x}^{4}+948290\,B{x}^{4}{b}^{2}c+1113210\,A{b}^{2}c{x}^{2}+371070\,B{x}^{2}{b}^{3}+449190\,A{b}^{3}}{4267305}{x}^{{\frac{19}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x)

[Out]

2/4267305*x^(19/2)*(121923*B*c^3*x^8+137655*A*c^3*x^6+412965*B*b*c^2*x^6+474145*A*b*c^2*x^4+474145*B*b^2*c*x^4
+556605*A*b^2*c*x^2+185535*B*b^3*x^2+224595*A*b^3)

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Maxima [A]  time = 1.1156, size = 99, normalized size = 1.16 \begin{align*} \frac{2}{35} \, B c^{3} x^{\frac{35}{2}} + \frac{2}{31} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac{31}{2}} + \frac{2}{9} \,{\left (B b^{2} c + A b c^{2}\right )} x^{\frac{27}{2}} + \frac{2}{19} \, A b^{3} x^{\frac{19}{2}} + \frac{2}{23} \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{\frac{23}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x, algorithm="maxima")

[Out]

2/35*B*c^3*x^(35/2) + 2/31*(3*B*b*c^2 + A*c^3)*x^(31/2) + 2/9*(B*b^2*c + A*b*c^2)*x^(27/2) + 2/19*A*b^3*x^(19/
2) + 2/23*(B*b^3 + 3*A*b^2*c)*x^(23/2)

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Fricas [A]  time = 1.6826, size = 216, normalized size = 2.54 \begin{align*} \frac{2}{4267305} \,{\left (121923 \, B c^{3} x^{17} + 137655 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{15} + 474145 \,{\left (B b^{2} c + A b c^{2}\right )} x^{13} + 224595 \, A b^{3} x^{9} + 185535 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{11}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x, algorithm="fricas")

[Out]

2/4267305*(121923*B*c^3*x^17 + 137655*(3*B*b*c^2 + A*c^3)*x^15 + 474145*(B*b^2*c + A*b*c^2)*x^13 + 224595*A*b^
3*x^9 + 185535*(B*b^3 + 3*A*b^2*c)*x^11)*sqrt(x)

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Sympy [A]  time = 165.782, size = 114, normalized size = 1.34 \begin{align*} \frac{2 A b^{3} x^{\frac{19}{2}}}{19} + \frac{6 A b^{2} c x^{\frac{23}{2}}}{23} + \frac{2 A b c^{2} x^{\frac{27}{2}}}{9} + \frac{2 A c^{3} x^{\frac{31}{2}}}{31} + \frac{2 B b^{3} x^{\frac{23}{2}}}{23} + \frac{2 B b^{2} c x^{\frac{27}{2}}}{9} + \frac{6 B b c^{2} x^{\frac{31}{2}}}{31} + \frac{2 B c^{3} x^{\frac{35}{2}}}{35} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(B*x**2+A)*(c*x**4+b*x**2)**3,x)

[Out]

2*A*b**3*x**(19/2)/19 + 6*A*b**2*c*x**(23/2)/23 + 2*A*b*c**2*x**(27/2)/9 + 2*A*c**3*x**(31/2)/31 + 2*B*b**3*x*
*(23/2)/23 + 2*B*b**2*c*x**(27/2)/9 + 6*B*b*c**2*x**(31/2)/31 + 2*B*c**3*x**(35/2)/35

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Giac [A]  time = 1.12029, size = 104, normalized size = 1.22 \begin{align*} \frac{2}{35} \, B c^{3} x^{\frac{35}{2}} + \frac{6}{31} \, B b c^{2} x^{\frac{31}{2}} + \frac{2}{31} \, A c^{3} x^{\frac{31}{2}} + \frac{2}{9} \, B b^{2} c x^{\frac{27}{2}} + \frac{2}{9} \, A b c^{2} x^{\frac{27}{2}} + \frac{2}{23} \, B b^{3} x^{\frac{23}{2}} + \frac{6}{23} \, A b^{2} c x^{\frac{23}{2}} + \frac{2}{19} \, A b^{3} x^{\frac{19}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x^2+A)*(c*x^4+b*x^2)^3,x, algorithm="giac")

[Out]

2/35*B*c^3*x^(35/2) + 6/31*B*b*c^2*x^(31/2) + 2/31*A*c^3*x^(31/2) + 2/9*B*b^2*c*x^(27/2) + 2/9*A*b*c^2*x^(27/2
) + 2/23*B*b^3*x^(23/2) + 6/23*A*b^2*c*x^(23/2) + 2/19*A*b^3*x^(19/2)