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

Optimal. Leaf size=85 \[ \frac{2}{21} b^2 x^{21/2} (3 A c+b B)+\frac{2}{17} A b^3 x^{17/2}+\frac{2}{29} c^2 x^{29/2} (A c+3 b B)+\frac{6}{25} b c x^{25/2} (A c+b B)+\frac{2}{33} B c^3 x^{33/2} \]

[Out]

(2*A*b^3*x^(17/2))/17 + (2*b^2*(b*B + 3*A*c)*x^(21/2))/21 + (6*b*c*(b*B + A*c)*x^(25/2))/25 + (2*c^2*(3*b*B +
A*c)*x^(29/2))/29 + (2*B*c^3*x^(33/2))/33

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Rubi [A]  time = 0.0501958, 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} \[ \frac{2}{21} b^2 x^{21/2} (3 A c+b B)+\frac{2}{17} A b^3 x^{17/2}+\frac{2}{29} c^2 x^{29/2} (A c+3 b B)+\frac{6}{25} b c x^{25/2} (A c+b B)+\frac{2}{33} B c^3 x^{33/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*b^3*x^(17/2))/17 + (2*b^2*(b*B + 3*A*c)*x^(21/2))/21 + (6*b*c*(b*B + A*c)*x^(25/2))/25 + (2*c^2*(3*b*B +
A*c)*x^(29/2))/29 + (2*B*c^3*x^(33/2))/33

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^{3/2} \left (A+B x^2\right ) \left (b x^2+c x^4\right )^3 \, dx &=\int x^{15/2} \left (A+B x^2\right ) \left (b+c x^2\right )^3 \, dx\\ &=\int \left (A b^3 x^{15/2}+b^2 (b B+3 A c) x^{19/2}+3 b c (b B+A c) x^{23/2}+c^2 (3 b B+A c) x^{27/2}+B c^3 x^{31/2}\right ) \, dx\\ &=\frac{2}{17} A b^3 x^{17/2}+\frac{2}{21} b^2 (b B+3 A c) x^{21/2}+\frac{6}{25} b c (b B+A c) x^{25/2}+\frac{2}{29} c^2 (3 b B+A c) x^{29/2}+\frac{2}{33} B c^3 x^{33/2}\\ \end{align*}

Mathematica [A]  time = 0.0409699, size = 85, normalized size = 1. \[ \frac{2}{21} b^2 x^{21/2} (3 A c+b B)+\frac{2}{17} A b^3 x^{17/2}+\frac{2}{29} c^2 x^{29/2} (A c+3 b B)+\frac{6}{25} b c x^{25/2} (A c+b B)+\frac{2}{33} B c^3 x^{33/2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*A*b^3*x^(17/2))/17 + (2*b^2*(b*B + 3*A*c)*x^(21/2))/21 + (6*b*c*(b*B + A*c)*x^(25/2))/25 + (2*c^2*(3*b*B +
A*c)*x^(29/2))/29 + (2*B*c^3*x^(33/2))/33

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Maple [A]  time = 0.005, size = 80, normalized size = 0.9 \begin{align*}{\frac{172550\,B{c}^{3}{x}^{8}+196350\,A{c}^{3}{x}^{6}+589050\,B{x}^{6}b{c}^{2}+683298\,Ab{c}^{2}{x}^{4}+683298\,B{x}^{4}{b}^{2}c+813450\,A{b}^{2}c{x}^{2}+271150\,B{x}^{2}{b}^{3}+334950\,A{b}^{3}}{2847075}{x}^{{\frac{17}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/2847075*x^(17/2)*(86275*B*c^3*x^8+98175*A*c^3*x^6+294525*B*b*c^2*x^6+341649*A*b*c^2*x^4+341649*B*b^2*c*x^4+4
06725*A*b^2*c*x^2+135575*B*b^3*x^2+167475*A*b^3)

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Maxima [A]  time = 1.19563, size = 99, normalized size = 1.16 \begin{align*} \frac{2}{33} \, B c^{3} x^{\frac{33}{2}} + \frac{2}{29} \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{\frac{29}{2}} + \frac{6}{25} \,{\left (B b^{2} c + A b c^{2}\right )} x^{\frac{25}{2}} + \frac{2}{17} \, A b^{3} x^{\frac{17}{2}} + \frac{2}{21} \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{\frac{21}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/33*B*c^3*x^(33/2) + 2/29*(3*B*b*c^2 + A*c^3)*x^(29/2) + 6/25*(B*b^2*c + A*b*c^2)*x^(25/2) + 2/17*A*b^3*x^(17
/2) + 2/21*(B*b^3 + 3*A*b^2*c)*x^(21/2)

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Fricas [A]  time = 1.84023, size = 213, normalized size = 2.51 \begin{align*} \frac{2}{2847075} \,{\left (86275 \, B c^{3} x^{16} + 98175 \,{\left (3 \, B b c^{2} + A c^{3}\right )} x^{14} + 341649 \,{\left (B b^{2} c + A b c^{2}\right )} x^{12} + 167475 \, A b^{3} x^{8} + 135575 \,{\left (B b^{3} + 3 \, A b^{2} c\right )} x^{10}\right )} \sqrt{x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/2847075*(86275*B*c^3*x^16 + 98175*(3*B*b*c^2 + A*c^3)*x^14 + 341649*(B*b^2*c + A*b*c^2)*x^12 + 167475*A*b^3*
x^8 + 135575*(B*b^3 + 3*A*b^2*c)*x^10)*sqrt(x)

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Sympy [A]  time = 113.16, size = 114, normalized size = 1.34 \begin{align*} \frac{2 A b^{3} x^{\frac{17}{2}}}{17} + \frac{2 A b^{2} c x^{\frac{21}{2}}}{7} + \frac{6 A b c^{2} x^{\frac{25}{2}}}{25} + \frac{2 A c^{3} x^{\frac{29}{2}}}{29} + \frac{2 B b^{3} x^{\frac{21}{2}}}{21} + \frac{6 B b^{2} c x^{\frac{25}{2}}}{25} + \frac{6 B b c^{2} x^{\frac{29}{2}}}{29} + \frac{2 B c^{3} x^{\frac{33}{2}}}{33} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*A*b**3*x**(17/2)/17 + 2*A*b**2*c*x**(21/2)/7 + 6*A*b*c**2*x**(25/2)/25 + 2*A*c**3*x**(29/2)/29 + 2*B*b**3*x*
*(21/2)/21 + 6*B*b**2*c*x**(25/2)/25 + 6*B*b*c**2*x**(29/2)/29 + 2*B*c**3*x**(33/2)/33

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Giac [A]  time = 1.14716, size = 104, normalized size = 1.22 \begin{align*} \frac{2}{33} \, B c^{3} x^{\frac{33}{2}} + \frac{6}{29} \, B b c^{2} x^{\frac{29}{2}} + \frac{2}{29} \, A c^{3} x^{\frac{29}{2}} + \frac{6}{25} \, B b^{2} c x^{\frac{25}{2}} + \frac{6}{25} \, A b c^{2} x^{\frac{25}{2}} + \frac{2}{21} \, B b^{3} x^{\frac{21}{2}} + \frac{2}{7} \, A b^{2} c x^{\frac{21}{2}} + \frac{2}{17} \, A b^{3} x^{\frac{17}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/33*B*c^3*x^(33/2) + 6/29*B*b*c^2*x^(29/2) + 2/29*A*c^3*x^(29/2) + 6/25*B*b^2*c*x^(25/2) + 6/25*A*b*c^2*x^(25
/2) + 2/21*B*b^3*x^(21/2) + 2/7*A*b^2*c*x^(21/2) + 2/17*A*b^3*x^(17/2)