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

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

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Rubi [A]  time = 0.05, antiderivative size = 85, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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]

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]

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*}

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Mathematica [A]  time = 0.05, size = 85, normalized size = 1.00 \begin {gather*} \frac {2}{17} A b^3 x^{17/2}+\frac {2}{21} b^2 x^{21/2} (3 A c+b B)+\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} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.06, size = 97, normalized size = 1.14 \begin {gather*} \frac {2 \left (167475 A b^3 x^{17/2}+406725 A b^2 c x^{21/2}+341649 A b c^2 x^{25/2}+98175 A c^3 x^{29/2}+135575 b^3 B x^{21/2}+341649 b^2 B c x^{25/2}+294525 b B c^2 x^{29/2}+86275 B c^3 x^{33/2}\right )}{2847075} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(167475*A*b^3*x^(17/2) + 135575*b^3*B*x^(21/2) + 406725*A*b^2*c*x^(21/2) + 341649*b^2*B*c*x^(25/2) + 341649
*A*b*c^2*x^(25/2) + 294525*b*B*c^2*x^(29/2) + 98175*A*c^3*x^(29/2) + 86275*B*c^3*x^(33/2)))/2847075

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fricas [A]  time = 0.40, size = 78, normalized size = 0.92 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.15, size = 77, normalized size = 0.91 \begin {gather*} \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 {gather*}

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)

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maple [A]  time = 0.05, size = 80, normalized size = 0.94 \begin {gather*} \frac {2 \left (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}+406725 A \,b^{2} c \,x^{2}+135575 B \,b^{3} x^{2}+167475 A \,b^{3}\right ) x^{\frac {17}{2}}}{2847075} \end {gather*}

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 = 1.38, size = 73, normalized size = 0.86 \begin {gather*} \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 {gather*}

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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mupad [B]  time = 0.04, size = 69, normalized size = 0.81 \begin {gather*} x^{21/2}\,\left (\frac {2\,B\,b^3}{21}+\frac {2\,A\,c\,b^2}{7}\right )+x^{29/2}\,\left (\frac {2\,A\,c^3}{29}+\frac {6\,B\,b\,c^2}{29}\right )+\frac {2\,A\,b^3\,x^{17/2}}{17}+\frac {2\,B\,c^3\,x^{33/2}}{33}+\frac {6\,b\,c\,x^{25/2}\,\left (A\,c+B\,b\right )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 54.01, size = 114, normalized size = 1.34 \begin {gather*} \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 {gather*}

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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