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

Optimal. Leaf size=51 \[ \frac {2}{17} b^3 x^{17/2}+\frac {2}{7} b^2 c x^{21/2}+\frac {6}{25} b c^2 x^{25/2}+\frac {2}{29} c^3 x^{29/2} \]

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Rubi [A]  time = 0.02, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1584, 270} \begin {gather*} \frac {2}{7} b^2 c x^{21/2}+\frac {2}{17} b^3 x^{17/2}+\frac {6}{25} b c^2 x^{25/2}+\frac {2}{29} c^3 x^{29/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(b*x^2 + c*x^4)^3,x]

[Out]

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

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 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 (b x^2+c x^4\right )^3 \, dx &=\int x^{15/2} \left (b+c x^2\right )^3 \, dx\\ &=\int \left (b^3 x^{15/2}+3 b^2 c x^{19/2}+3 b c^2 x^{23/2}+c^3 x^{27/2}\right ) \, dx\\ &=\frac {2}{17} b^3 x^{17/2}+\frac {2}{7} b^2 c x^{21/2}+\frac {6}{25} b c^2 x^{25/2}+\frac {2}{29} c^3 x^{29/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 51, normalized size = 1.00 \begin {gather*} \frac {2}{17} b^3 x^{17/2}+\frac {2}{7} b^2 c x^{21/2}+\frac {6}{25} b c^2 x^{25/2}+\frac {2}{29} c^3 x^{29/2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 0.03, size = 47, normalized size = 0.92 \begin {gather*} \frac {2 \left (5075 b^3 x^{17/2}+12325 b^2 c x^{21/2}+10353 b c^2 x^{25/2}+2975 c^3 x^{29/2}\right )}{86275} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5075*b^3*x^(17/2) + 12325*b^2*c*x^(21/2) + 10353*b*c^2*x^(25/2) + 2975*c^3*x^(29/2)))/86275

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fricas [A]  time = 0.94, size = 40, normalized size = 0.78 \begin {gather*} \frac {2}{86275} \, {\left (2975 \, c^{3} x^{14} + 10353 \, b c^{2} x^{12} + 12325 \, b^{2} c x^{10} + 5075 \, b^{3} x^{8}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/86275*(2975*c^3*x^14 + 10353*b*c^2*x^12 + 12325*b^2*c*x^10 + 5075*b^3*x^8)*sqrt(x)

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giac [A]  time = 0.15, size = 35, normalized size = 0.69 \begin {gather*} \frac {2}{29} \, c^{3} x^{\frac {29}{2}} + \frac {6}{25} \, b c^{2} x^{\frac {25}{2}} + \frac {2}{7} \, b^{2} c x^{\frac {21}{2}} + \frac {2}{17} \, b^{3} x^{\frac {17}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 38, normalized size = 0.75 \begin {gather*} \frac {2 \left (2975 c^{3} x^{6}+10353 b \,c^{2} x^{4}+12325 b^{2} c \,x^{2}+5075 b^{3}\right ) x^{\frac {17}{2}}}{86275} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/86275*x^(17/2)*(2975*c^3*x^6+10353*b*c^2*x^4+12325*b^2*c*x^2+5075*b^3)

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maxima [A]  time = 1.31, size = 35, normalized size = 0.69 \begin {gather*} \frac {2}{29} \, c^{3} x^{\frac {29}{2}} + \frac {6}{25} \, b c^{2} x^{\frac {25}{2}} + \frac {2}{7} \, b^{2} c x^{\frac {21}{2}} + \frac {2}{17} \, b^{3} x^{\frac {17}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 35, normalized size = 0.69 \begin {gather*} \frac {2\,b^3\,x^{17/2}}{17}+\frac {2\,c^3\,x^{29/2}}{29}+\frac {2\,b^2\,c\,x^{21/2}}{7}+\frac {6\,b\,c^2\,x^{25/2}}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(b*x^2 + c*x^4)^3,x)

[Out]

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

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sympy [A]  time = 33.34, size = 49, normalized size = 0.96 \begin {gather*} \frac {2 b^{3} x^{\frac {17}{2}}}{17} + \frac {2 b^{2} c x^{\frac {21}{2}}}{7} + \frac {6 b c^{2} x^{\frac {25}{2}}}{25} + \frac {2 c^{3} x^{\frac {29}{2}}}{29} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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