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

Optimal. Leaf size=51 \[ \frac {2}{11} b^3 x^{11/2}+\frac {2}{5} b^2 c x^{15/2}+\frac {6}{19} b c^2 x^{19/2}+\frac {2}{23} c^3 x^{23/2} \]

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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}{5} b^2 c x^{15/2}+\frac {2}{11} b^3 x^{11/2}+\frac {6}{19} b c^2 x^{19/2}+\frac {2}{23} c^3 x^{23/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^3/x^(3/2),x]

[Out]

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

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 \frac {\left (b x^2+c x^4\right )^3}{x^{3/2}} \, dx &=\int x^{9/2} \left (b+c x^2\right )^3 \, dx\\ &=\int \left (b^3 x^{9/2}+3 b^2 c x^{13/2}+3 b c^2 x^{17/2}+c^3 x^{21/2}\right ) \, dx\\ &=\frac {2}{11} b^3 x^{11/2}+\frac {2}{5} b^2 c x^{15/2}+\frac {6}{19} b c^2 x^{19/2}+\frac {2}{23} c^3 x^{23/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 41, normalized size = 0.80 \begin {gather*} \frac {2 x^{11/2} \left (2185 b^3+4807 b^2 c x^2+3795 b c^2 x^4+1045 c^3 x^6\right )}{24035} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(11/2)*(2185*b^3 + 4807*b^2*c*x^2 + 3795*b*c^2*x^4 + 1045*c^3*x^6))/24035

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IntegrateAlgebraic [A]  time = 0.03, size = 47, normalized size = 0.92 \begin {gather*} \frac {2 \left (2185 b^3 x^{11/2}+4807 b^2 c x^{15/2}+3795 b c^2 x^{19/2}+1045 c^3 x^{23/2}\right )}{24035} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(2185*b^3*x^(11/2) + 4807*b^2*c*x^(15/2) + 3795*b*c^2*x^(19/2) + 1045*c^3*x^(23/2)))/24035

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fricas [A]  time = 0.69, size = 40, normalized size = 0.78 \begin {gather*} \frac {2}{24035} \, {\left (1045 \, c^{3} x^{11} + 3795 \, b c^{2} x^{9} + 4807 \, b^{2} c x^{7} + 2185 \, b^{3} x^{5}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/24035*(1045*c^3*x^11 + 3795*b*c^2*x^9 + 4807*b^2*c*x^7 + 2185*b^3*x^5)*sqrt(x)

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giac [A]  time = 0.15, size = 35, normalized size = 0.69 \begin {gather*} \frac {2}{23} \, c^{3} x^{\frac {23}{2}} + \frac {6}{19} \, b c^{2} x^{\frac {19}{2}} + \frac {2}{5} \, b^{2} c x^{\frac {15}{2}} + \frac {2}{11} \, b^{3} x^{\frac {11}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 38, normalized size = 0.75 \begin {gather*} \frac {2 \left (1045 c^{3} x^{6}+3795 b \,c^{2} x^{4}+4807 b^{2} c \,x^{2}+2185 b^{3}\right ) x^{\frac {11}{2}}}{24035} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/24035*x^(11/2)*(1045*c^3*x^6+3795*b*c^2*x^4+4807*b^2*c*x^2+2185*b^3)

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maxima [A]  time = 1.31, size = 35, normalized size = 0.69 \begin {gather*} \frac {2}{23} \, c^{3} x^{\frac {23}{2}} + \frac {6}{19} \, b c^{2} x^{\frac {19}{2}} + \frac {2}{5} \, b^{2} c x^{\frac {15}{2}} + \frac {2}{11} \, b^{3} x^{\frac {11}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 35, normalized size = 0.69 \begin {gather*} \frac {2\,b^3\,x^{11/2}}{11}+\frac {2\,c^3\,x^{23/2}}{23}+\frac {2\,b^2\,c\,x^{15/2}}{5}+\frac {6\,b\,c^2\,x^{19/2}}{19} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^3/x^(3/2),x)

[Out]

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

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sympy [A]  time = 19.94, size = 49, normalized size = 0.96 \begin {gather*} \frac {2 b^{3} x^{\frac {11}{2}}}{11} + \frac {2 b^{2} c x^{\frac {15}{2}}}{5} + \frac {6 b c^{2} x^{\frac {19}{2}}}{19} + \frac {2 c^{3} x^{\frac {23}{2}}}{23} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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