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

Optimal. Leaf size=36 \[ \frac {2}{7} b^2 x^{7/2}+\frac {4}{11} b c x^{11/2}+\frac {2}{15} c^2 x^{15/2} \]

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Rubi [A]  time = 0.02, antiderivative size = 36, 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 x^{7/2}+\frac {4}{11} b c x^{11/2}+\frac {2}{15} c^2 x^{15/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^2/x^(3/2),x]

[Out]

(2*b^2*x^(7/2))/7 + (4*b*c*x^(11/2))/11 + (2*c^2*x^(15/2))/15

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 )^2}{x^{3/2}} \, dx &=\int x^{5/2} \left (b+c x^2\right )^2 \, dx\\ &=\int \left (b^2 x^{5/2}+2 b c x^{9/2}+c^2 x^{13/2}\right ) \, dx\\ &=\frac {2}{7} b^2 x^{7/2}+\frac {4}{11} b c x^{11/2}+\frac {2}{15} c^2 x^{15/2}\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 30, normalized size = 0.83 \begin {gather*} \frac {2 x^{7/2} \left (165 b^2+210 b c x^2+77 c^2 x^4\right )}{1155} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(7/2)*(165*b^2 + 210*b*c*x^2 + 77*c^2*x^4))/1155

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IntegrateAlgebraic [A]  time = 0.02, size = 34, normalized size = 0.94 \begin {gather*} \frac {2 \left (165 b^2 x^{7/2}+210 b c x^{11/2}+77 c^2 x^{15/2}\right )}{1155} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(165*b^2*x^(7/2) + 210*b*c*x^(11/2) + 77*c^2*x^(15/2)))/1155

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fricas [A]  time = 1.59, size = 29, normalized size = 0.81 \begin {gather*} \frac {2}{1155} \, {\left (77 \, c^{2} x^{7} + 210 \, b c x^{5} + 165 \, b^{2} x^{3}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155*(77*c^2*x^7 + 210*b*c*x^5 + 165*b^2*x^3)*sqrt(x)

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giac [A]  time = 0.15, size = 24, normalized size = 0.67 \begin {gather*} \frac {2}{15} \, c^{2} x^{\frac {15}{2}} + \frac {4}{11} \, b c x^{\frac {11}{2}} + \frac {2}{7} \, b^{2} x^{\frac {7}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*c^2*x^(15/2) + 4/11*b*c*x^(11/2) + 2/7*b^2*x^(7/2)

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maple [A]  time = 0.01, size = 27, normalized size = 0.75 \begin {gather*} \frac {2 \left (77 c^{2} x^{4}+210 b c \,x^{2}+165 b^{2}\right ) x^{\frac {7}{2}}}{1155} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155*x^(7/2)*(77*c^2*x^4+210*b*c*x^2+165*b^2)

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maxima [A]  time = 1.33, size = 24, normalized size = 0.67 \begin {gather*} \frac {2}{15} \, c^{2} x^{\frac {15}{2}} + \frac {4}{11} \, b c x^{\frac {11}{2}} + \frac {2}{7} \, b^{2} x^{\frac {7}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*c^2*x^(15/2) + 4/11*b*c*x^(11/2) + 2/7*b^2*x^(7/2)

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mupad [B]  time = 4.44, size = 26, normalized size = 0.72 \begin {gather*} \frac {2\,x^{7/2}\,\left (165\,b^2+210\,b\,c\,x^2+77\,c^2\,x^4\right )}{1155} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^2/x^(3/2),x)

[Out]

(2*x^(7/2)*(165*b^2 + 77*c^2*x^4 + 210*b*c*x^2))/1155

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sympy [A]  time = 5.22, size = 34, normalized size = 0.94 \begin {gather*} \frac {2 b^{2} x^{\frac {7}{2}}}{7} + \frac {4 b c x^{\frac {11}{2}}}{11} + \frac {2 c^{2} x^{\frac {15}{2}}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b**2*x**(7/2)/7 + 4*b*c*x**(11/2)/11 + 2*c**2*x**(15/2)/15

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