3.2.75 \(\int x^{5/2} (b x^2+c x^4) \, dx\)

Optimal. Leaf size=21 \[ \frac {2}{11} b x^{11/2}+\frac {2}{15} c x^{15/2} \]

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Rubi [A]  time = 0.01, antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {14} \begin {gather*} \frac {2}{11} b x^{11/2}+\frac {2}{15} c x^{15/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(b*x^2 + c*x^4),x]

[Out]

(2*b*x^(11/2))/11 + (2*c*x^(15/2))/15

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rubi steps

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

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Mathematica [A]  time = 0.00, size = 21, normalized size = 1.00 \begin {gather*} \frac {2}{11} b x^{11/2}+\frac {2}{15} c x^{15/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(b*x^2 + c*x^4),x]

[Out]

(2*b*x^(11/2))/11 + (2*c*x^(15/2))/15

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IntegrateAlgebraic [A]  time = 0.02, size = 21, normalized size = 1.00 \begin {gather*} \frac {2}{165} \left (15 b x^{11/2}+11 c x^{15/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)*(b*x^2 + c*x^4),x]

[Out]

(2*(15*b*x^(11/2) + 11*c*x^(15/2)))/165

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fricas [A]  time = 1.02, size = 18, normalized size = 0.86 \begin {gather*} \frac {2}{165} \, {\left (11 \, c x^{7} + 15 \, b x^{5}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^4+b*x^2),x, algorithm="fricas")

[Out]

2/165*(11*c*x^7 + 15*b*x^5)*sqrt(x)

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giac [A]  time = 0.15, size = 13, normalized size = 0.62 \begin {gather*} \frac {2}{15} \, c x^{\frac {15}{2}} + \frac {2}{11} \, b x^{\frac {11}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^4+b*x^2),x, algorithm="giac")

[Out]

2/15*c*x^(15/2) + 2/11*b*x^(11/2)

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maple [A]  time = 0.00, size = 16, normalized size = 0.76 \begin {gather*} \frac {2 \left (11 c \,x^{2}+15 b \right ) x^{\frac {11}{2}}}{165} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(c*x^4+b*x^2),x)

[Out]

2/165*x^(11/2)*(11*c*x^2+15*b)

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maxima [A]  time = 1.36, size = 13, normalized size = 0.62 \begin {gather*} \frac {2}{15} \, c x^{\frac {15}{2}} + \frac {2}{11} \, b x^{\frac {11}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(c*x^4+b*x^2),x, algorithm="maxima")

[Out]

2/15*c*x^(15/2) + 2/11*b*x^(11/2)

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mupad [B]  time = 0.03, size = 15, normalized size = 0.71 \begin {gather*} \frac {2\,x^{11/2}\,\left (11\,c\,x^2+15\,b\right )}{165} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(b*x^2 + c*x^4),x)

[Out]

(2*x^(11/2)*(15*b + 11*c*x^2))/165

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sympy [A]  time = 5.57, size = 19, normalized size = 0.90 \begin {gather*} \frac {2 b x^{\frac {11}{2}}}{11} + \frac {2 c x^{\frac {15}{2}}}{15} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)*(c*x**4+b*x**2),x)

[Out]

2*b*x**(11/2)/11 + 2*c*x**(15/2)/15

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