3.2.85 \(\int \sqrt {x} (b x^2+c x^4)^2 \, dx\)

Optimal. Leaf size=36 \[ \frac {2}{11} b^2 x^{11/2}+\frac {4}{15} b c x^{15/2}+\frac {2}{19} c^2 x^{19/2} \]

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Rubi [A]  time = 0.01, 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}{11} b^2 x^{11/2}+\frac {4}{15} b c x^{15/2}+\frac {2}{19} c^2 x^{19/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(b*x^2 + c*x^4)^2,x]

[Out]

(2*b^2*x^(11/2))/11 + (4*b*c*x^(15/2))/15 + (2*c^2*x^(19/2))/19

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

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Mathematica [A]  time = 0.01, size = 30, normalized size = 0.83 \begin {gather*} \frac {2 x^{11/2} \left (285 b^2+418 b c x^2+165 c^2 x^4\right )}{3135} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(11/2)*(285*b^2 + 418*b*c*x^2 + 165*c^2*x^4))/3135

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IntegrateAlgebraic [A]  time = 0.02, size = 34, normalized size = 0.94 \begin {gather*} \frac {2 \left (285 b^2 x^{11/2}+418 b c x^{15/2}+165 c^2 x^{19/2}\right )}{3135} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(285*b^2*x^(11/2) + 418*b*c*x^(15/2) + 165*c^2*x^(19/2)))/3135

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fricas [A]  time = 1.51, size = 29, normalized size = 0.81 \begin {gather*} \frac {2}{3135} \, {\left (165 \, c^{2} x^{9} + 418 \, b c x^{7} + 285 \, b^{2} x^{5}\right )} \sqrt {x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)*(c*x^4+b*x^2)^2,x, algorithm="fricas")

[Out]

2/3135*(165*c^2*x^9 + 418*b*c*x^7 + 285*b^2*x^5)*sqrt(x)

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giac [A]  time = 0.14, size = 24, normalized size = 0.67 \begin {gather*} \frac {2}{19} \, c^{2} x^{\frac {19}{2}} + \frac {4}{15} \, b c x^{\frac {15}{2}} + \frac {2}{11} \, b^{2} x^{\frac {11}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)*(c*x^4+b*x^2)^2,x, algorithm="giac")

[Out]

2/19*c^2*x^(19/2) + 4/15*b*c*x^(15/2) + 2/11*b^2*x^(11/2)

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maple [A]  time = 0.00, size = 27, normalized size = 0.75 \begin {gather*} \frac {2 \left (165 c^{2} x^{4}+418 b c \,x^{2}+285 b^{2}\right ) x^{\frac {11}{2}}}{3135} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(c*x^4+b*x^2)^2,x)

[Out]

2/3135*x^(11/2)*(165*c^2*x^4+418*b*c*x^2+285*b^2)

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maxima [A]  time = 1.36, size = 24, normalized size = 0.67 \begin {gather*} \frac {2}{19} \, c^{2} x^{\frac {19}{2}} + \frac {4}{15} \, b c x^{\frac {15}{2}} + \frac {2}{11} \, b^{2} x^{\frac {11}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(1/2)*(c*x^4+b*x^2)^2,x, algorithm="maxima")

[Out]

2/19*c^2*x^(19/2) + 4/15*b*c*x^(15/2) + 2/11*b^2*x^(11/2)

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mupad [B]  time = 0.04, size = 25, normalized size = 0.69 \begin {gather*} x^{11/2}\,\left (\frac {2\,b^2}{11}+\frac {4\,b\,c\,x^2}{15}+\frac {2\,c^2\,x^4}{19}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(b*x^2 + c*x^4)^2,x)

[Out]

x^(11/2)*((2*b^2)/11 + (2*c^2*x^4)/19 + (4*b*c*x^2)/15)

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sympy [A]  time = 2.75, size = 34, normalized size = 0.94 \begin {gather*} \frac {2 b^{2} x^{\frac {11}{2}}}{11} + \frac {4 b c x^{\frac {15}{2}}}{15} + \frac {2 c^{2} x^{\frac {19}{2}}}{19} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(1/2)*(c*x**4+b*x**2)**2,x)

[Out]

2*b**2*x**(11/2)/11 + 4*b*c*x**(15/2)/15 + 2*c**2*x**(19/2)/19

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