3.4.1 \(\int x^{5/2} (b x^2+c x^4)^2 \, dx\) [301]

Optimal. Leaf size=36 \[ \frac {2}{15} b^2 x^{15/2}+\frac {4}{19} b c x^{19/2}+\frac {2}{23} c^2 x^{23/2} \]

[Out]

2/15*b^2*x^(15/2)+4/19*b*c*x^(19/2)+2/23*c^2*x^(23/2)

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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 = {1598, 276} \begin {gather*} \frac {2}{15} b^2 x^{15/2}+\frac {4}{19} b c x^{19/2}+\frac {2}{23} c^2 x^{23/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(b*x^2 + c*x^4)^2,x]

[Out]

(2*b^2*x^(15/2))/15 + (4*b*c*x^(19/2))/19 + (2*c^2*x^(23/2))/23

Rule 276

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 1598

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

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Mathematica [A]
time = 0.02, size = 30, normalized size = 0.83 \begin {gather*} \frac {2 x^{15/2} \left (437 b^2+690 b c x^2+285 c^2 x^4\right )}{6555} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x^(15/2)*(437*b^2 + 690*b*c*x^2 + 285*c^2*x^4))/6555

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Maple [A]
time = 0.10, size = 25, normalized size = 0.69

method	result	size
derivativedivides	\(\frac {2 b^{2} x^{\frac {15}{2}}}{15}+\frac {4 b c \,x^{\frac {19}{2}}}{19}+\frac {2 c^{2} x^{\frac {23}{2}}}{23}\)	\(25\)
default	\(\frac {2 b^{2} x^{\frac {15}{2}}}{15}+\frac {4 b c \,x^{\frac {19}{2}}}{19}+\frac {2 c^{2} x^{\frac {23}{2}}}{23}\)	\(25\)
gosper	\(\frac {2 x^{\frac {15}{2}} \left (285 c^{2} x^{4}+690 b c \,x^{2}+437 b^{2}\right )}{6555}\)	\(27\)
trager	\(\frac {2 x^{\frac {15}{2}} \left (285 c^{2} x^{4}+690 b c \,x^{2}+437 b^{2}\right )}{6555}\)	\(27\)
risch	\(\frac {2 x^{\frac {15}{2}} \left (285 c^{2} x^{4}+690 b c \,x^{2}+437 b^{2}\right )}{6555}\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15*b^2*x^(15/2)+4/19*b*c*x^(19/2)+2/23*c^2*x^(23/2)

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Maxima [A]
time = 0.29, size = 24, normalized size = 0.67 \begin {gather*} \frac {2}{23} \, c^{2} x^{\frac {23}{2}} + \frac {4}{19} \, b c x^{\frac {19}{2}} + \frac {2}{15} \, b^{2} x^{\frac {15}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/23*c^2*x^(23/2) + 4/19*b*c*x^(19/2) + 2/15*b^2*x^(15/2)

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Fricas [A]
time = 0.35, size = 29, normalized size = 0.81 \begin {gather*} \frac {2}{6555} \, {\left (285 \, c^{2} x^{11} + 690 \, b c x^{9} + 437 \, b^{2} x^{7}\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)^2,x, algorithm="fricas")

[Out]

2/6555*(285*c^2*x^11 + 690*b*c*x^9 + 437*b^2*x^7)*sqrt(x)

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Sympy [A]
time = 1.18, size = 34, normalized size = 0.94 \begin {gather*} \frac {2 b^{2} x^{\frac {15}{2}}}{15} + \frac {4 b c x^{\frac {19}{2}}}{19} + \frac {2 c^{2} x^{\frac {23}{2}}}{23} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b**2*x**(15/2)/15 + 4*b*c*x**(19/2)/19 + 2*c**2*x**(23/2)/23

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Giac [A]
time = 3.19, size = 24, normalized size = 0.67 \begin {gather*} \frac {2}{23} \, c^{2} x^{\frac {23}{2}} + \frac {4}{19} \, b c x^{\frac {19}{2}} + \frac {2}{15} \, b^{2} x^{\frac {15}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/23*c^2*x^(23/2) + 4/19*b*c*x^(19/2) + 2/15*b^2*x^(15/2)

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Mupad [B]
time = 0.04, size = 25, normalized size = 0.69 \begin {gather*} x^{15/2}\,\left (\frac {2\,b^2}{15}+\frac {4\,b\,c\,x^2}{19}+\frac {2\,c^2\,x^4}{23}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(b*x^2 + c*x^4)^2,x)

[Out]

x^(15/2)*((2*b^2)/15 + (2*c^2*x^4)/23 + (4*b*c*x^2)/19)

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