3.3.31 \(\int x^4 \sqrt {b x^2+c x^4} \, dx\) [231]

Optimal. Leaf size=78 \[ \frac {8 b^2 \left (b x^2+c x^4\right )^{3/2}}{105 c^3 x^3}-\frac {4 b \left (b x^2+c x^4\right )^{3/2}}{35 c^2 x}+\frac {x \left (b x^2+c x^4\right )^{3/2}}{7 c} \]

[Out]

8/105*b^2*(c*x^4+b*x^2)^(3/2)/c^3/x^3-4/35*b*(c*x^4+b*x^2)^(3/2)/c^2/x+1/7*x*(c*x^4+b*x^2)^(3/2)/c

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Rubi [A]
time = 0.06, antiderivative size = 78, 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 = {2041, 2025} \begin {gather*} \frac {8 b^2 \left (b x^2+c x^4\right )^{3/2}}{105 c^3 x^3}-\frac {4 b \left (b x^2+c x^4\right )^{3/2}}{35 c^2 x}+\frac {x \left (b x^2+c x^4\right )^{3/2}}{7 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[b*x^2 + c*x^4],x]

[Out]

(8*b^2*(b*x^2 + c*x^4)^(3/2))/(105*c^3*x^3) - (4*b*(b*x^2 + c*x^4)^(3/2))/(35*c^2*x) + (x*(b*x^2 + c*x^4)^(3/2
))/(7*c)

Rule 2025

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(b*(n - j)*(p + 1)*x
^(n - 1)), x] /; FreeQ[{a, b, j, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && EqQ[j*p - n + j + 1, 0]

Rule 2041

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[c^(j - 1)*(c*x)^(m - j +
1)*((a*x^j + b*x^n)^(p + 1)/(a*(m + j*p + 1))), x] - Dist[b*((m + n*p + n - j + 1)/(a*c^(n - j)*(m + j*p + 1))
), Int[(c*x)^(m + n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[
n, j] && ILtQ[Simplify[(m + n*p + n - j + 1)/(n - j)], 0] && NeQ[m + j*p + 1, 0] && (IntegersQ[j, n] || GtQ[c,
0])

Rubi steps

\begin {align*} \int x^4 \sqrt {b x^2+c x^4} \, dx &=\frac {x \left (b x^2+c x^4\right )^{3/2}}{7 c}-\frac {(4 b) \int x^2 \sqrt {b x^2+c x^4} \, dx}{7 c}\\ &=-\frac {4 b \left (b x^2+c x^4\right )^{3/2}}{35 c^2 x}+\frac {x \left (b x^2+c x^4\right )^{3/2}}{7 c}+\frac {\left (8 b^2\right ) \int \sqrt {b x^2+c x^4} \, dx}{35 c^2}\\ &=\frac {8 b^2 \left (b x^2+c x^4\right )^{3/2}}{105 c^3 x^3}-\frac {4 b \left (b x^2+c x^4\right )^{3/2}}{35 c^2 x}+\frac {x \left (b x^2+c x^4\right )^{3/2}}{7 c}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 57, normalized size = 0.73 \begin {gather*} \frac {\sqrt {x^2 \left (b+c x^2\right )} \left (8 b^3-4 b^2 c x^2+3 b c^2 x^4+15 c^3 x^6\right )}{105 c^3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*Sqrt[b*x^2 + c*x^4],x]

[Out]

(Sqrt[x^2*(b + c*x^2)]*(8*b^3 - 4*b^2*c*x^2 + 3*b*c^2*x^4 + 15*c^3*x^6))/(105*c^3*x)

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Maple [A]
time = 0.10, size = 50, normalized size = 0.64

method	result	size
gosper	\(\frac {\left (c \,x^{2}+b \right ) \left (15 c^{2} x^{4}-12 b c \,x^{2}+8 b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{105 c^{3} x}\)	\(50\)
default	\(\frac {\left (c \,x^{2}+b \right ) \left (15 c^{2} x^{4}-12 b c \,x^{2}+8 b^{2}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{105 c^{3} x}\)	\(50\)
trager	\(\frac {\left (15 c^{3} x^{6}+3 b \,c^{2} x^{4}-4 b^{2} c \,x^{2}+8 b^{3}\right ) \sqrt {c \,x^{4}+b \,x^{2}}}{105 c^{3} x}\)	\(54\)
risch	\(\frac {\sqrt {x^{2} \left (c \,x^{2}+b \right )}\, \left (15 c^{3} x^{6}+3 b \,c^{2} x^{4}-4 b^{2} c \,x^{2}+8 b^{3}\right )}{105 x \,c^{3}}\)	\(54\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(c*x^2+b)*(15*c^2*x^4-12*b*c*x^2+8*b^2)*(c*x^4+b*x^2)^(1/2)/c^3/x

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Maxima [A]
time = 0.30, size = 46, normalized size = 0.59 \begin {gather*} \frac {{\left (15 \, c^{3} x^{6} + 3 \, b c^{2} x^{4} - 4 \, b^{2} c x^{2} + 8 \, b^{3}\right )} \sqrt {c x^{2} + b}}{105 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(15*c^3*x^6 + 3*b*c^2*x^4 - 4*b^2*c*x^2 + 8*b^3)*sqrt(c*x^2 + b)/c^3

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Fricas [A]
time = 0.36, size = 53, normalized size = 0.68 \begin {gather*} \frac {{\left (15 \, c^{3} x^{6} + 3 \, b c^{2} x^{4} - 4 \, b^{2} c x^{2} + 8 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{105 \, c^{3} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/105*(15*c^3*x^6 + 3*b*c^2*x^4 - 4*b^2*c*x^2 + 8*b^3)*sqrt(c*x^4 + b*x^2)/(c^3*x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int x^{4} \sqrt {x^{2} \left (b + c x^{2}\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4*sqrt(x**2*(b + c*x**2)), x)

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Giac [A]
time = 5.90, size = 60, normalized size = 0.77 \begin {gather*} -\frac {8 \, b^{\frac {7}{2}} \mathrm {sgn}\left (x\right )}{105 \, c^{3}} + \frac {15 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} \mathrm {sgn}\left (x\right ) - 42 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} b \mathrm {sgn}\left (x\right ) + 35 \, {\left (c x^{2} + b\right )}^{\frac {3}{2}} b^{2} \mathrm {sgn}\left (x\right )}{105 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/105*b^(7/2)*sgn(x)/c^3 + 1/105*(15*(c*x^2 + b)^(7/2)*sgn(x) - 42*(c*x^2 + b)^(5/2)*b*sgn(x) + 35*(c*x^2 + b
)^(3/2)*b^2*sgn(x))/c^3

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Mupad [B]
time = 4.23, size = 53, normalized size = 0.68 \begin {gather*} \frac {\sqrt {c\,x^4+b\,x^2}\,\left (8\,b^3-4\,b^2\,c\,x^2+3\,b\,c^2\,x^4+15\,c^3\,x^6\right )}{105\,c^3\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(b*x^2 + c*x^4)^(1/2),x)

[Out]

((b*x^2 + c*x^4)^(1/2)*(8*b^3 + 15*c^3*x^6 - 4*b^2*c*x^2 + 3*b*c^2*x^4))/(105*c^3*x)

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