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

Optimal. Leaf size=52 \[ \frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{35 c^2 x^5} \]

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Rubi [A]  time = 0.05, antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {2002, 2014} \begin {gather*} \frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{35 c^2 x^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2),x]

[Out]

(-2*b*(b*x^2 + c*x^4)^(5/2))/(35*c^2*x^5) + (b*x^2 + c*x^4)^(5/2)/(7*c*x^3)

Rule 2002

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

Rule 2014

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*(n - j)*(p + 1)), x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && N
eQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rubi steps

\begin {align*} \int \left (b x^2+c x^4\right )^{3/2} \, dx &=\frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}-\frac {(2 b) \int \frac {\left (b x^2+c x^4\right )^{3/2}}{x^2} \, dx}{7 c}\\ &=-\frac {2 b \left (b x^2+c x^4\right )^{5/2}}{35 c^2 x^5}+\frac {\left (b x^2+c x^4\right )^{5/2}}{7 c x^3}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 42, normalized size = 0.81 \begin {gather*} \frac {x \left (b+c x^2\right )^3 \left (5 c x^2-2 b\right )}{35 c^2 \sqrt {x^2 \left (b+c x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*x^2 + c*x^4)^(3/2),x]

[Out]

(x*(b + c*x^2)^3*(-2*b + 5*c*x^2))/(35*c^2*Sqrt[x^2*(b + c*x^2)])

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IntegrateAlgebraic [A]  time = 0.30, size = 56, normalized size = 1.08 \begin {gather*} \frac {\sqrt {b x^2+c x^4} \left (-2 b^3+b^2 c x^2+8 b c^2 x^4+5 c^3 x^6\right )}{35 c^2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[b*x^2 + c*x^4]*(-2*b^3 + b^2*c*x^2 + 8*b*c^2*x^4 + 5*c^3*x^6))/(35*c^2*x)

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fricas [A]  time = 1.04, size = 52, normalized size = 1.00 \begin {gather*} \frac {{\left (5 \, c^{3} x^{6} + 8 \, b c^{2} x^{4} + b^{2} c x^{2} - 2 \, b^{3}\right )} \sqrt {c x^{4} + b x^{2}}}{35 \, c^{2} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(5*c^3*x^6 + 8*b*c^2*x^4 + b^2*c*x^2 - 2*b^3)*sqrt(c*x^4 + b*x^2)/(c^2*x)

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giac [A]  time = 0.16, size = 44, normalized size = 0.85 \begin {gather*} \frac {2 \, b^{\frac {7}{2}} \mathrm {sgn}\relax (x)}{35 \, c^{2}} + \frac {5 \, {\left (c x^{2} + b\right )}^{\frac {7}{2}} \mathrm {sgn}\relax (x) - 7 \, {\left (c x^{2} + b\right )}^{\frac {5}{2}} b \mathrm {sgn}\relax (x)}{35 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/35*b^(7/2)*sgn(x)/c^2 + 1/35*(5*(c*x^2 + b)^(7/2)*sgn(x) - 7*(c*x^2 + b)^(5/2)*b*sgn(x))/c^2

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maple [A]  time = 0.01, size = 39, normalized size = 0.75 \begin {gather*} -\frac {\left (c \,x^{2}+b \right ) \left (-5 c \,x^{2}+2 b \right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{35 c^{2} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35*(c*x^2+b)*(-5*c*x^2+2*b)*(c*x^4+b*x^2)^(3/2)/c^2/x^3

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maxima [A]  time = 1.50, size = 45, normalized size = 0.87 \begin {gather*} \frac {{\left (5 \, c^{3} x^{6} + 8 \, b c^{2} x^{4} + b^{2} c x^{2} - 2 \, b^{3}\right )} \sqrt {c x^{2} + b}}{35 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(5*c^3*x^6 + 8*b*c^2*x^4 + b^2*c*x^2 - 2*b^3)*sqrt(c*x^2 + b)/c^2

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mupad [B]  time = 4.16, size = 40, normalized size = 0.77 \begin {gather*} -\frac {{\left (c\,x^2+b\right )}^2\,\sqrt {c\,x^4+b\,x^2}\,\left (2\,b-5\,c\,x^2\right )}{35\,c^2\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(3/2),x)

[Out]

-((b + c*x^2)^2*(b*x^2 + c*x^4)^(1/2)*(2*b - 5*c*x^2))/(35*c^2*x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (b x^{2} + c x^{4}\right )^{\frac {3}{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*x**2 + c*x**4)**(3/2), x)

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