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

Optimal. Leaf size=25 \[ \frac {\left (b x^2+c x^4\right )^{5/2}}{5 c x^5} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 25, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {2014} \[ \frac {\left (b x^2+c x^4\right )^{5/2}}{5 c x^5} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^2 + c*x^4)^(3/2)/x^2,x]

[Out]

(b*x^2 + c*x^4)^(5/2)/(5*c*x^5)

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

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Mathematica [A]  time = 0.01, size = 25, normalized size = 1.00 \[ \frac {\left (x^2 \left (b+c x^2\right )\right )^{5/2}}{5 c x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^2*(b + c*x^2))^(5/2)/(5*c*x^5)

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fricas [A]  time = 0.65, size = 39, normalized size = 1.56 \[ \frac {{\left (c^{2} x^{4} + 2 \, b c x^{2} + b^{2}\right )} \sqrt {c x^{4} + b x^{2}}}{5 \, c x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 27, normalized size = 1.08 \[ \frac {{\left (c x^{2} + b\right )}^{\frac {5}{2}} \mathrm {sgn}\relax (x)}{5 \, c} - \frac {b^{\frac {5}{2}} \mathrm {sgn}\relax (x)}{5 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 29, normalized size = 1.16 \[ \frac {\left (c \,x^{2}+b \right ) \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}}}{5 c \,x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.48, size = 32, normalized size = 1.28 \[ \frac {{\left (c^{2} x^{4} + 2 \, b c x^{2} + b^{2}\right )} \sqrt {c x^{2} + b}}{5 \, c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.15, size = 30, normalized size = 1.20 \[ \frac {{\left (c\,x^2+b\right )}^2\,\sqrt {c\,x^4+b\,x^2}}{5\,c\,x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2 + c*x^4)^(3/2)/x^2,x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x**2*(b + c*x**2))**(3/2)/x**2, x)

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