3.940 \(\int (a+\frac {b}{x^2}) \sqrt {c+\frac {d}{x^2}} x^4 \, dx\)

Optimal. Leaf size=53 \[ \frac {x^3 \left (c+\frac {d}{x^2}\right )^{3/2} (5 b c-2 a d)}{15 c^2}+\frac {a x^5 \left (c+\frac {d}{x^2}\right )^{3/2}}{5 c} \]

[Out]

1/15*(-2*a*d+5*b*c)*(c+d/x^2)^(3/2)*x^3/c^2+1/5*a*(c+d/x^2)^(3/2)*x^5/c

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Rubi [A]  time = 0.02, antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {453, 264} \[ \frac {x^3 \left (c+\frac {d}{x^2}\right )^{3/2} (5 b c-2 a d)}{15 c^2}+\frac {a x^5 \left (c+\frac {d}{x^2}\right )^{3/2}}{5 c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)*Sqrt[c + d/x^2]*x^4,x]

[Out]

((5*b*c - 2*a*d)*(c + d/x^2)^(3/2)*x^3)/(15*c^2) + (a*(c + d/x^2)^(3/2)*x^5)/(5*c)

Rule 264

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

Rule 453

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

Rubi steps

\begin {align*} \int \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^4 \, dx &=\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x^5}{5 c}+\frac {(5 b c-2 a d) \int \sqrt {c+\frac {d}{x^2}} x^2 \, dx}{5 c}\\ &=\frac {(5 b c-2 a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^3}{15 c^2}+\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x^5}{5 c}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 42, normalized size = 0.79 \[ \frac {x \sqrt {c+\frac {d}{x^2}} \left (c x^2+d\right ) \left (3 a c x^2-2 a d+5 b c\right )}{15 c^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d/x^2]*x*(d + c*x^2)*(5*b*c - 2*a*d + 3*a*c*x^2))/(15*c^2)

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fricas [A]  time = 0.53, size = 57, normalized size = 1.08 \[ \frac {{\left (3 \, a c^{2} x^{5} + {\left (5 \, b c^{2} + a c d\right )} x^{3} + {\left (5 \, b c d - 2 \, a d^{2}\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{15 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(3*a*c^2*x^5 + (5*b*c^2 + a*c*d)*x^3 + (5*b*c*d - 2*a*d^2)*x)*sqrt((c*x^2 + d)/x^2)/c^2

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giac [A]  time = 0.21, size = 72, normalized size = 1.36 \[ -\frac {{\left (5 \, b c d^{\frac {3}{2}} - 2 \, a d^{\frac {5}{2}}\right )} \mathrm {sgn}\relax (x)}{15 \, c^{2}} + \frac {3 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} a \mathrm {sgn}\relax (x) + 5 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} b c \mathrm {sgn}\relax (x) - 5 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} a d \mathrm {sgn}\relax (x)}{15 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(5*b*c*d^(3/2) - 2*a*d^(5/2))*sgn(x)/c^2 + 1/15*(3*(c*x^2 + d)^(5/2)*a*sgn(x) + 5*(c*x^2 + d)^(3/2)*b*c*
sgn(x) - 5*(c*x^2 + d)^(3/2)*a*d*sgn(x))/c^2

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maple [A]  time = 0.05, size = 43, normalized size = 0.81 \[ \frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (3 a \,x^{2} c -2 a d +5 b c \right ) \left (c \,x^{2}+d \right ) x}{15 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*((c*x^2+d)/x^2)^(1/2)*x*(3*a*c*x^2-2*a*d+5*b*c)*(c*x^2+d)/c^2

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maxima [A]  time = 0.66, size = 55, normalized size = 1.04 \[ \frac {b {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} x^{3}}{3 \, c} + \frac {{\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} x^{5} - 5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d x^{3}\right )} a}{15 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*b*(c + d/x^2)^(3/2)*x^3/c + 1/15*(3*(c + d/x^2)^(5/2)*x^5 - 5*(c + d/x^2)^(3/2)*d*x^3)*a/c^2

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mupad [B]  time = 4.44, size = 54, normalized size = 1.02 \[ \sqrt {c+\frac {d}{x^2}}\,\left (\frac {a\,x^5}{5}-\frac {x\,\left (2\,a\,d^2-5\,b\,c\,d\right )}{15\,c^2}+\frac {x^3\,\left (5\,b\,c^2+a\,d\,c\right )}{15\,c^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c + d/x^2)^(1/2)*((a*x^5)/5 - (x*(2*a*d^2 - 5*b*c*d))/(15*c^2) + (x^3*(5*b*c^2 + a*c*d))/(15*c^2))

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sympy [B]  time = 3.04, size = 119, normalized size = 2.25 \[ \frac {a \sqrt {d} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {a d^{\frac {3}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{15 c} - \frac {2 a d^{\frac {5}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{15 c^{2}} + \frac {b \sqrt {d} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{3} + \frac {b d^{\frac {3}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{3 c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*sqrt(d)*x**4*sqrt(c*x**2/d + 1)/5 + a*d**(3/2)*x**2*sqrt(c*x**2/d + 1)/(15*c) - 2*a*d**(5/2)*sqrt(c*x**2/d +
1)/(15*c**2) + b*sqrt(d)*x**2*sqrt(c*x**2/d + 1)/3 + b*d**(3/2)*sqrt(c*x**2/d + 1)/(3*c)

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