3.7.35 \(\int \frac {(a+\frac {b}{x^2}) x^2}{(c+\frac {d}{x^2})^{3/2}} \, dx\)

Optimal. Leaf size=79 \[ \frac {2 x \sqrt {c+\frac {d}{x^2}} (3 b c-4 a d)}{3 c^3}-\frac {x (3 b c-4 a d)}{3 c^2 \sqrt {c+\frac {d}{x^2}}}+\frac {a x^3}{3 c \sqrt {c+\frac {d}{x^2}}} \]

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Rubi [A]  time = 0.03, antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {453, 192, 191} \begin {gather*} \frac {2 x \sqrt {c+\frac {d}{x^2}} (3 b c-4 a d)}{3 c^3}-\frac {x (3 b c-4 a d)}{3 c^2 \sqrt {c+\frac {d}{x^2}}}+\frac {a x^3}{3 c \sqrt {c+\frac {d}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b/x^2)*x^2)/(c + d/x^2)^(3/2),x]

[Out]

-((3*b*c - 4*a*d)*x)/(3*c^2*Sqrt[c + d/x^2]) + (2*(3*b*c - 4*a*d)*Sqrt[c + d/x^2]*x)/(3*c^3) + (a*x^3)/(3*c*Sq
rt[c + d/x^2])

Rule 191

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

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -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 \frac {\left (a+\frac {b}{x^2}\right ) x^2}{\left (c+\frac {d}{x^2}\right )^{3/2}} \, dx &=\frac {a x^3}{3 c \sqrt {c+\frac {d}{x^2}}}+\frac {(3 b c-4 a d) \int \frac {1}{\left (c+\frac {d}{x^2}\right )^{3/2}} \, dx}{3 c}\\ &=-\frac {(3 b c-4 a d) x}{3 c^2 \sqrt {c+\frac {d}{x^2}}}+\frac {a x^3}{3 c \sqrt {c+\frac {d}{x^2}}}+\frac {(2 (3 b c-4 a d)) \int \frac {1}{\sqrt {c+\frac {d}{x^2}}} \, dx}{3 c^2}\\ &=-\frac {(3 b c-4 a d) x}{3 c^2 \sqrt {c+\frac {d}{x^2}}}+\frac {2 (3 b c-4 a d) \sqrt {c+\frac {d}{x^2}} x}{3 c^3}+\frac {a x^3}{3 c \sqrt {c+\frac {d}{x^2}}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 57, normalized size = 0.72 \begin {gather*} \frac {a \left (c^2 x^4-4 c d x^2-8 d^2\right )+3 b c \left (c x^2+2 d\right )}{3 c^3 x \sqrt {c+\frac {d}{x^2}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*b*c*(2*d + c*x^2) + a*(-8*d^2 - 4*c*d*x^2 + c^2*x^4))/(3*c^3*Sqrt[c + d/x^2]*x)

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IntegrateAlgebraic [A]  time = 0.09, size = 65, normalized size = 0.82 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (a c^2 x^4-4 a c d x^2-8 a d^2+3 b c^2 x^2+6 b c d\right )}{3 c^3 \left (c x^2+d\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 70, normalized size = 0.89 \begin {gather*} \frac {{\left (a c^{2} x^{5} + {\left (3 \, b c^{2} - 4 \, a c d\right )} x^{3} + 2 \, {\left (3 \, b c d - 4 \, a d^{2}\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{3 \, {\left (c^{4} x^{2} + c^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [ab
s(t_nostep)]Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is
real):Check [sign(t_nostep),sign(t_nostep+sqrt(d)/c*sign(t_nostep))]sym2poly/r2sym(const gen & e,const index_
m & i,const vecteur & l) Error: Bad Argument Value

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maple [A]  time = 0.06, size = 66, normalized size = 0.84 \begin {gather*} \frac {\left (a \,x^{4} c^{2}-4 a c d \,x^{2}+3 b \,c^{2} x^{2}-8 a \,d^{2}+6 b c d \right ) \left (c \,x^{2}+d \right )}{3 \left (\frac {c \,x^{2}+d}{x^{2}}\right )^{\frac {3}{2}} c^{3} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.53, size = 90, normalized size = 1.14 \begin {gather*} b {\left (\frac {\sqrt {c + \frac {d}{x^{2}}} x}{c^{2}} + \frac {d}{\sqrt {c + \frac {d}{x^{2}}} c^{2} x}\right )} + \frac {1}{3} \, a {\left (\frac {{\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} x^{3} - 6 \, \sqrt {c + \frac {d}{x^{2}}} d x}{c^{3}} - \frac {3 \, d^{2}}{\sqrt {c + \frac {d}{x^{2}}} c^{3} x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*(sqrt(c + d/x^2)*x/c^2 + d/(sqrt(c + d/x^2)*c^2*x)) + 1/3*a*(((c + d/x^2)^(3/2)*x^3 - 6*sqrt(c + d/x^2)*d*x)
/c^3 - 3*d^2/(sqrt(c + d/x^2)*c^3*x))

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mupad [B]  time = 5.20, size = 81, normalized size = 1.03 \begin {gather*} \frac {b\,c^2\,x^4+3\,b\,c\,d\,x^2+2\,b\,d^2}{c^2\,x^3\,{\left (c+\frac {d}{x^2}\right )}^{3/2}}-\frac {-a\,c^2\,x^4+4\,a\,c\,d\,x^2+8\,a\,d^2}{3\,c^3\,x\,\sqrt {c+\frac {d}{x^2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*b*d^2 + b*c^2*x^4 + 3*b*c*d*x^2)/(c^2*x^3*(c + d/x^2)^(3/2)) - (8*a*d^2 - a*c^2*x^4 + 4*a*c*d*x^2)/(3*c^3*x
*(c + d/x^2)^(1/2))

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sympy [B]  time = 7.31, size = 267, normalized size = 3.38 \begin {gather*} a \left (\frac {c^{3} d^{\frac {9}{2}} x^{6} \sqrt {\frac {c x^{2}}{d} + 1}}{3 c^{5} d^{4} x^{4} + 6 c^{4} d^{5} x^{2} + 3 c^{3} d^{6}} - \frac {3 c^{2} d^{\frac {11}{2}} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{3 c^{5} d^{4} x^{4} + 6 c^{4} d^{5} x^{2} + 3 c^{3} d^{6}} - \frac {12 c d^{\frac {13}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{3 c^{5} d^{4} x^{4} + 6 c^{4} d^{5} x^{2} + 3 c^{3} d^{6}} - \frac {8 d^{\frac {15}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{3 c^{5} d^{4} x^{4} + 6 c^{4} d^{5} x^{2} + 3 c^{3} d^{6}}\right ) + b \left (\frac {x^{2}}{c \sqrt {d} \sqrt {\frac {c x^{2}}{d} + 1}} + \frac {2 \sqrt {d}}{c^{2} \sqrt {\frac {c x^{2}}{d} + 1}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*(c**3*d**(9/2)*x**6*sqrt(c*x**2/d + 1)/(3*c**5*d**4*x**4 + 6*c**4*d**5*x**2 + 3*c**3*d**6) - 3*c**2*d**(11/2
)*x**4*sqrt(c*x**2/d + 1)/(3*c**5*d**4*x**4 + 6*c**4*d**5*x**2 + 3*c**3*d**6) - 12*c*d**(13/2)*x**2*sqrt(c*x**
2/d + 1)/(3*c**5*d**4*x**4 + 6*c**4*d**5*x**2 + 3*c**3*d**6) - 8*d**(15/2)*sqrt(c*x**2/d + 1)/(3*c**5*d**4*x**
4 + 6*c**4*d**5*x**2 + 3*c**3*d**6)) + b*(x**2/(c*sqrt(d)*sqrt(c*x**2/d + 1)) + 2*sqrt(d)/(c**2*sqrt(c*x**2/d
+ 1)))

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