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

Optimal. Leaf size=82 \[ -\frac {2 d x \sqrt {c+\frac {d}{x^2}} (5 b c-4 a d)}{15 c^3}+\frac {x^3 \sqrt {c+\frac {d}{x^2}} (5 b c-4 a d)}{15 c^2}+\frac {a x^5 \sqrt {c+\frac {d}{x^2}}}{5 c} \]

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Rubi [A]  time = 0.03, antiderivative size = 82, 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, 271, 191} \begin {gather*} \frac {x^3 \sqrt {c+\frac {d}{x^2}} (5 b c-4 a d)}{15 c^2}-\frac {2 d x \sqrt {c+\frac {d}{x^2}} (5 b c-4 a d)}{15 c^3}+\frac {a x^5 \sqrt {c+\frac {d}{x^2}}}{5 c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*d*(5*b*c - 4*a*d)*Sqrt[c + d/x^2]*x)/(15*c^3) + ((5*b*c - 4*a*d)*Sqrt[c + d/x^2]*x^3)/(15*c^2) + (a*Sqrt[c
+ d/x^2]*x^5)/(5*c)

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 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(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 \frac {\left (a+\frac {b}{x^2}\right ) x^4}{\sqrt {c+\frac {d}{x^2}}} \, dx &=\frac {a \sqrt {c+\frac {d}{x^2}} x^5}{5 c}+\frac {(5 b c-4 a d) \int \frac {x^2}{\sqrt {c+\frac {d}{x^2}}} \, dx}{5 c}\\ &=\frac {(5 b c-4 a d) \sqrt {c+\frac {d}{x^2}} x^3}{15 c^2}+\frac {a \sqrt {c+\frac {d}{x^2}} x^5}{5 c}-\frac {(2 d (5 b c-4 a d)) \int \frac {1}{\sqrt {c+\frac {d}{x^2}}} \, dx}{15 c^2}\\ &=-\frac {2 d (5 b c-4 a d) \sqrt {c+\frac {d}{x^2}} x}{15 c^3}+\frac {(5 b c-4 a d) \sqrt {c+\frac {d}{x^2}} x^3}{15 c^2}+\frac {a \sqrt {c+\frac {d}{x^2}} x^5}{5 c}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 56, normalized size = 0.68 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (a \left (3 c^2 x^4-4 c d x^2+8 d^2\right )+5 b c \left (c x^2-2 d\right )\right )}{15 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d/x^2]*x*(5*b*c*(-2*d + c*x^2) + a*(8*d^2 - 4*c*d*x^2 + 3*c^2*x^4)))/(15*c^3)

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IntegrateAlgebraic [A]  time = 0.08, size = 57, normalized size = 0.70 \begin {gather*} \frac {x \sqrt {c+\frac {d}{x^2}} \left (3 a c^2 x^4-4 a c d x^2+8 a d^2+5 b c^2 x^2-10 b c d\right )}{15 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c + d/x^2]*x*(-10*b*c*d + 8*a*d^2 + 5*b*c^2*x^2 - 4*a*c*d*x^2 + 3*a*c^2*x^4))/(15*c^3)

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fricas [A]  time = 0.41, size = 59, normalized size = 0.72 \begin {gather*} \frac {{\left (3 \, a c^{2} x^{5} + {\left (5 \, b c^{2} - 4 \, a c d\right )} x^{3} - 2 \, {\left (5 \, b c d - 4 \, a d^{2}\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{15 \, c^{3}} \end {gather*}

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

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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^4/(c+d/x^2)^(1/2),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 = 0.05, size = 67, normalized size = 0.82 \begin {gather*} \frac {\left (3 a \,x^{4} c^{2}-4 a c d \,x^{2}+5 b \,c^{2} x^{2}+8 a \,d^{2}-10 b c d \right ) \left (c \,x^{2}+d \right )}{15 \sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, c^{3} x} \end {gather*}

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/x*(3*a*c^2*x^4-4*a*c*d*x^2+5*b*c^2*x^2+8*a*d^2-10*b*c*d)*(c*x^2+d)/((c*x^2+d)/x^2)^(1/2)/c^3

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maxima [A]  time = 0.49, size = 85, normalized size = 1.04 \begin {gather*} \frac {{\left ({\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} x^{3} - 3 \, \sqrt {c + \frac {d}{x^{2}}} d x\right )} b}{3 \, c^{2}} + \frac {{\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} x^{5} - 10 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d x^{3} + 15 \, \sqrt {c + \frac {d}{x^{2}}} d^{2} x\right )} a}{15 \, c^{3}} \end {gather*}

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*((c + d/x^2)^(3/2)*x^3 - 3*sqrt(c + d/x^2)*d*x)*b/c^2 + 1/15*(3*(c + d/x^2)^(5/2)*x^5 - 10*(c + d/x^2)^(3/
2)*d*x^3 + 15*sqrt(c + d/x^2)*d^2*x)*a/c^3

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mupad [B]  time = 5.31, size = 53, normalized size = 0.65 \begin {gather*} \frac {x\,\sqrt {c+\frac {d}{x^2}}\,\left (3\,a\,c^2\,x^4+5\,b\,c^2\,x^2-4\,a\,c\,d\,x^2-10\,b\,c\,d+8\,a\,d^2\right )}{15\,c^3} \end {gather*}

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]

(x*(c + d/x^2)^(1/2)*(8*a*d^2 + 3*a*c^2*x^4 + 5*b*c^2*x^2 - 10*b*c*d - 4*a*c*d*x^2))/(15*c^3)

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sympy [B]  time = 3.55, size = 338, normalized size = 4.12 \begin {gather*} \frac {3 a c^{4} d^{\frac {9}{2}} x^{8} \sqrt {\frac {c x^{2}}{d} + 1}}{15 c^{5} d^{4} x^{4} + 30 c^{4} d^{5} x^{2} + 15 c^{3} d^{6}} + \frac {2 a c^{3} d^{\frac {11}{2}} x^{6} \sqrt {\frac {c x^{2}}{d} + 1}}{15 c^{5} d^{4} x^{4} + 30 c^{4} d^{5} x^{2} + 15 c^{3} d^{6}} + \frac {3 a c^{2} d^{\frac {13}{2}} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{15 c^{5} d^{4} x^{4} + 30 c^{4} d^{5} x^{2} + 15 c^{3} d^{6}} + \frac {12 a c d^{\frac {15}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{15 c^{5} d^{4} x^{4} + 30 c^{4} d^{5} x^{2} + 15 c^{3} d^{6}} + \frac {8 a d^{\frac {17}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{15 c^{5} d^{4} x^{4} + 30 c^{4} d^{5} x^{2} + 15 c^{3} d^{6}} + \frac {b \sqrt {d} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{3 c} - \frac {2 b d^{\frac {3}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{3 c^{2}} \end {gather*}

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]

3*a*c**4*d**(9/2)*x**8*sqrt(c*x**2/d + 1)/(15*c**5*d**4*x**4 + 30*c**4*d**5*x**2 + 15*c**3*d**6) + 2*a*c**3*d*
*(11/2)*x**6*sqrt(c*x**2/d + 1)/(15*c**5*d**4*x**4 + 30*c**4*d**5*x**2 + 15*c**3*d**6) + 3*a*c**2*d**(13/2)*x*
*4*sqrt(c*x**2/d + 1)/(15*c**5*d**4*x**4 + 30*c**4*d**5*x**2 + 15*c**3*d**6) + 12*a*c*d**(15/2)*x**2*sqrt(c*x*
*2/d + 1)/(15*c**5*d**4*x**4 + 30*c**4*d**5*x**2 + 15*c**3*d**6) + 8*a*d**(17/2)*sqrt(c*x**2/d + 1)/(15*c**5*d
**4*x**4 + 30*c**4*d**5*x**2 + 15*c**3*d**6) + b*sqrt(d)*x**2*sqrt(c*x**2/d + 1)/(3*c) - 2*b*d**(3/2)*sqrt(c*x
**2/d + 1)/(3*c**2)

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