3.939 \(\int (a+\frac {b}{x^2}) \sqrt {c+\frac {d}{x^2}} x^6 \, dx\)

Optimal. Leaf size=84 \[ -\frac {2 d x^3 \left (c+\frac {d}{x^2}\right )^{3/2} (7 b c-4 a d)}{105 c^3}+\frac {x^5 \left (c+\frac {d}{x^2}\right )^{3/2} (7 b c-4 a d)}{35 c^2}+\frac {a x^7 \left (c+\frac {d}{x^2}\right )^{3/2}}{7 c} \]

[Out]

-2/105*d*(-4*a*d+7*b*c)*(c+d/x^2)^(3/2)*x^3/c^3+1/35*(-4*a*d+7*b*c)*(c+d/x^2)^(3/2)*x^5/c^2+1/7*a*(c+d/x^2)^(3
/2)*x^7/c

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Rubi [A]  time = 0.04, antiderivative size = 84, 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, 264} \[ \frac {x^5 \left (c+\frac {d}{x^2}\right )^{3/2} (7 b c-4 a d)}{35 c^2}-\frac {2 d x^3 \left (c+\frac {d}{x^2}\right )^{3/2} (7 b c-4 a d)}{105 c^3}+\frac {a x^7 \left (c+\frac {d}{x^2}\right )^{3/2}}{7 c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x^2)*Sqrt[c + d/x^2]*x^6,x]

[Out]

(-2*d*(7*b*c - 4*a*d)*(c + d/x^2)^(3/2)*x^3)/(105*c^3) + ((7*b*c - 4*a*d)*(c + d/x^2)^(3/2)*x^5)/(35*c^2) + (a
*(c + d/x^2)^(3/2)*x^7)/(7*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 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 \left (a+\frac {b}{x^2}\right ) \sqrt {c+\frac {d}{x^2}} x^6 \, dx &=\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x^7}{7 c}+\frac {(7 b c-4 a d) \int \sqrt {c+\frac {d}{x^2}} x^4 \, dx}{7 c}\\ &=\frac {(7 b c-4 a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^5}{35 c^2}+\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x^7}{7 c}-\frac {(2 d (7 b c-4 a d)) \int \sqrt {c+\frac {d}{x^2}} x^2 \, dx}{35 c^2}\\ &=-\frac {2 d (7 b c-4 a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^3}{105 c^3}+\frac {(7 b c-4 a d) \left (c+\frac {d}{x^2}\right )^{3/2} x^5}{35 c^2}+\frac {a \left (c+\frac {d}{x^2}\right )^{3/2} x^7}{7 c}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 64, normalized size = 0.76 \[ \frac {x \sqrt {c+\frac {d}{x^2}} \left (c x^2+d\right ) \left (a \left (15 c^2 x^4-12 c d x^2+8 d^2\right )+7 b c \left (3 c x^2-2 d\right )\right )}{105 c^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x^2)*Sqrt[c + d/x^2]*x^6,x]

[Out]

(Sqrt[c + d/x^2]*x*(d + c*x^2)*(7*b*c*(-2*d + 3*c*x^2) + a*(8*d^2 - 12*c*d*x^2 + 15*c^2*x^4)))/(105*c^3)

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fricas [A]  time = 0.58, size = 82, normalized size = 0.98 \[ \frac {{\left (15 \, a c^{3} x^{7} + 3 \, {\left (7 \, b c^{3} + a c^{2} d\right )} x^{5} + {\left (7 \, b c^{2} d - 4 \, a c d^{2}\right )} x^{3} - 2 \, {\left (7 \, b c d^{2} - 4 \, a d^{3}\right )} x\right )} \sqrt {\frac {c x^{2} + d}{x^{2}}}}{105 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*x^6*(c+d/x^2)^(1/2),x, algorithm="fricas")

[Out]

1/105*(15*a*c^3*x^7 + 3*(7*b*c^3 + a*c^2*d)*x^5 + (7*b*c^2*d - 4*a*c*d^2)*x^3 - 2*(7*b*c*d^2 - 4*a*d^3)*x)*sqr
t((c*x^2 + d)/x^2)/c^3

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giac [A]  time = 0.17, size = 105, normalized size = 1.25 \[ \frac {2 \, {\left (7 \, b c d^{\frac {5}{2}} - 4 \, a d^{\frac {7}{2}}\right )} \mathrm {sgn}\relax (x)}{105 \, c^{3}} + \frac {15 \, {\left (c x^{2} + d\right )}^{\frac {7}{2}} a \mathrm {sgn}\relax (x) + 21 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} b c \mathrm {sgn}\relax (x) - 42 \, {\left (c x^{2} + d\right )}^{\frac {5}{2}} a d \mathrm {sgn}\relax (x) - 35 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} b c d \mathrm {sgn}\relax (x) + 35 \, {\left (c x^{2} + d\right )}^{\frac {3}{2}} a d^{2} \mathrm {sgn}\relax (x)}{105 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*x^6*(c+d/x^2)^(1/2),x, algorithm="giac")

[Out]

2/105*(7*b*c*d^(5/2) - 4*a*d^(7/2))*sgn(x)/c^3 + 1/105*(15*(c*x^2 + d)^(7/2)*a*sgn(x) + 21*(c*x^2 + d)^(5/2)*b
*c*sgn(x) - 42*(c*x^2 + d)^(5/2)*a*d*sgn(x) - 35*(c*x^2 + d)^(3/2)*b*c*d*sgn(x) + 35*(c*x^2 + d)^(3/2)*a*d^2*s
gn(x))/c^3

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maple [A]  time = 0.06, size = 65, normalized size = 0.77 \[ \frac {\sqrt {\frac {c \,x^{2}+d}{x^{2}}}\, \left (15 a \,x^{4} c^{2}-12 a c d \,x^{2}+21 b \,c^{2} x^{2}+8 a \,d^{2}-14 b c d \right ) \left (c \,x^{2}+d \right ) x}{105 c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x^2)*x^6*(c+d/x^2)^(1/2),x)

[Out]

1/105*((c*x^2+d)/x^2)^(1/2)*x*(15*a*c^2*x^4-12*a*c*d*x^2+21*b*c^2*x^2+8*a*d^2-14*b*c*d)*(c*x^2+d)/c^3

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maxima [A]  time = 0.53, size = 90, normalized size = 1.07 \[ \frac {{\left (3 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} x^{5} - 5 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d x^{3}\right )} b}{15 \, c^{2}} + \frac {{\left (15 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {7}{2}} x^{7} - 42 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {5}{2}} d x^{5} + 35 \, {\left (c + \frac {d}{x^{2}}\right )}^{\frac {3}{2}} d^{2} x^{3}\right )} a}{105 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x^2)*x^6*(c+d/x^2)^(1/2),x, algorithm="maxima")

[Out]

1/15*(3*(c + d/x^2)^(5/2)*x^5 - 5*(c + d/x^2)^(3/2)*d*x^3)*b/c^2 + 1/105*(15*(c + d/x^2)^(7/2)*x^7 - 42*(c + d
/x^2)^(5/2)*d*x^5 + 35*(c + d/x^2)^(3/2)*d^2*x^3)*a/c^3

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mupad [B]  time = 4.49, size = 77, normalized size = 0.92 \[ \sqrt {c+\frac {d}{x^2}}\,\left (\frac {a\,x^7}{7}+\frac {x\,\left (8\,a\,d^3-14\,b\,c\,d^2\right )}{105\,c^3}+\frac {x^5\,\left (21\,b\,c^3+3\,a\,d\,c^2\right )}{105\,c^3}-\frac {d\,x^3\,\left (4\,a\,d-7\,b\,c\right )}{105\,c^2}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^6*(a + b/x^2)*(c + d/x^2)^(1/2),x)

[Out]

(c + d/x^2)^(1/2)*((a*x^7)/7 + (x*(8*a*d^3 - 14*b*c*d^2))/(105*c^3) + (x^5*(21*b*c^3 + 3*a*c^2*d))/(105*c^3) -
(d*x^3*(4*a*d - 7*b*c))/(105*c^2))

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sympy [B]  time = 3.99, size = 422, normalized size = 5.02 \[ \frac {15 a c^{5} d^{\frac {9}{2}} x^{10} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {33 a c^{4} d^{\frac {11}{2}} x^{8} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {17 a c^{3} d^{\frac {13}{2}} x^{6} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {3 a c^{2} d^{\frac {15}{2}} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {12 a c d^{\frac {17}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {8 a d^{\frac {19}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{105 c^{5} d^{4} x^{4} + 210 c^{4} d^{5} x^{2} + 105 c^{3} d^{6}} + \frac {b \sqrt {d} x^{4} \sqrt {\frac {c x^{2}}{d} + 1}}{5} + \frac {b d^{\frac {3}{2}} x^{2} \sqrt {\frac {c x^{2}}{d} + 1}}{15 c} - \frac {2 b d^{\frac {5}{2}} \sqrt {\frac {c x^{2}}{d} + 1}}{15 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x**2)*x**6*(c+d/x**2)**(1/2),x)

[Out]

15*a*c**5*d**(9/2)*x**10*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 33*a*c
**4*d**(11/2)*x**8*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 17*a*c**3*d*
*(13/2)*x**6*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 3*a*c**2*d**(15/2)
*x**4*sqrt(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 12*a*c*d**(17/2)*x**2*sqr
t(c*x**2/d + 1)/(105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + 8*a*d**(19/2)*sqrt(c*x**2/d + 1)/(
105*c**5*d**4*x**4 + 210*c**4*d**5*x**2 + 105*c**3*d**6) + b*sqrt(d)*x**4*sqrt(c*x**2/d + 1)/5 + b*d**(3/2)*x*
*2*sqrt(c*x**2/d + 1)/(15*c) - 2*b*d**(5/2)*sqrt(c*x**2/d + 1)/(15*c**2)

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