3.568 \(\int \frac {A+B x^2}{x^8 \sqrt {a+b x^2}} \, dx\)

Optimal. Leaf size=117 \[ \frac {8 b^2 \sqrt {a+b x^2} (6 A b-7 a B)}{105 a^4 x}-\frac {4 b \sqrt {a+b x^2} (6 A b-7 a B)}{105 a^3 x^3}+\frac {\sqrt {a+b x^2} (6 A b-7 a B)}{35 a^2 x^5}-\frac {A \sqrt {a+b x^2}}{7 a x^7} \]

[Out]

-1/7*A*(b*x^2+a)^(1/2)/a/x^7+1/35*(6*A*b-7*B*a)*(b*x^2+a)^(1/2)/a^2/x^5-4/105*b*(6*A*b-7*B*a)*(b*x^2+a)^(1/2)/
a^3/x^3+8/105*b^2*(6*A*b-7*B*a)*(b*x^2+a)^(1/2)/a^4/x

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Rubi [A]  time = 0.05, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {453, 271, 264} \[ \frac {8 b^2 \sqrt {a+b x^2} (6 A b-7 a B)}{105 a^4 x}-\frac {4 b \sqrt {a+b x^2} (6 A b-7 a B)}{105 a^3 x^3}+\frac {\sqrt {a+b x^2} (6 A b-7 a B)}{35 a^2 x^5}-\frac {A \sqrt {a+b x^2}}{7 a x^7} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^8*Sqrt[a + b*x^2]),x]

[Out]

-(A*Sqrt[a + b*x^2])/(7*a*x^7) + ((6*A*b - 7*a*B)*Sqrt[a + b*x^2])/(35*a^2*x^5) - (4*b*(6*A*b - 7*a*B)*Sqrt[a
+ b*x^2])/(105*a^3*x^3) + (8*b^2*(6*A*b - 7*a*B)*Sqrt[a + b*x^2])/(105*a^4*x)

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 \frac {A+B x^2}{x^8 \sqrt {a+b x^2}} \, dx &=-\frac {A \sqrt {a+b x^2}}{7 a x^7}-\frac {(6 A b-7 a B) \int \frac {1}{x^6 \sqrt {a+b x^2}} \, dx}{7 a}\\ &=-\frac {A \sqrt {a+b x^2}}{7 a x^7}+\frac {(6 A b-7 a B) \sqrt {a+b x^2}}{35 a^2 x^5}+\frac {(4 b (6 A b-7 a B)) \int \frac {1}{x^4 \sqrt {a+b x^2}} \, dx}{35 a^2}\\ &=-\frac {A \sqrt {a+b x^2}}{7 a x^7}+\frac {(6 A b-7 a B) \sqrt {a+b x^2}}{35 a^2 x^5}-\frac {4 b (6 A b-7 a B) \sqrt {a+b x^2}}{105 a^3 x^3}-\frac {\left (8 b^2 (6 A b-7 a B)\right ) \int \frac {1}{x^2 \sqrt {a+b x^2}} \, dx}{105 a^3}\\ &=-\frac {A \sqrt {a+b x^2}}{7 a x^7}+\frac {(6 A b-7 a B) \sqrt {a+b x^2}}{35 a^2 x^5}-\frac {4 b (6 A b-7 a B) \sqrt {a+b x^2}}{105 a^3 x^3}+\frac {8 b^2 (6 A b-7 a B) \sqrt {a+b x^2}}{105 a^4 x}\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 84, normalized size = 0.72 \[ \frac {\left (\frac {b x^2}{a}+1\right ) \left (3 a^2-4 a b x^2+8 b^2 x^4\right ) (6 A b-7 a B)}{105 a^3 x^5 \sqrt {a+b x^2}}-\frac {A \sqrt {a+b x^2}}{7 a x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^8*Sqrt[a + b*x^2]),x]

[Out]

-1/7*(A*Sqrt[a + b*x^2])/(a*x^7) + ((6*A*b - 7*a*B)*(1 + (b*x^2)/a)*(3*a^2 - 4*a*b*x^2 + 8*b^2*x^4))/(105*a^3*
x^5*Sqrt[a + b*x^2])

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fricas [A]  time = 1.21, size = 82, normalized size = 0.70 \[ -\frac {{\left (8 \, {\left (7 \, B a b^{2} - 6 \, A b^{3}\right )} x^{6} - 4 \, {\left (7 \, B a^{2} b - 6 \, A a b^{2}\right )} x^{4} + 15 \, A a^{3} + 3 \, {\left (7 \, B a^{3} - 6 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{105 \, a^{4} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^8/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

-1/105*(8*(7*B*a*b^2 - 6*A*b^3)*x^6 - 4*(7*B*a^2*b - 6*A*a*b^2)*x^4 + 15*A*a^3 + 3*(7*B*a^3 - 6*A*a^2*b)*x^2)*
sqrt(b*x^2 + a)/(a^4*x^7)

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giac [B]  time = 0.42, size = 232, normalized size = 1.98 \[ \frac {16 \, {\left (70 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} B b^{\frac {5}{2}} - 175 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} B a b^{\frac {5}{2}} + 210 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} A b^{\frac {7}{2}} + 147 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a^{2} b^{\frac {5}{2}} - 126 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a b^{\frac {7}{2}} - 49 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{3} b^{\frac {5}{2}} + 42 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a^{2} b^{\frac {7}{2}} + 7 \, B a^{4} b^{\frac {5}{2}} - 6 \, A a^{3} b^{\frac {7}{2}}\right )}}{105 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^8/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

16/105*(70*(sqrt(b)*x - sqrt(b*x^2 + a))^8*B*b^(5/2) - 175*(sqrt(b)*x - sqrt(b*x^2 + a))^6*B*a*b^(5/2) + 210*(
sqrt(b)*x - sqrt(b*x^2 + a))^6*A*b^(7/2) + 147*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a^2*b^(5/2) - 126*(sqrt(b)*x
- sqrt(b*x^2 + a))^4*A*a*b^(7/2) - 49*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^3*b^(5/2) + 42*(sqrt(b)*x - sqrt(b*x
^2 + a))^2*A*a^2*b^(7/2) + 7*B*a^4*b^(5/2) - 6*A*a^3*b^(7/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^7

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maple [A]  time = 0.01, size = 83, normalized size = 0.71 \[ -\frac {\sqrt {b \,x^{2}+a}\, \left (-48 A \,b^{3} x^{6}+56 B a \,b^{2} x^{6}+24 x^{4} A a \,b^{2}-28 B \,a^{2} b \,x^{4}-18 A \,a^{2} b \,x^{2}+21 B \,a^{3} x^{2}+15 A \,a^{3}\right )}{105 a^{4} x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^8/(b*x^2+a)^(1/2),x)

[Out]

-1/105*(b*x^2+a)^(1/2)*(-48*A*b^3*x^6+56*B*a*b^2*x^6+24*A*a*b^2*x^4-28*B*a^2*b*x^4-18*A*a^2*b*x^2+21*B*a^3*x^2
+15*A*a^3)/x^7/a^4

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maxima [A]  time = 1.10, size = 138, normalized size = 1.18 \[ -\frac {8 \, \sqrt {b x^{2} + a} B b^{2}}{15 \, a^{3} x} + \frac {16 \, \sqrt {b x^{2} + a} A b^{3}}{35 \, a^{4} x} + \frac {4 \, \sqrt {b x^{2} + a} B b}{15 \, a^{2} x^{3}} - \frac {8 \, \sqrt {b x^{2} + a} A b^{2}}{35 \, a^{3} x^{3}} - \frac {\sqrt {b x^{2} + a} B}{5 \, a x^{5}} + \frac {6 \, \sqrt {b x^{2} + a} A b}{35 \, a^{2} x^{5}} - \frac {\sqrt {b x^{2} + a} A}{7 \, a x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^8/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

-8/15*sqrt(b*x^2 + a)*B*b^2/(a^3*x) + 16/35*sqrt(b*x^2 + a)*A*b^3/(a^4*x) + 4/15*sqrt(b*x^2 + a)*B*b/(a^2*x^3)
- 8/35*sqrt(b*x^2 + a)*A*b^2/(a^3*x^3) - 1/5*sqrt(b*x^2 + a)*B/(a*x^5) + 6/35*sqrt(b*x^2 + a)*A*b/(a^2*x^5) -
1/7*sqrt(b*x^2 + a)*A/(a*x^7)

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mupad [B]  time = 0.73, size = 105, normalized size = 0.90 \[ \frac {\sqrt {b\,x^2+a}\,\left (6\,A\,b-7\,B\,a\right )}{35\,a^2\,x^5}+\frac {\sqrt {b\,x^2+a}\,\left (48\,A\,b^3-56\,B\,a\,b^2\right )}{105\,a^4\,x}-\frac {\left (24\,A\,b^2-28\,B\,a\,b\right )\,\sqrt {b\,x^2+a}}{105\,a^3\,x^3}-\frac {A\,\sqrt {b\,x^2+a}}{7\,a\,x^7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^8*(a + b*x^2)^(1/2)),x)

[Out]

((a + b*x^2)^(1/2)*(6*A*b - 7*B*a))/(35*a^2*x^5) + ((a + b*x^2)^(1/2)*(48*A*b^3 - 56*B*a*b^2))/(105*a^4*x) - (
(24*A*b^2 - 28*B*a*b)*(a + b*x^2)^(1/2))/(105*a^3*x^3) - (A*(a + b*x^2)^(1/2))/(7*a*x^7)

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sympy [B]  time = 3.68, size = 819, normalized size = 7.00 \[ - \frac {5 A a^{6} b^{\frac {19}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{35 a^{7} b^{9} x^{6} + 105 a^{6} b^{10} x^{8} + 105 a^{5} b^{11} x^{10} + 35 a^{4} b^{12} x^{12}} - \frac {9 A a^{5} b^{\frac {21}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{35 a^{7} b^{9} x^{6} + 105 a^{6} b^{10} x^{8} + 105 a^{5} b^{11} x^{10} + 35 a^{4} b^{12} x^{12}} - \frac {5 A a^{4} b^{\frac {23}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{35 a^{7} b^{9} x^{6} + 105 a^{6} b^{10} x^{8} + 105 a^{5} b^{11} x^{10} + 35 a^{4} b^{12} x^{12}} + \frac {5 A a^{3} b^{\frac {25}{2}} x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{35 a^{7} b^{9} x^{6} + 105 a^{6} b^{10} x^{8} + 105 a^{5} b^{11} x^{10} + 35 a^{4} b^{12} x^{12}} + \frac {30 A a^{2} b^{\frac {27}{2}} x^{8} \sqrt {\frac {a}{b x^{2}} + 1}}{35 a^{7} b^{9} x^{6} + 105 a^{6} b^{10} x^{8} + 105 a^{5} b^{11} x^{10} + 35 a^{4} b^{12} x^{12}} + \frac {40 A a b^{\frac {29}{2}} x^{10} \sqrt {\frac {a}{b x^{2}} + 1}}{35 a^{7} b^{9} x^{6} + 105 a^{6} b^{10} x^{8} + 105 a^{5} b^{11} x^{10} + 35 a^{4} b^{12} x^{12}} + \frac {16 A b^{\frac {31}{2}} x^{12} \sqrt {\frac {a}{b x^{2}} + 1}}{35 a^{7} b^{9} x^{6} + 105 a^{6} b^{10} x^{8} + 105 a^{5} b^{11} x^{10} + 35 a^{4} b^{12} x^{12}} - \frac {3 B a^{4} b^{\frac {9}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {2 B a^{3} b^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {3 B a^{2} b^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {12 B a b^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {8 B b^{\frac {17}{2}} x^{8} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**8/(b*x**2+a)**(1/2),x)

[Out]

-5*A*a**6*b**(19/2)*sqrt(a/(b*x**2) + 1)/(35*a**7*b**9*x**6 + 105*a**6*b**10*x**8 + 105*a**5*b**11*x**10 + 35*
a**4*b**12*x**12) - 9*A*a**5*b**(21/2)*x**2*sqrt(a/(b*x**2) + 1)/(35*a**7*b**9*x**6 + 105*a**6*b**10*x**8 + 10
5*a**5*b**11*x**10 + 35*a**4*b**12*x**12) - 5*A*a**4*b**(23/2)*x**4*sqrt(a/(b*x**2) + 1)/(35*a**7*b**9*x**6 +
105*a**6*b**10*x**8 + 105*a**5*b**11*x**10 + 35*a**4*b**12*x**12) + 5*A*a**3*b**(25/2)*x**6*sqrt(a/(b*x**2) +
1)/(35*a**7*b**9*x**6 + 105*a**6*b**10*x**8 + 105*a**5*b**11*x**10 + 35*a**4*b**12*x**12) + 30*A*a**2*b**(27/2
)*x**8*sqrt(a/(b*x**2) + 1)/(35*a**7*b**9*x**6 + 105*a**6*b**10*x**8 + 105*a**5*b**11*x**10 + 35*a**4*b**12*x*
*12) + 40*A*a*b**(29/2)*x**10*sqrt(a/(b*x**2) + 1)/(35*a**7*b**9*x**6 + 105*a**6*b**10*x**8 + 105*a**5*b**11*x
**10 + 35*a**4*b**12*x**12) + 16*A*b**(31/2)*x**12*sqrt(a/(b*x**2) + 1)/(35*a**7*b**9*x**6 + 105*a**6*b**10*x*
*8 + 105*a**5*b**11*x**10 + 35*a**4*b**12*x**12) - 3*B*a**4*b**(9/2)*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 +
30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 2*B*a**3*b**(11/2)*x**2*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30
*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 3*B*a**2*b**(13/2)*x**4*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a*
*4*b**5*x**6 + 15*a**3*b**6*x**8) - 12*B*a*b**(15/2)*x**6*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b*
*5*x**6 + 15*a**3*b**6*x**8) - 8*B*b**(17/2)*x**8*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6
+ 15*a**3*b**6*x**8)

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