\(\int \frac {(a+b x^2)^{3/2} (A+B x^2)}{x^8} \, dx\) [534]

Optimal result

Integrand size = 22, antiderivative size = 53 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^8} \, dx=-\frac {A \left (a+b x^2\right )^{5/2}}{7 a x^7}+\frac {(2 A b-7 a B) \left (a+b x^2\right )^{5/2}}{35 a^2 x^5} \]

[Out]

-1/7*A*(b*x^2+a)^(5/2)/a/x^7+1/35*(2*A*b-7*B*a)*(b*x^2+a)^(5/2)/a^2/x^5

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {464, 270} \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^8} \, dx=\frac {\left (a+b x^2\right )^{5/2} (2 A b-7 a B)}{35 a^2 x^5}-\frac {A \left (a+b x^2\right )^{5/2}}{7 a x^7} \]

[In]

Int[((a + b*x^2)^(3/2)*(A + B*x^2))/x^8,x]

[Out]

-1/7*(A*(a + b*x^2)^(5/2))/(a*x^7) + ((2*A*b - 7*a*B)*(a + b*x^2)^(5/2))/(35*a^2*x^5)

Rule 270

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 464

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*} \text {integral}& = -\frac {A \left (a+b x^2\right )^{5/2}}{7 a x^7}-\frac {(2 A b-7 a B) \int \frac {\left (a+b x^2\right )^{3/2}}{x^6} \, dx}{7 a} \\ & = -\frac {A \left (a+b x^2\right )^{5/2}}{7 a x^7}+\frac {(2 A b-7 a B) \left (a+b x^2\right )^{5/2}}{35 a^2 x^5} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^8} \, dx=\frac {\left (a+b x^2\right )^{5/2} \left (-5 a A+2 A b x^2-7 a B x^2\right )}{35 a^2 x^7} \]

[In]

Integrate[((a + b*x^2)^(3/2)*(A + B*x^2))/x^8,x]

[Out]

((a + b*x^2)^(5/2)*(-5*a*A + 2*A*b*x^2 - 7*a*B*x^2))/(35*a^2*x^7)

Maple [A] (verified)

Time = 2.78 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.68

method	result	size
pseudoelliptic	\(-\frac {\left (\left (\frac {7 x^{2} B}{5}+A \right ) a -\frac {2 A b \,x^{2}}{5}\right ) \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{7 x^{7} a^{2}}\)	\(36\)
gosper	\(-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}} \left (-2 A b \,x^{2}+7 B a \,x^{2}+5 A a \right )}{35 x^{7} a^{2}}\)	\(37\)
default	\(A \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {5}{2}}}{7 a \,x^{7}}+\frac {2 b \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{35 a^{2} x^{5}}\right )-\frac {B \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5 a \,x^{5}}\)	\(58\)
trager	\(-\frac {\left (-2 x^{6} b^{3} A +7 x^{6} a \,b^{2} B +A a \,b^{2} x^{4}+14 B \,a^{2} b \,x^{4}+8 A \,a^{2} b \,x^{2}+7 B \,a^{3} x^{2}+5 a^{3} A \right ) \sqrt {b \,x^{2}+a}}{35 x^{7} a^{2}}\)	\(82\)
risch	\(-\frac {\left (-2 x^{6} b^{3} A +7 x^{6} a \,b^{2} B +A a \,b^{2} x^{4}+14 B \,a^{2} b \,x^{4}+8 A \,a^{2} b \,x^{2}+7 B \,a^{3} x^{2}+5 a^{3} A \right ) \sqrt {b \,x^{2}+a}}{35 x^{7} a^{2}}\)	\(82\)

[In]

int((b*x^2+a)^(3/2)*(B*x^2+A)/x^8,x,method=_RETURNVERBOSE)

[Out]

-1/7*((7/5*x^2*B+A)*a-2/5*A*b*x^2)*(b*x^2+a)^(5/2)/x^7/a^2

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.47 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^8} \, dx=-\frac {{\left ({\left (7 \, B a b^{2} - 2 \, A b^{3}\right )} x^{6} + {\left (14 \, B a^{2} b + A a b^{2}\right )} x^{4} + 5 \, A a^{3} + {\left (7 \, B a^{3} + 8 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{35 \, a^{2} x^{7}} \]

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^8,x, algorithm="fricas")

[Out]

-1/35*((7*B*a*b^2 - 2*A*b^3)*x^6 + (14*B*a^2*b + A*a*b^2)*x^4 + 5*A*a^3 + (7*B*a^3 + 8*A*a^2*b)*x^2)*sqrt(b*x^
2 + a)/(a^2*x^7)

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 518 vs. \(2 (46) = 92\).

Time = 2.34 (sec) , antiderivative size = 518, normalized size of antiderivative = 9.77 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^8} \, dx=- \frac {15 A a^{6} b^{\frac {9}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {33 A a^{5} b^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {17 A a^{4} b^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {3 A a^{3} b^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {12 A a^{2} b^{\frac {17}{2}} x^{8} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {8 A a b^{\frac {19}{2}} x^{10} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {A b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {A b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a x^{2}} + \frac {2 A b^{\frac {7}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{2}} - \frac {B a \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {2 B b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{2}} - \frac {B b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{5 a} \]

[In]

integrate((b*x**2+a)**(3/2)*(B*x**2+A)/x**8,x)

[Out]

-15*A*a**6*b**(9/2)*sqrt(a/(b*x**2) + 1)/(105*a**5*b**4*x**6 + 210*a**4*b**5*x**8 + 105*a**3*b**6*x**10) - 33*
A*a**5*b**(11/2)*x**2*sqrt(a/(b*x**2) + 1)/(105*a**5*b**4*x**6 + 210*a**4*b**5*x**8 + 105*a**3*b**6*x**10) - 1
7*A*a**4*b**(13/2)*x**4*sqrt(a/(b*x**2) + 1)/(105*a**5*b**4*x**6 + 210*a**4*b**5*x**8 + 105*a**3*b**6*x**10) -
3*A*a**3*b**(15/2)*x**6*sqrt(a/(b*x**2) + 1)/(105*a**5*b**4*x**6 + 210*a**4*b**5*x**8 + 105*a**3*b**6*x**10)
- 12*A*a**2*b**(17/2)*x**8*sqrt(a/(b*x**2) + 1)/(105*a**5*b**4*x**6 + 210*a**4*b**5*x**8 + 105*a**3*b**6*x**10
) - 8*A*a*b**(19/2)*x**10*sqrt(a/(b*x**2) + 1)/(105*a**5*b**4*x**6 + 210*a**4*b**5*x**8 + 105*a**3*b**6*x**10)
- A*b**(3/2)*sqrt(a/(b*x**2) + 1)/(5*x**4) - A*b**(5/2)*sqrt(a/(b*x**2) + 1)/(15*a*x**2) + 2*A*b**(7/2)*sqrt(
a/(b*x**2) + 1)/(15*a**2) - B*a*sqrt(b)*sqrt(a/(b*x**2) + 1)/(5*x**4) - 2*B*b**(3/2)*sqrt(a/(b*x**2) + 1)/(5*x
**2) - B*b**(5/2)*sqrt(a/(b*x**2) + 1)/(5*a)

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.06 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^8} \, dx=-\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B}{5 \, a x^{5}} + \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} A b}{35 \, a^{2} x^{5}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{7 \, a x^{7}} \]

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^8,x, algorithm="maxima")

[Out]

-1/5*(b*x^2 + a)^(5/2)*B/(a*x^5) + 2/35*(b*x^2 + a)^(5/2)*A*b/(a^2*x^5) - 1/7*(b*x^2 + a)^(5/2)*A/(a*x^7)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 344 vs. \(2 (45) = 90\).

Time = 0.32 (sec) , antiderivative size = 344, normalized size of antiderivative = 6.49 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^8} \, dx=\frac {2 \, {\left (35 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{12} B b^{\frac {5}{2}} - 70 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{10} B a b^{\frac {5}{2}} + 70 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{10} A b^{\frac {7}{2}} + 105 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} B a^{2} b^{\frac {5}{2}} + 70 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} A a b^{\frac {7}{2}} - 140 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} B a^{3} b^{\frac {5}{2}} + 140 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} A a^{2} b^{\frac {7}{2}} + 77 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a^{4} b^{\frac {5}{2}} + 28 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a^{3} b^{\frac {7}{2}} - 14 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{5} b^{\frac {5}{2}} + 14 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a^{4} b^{\frac {7}{2}} + 7 \, B a^{6} b^{\frac {5}{2}} - 2 \, A a^{5} b^{\frac {7}{2}}\right )}}{35 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{7}} \]

[In]

integrate((b*x^2+a)^(3/2)*(B*x^2+A)/x^8,x, algorithm="giac")

[Out]

2/35*(35*(sqrt(b)*x - sqrt(b*x^2 + a))^12*B*b^(5/2) - 70*(sqrt(b)*x - sqrt(b*x^2 + a))^10*B*a*b^(5/2) + 70*(sq
rt(b)*x - sqrt(b*x^2 + a))^10*A*b^(7/2) + 105*(sqrt(b)*x - sqrt(b*x^2 + a))^8*B*a^2*b^(5/2) + 70*(sqrt(b)*x -
sqrt(b*x^2 + a))^8*A*a*b^(7/2) - 140*(sqrt(b)*x - sqrt(b*x^2 + a))^6*B*a^3*b^(5/2) + 140*(sqrt(b)*x - sqrt(b*x
^2 + a))^6*A*a^2*b^(7/2) + 77*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a^4*b^(5/2) + 28*(sqrt(b)*x - sqrt(b*x^2 + a))
^4*A*a^3*b^(7/2) - 14*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^5*b^(5/2) + 14*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*a^4
*b^(7/2) + 7*B*a^6*b^(5/2) - 2*A*a^5*b^(7/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^7

Mupad [B] (verification not implemented)

Time = 6.36 (sec) , antiderivative size = 128, normalized size of antiderivative = 2.42 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2\right )}{x^8} \, dx=\frac {2\,A\,b^3\,\sqrt {b\,x^2+a}}{35\,a^2\,x}-\frac {8\,A\,b\,\sqrt {b\,x^2+a}}{35\,x^5}-\frac {B\,a\,\sqrt {b\,x^2+a}}{5\,x^5}-\frac {2\,B\,b\,\sqrt {b\,x^2+a}}{5\,x^3}-\frac {A\,b^2\,\sqrt {b\,x^2+a}}{35\,a\,x^3}-\frac {A\,a\,\sqrt {b\,x^2+a}}{7\,x^7}-\frac {B\,b^2\,\sqrt {b\,x^2+a}}{5\,a\,x} \]

[In]

int(((A + B*x^2)*(a + b*x^2)^(3/2))/x^8,x)

[Out]

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