3.551 \(\int \frac {(a+b x^2)^{5/2} (A+B x^2)}{x^8} \, dx\)

Optimal. Leaf size=108 \[ -\frac {A \left (a+b x^2\right )^{7/2}}{7 a x^7}+b^{5/2} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {b^2 B \sqrt {a+b x^2}}{x}-\frac {B \left (a+b x^2\right )^{5/2}}{5 x^5}-\frac {b B \left (a+b x^2\right )^{3/2}}{3 x^3} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {451, 277, 217, 206} \[ -\frac {A \left (a+b x^2\right )^{7/2}}{7 a x^7}-\frac {b^2 B \sqrt {a+b x^2}}{x}+b^{5/2} B \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {B \left (a+b x^2\right )^{5/2}}{5 x^5}-\frac {b B \left (a+b x^2\right )^{3/2}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^(5/2)*(A + B*x^2))/x^8,x]

[Out]

-((b^2*B*Sqrt[a + b*x^2])/x) - (b*B*(a + b*x^2)^(3/2))/(3*x^3) - (B*(a + b*x^2)^(5/2))/(5*x^5) - (A*(a + b*x^2
)^(7/2))/(7*a*x^7) + b^(5/2)*B*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 277

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 451

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[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1]))

Rubi steps

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

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Mathematica [C]  time = 0.09, size = 78, normalized size = 0.72 \[ -\frac {a^2 B \sqrt {a+b x^2} \, _2F_1\left (-\frac {5}{2},-\frac {5}{2};-\frac {3}{2};-\frac {b x^2}{a}\right )}{5 x^5 \sqrt {\frac {b x^2}{a}+1}}-\frac {A \left (a+b x^2\right )^{7/2}}{7 a x^7} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^(5/2)*(A + B*x^2))/x^8,x]

[Out]

-1/7*(A*(a + b*x^2)^(7/2))/(a*x^7) - (a^2*B*Sqrt[a + b*x^2]*Hypergeometric2F1[-5/2, -5/2, -3/2, -((b*x^2)/a)])
/(5*x^5*Sqrt[1 + (b*x^2)/a])

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fricas [A]  time = 0.74, size = 234, normalized size = 2.17 \[ \left [\frac {105 \, B a b^{\frac {5}{2}} x^{7} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left ({\left (161 \, B a b^{2} + 15 \, A b^{3}\right )} x^{6} + {\left (77 \, B a^{2} b + 45 \, A a b^{2}\right )} x^{4} + 15 \, A a^{3} + 3 \, {\left (7 \, B a^{3} + 15 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{210 \, a x^{7}}, -\frac {105 \, B a \sqrt {-b} b^{2} x^{7} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left ({\left (161 \, B a b^{2} + 15 \, A b^{3}\right )} x^{6} + {\left (77 \, B a^{2} b + 45 \, A a b^{2}\right )} x^{4} + 15 \, A a^{3} + 3 \, {\left (7 \, B a^{3} + 15 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{105 \, a x^{7}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^8,x, algorithm="fricas")

[Out]

[1/210*(105*B*a*b^(5/2)*x^7*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*((161*B*a*b^2 + 15*A*b^3)*x^6
+ (77*B*a^2*b + 45*A*a*b^2)*x^4 + 15*A*a^3 + 3*(7*B*a^3 + 15*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a*x^7), -1/105*(1
05*B*a*sqrt(-b)*b^2*x^7*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + ((161*B*a*b^2 + 15*A*b^3)*x^6 + (77*B*a^2*b + 45*
A*a*b^2)*x^4 + 15*A*a^3 + 3*(7*B*a^3 + 15*A*a^2*b)*x^2)*sqrt(b*x^2 + a))/(a*x^7)]

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giac [B]  time = 0.58, size = 320, normalized size = 2.96 \[ -\frac {1}{2} \, B b^{\frac {5}{2}} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, {\left (315 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{12} B a b^{\frac {5}{2}} + 105 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{12} A b^{\frac {7}{2}} - 1260 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{10} B a^{2} b^{\frac {5}{2}} + 2555 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} B a^{3} b^{\frac {5}{2}} + 525 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} A a^{2} b^{\frac {7}{2}} - 3080 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} B a^{4} b^{\frac {5}{2}} + 2121 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a^{5} b^{\frac {5}{2}} + 315 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a^{4} b^{\frac {7}{2}} - 812 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{6} b^{\frac {5}{2}} + 161 \, B a^{7} b^{\frac {5}{2}} + 15 \, A a^{6} b^{\frac {7}{2}}\right )}}{105 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^8,x, algorithm="giac")

[Out]

-1/2*B*b^(5/2)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2) + 2/105*(315*(sqrt(b)*x - sqrt(b*x^2 + a))^12*B*a*b^(5/2)
+ 105*(sqrt(b)*x - sqrt(b*x^2 + a))^12*A*b^(7/2) - 1260*(sqrt(b)*x - sqrt(b*x^2 + a))^10*B*a^2*b^(5/2) + 2555*
(sqrt(b)*x - sqrt(b*x^2 + a))^8*B*a^3*b^(5/2) + 525*(sqrt(b)*x - sqrt(b*x^2 + a))^8*A*a^2*b^(7/2) - 3080*(sqrt
(b)*x - sqrt(b*x^2 + a))^6*B*a^4*b^(5/2) + 2121*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a^5*b^(5/2) + 315*(sqrt(b)*x
- sqrt(b*x^2 + a))^4*A*a^4*b^(7/2) - 812*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^6*b^(5/2) + 161*B*a^7*b^(5/2) +
15*A*a^6*b^(7/2))/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^7

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maple [A]  time = 0.02, size = 155, normalized size = 1.44 \[ B \,b^{\frac {5}{2}} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )+\frac {\sqrt {b \,x^{2}+a}\, B \,b^{3} x}{a}+\frac {2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} B \,b^{3} x}{3 a^{2}}+\frac {8 \left (b \,x^{2}+a \right )^{\frac {5}{2}} B \,b^{3} x}{15 a^{3}}-\frac {8 \left (b \,x^{2}+a \right )^{\frac {7}{2}} B \,b^{2}}{15 a^{3} x}-\frac {2 \left (b \,x^{2}+a \right )^{\frac {7}{2}} B b}{15 a^{2} x^{3}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} B}{5 a \,x^{5}}-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}} A}{7 a \,x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2)*(B*x^2+A)/x^8,x)

[Out]

-1/5*B/a/x^5*(b*x^2+a)^(7/2)-2/15*B*b/a^2/x^3*(b*x^2+a)^(7/2)-8/15*B*b^2/a^3/x*(b*x^2+a)^(7/2)+8/15*B*b^3/a^3*
x*(b*x^2+a)^(5/2)+2/3*B*b^3/a^2*x*(b*x^2+a)^(3/2)+B*b^3/a*x*(b*x^2+a)^(1/2)+B*b^(5/2)*ln(b^(1/2)*x+(b*x^2+a)^(
1/2))-1/7*A*(b*x^2+a)^(7/2)/a/x^7

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maxima [A]  time = 1.09, size = 128, normalized size = 1.19 \[ \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{3} x}{3 \, a^{2}} + \frac {\sqrt {b x^{2} + a} B b^{3} x}{a} + B b^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{2}}{15 \, a^{2} x} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B b}{15 \, a^{2} x^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B}{5 \, a x^{5}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{7 \, a x^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2)*(B*x^2+A)/x^8,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{5/2}}{x^8} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x^2)*(a + b*x^2)^(5/2))/x^8,x)

[Out]

int(((A + B*x^2)*(a + b*x^2)^(5/2))/x^8, x)

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sympy [B]  time = 14.27, size = 592, normalized size = 5.48 \[ - \frac {15 A a^{7} b^{\frac {9}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {33 A a^{6} b^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {17 A a^{5} b^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {3 A a^{4} b^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {12 A a^{3} b^{\frac {17}{2}} x^{8} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {8 A a^{2} b^{\frac {19}{2}} x^{10} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {2 A a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {7 A b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 x^{2}} - \frac {A b^{\frac {7}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a} - \frac {B \sqrt {a} b^{2}}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {11 B a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 x^{2}} - \frac {8 B b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15} + B b^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {B b^{3} x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2)*(B*x**2+A)/x**8,x)

[Out]

-15*A*a**7*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**6*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**5*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**4*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**3*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**2*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) - 2*A*a*b**(3/2)*sqrt(a/(b*x**2) + 1)/(5*x**4) - 7*A*b**(5/2)*sqrt(a/(b*x**2) + 1)/(15*x**2) - A*b**(7/2)*
sqrt(a/(b*x**2) + 1)/(15*a) - B*sqrt(a)*b**2/(x*sqrt(1 + b*x**2/a)) - B*a**2*sqrt(b)*sqrt(a/(b*x**2) + 1)/(5*x
**4) - 11*B*a*b**(3/2)*sqrt(a/(b*x**2) + 1)/(15*x**2) - 8*B*b**(5/2)*sqrt(a/(b*x**2) + 1)/15 + B*b**(5/2)*asin
h(sqrt(b)*x/sqrt(a)) - B*b**3*x/(sqrt(a)*sqrt(1 + b*x**2/a))

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