\(\int \frac {(a+b x^2)^{5/2} (A+B x^2)}{x^8} \, dx\) [551]

Optimal result

Integrand size = 22, antiderivative size = 108 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^8} \, dx=-\frac {b^2 B \sqrt {a+b x^2}}{x}-\frac {b B \left (a+b x^2\right )^{3/2}}{3 x^3}-\frac {B \left (a+b x^2\right )^{5/2}}{5 x^5}-\frac {A \left (a+b x^2\right )^{7/2}}{7 a x^7}+b^{5/2} B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \]

[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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {462, 283, 223, 212} \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^8} \, dx=-\frac {A \left (a+b x^2\right )^{7/2}}{7 a x^7}+b^{5/2} B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )-\frac {b^2 B \sqrt {a+b x^2}}{x}-\frac {B \left (a+b x^2\right )^{5/2}}{5 x^5}-\frac {b B \left (a+b x^2\right )^{3/2}}{3 x^3} \]

[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 212

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

Rule 223

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 283

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 462

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*} \text {integral}& = -\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 ) \text {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*}

Mathematica [A] (verified)

Time = 0.22 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.04 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^8} \, dx=-\frac {\sqrt {a+b x^2} \left (15 A b^3 x^6+3 a^3 \left (5 A+7 B x^2\right )+a^2 b x^2 \left (45 A+77 B x^2\right )+a b^2 x^4 \left (45 A+161 B x^2\right )\right )}{105 a x^7}-b^{5/2} B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right ) \]

[In]

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

[Out]

-1/105*(Sqrt[a + b*x^2]*(15*A*b^3*x^6 + 3*a^3*(5*A + 7*B*x^2) + a^2*b*x^2*(45*A + 77*B*x^2) + a*b^2*x^4*(45*A
+ 161*B*x^2)))/(a*x^7) - b^(5/2)*B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]]

Maple [A] (verified)

Time = 2.94 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.95

method	result	size
pseudoelliptic	\(\frac {7 B a \,b^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right ) x^{7}-\left (\left (\frac {7 x^{2} B}{5}+A \right ) a^{3}+3 x^{2} b \left (\frac {77 x^{2} B}{45}+A \right ) a^{2}+3 x^{4} \left (\frac {161 x^{2} B}{45}+A \right ) b^{2} a +x^{6} b^{3} A \right ) \sqrt {b \,x^{2}+a}}{7 a \,x^{7}}\)	\(103\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (15 x^{6} b^{3} A +161 x^{6} a \,b^{2} B +45 A a \,b^{2} x^{4}+77 B \,a^{2} b \,x^{4}+45 A \,a^{2} b \,x^{2}+21 B \,a^{3} x^{2}+15 a^{3} A \right )}{105 x^{7} a}+B \,b^{\frac {5}{2}} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )\)	\(105\)
default	\(-\frac {A \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{7 a \,x^{7}}+B \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{5 a \,x^{5}}+\frac {2 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{3 a \,x^{3}}+\frac {4 b \left (-\frac {\left (b \,x^{2}+a \right )^{\frac {7}{2}}}{a x}+\frac {6 b \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{a}\right )}{3 a}\right )}{5 a}\right )\)	\(161\)

[In]

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

[Out]

1/7*(7*B*a*b^(5/2)*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))*x^7-((7/5*x^2*B+A)*a^3+3*x^2*b*(77/45*x^2*B+A)*a^2+3*x^4
*(161/45*x^2*B+A)*b^2*a+x^6*b^3*A)*(b*x^2+a)^(1/2))/a/x^7

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 234, normalized size of antiderivative = 2.17 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^8} \, dx=\left [\frac {105 \, B a b^{\frac {5}{2}} x^{7} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left ({\left (161 \, B a b^{2} + 15 \, A b^{3}\right )} x^{6} + {\left (77 \, B a^{2} b + 45 \, A a b^{2}\right )} x^{4} + 15 \, A a^{3} + 3 \, {\left (7 \, B a^{3} + 15 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{210 \, a x^{7}}, -\frac {105 \, B a \sqrt {-b} b^{2} x^{7} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left ({\left (161 \, B a b^{2} + 15 \, A b^{3}\right )} x^{6} + {\left (77 \, B a^{2} b + 45 \, A a b^{2}\right )} x^{4} + 15 \, A a^{3} + 3 \, {\left (7 \, B a^{3} + 15 \, A a^{2} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{105 \, a x^{7}}\right ] \]

[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)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 592 vs. \(2 (95) = 190\).

Time = 3.41 (sec) , antiderivative size = 592, normalized size of antiderivative = 5.48 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^8} \, dx=- \frac {15 A a^{7} b^{\frac {9}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {33 A a^{6} b^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {17 A a^{5} b^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {3 A a^{4} b^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {12 A a^{3} b^{\frac {17}{2}} x^{8} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {8 A a^{2} b^{\frac {19}{2}} x^{10} \sqrt {\frac {a}{b x^{2}} + 1}}{105 a^{5} b^{4} x^{6} + 210 a^{4} b^{5} x^{8} + 105 a^{3} b^{6} x^{10}} - \frac {2 A a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {7 A b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 x^{2}} - \frac {A b^{\frac {7}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a} - \frac {B \sqrt {a} b^{2}}{x \sqrt {1 + \frac {b x^{2}}{a}}} - \frac {B a^{2} \sqrt {b} \sqrt {\frac {a}{b x^{2}} + 1}}{5 x^{4}} - \frac {11 B a b^{\frac {3}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 x^{2}} - \frac {8 B b^{\frac {5}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15} + B b^{\frac {5}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )} - \frac {B b^{3} x}{\sqrt {a} \sqrt {1 + \frac {b x^{2}}{a}}} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.19 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^8} \, dx=\frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} B b^{3} x}{3 \, a^{2}} + \frac {\sqrt {b x^{2} + a} B b^{3} x}{a} + B b^{\frac {5}{2}} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right ) - \frac {8 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} B b^{2}}{15 \, a^{2} x} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B b}{15 \, a^{2} x^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B}{5 \, a x^{5}} - \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A}{7 \, a x^{7}} \]

[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)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 320 vs. \(2 (88) = 176\).

Time = 0.32 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.96 \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^8} \, dx=-\frac {1}{2} \, B b^{\frac {5}{2}} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right ) + \frac {2 \, {\left (315 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{12} B a b^{\frac {5}{2}} + 105 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{12} A b^{\frac {7}{2}} - 1260 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{10} B a^{2} b^{\frac {5}{2}} + 2555 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} B a^{3} b^{\frac {5}{2}} + 525 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} A a^{2} b^{\frac {7}{2}} - 3080 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} B a^{4} b^{\frac {5}{2}} + 2121 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a^{5} b^{\frac {5}{2}} + 315 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a^{4} b^{\frac {7}{2}} - 812 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{6} b^{\frac {5}{2}} + 161 \, B a^{7} b^{\frac {5}{2}} + 15 \, A a^{6} b^{\frac {7}{2}}\right )}}{105 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{7}} \]

[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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{5/2} \left (A+B x^2\right )}{x^8} \, dx=\int \frac {\left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{5/2}}{x^8} \,d x \]

[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)