3.6.97 \(\int \frac {A+B x^2}{x^6 (a+b x^2)^{5/2}} \, dx\) [597]

Optimal. Leaf size=146 \[ -\frac {A}{5 a x^5 \left (a+b x^2\right )^{3/2}}+\frac {8 A b-5 a B}{15 a^2 x^3 \left (a+b x^2\right )^{3/2}}-\frac {2 b (8 A b-5 a B)}{5 a^3 x \left (a+b x^2\right )^{3/2}}-\frac {8 b^2 (8 A b-5 a B) x}{15 a^4 \left (a+b x^2\right )^{3/2}}-\frac {16 b^2 (8 A b-5 a B) x}{15 a^5 \sqrt {a+b x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.04, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {464, 277, 198, 197} \begin {gather*} -\frac {16 b^2 x (8 A b-5 a B)}{15 a^5 \sqrt {a+b x^2}}-\frac {8 b^2 x (8 A b-5 a B)}{15 a^4 \left (a+b x^2\right )^{3/2}}-\frac {2 b (8 A b-5 a B)}{5 a^3 x \left (a+b x^2\right )^{3/2}}+\frac {8 A b-5 a B}{15 a^2 x^3 \left (a+b x^2\right )^{3/2}}-\frac {A}{5 a x^5 \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x^2)/(x^6*(a + b*x^2)^(5/2)),x]

[Out]

-1/5*A/(a*x^5*(a + b*x^2)^(3/2)) + (8*A*b - 5*a*B)/(15*a^2*x^3*(a + b*x^2)^(3/2)) - (2*b*(8*A*b - 5*a*B))/(5*a
^3*x*(a + b*x^2)^(3/2)) - (8*b^2*(8*A*b - 5*a*B)*x)/(15*a^4*(a + b*x^2)^(3/2)) - (16*b^2*(8*A*b - 5*a*B)*x)/(1
5*a^5*Sqrt[a + b*x^2])

Rule 197

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[x*((a + b*x^n)^(p + 1)/a), x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 198

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p +
1], 0] && NeQ[p, -1]

Rule 277

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 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*} \int \frac {A+B x^2}{x^6 \left (a+b x^2\right )^{5/2}} \, dx &=-\frac {A}{5 a x^5 \left (a+b x^2\right )^{3/2}}-\frac {(8 A b-5 a B) \int \frac {1}{x^4 \left (a+b x^2\right )^{5/2}} \, dx}{5 a}\\ &=-\frac {A}{5 a x^5 \left (a+b x^2\right )^{3/2}}+\frac {8 A b-5 a B}{15 a^2 x^3 \left (a+b x^2\right )^{3/2}}+\frac {(2 b (8 A b-5 a B)) \int \frac {1}{x^2 \left (a+b x^2\right )^{5/2}} \, dx}{5 a^2}\\ &=-\frac {A}{5 a x^5 \left (a+b x^2\right )^{3/2}}+\frac {8 A b-5 a B}{15 a^2 x^3 \left (a+b x^2\right )^{3/2}}-\frac {2 b (8 A b-5 a B)}{5 a^3 x \left (a+b x^2\right )^{3/2}}-\frac {\left (8 b^2 (8 A b-5 a B)\right ) \int \frac {1}{\left (a+b x^2\right )^{5/2}} \, dx}{5 a^3}\\ &=-\frac {A}{5 a x^5 \left (a+b x^2\right )^{3/2}}+\frac {8 A b-5 a B}{15 a^2 x^3 \left (a+b x^2\right )^{3/2}}-\frac {2 b (8 A b-5 a B)}{5 a^3 x \left (a+b x^2\right )^{3/2}}-\frac {8 b^2 (8 A b-5 a B) x}{15 a^4 \left (a+b x^2\right )^{3/2}}-\frac {\left (16 b^2 (8 A b-5 a B)\right ) \int \frac {1}{\left (a+b x^2\right )^{3/2}} \, dx}{15 a^4}\\ &=-\frac {A}{5 a x^5 \left (a+b x^2\right )^{3/2}}+\frac {8 A b-5 a B}{15 a^2 x^3 \left (a+b x^2\right )^{3/2}}-\frac {2 b (8 A b-5 a B)}{5 a^3 x \left (a+b x^2\right )^{3/2}}-\frac {8 b^2 (8 A b-5 a B) x}{15 a^4 \left (a+b x^2\right )^{3/2}}-\frac {16 b^2 (8 A b-5 a B) x}{15 a^5 \sqrt {a+b x^2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 0.17, size = 105, normalized size = 0.72 \begin {gather*} \frac {-128 A b^4 x^8+16 a b^3 x^6 \left (-12 A+5 B x^2\right )+24 a^2 b^2 x^4 \left (-2 A+5 B x^2\right )-a^4 \left (3 A+5 B x^2\right )+a^3 \left (8 A b x^2+30 b B x^4\right )}{15 a^5 x^5 \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x^2)/(x^6*(a + b*x^2)^(5/2)),x]

[Out]

(-128*A*b^4*x^8 + 16*a*b^3*x^6*(-12*A + 5*B*x^2) + 24*a^2*b^2*x^4*(-2*A + 5*B*x^2) - a^4*(3*A + 5*B*x^2) + a^3
*(8*A*b*x^2 + 30*b*B*x^4))/(15*a^5*x^5*(a + b*x^2)^(3/2))

________________________________________________________________________________________

Maple [A]
time = 0.11, size = 188, normalized size = 1.29

method	result	size
gosper	\(-\frac {128 A \,b^{4} x^{8}-80 B a \,b^{3} x^{8}+192 A a \,b^{3} x^{6}-120 B \,a^{2} b^{2} x^{6}+48 A \,a^{2} b^{2} x^{4}-30 B \,a^{3} b \,x^{4}-8 A \,a^{3} b \,x^{2}+5 B \,a^{4} x^{2}+3 A \,a^{4}}{15 x^{5} \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{5}}\)	\(107\)
trager	\(-\frac {128 A \,b^{4} x^{8}-80 B a \,b^{3} x^{8}+192 A a \,b^{3} x^{6}-120 B \,a^{2} b^{2} x^{6}+48 A \,a^{2} b^{2} x^{4}-30 B \,a^{3} b \,x^{4}-8 A \,a^{3} b \,x^{2}+5 B \,a^{4} x^{2}+3 A \,a^{4}}{15 x^{5} \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{5}}\)	\(107\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (73 A \,b^{2} x^{4}-40 B a b \,x^{4}-14 a A b \,x^{2}+5 B \,a^{2} x^{2}+3 a^{2} A \right )}{15 a^{5} x^{5}}-\frac {\sqrt {b \,x^{2}+a}\, x \left (11 A \,b^{2} x^{2}-8 B a b \,x^{2}+12 a b A -9 a^{2} B \right ) b^{2}}{3 a^{5} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )}\)	\(127\)
default	\(A \left (-\frac {1}{5 a \,x^{5} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {8 b \left (-\frac {1}{3 a \,x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 b \left (-\frac {1}{a x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{a}\right )}{a}\right )}{5 a}\right )+B \left (-\frac {1}{3 a \,x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 b \left (-\frac {1}{a x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{a}\right )}{a}\right )\)	\(188\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x^2+A)/x^6/(b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.35, size = 172, normalized size = 1.18 \begin {gather*} \frac {16 \, B b^{2} x}{3 \, \sqrt {b x^{2} + a} a^{4}} + \frac {8 \, B b^{2} x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3}} - \frac {128 \, A b^{3} x}{15 \, \sqrt {b x^{2} + a} a^{5}} - \frac {64 \, A b^{3} x}{15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{4}} + \frac {2 \, B b}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} x} - \frac {16 \, A b^{2}}{5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3} x} - \frac {B}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a x^{3}} + \frac {8 \, A b}{15 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} x^{3}} - \frac {A}{5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a x^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^6/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

16/3*B*b^2*x/(sqrt(b*x^2 + a)*a^4) + 8/3*B*b^2*x/((b*x^2 + a)^(3/2)*a^3) - 128/15*A*b^3*x/(sqrt(b*x^2 + a)*a^5
) - 64/15*A*b^3*x/((b*x^2 + a)^(3/2)*a^4) + 2*B*b/((b*x^2 + a)^(3/2)*a^2*x) - 16/5*A*b^2/((b*x^2 + a)^(3/2)*a^
3*x) - 1/3*B/((b*x^2 + a)^(3/2)*a*x^3) + 8/15*A*b/((b*x^2 + a)^(3/2)*a^2*x^3) - 1/5*A/((b*x^2 + a)^(3/2)*a*x^5
)

________________________________________________________________________________________

Fricas [A]
time = 1.95, size = 129, normalized size = 0.88 \begin {gather*} \frac {{\left (16 \, {\left (5 \, B a b^{3} - 8 \, A b^{4}\right )} x^{8} + 24 \, {\left (5 \, B a^{2} b^{2} - 8 \, A a b^{3}\right )} x^{6} - 3 \, A a^{4} + 6 \, {\left (5 \, B a^{3} b - 8 \, A a^{2} b^{2}\right )} x^{4} - {\left (5 \, B a^{4} - 8 \, A a^{3} b\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, {\left (a^{5} b^{2} x^{9} + 2 \, a^{6} b x^{7} + a^{7} x^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^6/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

1/15*(16*(5*B*a*b^3 - 8*A*b^4)*x^8 + 24*(5*B*a^2*b^2 - 8*A*a*b^3)*x^6 - 3*A*a^4 + 6*(5*B*a^3*b - 8*A*a^2*b^2)*
x^4 - (5*B*a^4 - 8*A*a^3*b)*x^2)*sqrt(b*x^2 + a)/(a^5*b^2*x^9 + 2*a^6*b*x^7 + a^7*x^5)

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 944 vs. \(2 (141) = 282\).
time = 16.30, size = 944, normalized size = 6.47 \begin {gather*} A \left (- \frac {3 a^{6} b^{\frac {33}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{6} + 90 a^{7} b^{18} x^{8} + 60 a^{6} b^{19} x^{10} + 15 a^{5} b^{20} x^{12}} + \frac {2 a^{5} b^{\frac {35}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{6} + 90 a^{7} b^{18} x^{8} + 60 a^{6} b^{19} x^{10} + 15 a^{5} b^{20} x^{12}} - \frac {35 a^{4} b^{\frac {37}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{6} + 90 a^{7} b^{18} x^{8} + 60 a^{6} b^{19} x^{10} + 15 a^{5} b^{20} x^{12}} - \frac {280 a^{3} b^{\frac {39}{2}} x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{6} + 90 a^{7} b^{18} x^{8} + 60 a^{6} b^{19} x^{10} + 15 a^{5} b^{20} x^{12}} - \frac {560 a^{2} b^{\frac {41}{2}} x^{8} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{6} + 90 a^{7} b^{18} x^{8} + 60 a^{6} b^{19} x^{10} + 15 a^{5} b^{20} x^{12}} - \frac {448 a b^{\frac {43}{2}} x^{10} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{6} + 90 a^{7} b^{18} x^{8} + 60 a^{6} b^{19} x^{10} + 15 a^{5} b^{20} x^{12}} - \frac {128 b^{\frac {45}{2}} x^{12} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{9} b^{16} x^{4} + 60 a^{8} b^{17} x^{6} + 90 a^{7} b^{18} x^{8} + 60 a^{6} b^{19} x^{10} + 15 a^{5} b^{20} x^{12}}\right ) + B \left (- \frac {a^{4} b^{\frac {19}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{7} b^{9} x^{2} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{6} + 3 a^{4} b^{12} x^{8}} + \frac {5 a^{3} b^{\frac {21}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{7} b^{9} x^{2} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{6} + 3 a^{4} b^{12} x^{8}} + \frac {30 a^{2} b^{\frac {23}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{7} b^{9} x^{2} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{6} + 3 a^{4} b^{12} x^{8}} + \frac {40 a b^{\frac {25}{2}} x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{7} b^{9} x^{2} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{6} + 3 a^{4} b^{12} x^{8}} + \frac {16 b^{\frac {27}{2}} x^{8} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{7} b^{9} x^{2} + 9 a^{6} b^{10} x^{4} + 9 a^{5} b^{11} x^{6} + 3 a^{4} b^{12} x^{8}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x**2+A)/x**6/(b*x**2+a)**(5/2),x)

[Out]

A*(-3*a**6*b**(33/2)*sqrt(a/(b*x**2) + 1)/(15*a**9*b**16*x**4 + 60*a**8*b**17*x**6 + 90*a**7*b**18*x**8 + 60*a
**6*b**19*x**10 + 15*a**5*b**20*x**12) + 2*a**5*b**(35/2)*x**2*sqrt(a/(b*x**2) + 1)/(15*a**9*b**16*x**4 + 60*a
**8*b**17*x**6 + 90*a**7*b**18*x**8 + 60*a**6*b**19*x**10 + 15*a**5*b**20*x**12) - 35*a**4*b**(37/2)*x**4*sqrt
(a/(b*x**2) + 1)/(15*a**9*b**16*x**4 + 60*a**8*b**17*x**6 + 90*a**7*b**18*x**8 + 60*a**6*b**19*x**10 + 15*a**5
*b**20*x**12) - 280*a**3*b**(39/2)*x**6*sqrt(a/(b*x**2) + 1)/(15*a**9*b**16*x**4 + 60*a**8*b**17*x**6 + 90*a**
7*b**18*x**8 + 60*a**6*b**19*x**10 + 15*a**5*b**20*x**12) - 560*a**2*b**(41/2)*x**8*sqrt(a/(b*x**2) + 1)/(15*a
**9*b**16*x**4 + 60*a**8*b**17*x**6 + 90*a**7*b**18*x**8 + 60*a**6*b**19*x**10 + 15*a**5*b**20*x**12) - 448*a*
b**(43/2)*x**10*sqrt(a/(b*x**2) + 1)/(15*a**9*b**16*x**4 + 60*a**8*b**17*x**6 + 90*a**7*b**18*x**8 + 60*a**6*b
**19*x**10 + 15*a**5*b**20*x**12) - 128*b**(45/2)*x**12*sqrt(a/(b*x**2) + 1)/(15*a**9*b**16*x**4 + 60*a**8*b**
17*x**6 + 90*a**7*b**18*x**8 + 60*a**6*b**19*x**10 + 15*a**5*b**20*x**12)) + B*(-a**4*b**(19/2)*sqrt(a/(b*x**2
) + 1)/(3*a**7*b**9*x**2 + 9*a**6*b**10*x**4 + 9*a**5*b**11*x**6 + 3*a**4*b**12*x**8) + 5*a**3*b**(21/2)*x**2*
sqrt(a/(b*x**2) + 1)/(3*a**7*b**9*x**2 + 9*a**6*b**10*x**4 + 9*a**5*b**11*x**6 + 3*a**4*b**12*x**8) + 30*a**2*
b**(23/2)*x**4*sqrt(a/(b*x**2) + 1)/(3*a**7*b**9*x**2 + 9*a**6*b**10*x**4 + 9*a**5*b**11*x**6 + 3*a**4*b**12*x
**8) + 40*a*b**(25/2)*x**6*sqrt(a/(b*x**2) + 1)/(3*a**7*b**9*x**2 + 9*a**6*b**10*x**4 + 9*a**5*b**11*x**6 + 3*
a**4*b**12*x**8) + 16*b**(27/2)*x**8*sqrt(a/(b*x**2) + 1)/(3*a**7*b**9*x**2 + 9*a**6*b**10*x**4 + 9*a**5*b**11
*x**6 + 3*a**4*b**12*x**8))

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (126) = 252\).
time = 1.63, size = 336, normalized size = 2.30 \begin {gather*} \frac {x {\left (\frac {{\left (8 \, B a^{5} b^{4} - 11 \, A a^{4} b^{5}\right )} x^{2}}{a^{9} b} + \frac {3 \, {\left (3 \, B a^{6} b^{3} - 4 \, A a^{5} b^{4}\right )}}{a^{9} b}\right )}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} - \frac {2 \, {\left (30 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} B a b^{\frac {3}{2}} - 45 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} A b^{\frac {5}{2}} - 150 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} B a^{2} b^{\frac {3}{2}} + 240 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} A a b^{\frac {5}{2}} + 250 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a^{3} b^{\frac {3}{2}} - 490 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A a^{2} b^{\frac {5}{2}} - 170 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{4} b^{\frac {3}{2}} + 320 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a^{3} b^{\frac {5}{2}} + 40 \, B a^{5} b^{\frac {3}{2}} - 73 \, A a^{4} b^{\frac {5}{2}}\right )}}{15 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5} a^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x^2+A)/x^6/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

1/3*x*((8*B*a^5*b^4 - 11*A*a^4*b^5)*x^2/(a^9*b) + 3*(3*B*a^6*b^3 - 4*A*a^5*b^4)/(a^9*b))/(b*x^2 + a)^(3/2) - 2
/15*(30*(sqrt(b)*x - sqrt(b*x^2 + a))^8*B*a*b^(3/2) - 45*(sqrt(b)*x - sqrt(b*x^2 + a))^8*A*b^(5/2) - 150*(sqrt
(b)*x - sqrt(b*x^2 + a))^6*B*a^2*b^(3/2) + 240*(sqrt(b)*x - sqrt(b*x^2 + a))^6*A*a*b^(5/2) + 250*(sqrt(b)*x -
sqrt(b*x^2 + a))^4*B*a^3*b^(3/2) - 490*(sqrt(b)*x - sqrt(b*x^2 + a))^4*A*a^2*b^(5/2) - 170*(sqrt(b)*x - sqrt(b
*x^2 + a))^2*B*a^4*b^(3/2) + 320*(sqrt(b)*x - sqrt(b*x^2 + a))^2*A*a^3*b^(5/2) + 40*B*a^5*b^(3/2) - 73*A*a^4*b
^(5/2))/(((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^5*a^4)

________________________________________________________________________________________

Mupad [B]
time = 0.49, size = 231, normalized size = 1.58 \begin {gather*} \frac {\frac {a\,\left (\frac {b^2\,\left (73\,A\,b-40\,B\,a\right )}{18\,a^4}+\frac {b^2\,\left (86\,A\,b-35\,B\,a\right )}{30\,a^4}+\frac {a\,\left (\frac {28\,A\,b^4-10\,B\,a\,b^3}{45\,a^5}-\frac {b^3\,\left (86\,A\,b-35\,B\,a\right )}{18\,a^5}\right )}{b}\right )}{b}-\frac {b\,\left (73\,A\,b-40\,B\,a\right )}{30\,a^3}}{x\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {x^2\,\left (\frac {28\,A\,b^3-10\,B\,a\,b^2}{15\,a^5}-\frac {2\,b^2\,\left (26\,A\,b-15\,B\,a\right )}{5\,a^5}\right )-\frac {b\,\left (26\,A\,b-15\,B\,a\right )}{5\,a^4}}{x\,\sqrt {b\,x^2+a}}-\frac {\sqrt {b\,x^2+a}\,\left (5\,B\,a^3-14\,A\,a^2\,b\right )}{15\,a^6\,x^3}-\frac {A\,\sqrt {b\,x^2+a}}{5\,a^3\,x^5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x^2)/(x^6*(a + b*x^2)^(5/2)),x)

[Out]

((a*((b^2*(73*A*b - 40*B*a))/(18*a^4) + (b^2*(86*A*b - 35*B*a))/(30*a^4) + (a*((28*A*b^4 - 10*B*a*b^3)/(45*a^5
) - (b^3*(86*A*b - 35*B*a))/(18*a^5)))/b))/b - (b*(73*A*b - 40*B*a))/(30*a^3))/(x*(a + b*x^2)^(3/2)) + (x^2*((
28*A*b^3 - 10*B*a*b^2)/(15*a^5) - (2*b^2*(26*A*b - 15*B*a))/(5*a^5)) - (b*(26*A*b - 15*B*a))/(5*a^4))/(x*(a +
b*x^2)^(1/2)) - ((a + b*x^2)^(1/2)*(5*B*a^3 - 14*A*a^2*b))/(15*a^6*x^3) - (A*(a + b*x^2)^(1/2))/(5*a^3*x^5)

________________________________________________________________________________________