3.1.42 \(\int \frac {A+B x}{x^3 (a+b x^2)^{5/2}} \, dx\)

Optimal. Leaf size=129 \[ \frac {5 A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{7/2}}-\frac {5 A \sqrt {a+b x^2}}{2 a^3 x^2}-\frac {8 B \sqrt {a+b x^2}}{3 a^3 x}+\frac {5 A+4 B x}{3 a^2 x^2 \sqrt {a+b x^2}}+\frac {A+B x}{3 a x^2 \left (a+b x^2\right )^{3/2}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {823, 835, 807, 266, 63, 208} \begin {gather*} \frac {5 A+4 B x}{3 a^2 x^2 \sqrt {a+b x^2}}-\frac {5 A \sqrt {a+b x^2}}{2 a^3 x^2}+\frac {5 A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{7/2}}-\frac {8 B \sqrt {a+b x^2}}{3 a^3 x}+\frac {A+B x}{3 a x^2 \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^3*(a + b*x^2)^(5/2)),x]

[Out]

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 266

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

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rubi steps

\begin {align*} \int \frac {A+B x}{x^3 \left (a+b x^2\right )^{5/2}} \, dx &=\frac {A+B x}{3 a x^2 \left (a+b x^2\right )^{3/2}}-\frac {\int \frac {-5 a A b-4 a b B x}{x^3 \left (a+b x^2\right )^{3/2}} \, dx}{3 a^2 b}\\ &=\frac {A+B x}{3 a x^2 \left (a+b x^2\right )^{3/2}}+\frac {5 A+4 B x}{3 a^2 x^2 \sqrt {a+b x^2}}+\frac {\int \frac {15 a^2 A b^2+8 a^2 b^2 B x}{x^3 \sqrt {a+b x^2}} \, dx}{3 a^4 b^2}\\ &=\frac {A+B x}{3 a x^2 \left (a+b x^2\right )^{3/2}}+\frac {5 A+4 B x}{3 a^2 x^2 \sqrt {a+b x^2}}-\frac {5 A \sqrt {a+b x^2}}{2 a^3 x^2}-\frac {\int \frac {-16 a^3 b^2 B+15 a^2 A b^3 x}{x^2 \sqrt {a+b x^2}} \, dx}{6 a^5 b^2}\\ &=\frac {A+B x}{3 a x^2 \left (a+b x^2\right )^{3/2}}+\frac {5 A+4 B x}{3 a^2 x^2 \sqrt {a+b x^2}}-\frac {5 A \sqrt {a+b x^2}}{2 a^3 x^2}-\frac {8 B \sqrt {a+b x^2}}{3 a^3 x}-\frac {(5 A b) \int \frac {1}{x \sqrt {a+b x^2}} \, dx}{2 a^3}\\ &=\frac {A+B x}{3 a x^2 \left (a+b x^2\right )^{3/2}}+\frac {5 A+4 B x}{3 a^2 x^2 \sqrt {a+b x^2}}-\frac {5 A \sqrt {a+b x^2}}{2 a^3 x^2}-\frac {8 B \sqrt {a+b x^2}}{3 a^3 x}-\frac {(5 A b) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )}{4 a^3}\\ &=\frac {A+B x}{3 a x^2 \left (a+b x^2\right )^{3/2}}+\frac {5 A+4 B x}{3 a^2 x^2 \sqrt {a+b x^2}}-\frac {5 A \sqrt {a+b x^2}}{2 a^3 x^2}-\frac {8 B \sqrt {a+b x^2}}{3 a^3 x}-\frac {(5 A) \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{2 a^3}\\ &=\frac {A+B x}{3 a x^2 \left (a+b x^2\right )^{3/2}}+\frac {5 A+4 B x}{3 a^2 x^2 \sqrt {a+b x^2}}-\frac {5 A \sqrt {a+b x^2}}{2 a^3 x^2}-\frac {8 B \sqrt {a+b x^2}}{3 a^3 x}+\frac {5 A b \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{7/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.22, size = 106, normalized size = 0.82 \begin {gather*} \frac {-\frac {3 a^3 (A+2 B x)}{x^2}-4 a^2 b (5 A+6 B x)-a b^2 x^2 (15 A+16 B x)+\frac {15 A b \left (a+b x^2\right )^2 \tanh ^{-1}\left (\sqrt {\frac {b x^2}{a}+1}\right )}{\sqrt {\frac {b x^2}{a}+1}}}{6 a^4 \left (a+b x^2\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^3*(a + b*x^2)^(5/2)),x]

[Out]

((-3*a^3*(A + 2*B*x))/x^2 - 4*a^2*b*(5*A + 6*B*x) - a*b^2*x^2*(15*A + 16*B*x) + (15*A*b*(a + b*x^2)^2*ArcTanh[
Sqrt[1 + (b*x^2)/a]])/Sqrt[1 + (b*x^2)/a])/(6*a^4*(a + b*x^2)^(3/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.63, size = 111, normalized size = 0.86 \begin {gather*} \frac {-3 a^2 A-6 a^2 B x-20 a A b x^2-24 a b B x^3-15 A b^2 x^4-16 b^2 B x^5}{6 a^3 x^2 \left (a+b x^2\right )^{3/2}}-\frac {5 A b \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}-\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(A + B*x)/(x^3*(a + b*x^2)^(5/2)),x]

[Out]

(-3*a^2*A - 6*a^2*B*x - 20*a*A*b*x^2 - 24*a*b*B*x^3 - 15*A*b^2*x^4 - 16*b^2*B*x^5)/(6*a^3*x^2*(a + b*x^2)^(3/2
)) - (5*A*b*ArcTanh[(Sqrt[b]*x)/Sqrt[a] - Sqrt[a + b*x^2]/Sqrt[a]])/a^(7/2)

________________________________________________________________________________________

fricas [A]  time = 0.79, size = 307, normalized size = 2.38 \begin {gather*} \left [\frac {15 \, {\left (A b^{3} x^{6} + 2 \, A a b^{2} x^{4} + A a^{2} b x^{2}\right )} \sqrt {a} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (16 \, B a b^{2} x^{5} + 15 \, A a b^{2} x^{4} + 24 \, B a^{2} b x^{3} + 20 \, A a^{2} b x^{2} + 6 \, B a^{3} x + 3 \, A a^{3}\right )} \sqrt {b x^{2} + a}}{12 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}, -\frac {15 \, {\left (A b^{3} x^{6} + 2 \, A a b^{2} x^{4} + A a^{2} b x^{2}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (16 \, B a b^{2} x^{5} + 15 \, A a b^{2} x^{4} + 24 \, B a^{2} b x^{3} + 20 \, A a^{2} b x^{2} + 6 \, B a^{3} x + 3 \, A a^{3}\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{4} b^{2} x^{6} + 2 \, a^{5} b x^{4} + a^{6} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[1/12*(15*(A*b^3*x^6 + 2*A*a*b^2*x^4 + A*a^2*b*x^2)*sqrt(a)*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2
) - 2*(16*B*a*b^2*x^5 + 15*A*a*b^2*x^4 + 24*B*a^2*b*x^3 + 20*A*a^2*b*x^2 + 6*B*a^3*x + 3*A*a^3)*sqrt(b*x^2 + a
))/(a^4*b^2*x^6 + 2*a^5*b*x^4 + a^6*x^2), -1/6*(15*(A*b^3*x^6 + 2*A*a*b^2*x^4 + A*a^2*b*x^2)*sqrt(-a)*arctan(s
qrt(-a)/sqrt(b*x^2 + a)) + (16*B*a*b^2*x^5 + 15*A*a*b^2*x^4 + 24*B*a^2*b*x^3 + 20*A*a^2*b*x^2 + 6*B*a^3*x + 3*
A*a^3)*sqrt(b*x^2 + a))/(a^4*b^2*x^6 + 2*a^5*b*x^4 + a^6*x^2)]

________________________________________________________________________________________

giac [A]  time = 0.53, size = 197, normalized size = 1.53 \begin {gather*} -\frac {{\left ({\left (\frac {5 \, B b^{2} x}{a^{3}} + \frac {6 \, A b^{2}}{a^{3}}\right )} x + \frac {6 \, B b}{a^{2}}\right )} x + \frac {7 \, A b}{a^{2}}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} - \frac {5 \, A b \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{3}} + \frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{3} A b + 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a \sqrt {b} + {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} A a b - 2 \, B a^{2} \sqrt {b}}{{\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{2} a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

-1/3*(((5*B*b^2*x/a^3 + 6*A*b^2/a^3)*x + 6*B*b/a^2)*x + 7*A*b/a^2)/(b*x^2 + a)^(3/2) - 5*A*b*arctan(-(sqrt(b)*
x - sqrt(b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^3) + ((sqrt(b)*x - sqrt(b*x^2 + a))^3*A*b + 2*(sqrt(b)*x - sqrt(b*x
^2 + a))^2*B*a*sqrt(b) + (sqrt(b)*x - sqrt(b*x^2 + a))*A*a*b - 2*B*a^2*sqrt(b))/(((sqrt(b)*x - sqrt(b*x^2 + a)
)^2 - a)^2*a^3)

________________________________________________________________________________________

maple [A]  time = 0.01, size = 134, normalized size = 1.04 \begin {gather*} -\frac {4 B b x}{3 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2}}-\frac {5 A b}{6 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2}}-\frac {8 B b x}{3 \sqrt {b \,x^{2}+a}\, a^{3}}+\frac {5 A b \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{2 a^{\frac {7}{2}}}-\frac {5 A b}{2 \sqrt {b \,x^{2}+a}\, a^{3}}-\frac {B}{\left (b \,x^{2}+a \right )^{\frac {3}{2}} a x}-\frac {A}{2 \left (b \,x^{2}+a \right )^{\frac {3}{2}} a \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^3/(b*x^2+a)^(5/2),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 1.34, size = 122, normalized size = 0.95 \begin {gather*} -\frac {8 \, B b x}{3 \, \sqrt {b x^{2} + a} a^{3}} - \frac {4 \, B b x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} + \frac {5 \, A b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {7}{2}}} - \frac {5 \, A b}{2 \, \sqrt {b x^{2} + a} a^{3}} - \frac {5 \, A b}{6 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2}} - \frac {B}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} a x} - \frac {A}{2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^3/(b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 1.62, size = 123, normalized size = 0.95 \begin {gather*} \frac {B\,a^2-8\,B\,{\left (b\,x^2+a\right )}^2+4\,B\,a\,\left (b\,x^2+a\right )}{3\,a^3\,x\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {10\,A\,b}{3\,a^2\,{\left (b\,x^2+a\right )}^{3/2}}-\frac {A}{2\,a\,x^2\,{\left (b\,x^2+a\right )}^{3/2}}+\frac {5\,A\,b\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{7/2}}-\frac {5\,A\,b^2\,x^2}{2\,a^3\,{\left (b\,x^2+a\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^3*(a + b*x^2)^(5/2)),x)

[Out]

(B*a^2 - 8*B*(a + b*x^2)^2 + 4*B*a*(a + b*x^2))/(3*a^3*x*(a + b*x^2)^(3/2)) - (10*A*b)/(3*a^2*(a + b*x^2)^(3/2
)) - A/(2*a*x^2*(a + b*x^2)^(3/2)) + (5*A*b*atanh((a + b*x^2)^(1/2)/a^(1/2)))/(2*a^(7/2)) - (5*A*b^2*x^2)/(2*a
^3*(a + b*x^2)^(3/2))

________________________________________________________________________________________

sympy [B]  time = 24.76, size = 1034, normalized size = 8.02 \begin {gather*} A \left (- \frac {6 a^{17} \sqrt {1 + \frac {b x^{2}}{a}}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {46 a^{16} b x^{2} \sqrt {1 + \frac {b x^{2}}{a}}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {15 a^{16} b x^{2} \log {\left (\frac {b x^{2}}{a} \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} + \frac {30 a^{16} b x^{2} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {70 a^{15} b^{2} x^{4} \sqrt {1 + \frac {b x^{2}}{a}}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {45 a^{15} b^{2} x^{4} \log {\left (\frac {b x^{2}}{a} \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} + \frac {90 a^{15} b^{2} x^{4} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {30 a^{14} b^{3} x^{6} \sqrt {1 + \frac {b x^{2}}{a}}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {45 a^{14} b^{3} x^{6} \log {\left (\frac {b x^{2}}{a} \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} + \frac {90 a^{14} b^{3} x^{6} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} - \frac {15 a^{13} b^{4} x^{8} \log {\left (\frac {b x^{2}}{a} \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}} + \frac {30 a^{13} b^{4} x^{8} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{12 a^{\frac {39}{2}} x^{2} + 36 a^{\frac {37}{2}} b x^{4} + 36 a^{\frac {35}{2}} b^{2} x^{6} + 12 a^{\frac {33}{2}} b^{3} x^{8}}\right ) + B \left (- \frac {3 a^{2} b^{\frac {9}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} - \frac {12 a b^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}} - \frac {8 b^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} + 6 a^{4} b^{5} x^{2} + 3 a^{3} b^{6} x^{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**3/(b*x**2+a)**(5/2),x)

[Out]

A*(-6*a**17*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2
)*b**3*x**8) - 46*a**16*b*x**2*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2
*x**6 + 12*a**(33/2)*b**3*x**8) - 15*a**16*b*x**2*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*
a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) + 30*a**16*b*x**2*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2
+ 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 70*a**15*b**2*x**4*sqrt(1 + b*x**2
/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 45*a**15*b**
2*x**4*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**
8) + 90*a**15*b**2*x**4*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b*
*2*x**6 + 12*a**(33/2)*b**3*x**8) - 30*a**14*b**3*x**6*sqrt(1 + b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*
x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 45*a**14*b**3*x**6*log(b*x**2/a)/(12*a**(39/2)*x**2
+ 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) + 90*a**14*b**3*x**6*log(sqrt(1 + b*x
**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8) - 15*a
**13*b**4*x**8*log(b*x**2/a)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(35/2)*b**2*x**6 + 12*a**(33/2)*
b**3*x**8) + 30*a**13*b**4*x**8*log(sqrt(1 + b*x**2/a) + 1)/(12*a**(39/2)*x**2 + 36*a**(37/2)*b*x**4 + 36*a**(
35/2)*b**2*x**6 + 12*a**(33/2)*b**3*x**8)) + B*(-3*a**2*b**(9/2)*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4 + 6*a**4*b*
*5*x**2 + 3*a**3*b**6*x**4) - 12*a*b**(11/2)*x**2*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4 + 6*a**4*b**5*x**2 + 3*a**
3*b**6*x**4) - 8*b**(13/2)*x**4*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4 + 6*a**4*b**5*x**2 + 3*a**3*b**6*x**4))

________________________________________________________________________________________