3.1.42 \(\int \frac {A+B x}{x^3 (a+b x^2)^{5/2}} \, dx\) [42]

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

[Out]

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

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Rubi [A]
time = 0.08, 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 = {837, 849, 821, 272, 65, 214} \begin {gather*} \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}} \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 65

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 214

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

Rule 272

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 821

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 837

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] +
Dist[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 849

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) \text {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) \text {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*}

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Mathematica [A]
time = 0.47, size = 102, normalized size = 0.79 \begin {gather*} \frac {-3 a^2 (A+2 B x)-4 a b x^2 (5 A+6 B x)-b^2 x^4 (15 A+16 B x)}{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+b x^2}}{\sqrt {a}}\right )}{a^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.15, size = 146, normalized size = 1.13

method	result	size
default	\(A \left (-\frac {1}{2 a \,x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {5 b \left (\frac {1}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}}{a}\right )}{2 a}\right )+B \left (-\frac {1}{a x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {4 b \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{a}\right )\)	\(146\)
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (2 B x +A \right )}{2 a^{3} x^{2}}+\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, A}{12 a^{3} \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {\sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, B}{12 a^{2} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )^{2}}-\frac {13 \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, A b}{12 a^{3} \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}-\frac {5 \sqrt {\left (x -\frac {\sqrt {-a b}}{b}\right )^{2} b +2 \sqrt {-a b}\, \left (x -\frac {\sqrt {-a b}}{b}\right )}\, B}{6 a^{3} \left (x -\frac {\sqrt {-a b}}{b}\right )}+\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, A}{12 a^{3} \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}+\frac {\sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, B}{12 a^{2} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )^{2}}+\frac {13 \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, A b}{12 a^{3} \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}-\frac {5 \sqrt {\left (x +\frac {\sqrt {-a b}}{b}\right )^{2} b -2 \sqrt {-a b}\, \left (x +\frac {\sqrt {-a b}}{b}\right )}\, B}{6 a^{3} \left (x +\frac {\sqrt {-a b}}{b}\right )}+\frac {5 \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) A b}{2 a^{\frac {7}{2}}}\)	\(565\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, 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)

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Fricas [A]
time = 9.97, 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)]

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 1034 vs. \(2 (122) = 244\).
time = 8.43, 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))

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Giac [A]
time = 0.93, 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)

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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))

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