\(\int \frac {A+B x}{x^2 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [720]

Optimal result

Integrand size = 29, antiderivative size = 196 \[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {2 A b-a B}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{2 a^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(3 A b-a B) (a+b x) \log (x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(3 A b-a B) (a+b x) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {784, 78} \[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {A b-a B}{2 a^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\log (x) (a+b x) (3 A b-a B)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (3 A b-a B) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 A b-a B}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}} \]

[In]

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

[Out]

-((2*A*b - a*B)/(a^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - (A*b - a*B)/(2*a^2*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x
^2]) - (A*(a + b*x))/(a^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - ((3*A*b - a*B)*(a + b*x)*Log[x])/(a^4*Sqrt[a^2 +
2*a*b*x + b^2*x^2]) + ((3*A*b - a*B)*(a + b*x)*Log[a + b*x])/(a^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 784

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps \begin{align*} \text {integral}& = \frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac {A+B x}{x^2 \left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = \frac {\left (b^2 \left (a b+b^2 x\right )\right ) \int \left (\frac {A}{a^3 b^3 x^2}+\frac {-3 A b+a B}{a^4 b^3 x}+\frac {A b-a B}{a^2 b^2 (a+b x)^3}+\frac {2 A b-a B}{a^3 b^2 (a+b x)^2}+\frac {3 A b-a B}{a^4 b^2 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}} \\ & = -\frac {2 A b-a B}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{2 a^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(3 A b-a B) (a+b x) \log (x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(3 A b-a B) (a+b x) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.56 \[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {a \left (-6 A b^2 x^2+a b x (-9 A+2 B x)+a^2 (-2 A+3 B x)\right )+2 (-3 A b+a B) x (a+b x)^2 \log (x)+2 (3 A b-a B) x (a+b x)^2 \log (a+b x)}{2 a^4 x (a+b x) \sqrt {(a+b x)^2}} \]

[In]

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

[Out]

(a*(-6*A*b^2*x^2 + a*b*x*(-9*A + 2*B*x) + a^2*(-2*A + 3*B*x)) + 2*(-3*A*b + a*B)*x*(a + b*x)^2*Log[x] + 2*(3*A
*b - a*B)*x*(a + b*x)^2*Log[a + b*x])/(2*a^4*x*(a + b*x)*Sqrt[(a + b*x)^2])

Maple [A] (verified)

Time = 0.24 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.67

method	result	size
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {b \left (3 A b -B a \right ) x^{2}}{a^{3}}-\frac {3 \left (3 A b -B a \right ) x}{2 a^{2}}-\frac {A}{a}\right )}{\left (b x +a \right )^{3} x}+\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 A b -B a \right ) \ln \left (-b x -a \right )}{\left (b x +a \right ) a^{4}}-\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (3 A b -B a \right ) \ln \left (x \right )}{\left (b x +a \right ) a^{4}}\)	\(132\)
default	\(-\frac {\left (6 A \ln \left (x \right ) x^{3} b^{3}-6 A \ln \left (b x +a \right ) x^{3} b^{3}-2 B \ln \left (x \right ) x^{3} a \,b^{2}+2 B \ln \left (b x +a \right ) x^{3} a \,b^{2}+12 A \ln \left (x \right ) x^{2} a \,b^{2}-12 A \ln \left (b x +a \right ) x^{2} a \,b^{2}-4 B \ln \left (x \right ) x^{2} a^{2} b +4 B \ln \left (b x +a \right ) x^{2} a^{2} b +6 A \ln \left (x \right ) x \,a^{2} b -6 A \ln \left (b x +a \right ) x \,a^{2} b +6 A a \,b^{2} x^{2}-2 B \,a^{3} \ln \left (x \right ) x +2 B \ln \left (b x +a \right ) a^{3} x -2 B \,a^{2} b \,x^{2}+9 A \,a^{2} b x -3 a^{3} B x +2 A \,a^{3}\right ) \left (b x +a \right )}{2 x \,a^{4} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(221\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.41 (sec) , antiderivative size = 187, normalized size of antiderivative = 0.95 \[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {2 \, A a^{3} - 2 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} - 3 \, {\left (B a^{3} - 3 \, A a^{2} b\right )} x + 2 \, {\left ({\left (B a b^{2} - 3 \, A b^{3}\right )} x^{3} + 2 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} + {\left (B a^{3} - 3 \, A a^{2} b\right )} x\right )} \log \left (b x + a\right ) - 2 \, {\left ({\left (B a b^{2} - 3 \, A b^{3}\right )} x^{3} + 2 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} + {\left (B a^{3} - 3 \, A a^{2} b\right )} x\right )} \log \left (x\right )}{2 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {A + B x}{x^{2} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 191, normalized size of antiderivative = 0.97 \[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=-\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} B \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{3}} + \frac {3 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{4}} + \frac {B}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}} - \frac {3 \, A b}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3}} - \frac {A}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} x} + \frac {B}{2 \, a b^{2} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {A}{2 \, a^{2} b {\left (x + \frac {a}{b}\right )}^{2}} \]

[In]

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

[Out]

-(-1)^(2*a*b*x + 2*a^2)*B*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^3 + 3*(-1)^(2*a*b*x + 2*a^2)*A*b*log(2*a*b*x/ab
s(x) + 2*a^2/abs(x))/a^4 + B/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2) - 3*A*b/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^3) -
A/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^2*x) + 1/2*B/(a*b^2*(x + a/b)^2) - 1/2*A/(a^2*b*(x + a/b)^2)

Giac [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.63 \[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\frac {{\left (B a - 3 \, A b\right )} \log \left ({\left | x \right |}\right )}{a^{4} \mathrm {sgn}\left (b x + a\right )} - \frac {{\left (B a b - 3 \, A b^{2}\right )} \log \left ({\left | b x + a \right |}\right )}{a^{4} b \mathrm {sgn}\left (b x + a\right )} - \frac {2 \, A a^{3} - 2 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} - 3 \, {\left (B a^{3} - 3 \, A a^{2} b\right )} x}{2 \, {\left (b x + a\right )}^{2} a^{4} x \mathrm {sgn}\left (b x + a\right )} \]

[In]

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

[Out]

(B*a - 3*A*b)*log(abs(x))/(a^4*sgn(b*x + a)) - (B*a*b - 3*A*b^2)*log(abs(b*x + a))/(a^4*b*sgn(b*x + a)) - 1/2*
(2*A*a^3 - 2*(B*a^2*b - 3*A*a*b^2)*x^2 - 3*(B*a^3 - 3*A*a^2*b)*x)/((b*x + a)^2*a^4*x*sgn(b*x + a))

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B x}{x^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx=\int \frac {A+B\,x}{x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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