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

Optimal. Leaf size=196 \[ -\frac {A b-a B}{2 a^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\log (x) (a+b x) (3 A b-a B)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (3 A b-a B) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 A b-a B}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}} \]

[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)

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Rubi [A]  time = 0.14, antiderivative size = 196, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {770, 77} \[ -\frac {A b-a B}{2 a^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 A b-a B}{a^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\log (x) (a+b x) (3 A b-a B)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (3 A b-a B) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[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 77

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 770

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*} \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^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*}

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Mathematica [A]  time = 0.06, size = 110, normalized size = 0.56 \[ \frac {a \left (a^2 (3 B x-2 A)+a b x (2 B x-9 A)-6 A b^2 x^2\right )+2 x \log (x) (a+b x)^2 (a B-3 A b)+2 x (a+b x)^2 (3 A b-a B) \log (a+b x)}{2 a^4 x (a+b x) \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[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])

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fricas [A]  time = 0.66, size = 187, normalized size = 0.95 \[ -\frac {2 \, A a^{3} - 2 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} - 3 \, {\left (B a^{3} - 3 \, A a^{2} b\right )} x + 2 \, {\left ({\left (B a b^{2} - 3 \, A b^{3}\right )} x^{3} + 2 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} + {\left (B a^{3} - 3 \, A a^{2} b\right )} x\right )} \log \left (b x + a\right ) - 2 \, {\left ({\left (B a b^{2} - 3 \, A b^{3}\right )} x^{3} + 2 \, {\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} + {\left (B a^{3} - 3 \, A a^{2} b\right )} x\right )} \log \relax (x)}{2 \, {\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \mathit {sage}_{0} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.07, size = 221, normalized size = 1.13 \[ \frac {\left (-6 A \,b^{3} x^{3} \ln \relax (x )+6 A \,b^{3} x^{3} \ln \left (b x +a \right )+2 B a \,b^{2} x^{3} \ln \relax (x )-2 B a \,b^{2} x^{3} \ln \left (b x +a \right )-12 A a \,b^{2} x^{2} \ln \relax (x )+12 A a \,b^{2} x^{2} \ln \left (b x +a \right )+4 B \,a^{2} b \,x^{2} \ln \relax (x )-4 B \,a^{2} b \,x^{2} \ln \left (b x +a \right )-6 A \,a^{2} b x \ln \relax (x )+6 A \,a^{2} b x \ln \left (b x +a \right )-6 A a \,b^{2} x^{2}+2 B \,a^{3} x \ln \relax (x )-2 B \,a^{3} x \ln \left (b x +a \right )+2 B \,a^{2} b \,x^{2}-9 A \,a^{2} b x +3 B \,a^{3} x -2 A \,a^{3}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} a^{4} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(6*A*b^3*x^3*ln(b*x+a)-6*A*b^3*x^3*ln(x)-2*B*a*b^2*x^3*ln(b*x+a)+2*B*a*b^2*x^3*ln(x)+12*A*ln(b*x+a)*x^2*a*
b^2-12*A*a*b^2*x^2*ln(x)-4*B*ln(b*x+a)*x^2*a^2*b+4*B*a^2*b*x^2*ln(x)+6*A*ln(b*x+a)*x*a^2*b-6*A*ln(x)*x*a^2*b-6
*A*a*b^2*x^2-2*B*ln(b*x+a)*x*a^3+2*B*a^3*x*ln(x)+2*B*a^2*b*x^2-9*A*a^2*b*x+3*B*a^3*x-2*A*a^3)*(b*x+a)/x/a^4/((
b*x+a)^2)^(3/2)

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maxima [A]  time = 0.52, size = 191, normalized size = 0.97 \[ -\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} B \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{3}} + \frac {3 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{4}} + \frac {B}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2}} - \frac {3 \, A b}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{3}} - \frac {A}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} x} + \frac {B}{2 \, a b^{2} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {A}{2 \, a^{2} b {\left (x + \frac {a}{b}\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {A+B\,x}{x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x}{x^{2} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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