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

Optimal. Leaf size=282 \[ -\frac {A b-a B}{4 a^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\log (x) (a+b x) (5 A b-a B)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (5 A b-a B) \log (a+b x)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {4 A b-a B}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{a^5 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 A b-a B}{2 a^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 A b-a B}{3 a^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 282, 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} \[ -\frac {A b-a B}{4 a^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {4 A b-a B}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 A b-a B}{2 a^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 A b-a B}{3 a^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\log (x) (a+b x) (5 A b-a B)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (5 A b-a B) \log (a+b x)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{a^5 x \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((4*A*b - a*B)/(a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) - (A*b - a*B)/(4*a^2*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2
*x^2]) - (2*A*b - a*B)/(3*a^3*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*A*b - a*B)/(2*a^4*(a + b*x)*Sqrt
[a^2 + 2*a*b*x + b^2*x^2]) - (A*(a + b*x))/(a^5*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - ((5*A*b - a*B)*(a + b*x)*Lo
g[x])/(a^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((5*A*b - a*B)*(a + b*x)*Log[a + b*x])/(a^6*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 )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {A+B x}{x^2 \left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \left (\frac {A}{a^5 b^5 x^2}+\frac {-5 A b+a B}{a^6 b^5 x}+\frac {A b-a B}{a^2 b^4 (a+b x)^5}+\frac {2 A b-a B}{a^3 b^4 (a+b x)^4}+\frac {3 A b-a B}{a^4 b^4 (a+b x)^3}+\frac {4 A b-a B}{a^5 b^4 (a+b x)^2}+\frac {5 A b-a B}{a^6 b^4 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {4 A b-a B}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A b-a B}{4 a^2 (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 A b-a B}{3 a^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 A b-a B}{2 a^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{a^5 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(5 A b-a B) (a+b x) \log (x)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(5 A b-a B) (a+b x) \log (a+b x)}{a^6 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 148, normalized size = 0.52 \[ \frac {a \left (a^4 (25 B x-12 A)+a^3 b x (52 B x-125 A)+2 a^2 b^2 x^2 (21 B x-130 A)+6 a b^3 x^3 (2 B x-35 A)-60 A b^4 x^4\right )+12 x \log (x) (a+b x)^4 (a B-5 A b)+12 x (a+b x)^4 (5 A b-a B) \log (a+b x)}{12 a^6 x (a+b x)^3 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-60*A*b^4*x^4 + 6*a*b^3*x^3*(-35*A + 2*B*x) + 2*a^2*b^2*x^2*(-130*A + 21*B*x) + a^4*(-12*A + 25*B*x) + a^3
*b*x*(-125*A + 52*B*x)) + 12*(-5*A*b + a*B)*x*(a + b*x)^4*Log[x] + 12*(5*A*b - a*B)*x*(a + b*x)^4*Log[a + b*x]
)/(12*a^6*x*(a + b*x)^3*Sqrt[(a + b*x)^2])

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fricas [A]  time = 0.83, size = 347, normalized size = 1.23 \[ -\frac {12 \, A a^{5} - 12 \, {\left (B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4} - 42 \, {\left (B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{3} - 52 \, {\left (B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{2} - 25 \, {\left (B a^{5} - 5 \, A a^{4} b\right )} x + 12 \, {\left ({\left (B a b^{4} - 5 \, A b^{5}\right )} x^{5} + 4 \, {\left (B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4} + 6 \, {\left (B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{3} + 4 \, {\left (B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{2} + {\left (B a^{5} - 5 \, A a^{4} b\right )} x\right )} \log \left (b x + a\right ) - 12 \, {\left ({\left (B a b^{4} - 5 \, A b^{5}\right )} x^{5} + 4 \, {\left (B a^{2} b^{3} - 5 \, A a b^{4}\right )} x^{4} + 6 \, {\left (B a^{3} b^{2} - 5 \, A a^{2} b^{3}\right )} x^{3} + 4 \, {\left (B a^{4} b - 5 \, A a^{3} b^{2}\right )} x^{2} + {\left (B a^{5} - 5 \, A a^{4} b\right )} x\right )} \log \relax (x)}{12 \, {\left (a^{6} b^{4} x^{5} + 4 \, a^{7} b^{3} x^{4} + 6 \, a^{8} b^{2} x^{3} + 4 \, a^{9} b x^{2} + a^{10} x\right )}} \]

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)^(5/2),x, algorithm="fricas")

[Out]

-1/12*(12*A*a^5 - 12*(B*a^2*b^3 - 5*A*a*b^4)*x^4 - 42*(B*a^3*b^2 - 5*A*a^2*b^3)*x^3 - 52*(B*a^4*b - 5*A*a^3*b^
2)*x^2 - 25*(B*a^5 - 5*A*a^4*b)*x + 12*((B*a*b^4 - 5*A*b^5)*x^5 + 4*(B*a^2*b^3 - 5*A*a*b^4)*x^4 + 6*(B*a^3*b^2
- 5*A*a^2*b^3)*x^3 + 4*(B*a^4*b - 5*A*a^3*b^2)*x^2 + (B*a^5 - 5*A*a^4*b)*x)*log(b*x + a) - 12*((B*a*b^4 - 5*A
*b^5)*x^5 + 4*(B*a^2*b^3 - 5*A*a*b^4)*x^4 + 6*(B*a^3*b^2 - 5*A*a^2*b^3)*x^3 + 4*(B*a^4*b - 5*A*a^3*b^2)*x^2 +
(B*a^5 - 5*A*a^4*b)*x)*log(x))/(a^6*b^4*x^5 + 4*a^7*b^3*x^4 + 6*a^8*b^2*x^3 + 4*a^9*b*x^2 + a^10*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)^(5/2),x, algorithm="giac")

[Out]

sage0*x

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maple [B]  time = 0.06, size = 397, normalized size = 1.41 \[ \frac {\left (-60 A \,b^{5} x^{5} \ln \relax (x )+60 A \,b^{5} x^{5} \ln \left (b x +a \right )+12 B a \,b^{4} x^{5} \ln \relax (x )-12 B a \,b^{4} x^{5} \ln \left (b x +a \right )-240 A a \,b^{4} x^{4} \ln \relax (x )+240 A a \,b^{4} x^{4} \ln \left (b x +a \right )+48 B \,a^{2} b^{3} x^{4} \ln \relax (x )-48 B \,a^{2} b^{3} x^{4} \ln \left (b x +a \right )-360 A \,a^{2} b^{3} x^{3} \ln \relax (x )+360 A \,a^{2} b^{3} x^{3} \ln \left (b x +a \right )-60 A a \,b^{4} x^{4}+72 B \,a^{3} b^{2} x^{3} \ln \relax (x )-72 B \,a^{3} b^{2} x^{3} \ln \left (b x +a \right )+12 B \,a^{2} b^{3} x^{4}-240 A \,a^{3} b^{2} x^{2} \ln \relax (x )+240 A \,a^{3} b^{2} x^{2} \ln \left (b x +a \right )-210 A \,a^{2} b^{3} x^{3}+48 B \,a^{4} b \,x^{2} \ln \relax (x )-48 B \,a^{4} b \,x^{2} \ln \left (b x +a \right )+42 B \,a^{3} b^{2} x^{3}-60 A \,a^{4} b x \ln \relax (x )+60 A \,a^{4} b x \ln \left (b x +a \right )-260 A \,a^{3} b^{2} x^{2}+12 B \,a^{5} x \ln \relax (x )-12 B \,a^{5} x \ln \left (b x +a \right )+52 B \,a^{4} b \,x^{2}-125 A \,a^{4} b x +25 B \,a^{5} x -12 A \,a^{5}\right ) \left (b x +a \right )}{12 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} a^{6} x} \]

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)^(5/2),x)

[Out]

1/12*(52*B*a^4*b*x^2-125*A*a^4*b*x-60*A*a*b^4*x^4+12*B*a^2*b^3*x^4-260*A*a^3*b^2*x^2-210*A*a^2*b^3*x^3+42*B*a^
3*b^2*x^3-60*A*b^5*x^5*ln(x)+12*B*a^5*x*ln(x)-12*A*a^5+12*B*a*b^4*x^5*ln(x)-240*A*a^3*b^2*x^2*ln(x)+48*B*a^4*b
*x^2*ln(x)-60*A*ln(x)*x*a^4*b-12*B*ln(b*x+a)*x^5*a*b^4+240*A*ln(b*x+a)*x^4*a*b^4+25*B*a^5*x+60*A*ln(b*x+a)*x^5
*b^5-12*B*ln(b*x+a)*x*a^5-48*B*ln(b*x+a)*x^4*a^2*b^3-240*A*ln(x)*x^4*a*b^4+48*B*ln(x)*x^4*a^2*b^3-360*A*ln(x)*
x^3*a^2*b^3+72*B*ln(x)*x^3*a^3*b^2+360*A*ln(b*x+a)*x^3*a^2*b^3-72*B*ln(b*x+a)*x^3*a^3*b^2+240*A*ln(b*x+a)*x^2*
a^3*b^2-48*B*ln(b*x+a)*x^2*a^4*b+60*A*ln(b*x+a)*x*a^4*b)*(b*x+a)/x/a^6/((b*x+a)^2)^(5/2)

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maxima [A]  time = 0.47, size = 276, normalized size = 0.98 \[ -\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} B \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{5}} + \frac {5 \, \left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{6}} + \frac {B}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2}} - \frac {5 \, A b}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{3}} + \frac {B}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4}} - \frac {5 \, A b}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{5}} - \frac {A}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2} x} + \frac {B}{2 \, a^{3} b^{2} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {5 \, A}{2 \, a^{4} b {\left (x + \frac {a}{b}\right )}^{2}} + \frac {B}{4 \, a b^{4} {\left (x + \frac {a}{b}\right )}^{4}} - \frac {A}{4 \, a^{2} b^{3} {\left (x + \frac {a}{b}\right )}^{4}} \]

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)^(5/2),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^5 + 5*(-1)^(2*a*b*x + 2*a^2)*A*b*log(2*a*b*x/ab
s(x) + 2*a^2/abs(x))/a^6 + 1/3*B/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*a^2) - 5/3*A*b/((b^2*x^2 + 2*a*b*x + a^2)^(3
/2)*a^3) + B/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^4) - 5*A*b/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*a^5) - A/((b^2*x^2 + 2
*a*b*x + a^2)^(3/2)*a^2*x) + 1/2*B/(a^3*b^2*(x + a/b)^2) - 5/2*A/(a^4*b*(x + a/b)^2) + 1/4*B/(a*b^4*(x + a/b)^
4) - 1/4*A/(a^2*b^3*(x + a/b)^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,x}{x^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \]

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)^(5/2)),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B x}{x^{2} \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \]

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)**(5/2),x)

[Out]

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

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