3.712 \(\int \frac {A+B x}{x^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=211 \[ \frac {(a+b x) (A b-a B)}{2 a^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{3 a x^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 \log (x) (a+b x) (A b-a B)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (a+b x) (A b-a B) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (a+b x) (A b-a B)}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

-1/3*A*(b*x+a)/a/x^3/((b*x+a)^2)^(1/2)+1/2*(A*b-B*a)*(b*x+a)/a^2/x^2/((b*x+a)^2)^(1/2)-b*(A*b-B*a)*(b*x+a)/a^3
/x/((b*x+a)^2)^(1/2)-b^2*(A*b-B*a)*(b*x+a)*ln(x)/a^4/((b*x+a)^2)^(1/2)+b^2*(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 = 0.10, antiderivative size = 211, 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} \[ -\frac {b (a+b x) (A b-a B)}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (A b-a B)}{2 a^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 \log (x) (a+b x) (A b-a B)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (a+b x) (A b-a B) \log (a+b x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{3 a x^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

-(A*(a + b*x))/(3*a*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - a*B)*(a + b*x))/(2*a^2*x^2*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]) - (b*(A*b - a*B)*(a + b*x))/(a^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (b^2*(A*b - a*B)*(a + b*x)*Lo
g[x])/(a^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (b^2*(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^4 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {\left (a b+b^2 x\right ) \int \frac {A+B x}{x^4 \left (a b+b^2 x\right )} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \left (\frac {A}{a b x^4}+\frac {-A b+a B}{a^2 b x^3}+\frac {A b-a B}{a^3 x^2}+\frac {b (-A b+a B)}{a^4 x}-\frac {b^2 (-A b+a B)}{a^4 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {A (a+b x)}{3 a x^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) (a+b x)}{2 a^2 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (A b-a B) (a+b x)}{a^3 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2 (A b-a B) (a+b x) \log (x)}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (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 = 0.05, size = 102, normalized size = 0.48 \[ -\frac {(a+b x) \left (a \left (a^2 (2 A+3 B x)-3 a b x (A+2 B x)+6 A b^2 x^2\right )+6 b^2 x^3 \log (x) (A b-a B)+6 b^2 x^3 (a B-A b) \log (a+b x)\right )}{6 a^4 x^3 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]),x]

[Out]

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

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fricas [A]  time = 0.97, size = 94, normalized size = 0.45 \[ -\frac {6 \, {\left (B a b^{2} - A b^{3}\right )} x^{3} \log \left (b x + a\right ) - 6 \, {\left (B a b^{2} - A b^{3}\right )} x^{3} \log \relax (x) + 2 \, A a^{3} - 6 \, {\left (B a^{2} b - A a b^{2}\right )} x^{2} + 3 \, {\left (B a^{3} - A a^{2} b\right )} x}{6 \, a^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.16, size = 153, normalized size = 0.73 \[ \frac {{\left (B a b^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right )}{a^{4}} - \frac {{\left (B a b^{3} \mathrm {sgn}\left (b x + a\right ) - A b^{4} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{a^{4} b} - \frac {2 \, A a^{3} \mathrm {sgn}\left (b x + a\right ) - 6 \, {\left (B a^{2} b \mathrm {sgn}\left (b x + a\right ) - A a b^{2} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 3 \, {\left (B a^{3} \mathrm {sgn}\left (b x + a\right ) - A a^{2} b \mathrm {sgn}\left (b x + a\right )\right )} x}{6 \, a^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 119, normalized size = 0.56 \[ \frac {\left (b x +a \right ) \left (-6 A \,b^{3} x^{3} \ln \relax (x )+6 A \,b^{3} x^{3} \ln \left (b x +a \right )+6 B a \,b^{2} x^{3} \ln \relax (x )-6 B a \,b^{2} x^{3} \ln \left (b x +a \right )-6 A a \,b^{2} x^{2}+6 B \,a^{2} b \,x^{2}+3 A \,a^{2} b x -3 B \,a^{3} x -2 A \,a^{3}\right )}{6 \sqrt {\left (b x +a \right )^{2}}\, a^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^4/((b*x+a)^2)^(1/2),x)

[Out]

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

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maxima [A]  time = 0.67, size = 224, normalized size = 1.06 \[ -\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} B b^{2} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{3}} + \frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{4}} + \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b}{2 \, a^{3} x} - \frac {11 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{2}}{6 \, a^{4} x} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B}{2 \, a^{2} x^{2}} + \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b}{6 \, a^{3} x^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A}{3 \, a^{2} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^4/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

-(-1)^(2*a*b*x + 2*a^2)*B*b^2*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^3 + (-1)^(2*a*b*x + 2*a^2)*A*b^3*log(2*a*b*
x/abs(x) + 2*a^2/abs(x))/a^4 + 3/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b/(a^3*x) - 11/6*sqrt(b^2*x^2 + 2*a*b*x + a
^2)*A*b^2/(a^4*x) - 1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B/(a^2*x^2) + 5/6*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b/(a^3
*x^2) - 1/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A/(a^2*x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+B\,x}{x^4\,\sqrt {{\left (a+b\,x\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^4*((a + b*x)^2)^(1/2)),x)

[Out]

int((A + B*x)/(x^4*((a + b*x)^2)^(1/2)), x)

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sympy [A]  time = 0.53, size = 165, normalized size = 0.78 \[ \frac {- 2 A a^{2} + x^{2} \left (- 6 A b^{2} + 6 B a b\right ) + x \left (3 A a b - 3 B a^{2}\right )}{6 a^{3} x^{3}} + \frac {b^{2} \left (- A b + B a\right ) \log {\left (x + \frac {- A a b^{3} + B a^{2} b^{2} - a b^{2} \left (- A b + B a\right )}{- 2 A b^{4} + 2 B a b^{3}} \right )}}{a^{4}} - \frac {b^{2} \left (- A b + B a\right ) \log {\left (x + \frac {- A a b^{3} + B a^{2} b^{2} + a b^{2} \left (- A b + B a\right )}{- 2 A b^{4} + 2 B a b^{3}} \right )}}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**4/((b*x+a)**2)**(1/2),x)

[Out]

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

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