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

Optimal. Leaf size=256 \[ \frac {(a+b x) (A b-a B)}{3 a^2 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{4 a x^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^3 \log (x) (a+b x) (A b-a B)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^3 (a+b x) (A b-a B) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^2 (a+b x) (A b-a B)}{a^4 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (a+b x) (A b-a B)}{2 a^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 256, 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} \[ \frac {b^2 (a+b x) (A b-a B)}{a^4 x \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (a+b x) (A b-a B)}{2 a^3 x^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (A b-a B)}{3 a^2 x^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b^3 \log (x) (a+b x) (A b-a B)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^3 (a+b x) (A b-a B) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x)}{4 a x^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.06, size = 121, normalized size = 0.47 \[ -\frac {(a+b x) \left (a \left (a^3 (3 A+4 B x)-2 a^2 b x (2 A+3 B x)+6 a b^2 x^2 (A+2 B x)-12 A b^3 x^3\right )-12 b^3 x^4 \log (x) (A b-a B)+12 b^3 x^4 (A b-a B) \log (a+b x)\right )}{12 a^5 x^4 \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*((a + b*x)*(a*(-12*A*b^3*x^3 + 6*a*b^2*x^2*(A + 2*B*x) - 2*a^2*b*x*(2*A + 3*B*x) + a^3*(3*A + 4*B*x)) -
12*b^3*(A*b - a*B)*x^4*Log[x] + 12*b^3*(A*b - a*B)*x^4*Log[a + b*x]))/(a^5*x^4*Sqrt[(a + b*x)^2])

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fricas [A]  time = 0.89, size = 117, normalized size = 0.46 \[ \frac {12 \, {\left (B a b^{3} - A b^{4}\right )} x^{4} \log \left (b x + a\right ) - 12 \, {\left (B a b^{3} - A b^{4}\right )} x^{4} \log \relax (x) - 3 \, A a^{4} - 12 \, {\left (B a^{2} b^{2} - A a b^{3}\right )} x^{3} + 6 \, {\left (B a^{3} b - A a^{2} b^{2}\right )} x^{2} - 4 \, {\left (B a^{4} - A a^{3} b\right )} x}{12 \, a^{5} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^5/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

1/12*(12*(B*a*b^3 - A*b^4)*x^4*log(b*x + a) - 12*(B*a*b^3 - A*b^4)*x^4*log(x) - 3*A*a^4 - 12*(B*a^2*b^2 - A*a*
b^3)*x^3 + 6*(B*a^3*b - A*a^2*b^2)*x^2 - 4*(B*a^4 - A*a^3*b)*x)/(a^5*x^4)

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giac [A]  time = 0.23, size = 188, normalized size = 0.73 \[ -\frac {{\left (B a b^{3} \mathrm {sgn}\left (b x + a\right ) - A b^{4} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x \right |}\right )}{a^{5}} + \frac {{\left (B a b^{4} \mathrm {sgn}\left (b x + a\right ) - A b^{5} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{a^{5} b} - \frac {3 \, A a^{4} \mathrm {sgn}\left (b x + a\right ) + 12 \, {\left (B a^{2} b^{2} \mathrm {sgn}\left (b x + a\right ) - A a b^{3} \mathrm {sgn}\left (b x + a\right )\right )} x^{3} - 6 \, {\left (B a^{3} b \mathrm {sgn}\left (b x + a\right ) - A a^{2} b^{2} \mathrm {sgn}\left (b x + a\right )\right )} x^{2} + 4 \, {\left (B a^{4} \mathrm {sgn}\left (b x + a\right ) - A a^{3} b \mathrm {sgn}\left (b x + a\right )\right )} x}{12 \, a^{5} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^5/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

-(B*a*b^3*sgn(b*x + a) - A*b^4*sgn(b*x + a))*log(abs(x))/a^5 + (B*a*b^4*sgn(b*x + a) - A*b^5*sgn(b*x + a))*log
(abs(b*x + a))/(a^5*b) - 1/12*(3*A*a^4*sgn(b*x + a) + 12*(B*a^2*b^2*sgn(b*x + a) - A*a*b^3*sgn(b*x + a))*x^3 -
6*(B*a^3*b*sgn(b*x + a) - A*a^2*b^2*sgn(b*x + a))*x^2 + 4*(B*a^4*sgn(b*x + a) - A*a^3*b*sgn(b*x + a))*x)/(a^5
*x^4)

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maple [A]  time = 0.07, size = 143, normalized size = 0.56 \[ -\frac {\left (b x +a \right ) \left (-12 A \,b^{4} x^{4} \ln \relax (x )+12 A \,b^{4} x^{4} \ln \left (b x +a \right )+12 B a \,b^{3} x^{4} \ln \relax (x )-12 B a \,b^{3} x^{4} \ln \left (b x +a \right )-12 A a \,b^{3} x^{3}+12 B \,a^{2} b^{2} x^{3}+6 A \,a^{2} b^{2} x^{2}-6 B \,a^{3} b \,x^{2}-4 A \,a^{3} b x +4 B \,a^{4} x +3 A \,a^{4}\right )}{12 \sqrt {\left (b x +a \right )^{2}}\, a^{5} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/x^5/((b*x+a)^2)^(1/2),x)

[Out]

-1/12*(b*x+a)*(12*A*ln(b*x+a)*x^4*b^4-12*A*ln(x)*x^4*b^4-12*B*ln(b*x+a)*x^4*a*b^3+12*B*ln(x)*x^4*a*b^3-12*A*a*
b^3*x^3+12*B*x^3*a^2*b^2+6*A*a^2*b^2*x^2-6*B*x^2*a^3*b-4*A*a^3*b*x+4*B*a^4*x+3*A*a^4)/((b*x+a)^2)^(1/2)/x^4/a^
5

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maxima [A]  time = 0.54, size = 284, normalized size = 1.11 \[ \frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} B b^{3} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{4}} - \frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} A b^{4} \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{5}} - \frac {11 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b^{2}}{6 \, a^{4} x} + \frac {25 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{3}}{12 \, a^{5} x} + \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B b}{6 \, a^{3} x^{2}} - \frac {13 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b^{2}}{12 \, a^{4} x^{2}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} B}{3 \, a^{2} x^{3}} + \frac {7 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A b}{12 \, a^{3} x^{3}} - \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} A}{4 \, a^{2} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^5/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

(-1)^(2*a*b*x + 2*a^2)*B*b^3*log(2*a*b*x/abs(x) + 2*a^2/abs(x))/a^4 - (-1)^(2*a*b*x + 2*a^2)*A*b^4*log(2*a*b*x
/abs(x) + 2*a^2/abs(x))/a^5 - 11/6*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b^2/(a^4*x) + 25/12*sqrt(b^2*x^2 + 2*a*b*x
+ a^2)*A*b^3/(a^5*x) + 5/6*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B*b/(a^3*x^2) - 13/12*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A
*b^2/(a^4*x^2) - 1/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*B/(a^2*x^3) + 7/12*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A*b/(a^3*x
^3) - 1/4*sqrt(b^2*x^2 + 2*a*b*x + a^2)*A/(a^2*x^4)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*x)/(x^5*((a + b*x)^2)^(1/2)),x)

[Out]

int((A + B*x)/(x^5*((a + b*x)^2)^(1/2)), x)

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sympy [A]  time = 0.59, size = 189, normalized size = 0.74 \[ \frac {- 3 A a^{3} + x^{3} \left (12 A b^{3} - 12 B a b^{2}\right ) + x^{2} \left (- 6 A a b^{2} + 6 B a^{2} b\right ) + x \left (4 A a^{2} b - 4 B a^{3}\right )}{12 a^{4} x^{4}} - \frac {b^{3} \left (- A b + B a\right ) \log {\left (x + \frac {- A a b^{4} + B a^{2} b^{3} - a b^{3} \left (- A b + B a\right )}{- 2 A b^{5} + 2 B a b^{4}} \right )}}{a^{5}} + \frac {b^{3} \left (- A b + B a\right ) \log {\left (x + \frac {- A a b^{4} + B a^{2} b^{3} + a b^{3} \left (- A b + B a\right )}{- 2 A b^{5} + 2 B a b^{4}} \right )}}{a^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x**5/((b*x+a)**2)**(1/2),x)

[Out]

(-3*A*a**3 + x**3*(12*A*b**3 - 12*B*a*b**2) + x**2*(-6*A*a*b**2 + 6*B*a**2*b) + x*(4*A*a**2*b - 4*B*a**3))/(12
*a**4*x**4) - b**3*(-A*b + B*a)*log(x + (-A*a*b**4 + B*a**2*b**3 - a*b**3*(-A*b + B*a))/(-2*A*b**5 + 2*B*a*b**
4))/a**5 + b**3*(-A*b + B*a)*log(x + (-A*a*b**4 + B*a**2*b**3 + a*b**3*(-A*b + B*a))/(-2*A*b**5 + 2*B*a*b**4))
/a**5

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