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

Optimal. Leaf size=210 \[ \frac {A b-a B}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A}{3 a^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A \log (x) (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.12, antiderivative size = 210, 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} \begin {gather*} \frac {A b-a B}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A}{3 a^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A \log (x) (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (a+b x) \log (a+b x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

A/(a^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (A*b - a*B)/(4*a*b*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + A/(3*a
^2*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + A/(2*a^3*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (A*(a + b*
x)*Log[x])/(a^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (A*(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 \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 \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}+\frac {-A b+a B}{a b^5 (a+b x)^5}-\frac {A}{a^2 b^4 (a+b x)^4}-\frac {A}{a^3 b^4 (a+b x)^3}-\frac {A}{a^4 b^4 (a+b x)^2}-\frac {A}{a^5 b^4 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {A}{a^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A b-a B}{4 a b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A}{3 a^2 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A}{2 a^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {A (a+b x) \log (x)}{a^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {A (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 = 104, normalized size = 0.50 \begin {gather*} \frac {a \left (-3 a^4 B+25 a^3 A b+52 a^2 A b^2 x+42 a A b^3 x^2+12 A b^4 x^3\right )+12 A b \log (x) (a+b x)^4-12 A b (a+b x)^4 \log (a+b x)}{12 a^5 b (a+b x)^3 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(25*a^3*A*b - 3*a^4*B + 52*a^2*A*b^2*x + 42*a*A*b^3*x^2 + 12*A*b^4*x^3) + 12*A*b*(a + b*x)^4*Log[x] - 12*A*
b*(a + b*x)^4*Log[a + b*x])/(12*a^5*b*(a + b*x)^3*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [A]  time = 1.17, size = 391, normalized size = 1.86 \begin {gather*} \frac {2 A \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 a b x+b^2 x^2}}{a}\right )}{a^5}+\frac {2 \left (3 a^8 b B-3 a^7 A b^2+3 a^4 b^5 B x^4-25 a^3 A b^6 x^4-52 a^2 A b^7 x^5-42 a A b^8 x^6-12 A b^9 x^7\right )+2 \sqrt {b^2} \sqrt {a^2+2 a b x+b^2 x^2} \left (3 a^7 B-3 a^6 A b-3 a^6 b B x+3 a^5 A b^2 x+3 a^5 b^2 B x^2-3 a^4 A b^3 x^2-3 a^4 b^3 B x^3+3 a^3 A b^4 x^3+22 a^2 A b^5 x^4+30 a A b^6 x^5+12 A b^7 x^6\right )}{3 a^4 b x^4 \sqrt {a^2+2 a b x+b^2 x^2} \left (-8 a^3 b^5-24 a^2 b^6 x-24 a b^7 x^2-8 b^8 x^3\right )+3 a^4 b \sqrt {b^2} x^4 \left (8 a^4 b^4+32 a^3 b^5 x+48 a^2 b^6 x^2+32 a b^7 x^3+8 b^8 x^4\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-3*a^6*A*b + 3*a^7*B + 3*a^5*A*b^2*x - 3*a^6*b*B*x - 3*a^4*A*b^3*x
^2 + 3*a^5*b^2*B*x^2 + 3*a^3*A*b^4*x^3 - 3*a^4*b^3*B*x^3 + 22*a^2*A*b^5*x^4 + 30*a*A*b^6*x^5 + 12*A*b^7*x^6) +
2*(-3*a^7*A*b^2 + 3*a^8*b*B - 25*a^3*A*b^6*x^4 + 3*a^4*b^5*B*x^4 - 52*a^2*A*b^7*x^5 - 42*a*A*b^8*x^6 - 12*A*b
^9*x^7))/(3*a^4*b*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-8*a^3*b^5 - 24*a^2*b^6*x - 24*a*b^7*x^2 - 8*b^8*x^3) + 3
*a^4*b*Sqrt[b^2]*x^4*(8*a^4*b^4 + 32*a^3*b^5*x + 48*a^2*b^6*x^2 + 32*a*b^7*x^3 + 8*b^8*x^4)) + (2*A*ArcTanh[(S
qrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a])/a^5

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fricas [A]  time = 0.41, size = 203, normalized size = 0.97 \begin {gather*} \frac {12 \, A a b^{4} x^{3} + 42 \, A a^{2} b^{3} x^{2} + 52 \, A a^{3} b^{2} x - 3 \, B a^{5} + 25 \, A a^{4} b - 12 \, {\left (A b^{5} x^{4} + 4 \, A a b^{4} x^{3} + 6 \, A a^{2} b^{3} x^{2} + 4 \, A a^{3} b^{2} x + A a^{4} b\right )} \log \left (b x + a\right ) + 12 \, {\left (A b^{5} x^{4} + 4 \, A a b^{4} x^{3} + 6 \, A a^{2} b^{3} x^{2} + 4 \, A a^{3} b^{2} x + A a^{4} b\right )} \log \relax (x)}{12 \, {\left (a^{5} b^{5} x^{4} + 4 \, a^{6} b^{4} x^{3} + 6 \, a^{7} b^{3} x^{2} + 4 \, a^{8} b^{2} x + a^{9} b\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.07, size = 205, normalized size = 0.98 \begin {gather*} -\frac {\left (-12 A \,b^{5} x^{4} \ln \relax (x )+12 A \,b^{5} x^{4} \ln \left (b x +a \right )-48 A a \,b^{4} x^{3} \ln \relax (x )+48 A a \,b^{4} x^{3} \ln \left (b x +a \right )-72 A \,a^{2} b^{3} x^{2} \ln \relax (x )+72 A \,a^{2} b^{3} x^{2} \ln \left (b x +a \right )-12 A a \,b^{4} x^{3}-48 A \,a^{3} b^{2} x \ln \relax (x )+48 A \,a^{3} b^{2} x \ln \left (b x +a \right )-42 A \,a^{2} b^{3} x^{2}-12 A \,a^{4} b \ln \relax (x )+12 A \,a^{4} b \ln \left (b x +a \right )-52 A \,a^{3} b^{2} x -25 A \,a^{4} b +3 B \,a^{5}\right ) \left (b x +a \right )}{12 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} a^{5} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(12*A*b^5*x^4*ln(b*x+a)-12*A*ln(x)*x^4*b^5+48*A*a*b^4*x^3*ln(b*x+a)-48*A*ln(x)*x^3*a*b^4+72*A*a^2*b^3*x^
2*ln(b*x+a)-72*A*ln(x)*x^2*a^2*b^3-12*A*a*b^4*x^3+48*A*a^3*b^2*x*ln(b*x+a)-48*A*ln(x)*x*a^3*b^2-42*A*a^2*b^3*x
^2+12*A*a^4*b*ln(b*x+a)-12*A*ln(x)*a^4*b-52*A*a^3*b^2*x-25*A*a^4*b+3*B*a^5)*(b*x+a)/b/a^5/((b*x+a)^2)^(5/2)

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maxima [A]  time = 0.49, size = 138, normalized size = 0.66 \begin {gather*} -\frac {\left (-1\right )^{2 \, a b x + 2 \, a^{2}} A \log \left (\frac {2 \, a b x}{{\left | x \right |}} + \frac {2 \, a^{2}}{{\left | x \right |}}\right )}{a^{5}} + \frac {A}{3 \, {\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} a^{2}} + \frac {A}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} a^{4}} + \frac {A}{2 \, a^{3} b^{2} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {B}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} + \frac {A}{4 \, a b^{4} {\left (x + \frac {a}{b}\right )}^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((A + B*x)/(x*(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 \begin {gather*} \int \frac {A + B x}{x \left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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