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 \[ -\frac{A b-a B}{4 a^2 (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{4 A b-a B}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 A b-a B}{2 a^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 A b-a B}{3 a^3 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\log (x) (a+b x) (5 A b-a B)}{a^6 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (5 A b-a B) \log (a+b x)}{a^6 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A (a+b x)}{a^5 x \sqrt{a^2+2 a b x+b^2 x^2}} \]

[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])

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Rubi [A]  time = 0.206947, antiderivative size = 282, normalized size of antiderivative = 1., 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 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]

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]))))

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*}

Mathematica [A]  time = 0.0997378, size = 148, normalized size = 0.52 \[ \frac{a \left (2 a^2 b^2 x^2 (21 B x-130 A)+a^3 b x (52 B x-125 A)+a^4 (25 B x-12 A)+6 a b^3 x^3 (2 B x-35 A)-60 A b^4 x^4\right )+12 x \log (x) (a+b x)^4 (a B-5 A b)+12 x (a+b x)^4 (5 A b-a B) \log (a+b x)}{12 a^6 x (a+b x)^3 \sqrt{(a+b x)^2}} \]

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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Maple [B]  time = 0.016, size = 397, normalized size = 1.4 \begin{align*} -{\frac{ \left ( -25\,B{a}^{5}x+12\,A{a}^{5}-72\,B\ln \left ( x \right ){x}^{3}{a}^{3}{b}^{2}+240\,A\ln \left ( x \right ){x}^{4}a{b}^{4}-48\,B\ln \left ( x \right ){x}^{4}{a}^{2}{b}^{3}+360\,A\ln \left ( x \right ){x}^{3}{a}^{2}{b}^{3}-12\,B\ln \left ( x \right ){x}^{5}a{b}^{4}+240\,A\ln \left ( x \right ){x}^{2}{a}^{3}{b}^{2}-48\,B\ln \left ( x \right ){x}^{2}{a}^{4}b+60\,A\ln \left ( x \right ) x{a}^{4}b-60\,A\ln \left ( bx+a \right ){x}^{5}{b}^{5}+12\,B\ln \left ( bx+a \right ) x{a}^{5}+60\,A\ln \left ( x \right ){x}^{5}{b}^{5}-12\,B\ln \left ( x \right ) x{a}^{5}+60\,A{x}^{4}a{b}^{4}-12\,B{x}^{4}{a}^{2}{b}^{3}+210\,A{x}^{3}{a}^{2}{b}^{3}-42\,B{x}^{3}{a}^{3}{b}^{2}+260\,A{x}^{2}{a}^{3}{b}^{2}-52\,B{x}^{2}{a}^{4}b+125\,A{a}^{4}bx-240\,A\ln \left ( bx+a \right ){x}^{2}{a}^{3}{b}^{2}+48\,B\ln \left ( bx+a \right ){x}^{2}{a}^{4}b-60\,A\ln \left ( bx+a \right ) x{a}^{4}b+12\,B\ln \left ( bx+a \right ){x}^{5}a{b}^{4}-240\,A\ln \left ( bx+a \right ){x}^{4}a{b}^{4}+48\,B\ln \left ( bx+a \right ){x}^{4}{a}^{2}{b}^{3}-360\,A\ln \left ( bx+a \right ){x}^{3}{a}^{2}{b}^{3}+72\,B\ln \left ( bx+a \right ){x}^{3}{a}^{3}{b}^{2} \right ) \left ( bx+a \right ) }{12\,x{a}^{6}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.58776, size = 730, normalized size = 2.59 \begin{align*} -\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 \left (x\right )}{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 )}} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x}{x^{2} \left (\left (a + b x\right )^{2}\right )^{\frac{5}{2}}}\, dx \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{0} x \end{align*}

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