3.720 \(\int \frac{A+B x}{x^2 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=196 \[ -\frac{A b-a B}{2 a^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 A b-a B}{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\log (x) (a+b x) (3 A b-a B)}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (3 A b-a B) \log (a+b x)}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A (a+b x)}{a^3 x \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.135734, antiderivative size = 196, 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} \[ -\frac{A b-a B}{2 a^2 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 A b-a B}{a^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\log (x) (a+b x) (3 A b-a B)}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (3 A b-a B) \log (a+b x)}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A (a+b x)}{a^3 x \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

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

Mathematica [A]  time = 0.063217, size = 110, normalized size = 0.56 \[ \frac{a \left (a^2 (3 B x-2 A)+a b x (2 B x-9 A)-6 A b^2 x^2\right )+2 x \log (x) (a+b x)^2 (a B-3 A b)+2 x (a+b x)^2 (3 A b-a B) \log (a+b x)}{2 a^4 x (a+b x) \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^2*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)),x]

[Out]

(a*(-6*A*b^2*x^2 + a*b*x*(-9*A + 2*B*x) + a^2*(-2*A + 3*B*x)) + 2*(-3*A*b + a*B)*x*(a + b*x)^2*Log[x] + 2*(3*A
*b - a*B)*x*(a + b*x)^2*Log[a + b*x])/(2*a^4*x*(a + b*x)*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.016, size = 221, normalized size = 1.1 \begin{align*} -{\frac{ \left ( 6\,A\ln \left ( x \right ){x}^{3}{b}^{3}-6\,A\ln \left ( bx+a \right ){x}^{3}{b}^{3}-2\,B\ln \left ( x \right ){x}^{3}a{b}^{2}+2\,B\ln \left ( bx+a \right ){x}^{3}a{b}^{2}+12\,A\ln \left ( x \right ){x}^{2}a{b}^{2}-12\,A\ln \left ( bx+a \right ){x}^{2}a{b}^{2}-4\,B\ln \left ( x \right ){x}^{2}{a}^{2}b+4\,B\ln \left ( bx+a \right ){x}^{2}{a}^{2}b+6\,A\ln \left ( x \right ) x{a}^{2}b-6\,A\ln \left ( bx+a \right ) x{a}^{2}b+6\,A{x}^{2}a{b}^{2}-2\,B\ln \left ( x \right ) x{a}^{3}+2\,B\ln \left ( bx+a \right ) x{a}^{3}-2\,B{x}^{2}{a}^{2}b+9\,A{a}^{2}bx-3\,{a}^{3}Bx+2\,A{a}^{3} \right ) \left ( bx+a \right ) }{2\,x{a}^{4}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{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)^(3/2),x)

[Out]

-1/2*(6*A*ln(x)*x^3*b^3-6*A*ln(b*x+a)*x^3*b^3-2*B*ln(x)*x^3*a*b^2+2*B*ln(b*x+a)*x^3*a*b^2+12*A*ln(x)*x^2*a*b^2
-12*A*ln(b*x+a)*x^2*a*b^2-4*B*ln(x)*x^2*a^2*b+4*B*ln(b*x+a)*x^2*a^2*b+6*A*ln(x)*x*a^2*b-6*A*ln(b*x+a)*x*a^2*b+
6*A*x^2*a*b^2-2*B*ln(x)*x*a^3+2*B*ln(b*x+a)*x*a^3-2*B*x^2*a^2*b+9*A*a^2*b*x-3*a^3*B*x+2*A*a^3)*(b*x+a)/x/a^4/(
(b*x+a)^2)^(3/2)

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.85385, size = 400, normalized size = 2.04 \begin{align*} -\frac{2 \, A a^{3} - 2 \,{\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} - 3 \,{\left (B a^{3} - 3 \, A a^{2} b\right )} x + 2 \,{\left ({\left (B a b^{2} - 3 \, A b^{3}\right )} x^{3} + 2 \,{\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} +{\left (B a^{3} - 3 \, A a^{2} b\right )} x\right )} \log \left (b x + a\right ) - 2 \,{\left ({\left (B a b^{2} - 3 \, A b^{3}\right )} x^{3} + 2 \,{\left (B a^{2} b - 3 \, A a b^{2}\right )} x^{2} +{\left (B a^{3} - 3 \, A a^{2} b\right )} x\right )} \log \left (x\right )}{2 \,{\left (a^{4} b^{2} x^{3} + 2 \, a^{5} b x^{2} + a^{6} 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)^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

Integral((A + B*x)/(x**2*((a + b*x)**2)**(3/2)), 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)^(3/2),x, algorithm="giac")

[Out]

sage0*x