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

Optimal. Leaf size=243 \[ \frac{b (3 A b-2 a B)}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (3 A b-a B)}{a^4 x \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{b (A b-a B)}{2 a^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 b \log (x) (a+b x) (2 A b-a B)}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 b (a+b x) (2 A b-a B) \log (a+b x)}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A (a+b x)}{2 a^3 x^2 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.161379, antiderivative size = 243, 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} \[ \frac{b (3 A b-2 a B)}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (3 A b-a B)}{a^4 x \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{b (A b-a B)}{2 a^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 b \log (x) (a+b x) (2 A b-a B)}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 b (a+b x) (2 A b-a B) \log (a+b x)}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A (a+b x)}{2 a^3 x^2 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0758207, size = 133, normalized size = 0.55 \[ \frac{-a \left (a^2 b x (9 B x-4 A)+a^3 (A+2 B x)+6 a b^2 x^2 (B x-3 A)-12 A b^3 x^3\right )+6 b x^2 \log (x) (a+b x)^2 (2 A b-a B)+6 b x^2 (a+b x)^2 (a B-2 A b) \log (a+b x)}{2 a^5 x^2 (a+b x) \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*(-12*A*b^3*x^3 + 6*a*b^2*x^2*(-3*A + B*x) + a^3*(A + 2*B*x) + a^2*b*x*(-4*A + 9*B*x))) + 6*b*(2*A*b - a*B
)*x^2*(a + b*x)^2*Log[x] + 6*b*(-2*A*b + a*B)*x^2*(a + b*x)^2*Log[a + b*x])/(2*a^5*x^2*(a + b*x)*Sqrt[(a + b*x
)^2])

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Maple [A]  time = 0.016, size = 263, normalized size = 1.1 \begin{align*}{\frac{ \left ( 12\,A\ln \left ( x \right ){x}^{4}{b}^{4}-12\,A\ln \left ( bx+a \right ){x}^{4}{b}^{4}-6\,B\ln \left ( x \right ){x}^{4}a{b}^{3}+6\,B\ln \left ( bx+a \right ){x}^{4}a{b}^{3}+24\,A\ln \left ( x \right ){x}^{3}a{b}^{3}-24\,A\ln \left ( bx+a \right ){x}^{3}a{b}^{3}-12\,B\ln \left ( x \right ){x}^{3}{a}^{2}{b}^{2}+12\,B\ln \left ( bx+a \right ){x}^{3}{a}^{2}{b}^{2}+12\,A\ln \left ( x \right ){x}^{2}{a}^{2}{b}^{2}-12\,A\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{2}+12\,aA{b}^{3}{x}^{3}-6\,B\ln \left ( x \right ){x}^{2}{a}^{3}b+6\,B\ln \left ( bx+a \right ){x}^{2}{a}^{3}b-6\,B{x}^{3}{a}^{2}{b}^{2}+18\,{a}^{2}A{b}^{2}{x}^{2}-9\,B{x}^{2}{a}^{3}b+4\,{a}^{3}Abx-2\,{a}^{4}Bx-A{a}^{4} \right ) \left ( bx+a \right ) }{2\,{x}^{2}{a}^{5}} \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^3/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

1/2*(12*A*ln(x)*x^4*b^4-12*A*ln(b*x+a)*x^4*b^4-6*B*ln(x)*x^4*a*b^3+6*B*ln(b*x+a)*x^4*a*b^3+24*A*ln(x)*x^3*a*b^
3-24*A*ln(b*x+a)*x^3*a*b^3-12*B*ln(x)*x^3*a^2*b^2+12*B*ln(b*x+a)*x^3*a^2*b^2+12*A*ln(x)*x^2*a^2*b^2-12*A*ln(b*
x+a)*x^2*a^2*b^2+12*a*A*b^3*x^3-6*B*ln(x)*x^2*a^3*b+6*B*ln(b*x+a)*x^2*a^3*b-6*B*x^3*a^2*b^2+18*a^2*A*b^2*x^2-9
*B*x^2*a^3*b+4*a^3*A*b*x-2*a^4*B*x-A*a^4)*(b*x+a)/x^2/a^5/((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^3/(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 = 1.92137, size = 467, normalized size = 1.92 \begin{align*} -\frac{A a^{4} + 6 \,{\left (B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{3} + 9 \,{\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{2} + 2 \,{\left (B a^{4} - 2 \, A a^{3} b\right )} x - 6 \,{\left ({\left (B a b^{3} - 2 \, A b^{4}\right )} x^{4} + 2 \,{\left (B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{3} +{\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{2}\right )} \log \left (b x + a\right ) + 6 \,{\left ({\left (B a b^{3} - 2 \, A b^{4}\right )} x^{4} + 2 \,{\left (B a^{2} b^{2} - 2 \, A a b^{3}\right )} x^{3} +{\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x^{2}\right )} \log \left (x\right )}{2 \,{\left (a^{5} b^{2} x^{4} + 2 \, a^{6} b x^{3} + a^{7} x^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B x}{x^{3} \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**3/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

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

[Out]

sage0*x