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

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

[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]
)

________________________________________________________________________________________

Rubi [A]  time = 0.120337, antiderivative size = 210, 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} \[ \frac{A b-a B}{4 a b (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A}{2 a^3 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A}{3 a^2 (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A}{a^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A \log (x) (a+b x)}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A (a+b x) \log (a+b x)}{a^5 \sqrt{a^2+2 a b x+b^2 x^2}} \]

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 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 \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*}

Mathematica [A]  time = 0.0514607, size = 104, normalized size = 0.5 \[ \frac{a \left (52 a^2 A b^2 x+25 a^3 A b-3 a^4 B+42 a A b^3 x^2+12 A b^4 x^3\right )+12 A b \log (x) (a+b x)^4-12 A b (a+b x)^4 \log (a+b x)}{12 a^5 b (a+b x)^3 \sqrt{(a+b x)^2}} \]

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])

________________________________________________________________________________________

Maple [A]  time = 0.014, size = 205, normalized size = 1. \begin{align*}{\frac{ \left ( 12\,A\ln \left ( x \right ){x}^{4}{b}^{5}-12\,A\ln \left ( bx+a \right ){x}^{4}{b}^{5}+48\,A\ln \left ( x \right ){x}^{3}a{b}^{4}-48\,A\ln \left ( bx+a \right ){x}^{3}a{b}^{4}+72\,A\ln \left ( x \right ){x}^{2}{a}^{2}{b}^{3}-72\,A\ln \left ( bx+a \right ){x}^{2}{a}^{2}{b}^{3}+12\,A{x}^{3}a{b}^{4}+48\,A\ln \left ( x \right ) x{a}^{3}{b}^{2}-48\,A\ln \left ( bx+a \right ) x{a}^{3}{b}^{2}+42\,A{x}^{2}{a}^{2}{b}^{3}+12\,A\ln \left ( x \right ){a}^{4}b-12\,A\ln \left ( bx+a \right ){a}^{4}b+52\,A{a}^{3}{b}^{2}x+25\,A{a}^{4}b-3\,B{a}^{5} \right ) \left ( bx+a \right ) }{12\,b{a}^{5}} \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/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.66049, size = 437, normalized size = 2.08 \begin{align*} \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 \left (x\right )}{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{align*}

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)

________________________________________________________________________________________

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

________________________________________________________________________________________

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

[Out]

sage0*x