3.663 \(\int \frac{(A+B x) \sqrt{a^2+2 a b x+b^2 x^2}}{x^6} \, dx\)

Optimal. Leaf size=114 \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{4 x^4 (a+b x)}-\frac{a A \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac{b B \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

[Out]

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

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Rubi [A]  time = 0.0458225, antiderivative size = 114, 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, 76} \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{4 x^4 (a+b x)}-\frac{a A \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac{b B \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^6,x]

[Out]

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

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 76

Int[((d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_))*((e_) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*
x)*(d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, d, e, f, n}, x] && IGtQ[p, 0] && (NeQ[n, -1] || EqQ[p, 1]) && N
eQ[b*e + a*f, 0] && ( !IntegerQ[n] || LtQ[9*p + 5*n, 0] || GeQ[n + p + 1, 0] || (GeQ[n + p + 2, 0] && Rational
Q[a, b, d, e, f])) && (NeQ[n + p + 3, 0] || EqQ[p, 1])

Rubi steps

\begin{align*} \int \frac{(A+B x) \sqrt{a^2+2 a b x+b^2 x^2}}{x^6} \, dx &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \frac{\left (a b+b^2 x\right ) (A+B x)}{x^6} \, dx}{a b+b^2 x}\\ &=\frac{\sqrt{a^2+2 a b x+b^2 x^2} \int \left (\frac{a A b}{x^6}+\frac{b (A b+a B)}{x^5}+\frac{b^2 B}{x^4}\right ) \, dx}{a b+b^2 x}\\ &=-\frac{a A \sqrt{a^2+2 a b x+b^2 x^2}}{5 x^5 (a+b x)}-\frac{(A b+a B) \sqrt{a^2+2 a b x+b^2 x^2}}{4 x^4 (a+b x)}-\frac{b B \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^3 (a+b x)}\\ \end{align*}

Mathematica [A]  time = 0.015919, size = 49, normalized size = 0.43 \[ -\frac{\sqrt{(a+b x)^2} (3 a (4 A+5 B x)+5 b x (3 A+4 B x))}{60 x^5 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^6,x]

[Out]

-(Sqrt[(a + b*x)^2]*(5*b*x*(3*A + 4*B*x) + 3*a*(4*A + 5*B*x)))/(60*x^5*(a + b*x))

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Maple [A]  time = 0.003, size = 44, normalized size = 0.4 \begin{align*} -{\frac{20\,Bb{x}^{2}+15\,Abx+15\,aBx+12\,aA}{60\,{x}^{5} \left ( bx+a \right ) }\sqrt{ \left ( bx+a \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*((b*x+a)^2)^(1/2)/x^6,x)

[Out]

-1/60*(20*B*b*x^2+15*A*b*x+15*B*a*x+12*A*a)*((b*x+a)^2)^(1/2)/x^5/(b*x+a)

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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)*((b*x+a)^2)^(1/2)/x^6,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.57358, size = 70, normalized size = 0.61 \begin{align*} -\frac{20 \, B b x^{2} + 12 \, A a + 15 \,{\left (B a + A b\right )} x}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^6,x, algorithm="fricas")

[Out]

-1/60*(20*B*b*x^2 + 12*A*a + 15*(B*a + A*b)*x)/x^5

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Sympy [A]  time = 0.913517, size = 31, normalized size = 0.27 \begin{align*} - \frac{12 A a + 20 B b x^{2} + x \left (15 A b + 15 B a\right )}{60 x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/x**6,x)

[Out]

-(12*A*a + 20*B*b*x**2 + x*(15*A*b + 15*B*a))/(60*x**5)

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Giac [A]  time = 1.15266, size = 104, normalized size = 0.91 \begin{align*} -\frac{{\left (5 \, B a b^{4} - 3 \, A b^{5}\right )} \mathrm{sgn}\left (b x + a\right )}{60 \, a^{4}} - \frac{20 \, B b x^{2} \mathrm{sgn}\left (b x + a\right ) + 15 \, B a x \mathrm{sgn}\left (b x + a\right ) + 15 \, A b x \mathrm{sgn}\left (b x + a\right ) + 12 \, A a \mathrm{sgn}\left (b x + a\right )}{60 \, x^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^6,x, algorithm="giac")

[Out]

-1/60*(5*B*a*b^4 - 3*A*b^5)*sgn(b*x + a)/a^4 - 1/60*(20*B*b*x^2*sgn(b*x + a) + 15*B*a*x*sgn(b*x + a) + 15*A*b*
x*sgn(b*x + a) + 12*A*a*sgn(b*x + a))/x^5