3.661 \(\int \frac{(A+B x) \sqrt{a^2+2 a b x+b^2 x^2}}{x^4} \, dx\)

Optimal. Leaf size=75 \[ \frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (A b-a B)}{2 a^2 x^2}-\frac{A \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 a^2 x^3} \]

[Out]

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

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Rubi [A]  time = 0.0448886, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {769, 646, 37} \[ \frac{(a+b x) \sqrt{a^2+2 a b x+b^2 x^2} (A b-a B)}{2 a^2 x^2}-\frac{A \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 a^2 x^3} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^4,x]

[Out]

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

Rule 769

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)^2), x] + Dist[(2*c*f -
b*g)/(2*c*d - b*e), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x]
&& EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 0]

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps

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

Mathematica [A]  time = 0.0146603, size = 46, normalized size = 0.61 \[ -\frac{\sqrt{(a+b x)^2} (a (2 A+3 B x)+3 b x (A+2 B x))}{6 x^3 (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^4,x]

[Out]

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

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Maple [A]  time = 0.006, size = 44, normalized size = 0.6 \begin{align*} -{\frac{6\,Bb{x}^{2}+3\,Abx+3\,aBx+2\,aA}{6\,{x}^{3} \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^4,x)

[Out]

-1/6*(6*B*b*x^2+3*A*b*x+3*B*a*x+2*A*a)*((b*x+a)^2)^(1/2)/x^3/(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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.33395, size = 65, normalized size = 0.87 \begin{align*} -\frac{6 \, B b x^{2} + 2 \, A a + 3 \,{\left (B a + A b\right )} x}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^4,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.651614, size = 31, normalized size = 0.41 \begin{align*} - \frac{2 A a + 6 B b x^{2} + x \left (3 A b + 3 B a\right )}{6 x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/x**4,x)

[Out]

-(2*A*a + 6*B*b*x**2 + x*(3*A*b + 3*B*a))/(6*x**3)

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Giac [A]  time = 1.1849, size = 104, normalized size = 1.39 \begin{align*} -\frac{{\left (3 \, B a b^{2} - A b^{3}\right )} \mathrm{sgn}\left (b x + a\right )}{6 \, a^{2}} - \frac{6 \, B b x^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, B a x \mathrm{sgn}\left (b x + a\right ) + 3 \, A b x \mathrm{sgn}\left (b x + a\right ) + 2 \, A a \mathrm{sgn}\left (b x + a\right )}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^4,x, algorithm="giac")

[Out]

-1/6*(3*B*a*b^2 - A*b^3)*sgn(b*x + a)/a^2 - 1/6*(6*B*b*x^2*sgn(b*x + a) + 3*B*a*x*sgn(b*x + a) + 3*A*b*x*sgn(b
*x + a) + 2*A*a*sgn(b*x + a))/x^3