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

Optimal. Leaf size=116 \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{\sqrt{x} (a+b x)}-\frac{2 a A \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}+\frac{2 b B \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

[Out]

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

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Rubi [A]  time = 0.0435198, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {770, 76} \[ -\frac{2 \sqrt{a^2+2 a b x+b^2 x^2} (a B+A b)}{\sqrt{x} (a+b x)}-\frac{2 a A \sqrt{a^2+2 a b x+b^2 x^2}}{3 x^{3/2} (a+b x)}+\frac{2 b B \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}{a+b x} \]

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^(5/2),x]

[Out]

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

Mathematica [A]  time = 0.028561, size = 46, normalized size = 0.4 \[ -\frac{2 \sqrt{(a+b x)^2} (a (A+3 B x)+3 b x (A-B x))}{3 x^{3/2} (a+b x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/x^(5/2),x]

[Out]

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

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Maple [A]  time = 0.004, size = 43, normalized size = 0.4 \begin{align*} -{\frac{-6\,Bb{x}^{2}+6\,Abx+6\,aBx+2\,aA}{3\,bx+3\,a}\sqrt{ \left ( bx+a \right ) ^{2}}{x}^{-{\frac{3}{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^(5/2),x)

[Out]

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

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Maxima [A]  time = 1.12713, size = 45, normalized size = 0.39 \begin{align*} \frac{2 \,{\left (b x^{2} - a x\right )} B}{x^{\frac{3}{2}}} - \frac{2 \,{\left (3 \, b x^{2} + a x\right )} A}{3 \, x^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(5/2),x, algorithm="maxima")

[Out]

2*(b*x^2 - a*x)*B/x^(3/2) - 2/3*(3*b*x^2 + a*x)*A/x^(5/2)

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Fricas [A]  time = 1.53443, size = 66, normalized size = 0.57 \begin{align*} \frac{2 \,{\left (3 \, B b x^{2} - A a - 3 \,{\left (B a + A b\right )} x\right )}}{3 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)**2)**(1/2)/x**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.17982, size = 69, normalized size = 0.59 \begin{align*} 2 \, B b \sqrt{x} \mathrm{sgn}\left (b x + a\right ) - \frac{2 \,{\left (3 \, B a x \mathrm{sgn}\left (b x + a\right ) + 3 \, A b x \mathrm{sgn}\left (b x + a\right ) + A a \mathrm{sgn}\left (b x + a\right )\right )}}{3 \, x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*((b*x+a)^2)^(1/2)/x^(5/2),x, algorithm="giac")

[Out]

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