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

Optimal. Leaf size=144 \[ \frac{2 (a+b x) (A b-a B)}{a^2 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 \sqrt{b} (a+b x) (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{a^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 A (a+b x)}{3 a x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0749992, antiderivative size = 144, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.161, Rules used = {770, 78, 51, 63, 205} \[ \frac{2 (a+b x) (A b-a B)}{a^2 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 \sqrt{b} (a+b x) (A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{a^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 A (a+b x)}{3 a x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*(a + b*x))/(3*a*x^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (2*(A*b - a*B)*(a + b*x))/(a^2*Sqrt[x]*Sqrt[a^2
+ 2*a*b*x + b^2*x^2]) + (2*Sqrt[b]*(A*b - a*B)*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(a^(5/2)*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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 51

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] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

\begin{align*} \int \frac{A+B x}{x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{A+B x}{x^{5/2} \left (a b+b^2 x\right )} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{2 A (a+b x)}{3 a x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 \left (-\frac{3 A b^2}{2}+\frac{3 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{3/2} \left (a b+b^2 x\right )} \, dx}{3 a b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{2 A (a+b x)}{3 a x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (a+b x)}{a^2 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (2 \left (-\frac{3 A b^2}{2}+\frac{3 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \int \frac{1}{\sqrt{x} \left (a b+b^2 x\right )} \, dx}{3 a^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{2 A (a+b x)}{3 a x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (a+b x)}{a^2 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (4 \left (-\frac{3 A b^2}{2}+\frac{3 a b B}{2}\right ) \left (a b+b^2 x\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a b+b^2 x^2} \, dx,x,\sqrt{x}\right )}{3 a^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{2 A (a+b x)}{3 a x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (a+b x)}{a^2 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 \sqrt{b} (A b-a B) (a+b x) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{a^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0216479, size = 59, normalized size = 0.41 \[ -\frac{2 (a+b x) \left (\, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\frac{b x}{a}\right ) (3 a B x-3 A b x)+a A\right )}{3 a^2 x^{3/2} \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a + b*x)*(a*A + (-3*A*b*x + 3*a*B*x)*Hypergeometric2F1[-1/2, 1, 1/2, -((b*x)/a)]))/(3*a^2*x^(3/2)*Sqrt[(a
+ b*x)^2])

________________________________________________________________________________________

Maple [A]  time = 0.016, size = 97, normalized size = 0.7 \begin{align*}{\frac{2\,bx+2\,a}{3\,{a}^{2}} \left ( 3\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{3/2}{b}^{2}-3\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{3/2}ab+3\,A\sqrt{ab}xb-3\,B\sqrt{ab}xa-Aa\sqrt{ab} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}{x}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.46934, size = 335, normalized size = 2.33 \begin{align*} \left [-\frac{3 \,{\left (B a - A b\right )} x^{2} \sqrt{-\frac{b}{a}} \log \left (\frac{b x + 2 \, a \sqrt{x} \sqrt{-\frac{b}{a}} - a}{b x + a}\right ) + 2 \,{\left (A a + 3 \,{\left (B a - A b\right )} x\right )} \sqrt{x}}{3 \, a^{2} x^{2}}, \frac{2 \,{\left (3 \,{\left (B a - A b\right )} x^{2} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) -{\left (A a + 3 \,{\left (B a - A b\right )} x\right )} \sqrt{x}\right )}}{3 \, a^{2} x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/3*(3*(B*a - A*b)*x^2*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 2*(A*a + 3*(B*a - A*b)
*x)*sqrt(x))/(a^2*x^2), 2/3*(3*(B*a - A*b)*x^2*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (A*a + 3*(B*a - A*b
)*x)*sqrt(x))/(a^2*x^2)]

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.14362, size = 115, normalized size = 0.8 \begin{align*} -\frac{2 \,{\left (B a b \mathrm{sgn}\left (b x + a\right ) - A b^{2} \mathrm{sgn}\left (b x + a\right )\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{\sqrt{a b} a^{2}} - \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 \, a^{2} x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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