3.771 \(\int \frac{A+B x}{x^{3/2} (a^2+2 a b x+b^2 x^2)^2} \, dx\)

Optimal. Leaf size=157 \[ -\frac{5 (7 A b-a B)}{8 a^4 b \sqrt{x}}+\frac{5 (7 A b-a B)}{24 a^3 b \sqrt{x} (a+b x)}+\frac{7 A b-a B}{12 a^2 b \sqrt{x} (a+b x)^2}-\frac{5 (7 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{8 a^{9/2} \sqrt{b}}+\frac{A b-a B}{3 a b \sqrt{x} (a+b x)^3} \]

[Out]

(-5*(7*A*b - a*B))/(8*a^4*b*Sqrt[x]) + (A*b - a*B)/(3*a*b*Sqrt[x]*(a + b*x)^3) + (7*A*b - a*B)/(12*a^2*b*Sqrt[
x]*(a + b*x)^2) + (5*(7*A*b - a*B))/(24*a^3*b*Sqrt[x]*(a + b*x)) - (5*(7*A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/S
qrt[a]])/(8*a^(9/2)*Sqrt[b])

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Rubi [A]  time = 0.0650374, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {27, 78, 51, 63, 205} \[ -\frac{5 (7 A b-a B)}{8 a^4 b \sqrt{x}}+\frac{5 (7 A b-a B)}{24 a^3 b \sqrt{x} (a+b x)}+\frac{7 A b-a B}{12 a^2 b \sqrt{x} (a+b x)^2}-\frac{5 (7 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{8 a^{9/2} \sqrt{b}}+\frac{A b-a B}{3 a b \sqrt{x} (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(3/2)*(a^2 + 2*a*b*x + b^2*x^2)^2),x]

[Out]

(-5*(7*A*b - a*B))/(8*a^4*b*Sqrt[x]) + (A*b - a*B)/(3*a*b*Sqrt[x]*(a + b*x)^3) + (7*A*b - a*B)/(12*a^2*b*Sqrt[
x]*(a + b*x)^2) + (5*(7*A*b - a*B))/(24*a^3*b*Sqrt[x]*(a + b*x)) - (5*(7*A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/S
qrt[a]])/(8*a^(9/2)*Sqrt[b])

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

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^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{A+B x}{x^{3/2} (a+b x)^4} \, dx\\ &=\frac{A b-a B}{3 a b \sqrt{x} (a+b x)^3}-\frac{\left (-\frac{7 A b}{2}+\frac{a B}{2}\right ) \int \frac{1}{x^{3/2} (a+b x)^3} \, dx}{3 a b}\\ &=\frac{A b-a B}{3 a b \sqrt{x} (a+b x)^3}+\frac{7 A b-a B}{12 a^2 b \sqrt{x} (a+b x)^2}+\frac{(5 (7 A b-a B)) \int \frac{1}{x^{3/2} (a+b x)^2} \, dx}{24 a^2 b}\\ &=\frac{A b-a B}{3 a b \sqrt{x} (a+b x)^3}+\frac{7 A b-a B}{12 a^2 b \sqrt{x} (a+b x)^2}+\frac{5 (7 A b-a B)}{24 a^3 b \sqrt{x} (a+b x)}+\frac{(5 (7 A b-a B)) \int \frac{1}{x^{3/2} (a+b x)} \, dx}{16 a^3 b}\\ &=-\frac{5 (7 A b-a B)}{8 a^4 b \sqrt{x}}+\frac{A b-a B}{3 a b \sqrt{x} (a+b x)^3}+\frac{7 A b-a B}{12 a^2 b \sqrt{x} (a+b x)^2}+\frac{5 (7 A b-a B)}{24 a^3 b \sqrt{x} (a+b x)}-\frac{(5 (7 A b-a B)) \int \frac{1}{\sqrt{x} (a+b x)} \, dx}{16 a^4}\\ &=-\frac{5 (7 A b-a B)}{8 a^4 b \sqrt{x}}+\frac{A b-a B}{3 a b \sqrt{x} (a+b x)^3}+\frac{7 A b-a B}{12 a^2 b \sqrt{x} (a+b x)^2}+\frac{5 (7 A b-a B)}{24 a^3 b \sqrt{x} (a+b x)}-\frac{(5 (7 A b-a B)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{x}\right )}{8 a^4}\\ &=-\frac{5 (7 A b-a B)}{8 a^4 b \sqrt{x}}+\frac{A b-a B}{3 a b \sqrt{x} (a+b x)^3}+\frac{7 A b-a B}{12 a^2 b \sqrt{x} (a+b x)^2}+\frac{5 (7 A b-a B)}{24 a^3 b \sqrt{x} (a+b x)}-\frac{5 (7 A b-a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{8 a^{9/2} \sqrt{b}}\\ \end{align*}

Mathematica [C]  time = 0.0275284, size = 59, normalized size = 0.38 \[ \frac{\frac{a^3 (A b-a B)}{(a+b x)^3}+(a B-7 A b) \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};-\frac{b x}{a}\right )}{3 a^4 b \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(3/2)*(a^2 + 2*a*b*x + b^2*x^2)^2),x]

[Out]

((a^3*(A*b - a*B))/(a + b*x)^3 + (-7*A*b + a*B)*Hypergeometric2F1[-1/2, 3, 1/2, -((b*x)/a)])/(3*a^4*b*Sqrt[x])

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Maple [A]  time = 0.022, size = 163, normalized size = 1. \begin{align*} -2\,{\frac{A}{{a}^{4}\sqrt{x}}}-{\frac{19\,A{b}^{3}}{8\,{a}^{4} \left ( bx+a \right ) ^{3}}{x}^{{\frac{5}{2}}}}+{\frac{5\,{b}^{2}B}{8\,{a}^{3} \left ( bx+a \right ) ^{3}}{x}^{{\frac{5}{2}}}}-{\frac{17\,A{b}^{2}}{3\,{a}^{3} \left ( bx+a \right ) ^{3}}{x}^{{\frac{3}{2}}}}+{\frac{5\,bB}{3\,{a}^{2} \left ( bx+a \right ) ^{3}}{x}^{{\frac{3}{2}}}}-{\frac{29\,Ab}{8\,{a}^{2} \left ( bx+a \right ) ^{3}}\sqrt{x}}+{\frac{11\,B}{8\,a \left ( bx+a \right ) ^{3}}\sqrt{x}}-{\frac{35\,Ab}{8\,{a}^{4}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{5\,B}{8\,{a}^{3}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*A/a^4/x^(1/2)-19/8/a^4/(b*x+a)^3*x^(5/2)*A*b^3+5/8/a^3/(b*x+a)^3*x^(5/2)*B*b^2-17/3/a^3/(b*x+a)^3*A*x^(3/2)
*b^2+5/3/a^2/(b*x+a)^3*B*x^(3/2)*b-29/8/a^2/(b*x+a)^3*x^(1/2)*A*b+11/8/a/(b*x+a)^3*x^(1/2)*B-35/8/a^4/(a*b)^(1
/2)*arctan(x^(1/2)*b/(a*b)^(1/2))*A*b+5/8/a^3/(a*b)^(1/2)*arctan(x^(1/2)*b/(a*b)^(1/2))*B

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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)/x^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.93682, size = 961, normalized size = 6.12 \begin{align*} \left [\frac{15 \,{\left ({\left (B a b^{3} - 7 \, A b^{4}\right )} x^{4} + 3 \,{\left (B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + 3 \,{\left (B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} +{\left (B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt{-a b} \log \left (\frac{b x - a + 2 \, \sqrt{-a b} \sqrt{x}}{b x + a}\right ) - 2 \,{\left (48 \, A a^{4} b - 15 \,{\left (B a^{2} b^{3} - 7 \, A a b^{4}\right )} x^{3} - 40 \,{\left (B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{2} - 33 \,{\left (B a^{4} b - 7 \, A a^{3} b^{2}\right )} x\right )} \sqrt{x}}{48 \,{\left (a^{5} b^{4} x^{4} + 3 \, a^{6} b^{3} x^{3} + 3 \, a^{7} b^{2} x^{2} + a^{8} b x\right )}}, -\frac{15 \,{\left ({\left (B a b^{3} - 7 \, A b^{4}\right )} x^{4} + 3 \,{\left (B a^{2} b^{2} - 7 \, A a b^{3}\right )} x^{3} + 3 \,{\left (B a^{3} b - 7 \, A a^{2} b^{2}\right )} x^{2} +{\left (B a^{4} - 7 \, A a^{3} b\right )} x\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{b \sqrt{x}}\right ) +{\left (48 \, A a^{4} b - 15 \,{\left (B a^{2} b^{3} - 7 \, A a b^{4}\right )} x^{3} - 40 \,{\left (B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{2} - 33 \,{\left (B a^{4} b - 7 \, A a^{3} b^{2}\right )} x\right )} \sqrt{x}}{24 \,{\left (a^{5} b^{4} x^{4} + 3 \, a^{6} b^{3} x^{3} + 3 \, a^{7} b^{2} x^{2} + a^{8} b x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")

[Out]

[1/48*(15*((B*a*b^3 - 7*A*b^4)*x^4 + 3*(B*a^2*b^2 - 7*A*a*b^3)*x^3 + 3*(B*a^3*b - 7*A*a^2*b^2)*x^2 + (B*a^4 -
7*A*a^3*b)*x)*sqrt(-a*b)*log((b*x - a + 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) - 2*(48*A*a^4*b - 15*(B*a^2*b^3 - 7*A
*a*b^4)*x^3 - 40*(B*a^3*b^2 - 7*A*a^2*b^3)*x^2 - 33*(B*a^4*b - 7*A*a^3*b^2)*x)*sqrt(x))/(a^5*b^4*x^4 + 3*a^6*b
^3*x^3 + 3*a^7*b^2*x^2 + a^8*b*x), -1/24*(15*((B*a*b^3 - 7*A*b^4)*x^4 + 3*(B*a^2*b^2 - 7*A*a*b^3)*x^3 + 3*(B*a
^3*b - 7*A*a^2*b^2)*x^2 + (B*a^4 - 7*A*a^3*b)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (48*A*a^4*b - 15*(B
*a^2*b^3 - 7*A*a*b^4)*x^3 - 40*(B*a^3*b^2 - 7*A*a^2*b^3)*x^2 - 33*(B*a^4*b - 7*A*a^3*b^2)*x)*sqrt(x))/(a^5*b^4
*x^4 + 3*a^6*b^3*x^3 + 3*a^7*b^2*x^2 + a^8*b*x)]

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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)/x**(3/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.20598, size = 149, normalized size = 0.95 \begin{align*} \frac{5 \,{\left (B a - 7 \, A b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} a^{4}} - \frac{2 \, A}{a^{4} \sqrt{x}} + \frac{15 \, B a b^{2} x^{\frac{5}{2}} - 57 \, A b^{3} x^{\frac{5}{2}} + 40 \, B a^{2} b x^{\frac{3}{2}} - 136 \, A a b^{2} x^{\frac{3}{2}} + 33 \, B a^{3} \sqrt{x} - 87 \, A a^{2} b \sqrt{x}}{24 \,{\left (b x + a\right )}^{3} a^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(3/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")

[Out]

5/8*(B*a - 7*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^4) - 2*A/(a^4*sqrt(x)) + 1/24*(15*B*a*b^2*x^(5/2) -
57*A*b^3*x^(5/2) + 40*B*a^2*b*x^(3/2) - 136*A*a*b^2*x^(3/2) + 33*B*a^3*sqrt(x) - 87*A*a^2*b*sqrt(x))/((b*x +
a)^3*a^4)