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

Optimal. Leaf size=357 \[ \frac{105 (a+b x) (11 A b-3 a B)}{64 a^6 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 (a+b x) (11 A b-3 a B)}{64 a^5 b x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 (11 A b-3 a B)}{32 a^3 b x^{3/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{21 (11 A b-3 a B)}{64 a^4 b x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 A b-3 a B}{24 a^2 b x^{3/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{3/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{105 \sqrt{b} (a+b x) (11 A b-3 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{64 a^{13/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(21*(11*A*b - 3*a*B))/(64*a^4*b*x^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (A*b - a*B)/(4*a*b*x^(3/2)*(a + b*x)^
3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (11*A*b - 3*a*B)/(24*a^2*b*x^(3/2)*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2
]) + (3*(11*A*b - 3*a*B))/(32*a^3*b*x^(3/2)*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (35*(11*A*b - 3*a*B)*(a
+ b*x))/(64*a^5*b*x^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (105*(11*A*b - 3*a*B)*(a + b*x))/(64*a^6*Sqrt[x]*S
qrt[a^2 + 2*a*b*x + b^2*x^2]) + (105*Sqrt[b]*(11*A*b - 3*a*B)*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(64
*a^(13/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.185963, antiderivative size = 357, normalized size of antiderivative = 1., number of steps used = 9, 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} \[ \frac{105 (a+b x) (11 A b-3 a B)}{64 a^6 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 (a+b x) (11 A b-3 a B)}{64 a^5 b x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 (11 A b-3 a B)}{32 a^3 b x^{3/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{21 (11 A b-3 a B)}{64 a^4 b x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 A b-3 a B}{24 a^2 b x^{3/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{3/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{105 \sqrt{b} (a+b x) (11 A b-3 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{64 a^{13/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(21*(11*A*b - 3*a*B))/(64*a^4*b*x^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (A*b - a*B)/(4*a*b*x^(3/2)*(a + b*x)^
3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (11*A*b - 3*a*B)/(24*a^2*b*x^(3/2)*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2
]) + (3*(11*A*b - 3*a*B))/(32*a^3*b*x^(3/2)*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (35*(11*A*b - 3*a*B)*(a
+ b*x))/(64*a^5*b*x^(3/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (105*(11*A*b - 3*a*B)*(a + b*x))/(64*a^6*Sqrt[x]*S
qrt[a^2 + 2*a*b*x + b^2*x^2]) + (105*Sqrt[b]*(11*A*b - 3*a*B)*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(64
*a^(13/2)*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} \left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac{\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac{A+B x}{x^{5/2} \left (a b+b^2 x\right )^5} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{A b-a B}{4 a b x^{3/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (b^2 (11 A b-3 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{5/2} \left (a b+b^2 x\right )^4} \, dx}{8 a \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{A b-a B}{4 a b x^{3/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 A b-3 a B}{24 a^2 b x^{3/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (3 b (11 A b-3 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{5/2} \left (a b+b^2 x\right )^3} \, dx}{16 a^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{A b-a B}{4 a b x^{3/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 A b-3 a B}{24 a^2 b x^{3/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 (11 A b-3 a B)}{32 a^3 b x^{3/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (21 (11 A b-3 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{5/2} \left (a b+b^2 x\right )^2} \, dx}{64 a^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{21 (11 A b-3 a B)}{64 a^4 b x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{3/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 A b-3 a B}{24 a^2 b x^{3/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 (11 A b-3 a B)}{32 a^3 b x^{3/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (105 (11 A b-3 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{5/2} \left (a b+b^2 x\right )} \, dx}{128 a^4 b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{21 (11 A b-3 a B)}{64 a^4 b x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{3/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 A b-3 a B}{24 a^2 b x^{3/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 (11 A b-3 a B)}{32 a^3 b x^{3/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 (11 A b-3 a B) (a+b x)}{64 a^5 b x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (105 (11 A b-3 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{3/2} \left (a b+b^2 x\right )} \, dx}{128 a^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{21 (11 A b-3 a B)}{64 a^4 b x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{3/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 A b-3 a B}{24 a^2 b x^{3/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 (11 A b-3 a B)}{32 a^3 b x^{3/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 (11 A b-3 a B) (a+b x)}{64 a^5 b x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{105 (11 A b-3 a B) (a+b x)}{64 a^6 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (105 b (11 A b-3 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{\sqrt{x} \left (a b+b^2 x\right )} \, dx}{128 a^6 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{21 (11 A b-3 a B)}{64 a^4 b x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{3/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 A b-3 a B}{24 a^2 b x^{3/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 (11 A b-3 a B)}{32 a^3 b x^{3/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 (11 A b-3 a B) (a+b x)}{64 a^5 b x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{105 (11 A b-3 a B) (a+b x)}{64 a^6 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (105 b (11 A b-3 a B) \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 )}{64 a^6 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{21 (11 A b-3 a B)}{64 a^4 b x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{3/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 A b-3 a B}{24 a^2 b x^{3/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 (11 A b-3 a B)}{32 a^3 b x^{3/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{35 (11 A b-3 a B) (a+b x)}{64 a^5 b x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{105 (11 A b-3 a B) (a+b x)}{64 a^6 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{105 \sqrt{b} (11 A b-3 a B) (a+b x) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{64 a^{13/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.03559, size = 80, normalized size = 0.22 \[ \frac{-3 a^4 (a B-A b)-(a+b x)^4 (11 A b-3 a B) \, _2F_1\left (-\frac{3}{2},4;-\frac{1}{2};-\frac{b x}{a}\right )}{12 a^5 b x^{3/2} (a+b x)^3 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a^4*(-(A*b) + a*B) - (11*A*b - 3*a*B)*(a + b*x)^4*Hypergeometric2F1[-3/2, 4, -1/2, -((b*x)/a)])/(12*a^5*b*
x^(3/2)*(a + b*x)^3*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.023, size = 413, normalized size = 1.2 \begin{align*}{\frac{bx+a}{192\,{a}^{6}} \left ( -3780\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{5/2}{a}^{4}{b}^{2}+13860\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{5/2}{a}^{3}{b}^{3}+20790\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{7/2}{a}^{2}{b}^{4}-5670\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{7/2}{a}^{3}{b}^{3}-3780\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{9/2}{a}^{2}{b}^{4}+13860\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{9/2}a{b}^{5}-945\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{11/2}a{b}^{5}+3465\,A\sqrt{ab}{x}^{5}{b}^{5}-384\,B\sqrt{ab}x{a}^{5}-128\,A\sqrt{ab}{a}^{5}+16863\,A\sqrt{ab}{x}^{3}{a}^{2}{b}^{3}-4599\,B\sqrt{ab}{x}^{3}{a}^{3}{b}^{2}+9207\,A\sqrt{ab}{x}^{2}{a}^{3}{b}^{2}+3465\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{3/2}{a}^{4}{b}^{2}-2511\,B\sqrt{ab}{x}^{2}{a}^{4}b-945\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{3/2}{a}^{5}b+1408\,A\sqrt{ab}x{a}^{4}b-945\,B\sqrt{ab}{x}^{5}a{b}^{4}+12705\,A\sqrt{ab}{x}^{4}a{b}^{4}+3465\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{11/2}{b}^{6}-3465\,B\sqrt{ab}{x}^{4}{a}^{2}{b}^{3} \right ){\frac{1}{\sqrt{ab}}}{x}^{-{\frac{3}{2}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(-3780*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(5/2)*a^4*b^2+13860*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(5/2)*a^3
*b^3+20790*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(7/2)*a^2*b^4-5670*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(7/2)*a^3*b^
3-3780*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(9/2)*a^2*b^4+13860*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(9/2)*a*b^5-945
*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(11/2)*a*b^5+3465*A*(a*b)^(1/2)*x^5*b^5-384*B*(a*b)^(1/2)*x*a^5-128*A*(a*b)
^(1/2)*a^5+16863*A*(a*b)^(1/2)*x^3*a^2*b^3-4599*B*(a*b)^(1/2)*x^3*a^3*b^2+9207*A*(a*b)^(1/2)*x^2*a^3*b^2+3465*
A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(3/2)*a^4*b^2-2511*B*(a*b)^(1/2)*x^2*a^4*b-945*B*arctan(x^(1/2)*b/(a*b)^(1/2
))*x^(3/2)*a^5*b+1408*A*(a*b)^(1/2)*x*a^4*b-945*B*(a*b)^(1/2)*x^5*a*b^4+12705*A*(a*b)^(1/2)*x^4*a*b^4+3465*A*a
rctan(x^(1/2)*b/(a*b)^(1/2))*x^(11/2)*b^6-3465*B*(a*b)^(1/2)*x^4*a^2*b^3)*(b*x+a)/(a*b)^(1/2)/x^(3/2)/a^6/((b*
x+a)^2)^(5/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.39563, size = 1364, normalized size = 3.82 \begin{align*} \left [-\frac{315 \,{\left ({\left (3 \, B a b^{4} - 11 \, A b^{5}\right )} x^{6} + 4 \,{\left (3 \, B a^{2} b^{3} - 11 \, A a b^{4}\right )} x^{5} + 6 \,{\left (3 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{4} + 4 \,{\left (3 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{3} +{\left (3 \, B a^{5} - 11 \, A a^{4} b\right )} x^{2}\right )} \sqrt{-\frac{b}{a}} \log \left (\frac{b x + 2 \, a \sqrt{x} \sqrt{-\frac{b}{a}} - a}{b x + a}\right ) + 2 \,{\left (128 \, A a^{5} + 315 \,{\left (3 \, B a b^{4} - 11 \, A b^{5}\right )} x^{5} + 1155 \,{\left (3 \, B a^{2} b^{3} - 11 \, A a b^{4}\right )} x^{4} + 1533 \,{\left (3 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{3} + 837 \,{\left (3 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{2} + 128 \,{\left (3 \, B a^{5} - 11 \, A a^{4} b\right )} x\right )} \sqrt{x}}{384 \,{\left (a^{6} b^{4} x^{6} + 4 \, a^{7} b^{3} x^{5} + 6 \, a^{8} b^{2} x^{4} + 4 \, a^{9} b x^{3} + a^{10} x^{2}\right )}}, \frac{315 \,{\left ({\left (3 \, B a b^{4} - 11 \, A b^{5}\right )} x^{6} + 4 \,{\left (3 \, B a^{2} b^{3} - 11 \, A a b^{4}\right )} x^{5} + 6 \,{\left (3 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{4} + 4 \,{\left (3 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{3} +{\left (3 \, B a^{5} - 11 \, A a^{4} b\right )} x^{2}\right )} \sqrt{\frac{b}{a}} \arctan \left (\frac{a \sqrt{\frac{b}{a}}}{b \sqrt{x}}\right ) -{\left (128 \, A a^{5} + 315 \,{\left (3 \, B a b^{4} - 11 \, A b^{5}\right )} x^{5} + 1155 \,{\left (3 \, B a^{2} b^{3} - 11 \, A a b^{4}\right )} x^{4} + 1533 \,{\left (3 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{3} + 837 \,{\left (3 \, B a^{4} b - 11 \, A a^{3} b^{2}\right )} x^{2} + 128 \,{\left (3 \, B a^{5} - 11 \, A a^{4} b\right )} x\right )} \sqrt{x}}{192 \,{\left (a^{6} b^{4} x^{6} + 4 \, a^{7} b^{3} x^{5} + 6 \, a^{8} b^{2} x^{4} + 4 \, a^{9} b x^{3} + a^{10} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/384*(315*((3*B*a*b^4 - 11*A*b^5)*x^6 + 4*(3*B*a^2*b^3 - 11*A*a*b^4)*x^5 + 6*(3*B*a^3*b^2 - 11*A*a^2*b^3)*x
^4 + 4*(3*B*a^4*b - 11*A*a^3*b^2)*x^3 + (3*B*a^5 - 11*A*a^4*b)*x^2)*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/
a) - a)/(b*x + a)) + 2*(128*A*a^5 + 315*(3*B*a*b^4 - 11*A*b^5)*x^5 + 1155*(3*B*a^2*b^3 - 11*A*a*b^4)*x^4 + 153
3*(3*B*a^3*b^2 - 11*A*a^2*b^3)*x^3 + 837*(3*B*a^4*b - 11*A*a^3*b^2)*x^2 + 128*(3*B*a^5 - 11*A*a^4*b)*x)*sqrt(x
))/(a^6*b^4*x^6 + 4*a^7*b^3*x^5 + 6*a^8*b^2*x^4 + 4*a^9*b*x^3 + a^10*x^2), 1/192*(315*((3*B*a*b^4 - 11*A*b^5)*
x^6 + 4*(3*B*a^2*b^3 - 11*A*a*b^4)*x^5 + 6*(3*B*a^3*b^2 - 11*A*a^2*b^3)*x^4 + 4*(3*B*a^4*b - 11*A*a^3*b^2)*x^3
+ (3*B*a^5 - 11*A*a^4*b)*x^2)*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - (128*A*a^5 + 315*(3*B*a*b^4 - 11*A*
b^5)*x^5 + 1155*(3*B*a^2*b^3 - 11*A*a*b^4)*x^4 + 1533*(3*B*a^3*b^2 - 11*A*a^2*b^3)*x^3 + 837*(3*B*a^4*b - 11*A
*a^3*b^2)*x^2 + 128*(3*B*a^5 - 11*A*a^4*b)*x)*sqrt(x))/(a^6*b^4*x^6 + 4*a^7*b^3*x^5 + 6*a^8*b^2*x^4 + 4*a^9*b*
x^3 + a^10*x^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)/x**(5/2)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.17978, size = 243, normalized size = 0.68 \begin{align*} -\frac{105 \,{\left (3 \, B a b - 11 \, A b^{2}\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{64 \, \sqrt{a b} a^{6} \mathrm{sgn}\left (b x + a\right )} - \frac{2 \,{\left (3 \, B a x - 15 \, A b x + A a\right )}}{3 \, a^{6} x^{\frac{3}{2}} \mathrm{sgn}\left (b x + a\right )} - \frac{561 \, B a b^{4} x^{\frac{7}{2}} - 1545 \, A b^{5} x^{\frac{7}{2}} + 1929 \, B a^{2} b^{3} x^{\frac{5}{2}} - 5153 \, A a b^{4} x^{\frac{5}{2}} + 2295 \, B a^{3} b^{2} x^{\frac{3}{2}} - 5855 \, A a^{2} b^{3} x^{\frac{3}{2}} + 975 \, B a^{4} b \sqrt{x} - 2295 \, A a^{3} b^{2} \sqrt{x}}{192 \,{\left (b x + a\right )}^{4} a^{6} \mathrm{sgn}\left (b x + a\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-105/64*(3*B*a*b - 11*A*b^2)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^6*sgn(b*x + a)) - 2/3*(3*B*a*x - 15*A*b*
x + A*a)/(a^6*x^(3/2)*sgn(b*x + a)) - 1/192*(561*B*a*b^4*x^(7/2) - 1545*A*b^5*x^(7/2) + 1929*B*a^2*b^3*x^(5/2)
- 5153*A*a*b^4*x^(5/2) + 2295*B*a^3*b^2*x^(3/2) - 5855*A*a^2*b^3*x^(3/2) + 975*B*a^4*b*sqrt(x) - 2295*A*a^3*b
^2*sqrt(x))/((b*x + a)^4*a^6*sgn(b*x + a))