3.835 \(\int \frac{A+B x}{x^{7/2} (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=404 \[ -\frac{231 b (a+b x) (13 A b-5 a B)}{64 a^7 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{77 (a+b x) (13 A b-5 a B)}{64 a^6 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 (a+b x) (13 A b-5 a B)}{320 a^5 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 (13 A b-5 a B)}{96 a^3 b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{13 A b-5 a B}{24 a^2 b x^{5/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{33 (13 A b-5 a B)}{64 a^4 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{5/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 b^{3/2} (a+b x) (13 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{64 a^{15/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(33*(13*A*b - 5*a*B))/(64*a^4*b*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (A*b - a*B)/(4*a*b*x^(5/2)*(a + b*x)^
3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (13*A*b - 5*a*B)/(24*a^2*b*x^(5/2)*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2
]) + (11*(13*A*b - 5*a*B))/(96*a^3*b*x^(5/2)*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (231*(13*A*b - 5*a*B)*
(a + b*x))/(320*a^5*b*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (77*(13*A*b - 5*a*B)*(a + b*x))/(64*a^6*x^(3/2)
*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (231*b*(13*A*b - 5*a*B)*(a + b*x))/(64*a^7*Sqrt[x]*Sqrt[a^2 + 2*a*b*x + b^2*
x^2]) - (231*b^(3/2)*(13*A*b - 5*a*B)*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(64*a^(15/2)*Sqrt[a^2 + 2*a
*b*x + b^2*x^2])

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Rubi [A]  time = 0.211377, antiderivative size = 404, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ -\frac{231 b (a+b x) (13 A b-5 a B)}{64 a^7 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{77 (a+b x) (13 A b-5 a B)}{64 a^6 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 (a+b x) (13 A b-5 a B)}{320 a^5 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 (13 A b-5 a B)}{96 a^3 b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{13 A b-5 a B}{24 a^2 b x^{5/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{33 (13 A b-5 a B)}{64 a^4 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{5/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 b^{3/2} (a+b x) (13 A b-5 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{64 a^{15/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/(x^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(33*(13*A*b - 5*a*B))/(64*a^4*b*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (A*b - a*B)/(4*a*b*x^(5/2)*(a + b*x)^
3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (13*A*b - 5*a*B)/(24*a^2*b*x^(5/2)*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2
]) + (11*(13*A*b - 5*a*B))/(96*a^3*b*x^(5/2)*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (231*(13*A*b - 5*a*B)*
(a + b*x))/(320*a^5*b*x^(5/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (77*(13*A*b - 5*a*B)*(a + b*x))/(64*a^6*x^(3/2)
*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (231*b*(13*A*b - 5*a*B)*(a + b*x))/(64*a^7*Sqrt[x]*Sqrt[a^2 + 2*a*b*x + b^2*
x^2]) - (231*b^(3/2)*(13*A*b - 5*a*B)*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(64*a^(15/2)*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^{7/2} \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^{7/2} \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^{5/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (b^2 (13 A b-5 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{7/2} \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^{5/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{13 A b-5 a B}{24 a^2 b x^{5/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (11 b (13 A b-5 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{7/2} \left (a b+b^2 x\right )^3} \, dx}{48 a^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{A b-a B}{4 a b x^{5/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{13 A b-5 a B}{24 a^2 b x^{5/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 (13 A b-5 a B)}{96 a^3 b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (33 (13 A b-5 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{7/2} \left (a b+b^2 x\right )^2} \, dx}{64 a^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{33 (13 A b-5 a B)}{64 a^4 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{5/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{13 A b-5 a B}{24 a^2 b x^{5/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 (13 A b-5 a B)}{96 a^3 b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (231 (13 A b-5 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{x^{7/2} \left (a b+b^2 x\right )} \, dx}{128 a^4 b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{33 (13 A b-5 a B)}{64 a^4 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{5/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{13 A b-5 a B}{24 a^2 b x^{5/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 (13 A b-5 a B)}{96 a^3 b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 (13 A b-5 a B) (a+b x)}{320 a^5 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (231 (13 A b-5 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^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{33 (13 A b-5 a B)}{64 a^4 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{5/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{13 A b-5 a B}{24 a^2 b x^{5/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 (13 A b-5 a B)}{96 a^3 b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 (13 A b-5 a B) (a+b x)}{320 a^5 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{77 (13 A b-5 a B) (a+b x)}{64 a^6 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (231 b (13 A b-5 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^6 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{33 (13 A b-5 a B)}{64 a^4 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{5/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{13 A b-5 a B}{24 a^2 b x^{5/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 (13 A b-5 a B)}{96 a^3 b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 (13 A b-5 a B) (a+b x)}{320 a^5 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{77 (13 A b-5 a B) (a+b x)}{64 a^6 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 b (13 A b-5 a B) (a+b x)}{64 a^7 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (231 b^2 (13 A b-5 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^7 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{33 (13 A b-5 a B)}{64 a^4 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{5/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{13 A b-5 a B}{24 a^2 b x^{5/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 (13 A b-5 a B)}{96 a^3 b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 (13 A b-5 a B) (a+b x)}{320 a^5 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{77 (13 A b-5 a B) (a+b x)}{64 a^6 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 b (13 A b-5 a B) (a+b x)}{64 a^7 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (231 b^2 (13 A b-5 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^7 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{33 (13 A b-5 a B)}{64 a^4 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{A b-a B}{4 a b x^{5/2} (a+b x)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{13 A b-5 a B}{24 a^2 b x^{5/2} (a+b x)^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{11 (13 A b-5 a B)}{96 a^3 b x^{5/2} (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 (13 A b-5 a B) (a+b x)}{320 a^5 b x^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{77 (13 A b-5 a B) (a+b x)}{64 a^6 x^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 b (13 A b-5 a B) (a+b x)}{64 a^7 \sqrt{x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{231 b^{3/2} (13 A b-5 a B) (a+b x) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{64 a^{15/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0366308, size = 80, normalized size = 0.2 \[ \frac{-5 a^4 (a B-A b)-(a+b x)^4 (13 A b-5 a B) \, _2F_1\left (-\frac{5}{2},4;-\frac{3}{2};-\frac{b x}{a}\right )}{20 a^5 b x^{5/2} (a+b x)^3 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x)/(x^(7/2)*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(-5*a^4*(-(A*b) + a*B) - (13*A*b - 5*a*B)*(a + b*x)^4*Hypergeometric2F1[-5/2, 4, -3/2, -((b*x)/a)])/(20*a^5*b*
x^(5/2)*(a + b*x)^3*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.023, size = 449, normalized size = 1.1 \begin{align*} -{\frac{bx+a}{960\,{a}^{7}} \left ( -103950\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{9/2}{a}^{3}{b}^{4}+180180\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{7/2}{a}^{3}{b}^{4}-69300\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{7/2}{a}^{4}{b}^{3}-17325\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{13/2}a{b}^{6}+180180\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{11/2}a{b}^{6}-69300\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{11/2}{a}^{2}{b}^{5}+270270\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{9/2}{a}^{2}{b}^{5}+45045\,A\sqrt{ab}{x}^{6}{b}^{6}+640\,B\sqrt{ab}x{a}^{6}+384\,A\sqrt{ab}{a}^{6}-17325\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{5/2}{a}^{5}{b}^{2}+18304\,A\sqrt{ab}{x}^{2}{a}^{4}{b}^{2}-7040\,B\sqrt{ab}{x}^{2}{a}^{5}b-1664\,A\sqrt{ab}x{a}^{5}b-17325\,B\sqrt{ab}{x}^{6}a{b}^{5}+165165\,A\sqrt{ab}{x}^{5}a{b}^{5}+45045\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{13/2}{b}^{7}-63525\,B\sqrt{ab}{x}^{5}{a}^{2}{b}^{4}+219219\,A\sqrt{ab}{x}^{4}{a}^{2}{b}^{4}-84315\,B\sqrt{ab}{x}^{4}{a}^{3}{b}^{3}+119691\,A\sqrt{ab}{x}^{3}{a}^{3}{b}^{3}+45045\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{5/2}{a}^{4}{b}^{3}-46035\,B\sqrt{ab}{x}^{3}{a}^{4}{b}^{2} \right ){\frac{1}{\sqrt{ab}}}{x}^{-{\frac{5}{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^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

-1/960*(-103950*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(9/2)*a^3*b^4+180180*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(7/2)
*a^3*b^4-69300*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(7/2)*a^4*b^3-17325*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(13/2)*
a*b^6+180180*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(11/2)*a*b^6-69300*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(11/2)*a^2
*b^5+270270*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(9/2)*a^2*b^5+45045*A*(a*b)^(1/2)*x^6*b^6+640*B*(a*b)^(1/2)*x*a^
6+384*A*(a*b)^(1/2)*a^6-17325*B*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(5/2)*a^5*b^2+18304*A*(a*b)^(1/2)*x^2*a^4*b^2-
7040*B*(a*b)^(1/2)*x^2*a^5*b-1664*A*(a*b)^(1/2)*x*a^5*b-17325*B*(a*b)^(1/2)*x^6*a*b^5+165165*A*(a*b)^(1/2)*x^5
*a*b^5+45045*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^(13/2)*b^7-63525*B*(a*b)^(1/2)*x^5*a^2*b^4+219219*A*(a*b)^(1/2)
*x^4*a^2*b^4-84315*B*(a*b)^(1/2)*x^4*a^3*b^3+119691*A*(a*b)^(1/2)*x^3*a^3*b^3+45045*A*arctan(x^(1/2)*b/(a*b)^(
1/2))*x^(5/2)*a^4*b^3-46035*B*(a*b)^(1/2)*x^3*a^4*b^2)*(b*x+a)/(a*b)^(1/2)/x^(5/2)/a^7/((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^(7/2)/(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 = 1.44835, size = 1507, normalized size = 3.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/x^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

[-1/1920*(3465*((5*B*a*b^5 - 13*A*b^6)*x^7 + 4*(5*B*a^2*b^4 - 13*A*a*b^5)*x^6 + 6*(5*B*a^3*b^3 - 13*A*a^2*b^4)
*x^5 + 4*(5*B*a^4*b^2 - 13*A*a^3*b^3)*x^4 + (5*B*a^5*b - 13*A*a^4*b^2)*x^3)*sqrt(-b/a)*log((b*x - 2*a*sqrt(x)*
sqrt(-b/a) - a)/(b*x + a)) + 2*(384*A*a^6 - 3465*(5*B*a*b^5 - 13*A*b^6)*x^6 - 12705*(5*B*a^2*b^4 - 13*A*a*b^5)
*x^5 - 16863*(5*B*a^3*b^3 - 13*A*a^2*b^4)*x^4 - 9207*(5*B*a^4*b^2 - 13*A*a^3*b^3)*x^3 - 1408*(5*B*a^5*b - 13*A
*a^4*b^2)*x^2 + 128*(5*B*a^6 - 13*A*a^5*b)*x)*sqrt(x))/(a^7*b^4*x^7 + 4*a^8*b^3*x^6 + 6*a^9*b^2*x^5 + 4*a^10*b
*x^4 + a^11*x^3), -1/960*(3465*((5*B*a*b^5 - 13*A*b^6)*x^7 + 4*(5*B*a^2*b^4 - 13*A*a*b^5)*x^6 + 6*(5*B*a^3*b^3
- 13*A*a^2*b^4)*x^5 + 4*(5*B*a^4*b^2 - 13*A*a^3*b^3)*x^4 + (5*B*a^5*b - 13*A*a^4*b^2)*x^3)*sqrt(b/a)*arctan(a
*sqrt(b/a)/(b*sqrt(x))) + (384*A*a^6 - 3465*(5*B*a*b^5 - 13*A*b^6)*x^6 - 12705*(5*B*a^2*b^4 - 13*A*a*b^5)*x^5
- 16863*(5*B*a^3*b^3 - 13*A*a^2*b^4)*x^4 - 9207*(5*B*a^4*b^2 - 13*A*a^3*b^3)*x^3 - 1408*(5*B*a^5*b - 13*A*a^4*
b^2)*x^2 + 128*(5*B*a^6 - 13*A*a^5*b)*x)*sqrt(x))/(a^7*b^4*x^7 + 4*a^8*b^3*x^6 + 6*a^9*b^2*x^5 + 4*a^10*b*x^4
+ a^11*x^3)]

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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**(7/2)/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Timed out

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Giac [A]  time = 1.20233, size = 279, normalized size = 0.69 \begin{align*} \frac{231 \,{\left (5 \, B a b^{2} - 13 \, A b^{3}\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{64 \, \sqrt{a b} a^{7} \mathrm{sgn}\left (b x + a\right )} + \frac{2 \,{\left (75 \, B a b x^{2} - 225 \, A b^{2} x^{2} - 5 \, B a^{2} x + 25 \, A a b x - 3 \, A a^{2}\right )}}{15 \, a^{7} x^{\frac{5}{2}} \mathrm{sgn}\left (b x + a\right )} + \frac{1545 \, B a b^{5} x^{\frac{7}{2}} - 3249 \, A b^{6} x^{\frac{7}{2}} + 5153 \, B a^{2} b^{4} x^{\frac{5}{2}} - 10633 \, A a b^{5} x^{\frac{5}{2}} + 5855 \, B a^{3} b^{3} x^{\frac{3}{2}} - 11767 \, A a^{2} b^{4} x^{\frac{3}{2}} + 2295 \, B a^{4} b^{2} \sqrt{x} - 4431 \, A a^{3} b^{3} \sqrt{x}}{192 \,{\left (b x + a\right )}^{4} a^{7} \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^(7/2)/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

231/64*(5*B*a*b^2 - 13*A*b^3)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^7*sgn(b*x + a)) + 2/15*(75*B*a*b*x^2 -
225*A*b^2*x^2 - 5*B*a^2*x + 25*A*a*b*x - 3*A*a^2)/(a^7*x^(5/2)*sgn(b*x + a)) + 1/192*(1545*B*a*b^5*x^(7/2) - 3
249*A*b^6*x^(7/2) + 5153*B*a^2*b^4*x^(5/2) - 10633*A*a*b^5*x^(5/2) + 5855*B*a^3*b^3*x^(3/2) - 11767*A*a^2*b^4*
x^(3/2) + 2295*B*a^4*b^2*sqrt(x) - 4431*A*a^3*b^3*sqrt(x))/((b*x + a)^4*a^7*sgn(b*x + a))