3.773 \(\int \frac{x^{11/2} (A+B x)}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=240 \[ \frac{x^{11/2} (3 A b-13 a B)}{40 a b^2 (a+b x)^4}+\frac{11 x^{9/2} (3 A b-13 a B)}{240 a b^3 (a+b x)^3}+\frac{33 x^{7/2} (3 A b-13 a B)}{320 a b^4 (a+b x)^2}+\frac{231 x^{5/2} (3 A b-13 a B)}{640 a b^5 (a+b x)}-\frac{77 x^{3/2} (3 A b-13 a B)}{128 a b^6}+\frac{231 \sqrt{x} (3 A b-13 a B)}{128 b^7}-\frac{231 \sqrt{a} (3 A b-13 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{128 b^{15/2}}+\frac{x^{13/2} (A b-a B)}{5 a b (a+b x)^5} \]

[Out]

(231*(3*A*b - 13*a*B)*Sqrt[x])/(128*b^7) - (77*(3*A*b - 13*a*B)*x^(3/2))/(128*a*b^6) + ((A*b - a*B)*x^(13/2))/
(5*a*b*(a + b*x)^5) + ((3*A*b - 13*a*B)*x^(11/2))/(40*a*b^2*(a + b*x)^4) + (11*(3*A*b - 13*a*B)*x^(9/2))/(240*
a*b^3*(a + b*x)^3) + (33*(3*A*b - 13*a*B)*x^(7/2))/(320*a*b^4*(a + b*x)^2) + (231*(3*A*b - 13*a*B)*x^(5/2))/(6
40*a*b^5*(a + b*x)) - (231*Sqrt[a]*(3*A*b - 13*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(128*b^(15/2))

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Rubi [A]  time = 0.117198, antiderivative size = 240, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {27, 78, 47, 50, 63, 205} \[ \frac{x^{11/2} (3 A b-13 a B)}{40 a b^2 (a+b x)^4}+\frac{11 x^{9/2} (3 A b-13 a B)}{240 a b^3 (a+b x)^3}+\frac{33 x^{7/2} (3 A b-13 a B)}{320 a b^4 (a+b x)^2}+\frac{231 x^{5/2} (3 A b-13 a B)}{640 a b^5 (a+b x)}-\frac{77 x^{3/2} (3 A b-13 a B)}{128 a b^6}+\frac{231 \sqrt{x} (3 A b-13 a B)}{128 b^7}-\frac{231 \sqrt{a} (3 A b-13 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{128 b^{15/2}}+\frac{x^{13/2} (A b-a B)}{5 a b (a+b x)^5} \]

Antiderivative was successfully verified.

[In]

Int[(x^(11/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

(231*(3*A*b - 13*a*B)*Sqrt[x])/(128*b^7) - (77*(3*A*b - 13*a*B)*x^(3/2))/(128*a*b^6) + ((A*b - a*B)*x^(13/2))/
(5*a*b*(a + b*x)^5) + ((3*A*b - 13*a*B)*x^(11/2))/(40*a*b^2*(a + b*x)^4) + (11*(3*A*b - 13*a*B)*x^(9/2))/(240*
a*b^3*(a + b*x)^3) + (33*(3*A*b - 13*a*B)*x^(7/2))/(320*a*b^4*(a + b*x)^2) + (231*(3*A*b - 13*a*B)*x^(5/2))/(6
40*a*b^5*(a + b*x)) - (231*Sqrt[a]*(3*A*b - 13*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(128*b^(15/2))

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 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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{x^{11/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac{x^{11/2} (A+B x)}{(a+b x)^6} \, dx\\ &=\frac{(A b-a B) x^{13/2}}{5 a b (a+b x)^5}-\frac{\left (\frac{3 A b}{2}-\frac{13 a B}{2}\right ) \int \frac{x^{11/2}}{(a+b x)^5} \, dx}{5 a b}\\ &=\frac{(A b-a B) x^{13/2}}{5 a b (a+b x)^5}+\frac{(3 A b-13 a B) x^{11/2}}{40 a b^2 (a+b x)^4}-\frac{(11 (3 A b-13 a B)) \int \frac{x^{9/2}}{(a+b x)^4} \, dx}{80 a b^2}\\ &=\frac{(A b-a B) x^{13/2}}{5 a b (a+b x)^5}+\frac{(3 A b-13 a B) x^{11/2}}{40 a b^2 (a+b x)^4}+\frac{11 (3 A b-13 a B) x^{9/2}}{240 a b^3 (a+b x)^3}-\frac{(33 (3 A b-13 a B)) \int \frac{x^{7/2}}{(a+b x)^3} \, dx}{160 a b^3}\\ &=\frac{(A b-a B) x^{13/2}}{5 a b (a+b x)^5}+\frac{(3 A b-13 a B) x^{11/2}}{40 a b^2 (a+b x)^4}+\frac{11 (3 A b-13 a B) x^{9/2}}{240 a b^3 (a+b x)^3}+\frac{33 (3 A b-13 a B) x^{7/2}}{320 a b^4 (a+b x)^2}-\frac{(231 (3 A b-13 a B)) \int \frac{x^{5/2}}{(a+b x)^2} \, dx}{640 a b^4}\\ &=\frac{(A b-a B) x^{13/2}}{5 a b (a+b x)^5}+\frac{(3 A b-13 a B) x^{11/2}}{40 a b^2 (a+b x)^4}+\frac{11 (3 A b-13 a B) x^{9/2}}{240 a b^3 (a+b x)^3}+\frac{33 (3 A b-13 a B) x^{7/2}}{320 a b^4 (a+b x)^2}+\frac{231 (3 A b-13 a B) x^{5/2}}{640 a b^5 (a+b x)}-\frac{(231 (3 A b-13 a B)) \int \frac{x^{3/2}}{a+b x} \, dx}{256 a b^5}\\ &=-\frac{77 (3 A b-13 a B) x^{3/2}}{128 a b^6}+\frac{(A b-a B) x^{13/2}}{5 a b (a+b x)^5}+\frac{(3 A b-13 a B) x^{11/2}}{40 a b^2 (a+b x)^4}+\frac{11 (3 A b-13 a B) x^{9/2}}{240 a b^3 (a+b x)^3}+\frac{33 (3 A b-13 a B) x^{7/2}}{320 a b^4 (a+b x)^2}+\frac{231 (3 A b-13 a B) x^{5/2}}{640 a b^5 (a+b x)}+\frac{(231 (3 A b-13 a B)) \int \frac{\sqrt{x}}{a+b x} \, dx}{256 b^6}\\ &=\frac{231 (3 A b-13 a B) \sqrt{x}}{128 b^7}-\frac{77 (3 A b-13 a B) x^{3/2}}{128 a b^6}+\frac{(A b-a B) x^{13/2}}{5 a b (a+b x)^5}+\frac{(3 A b-13 a B) x^{11/2}}{40 a b^2 (a+b x)^4}+\frac{11 (3 A b-13 a B) x^{9/2}}{240 a b^3 (a+b x)^3}+\frac{33 (3 A b-13 a B) x^{7/2}}{320 a b^4 (a+b x)^2}+\frac{231 (3 A b-13 a B) x^{5/2}}{640 a b^5 (a+b x)}-\frac{(231 a (3 A b-13 a B)) \int \frac{1}{\sqrt{x} (a+b x)} \, dx}{256 b^7}\\ &=\frac{231 (3 A b-13 a B) \sqrt{x}}{128 b^7}-\frac{77 (3 A b-13 a B) x^{3/2}}{128 a b^6}+\frac{(A b-a B) x^{13/2}}{5 a b (a+b x)^5}+\frac{(3 A b-13 a B) x^{11/2}}{40 a b^2 (a+b x)^4}+\frac{11 (3 A b-13 a B) x^{9/2}}{240 a b^3 (a+b x)^3}+\frac{33 (3 A b-13 a B) x^{7/2}}{320 a b^4 (a+b x)^2}+\frac{231 (3 A b-13 a B) x^{5/2}}{640 a b^5 (a+b x)}-\frac{(231 a (3 A b-13 a B)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{x}\right )}{128 b^7}\\ &=\frac{231 (3 A b-13 a B) \sqrt{x}}{128 b^7}-\frac{77 (3 A b-13 a B) x^{3/2}}{128 a b^6}+\frac{(A b-a B) x^{13/2}}{5 a b (a+b x)^5}+\frac{(3 A b-13 a B) x^{11/2}}{40 a b^2 (a+b x)^4}+\frac{11 (3 A b-13 a B) x^{9/2}}{240 a b^3 (a+b x)^3}+\frac{33 (3 A b-13 a B) x^{7/2}}{320 a b^4 (a+b x)^2}+\frac{231 (3 A b-13 a B) x^{5/2}}{640 a b^5 (a+b x)}-\frac{231 \sqrt{a} (3 A b-13 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{128 b^{15/2}}\\ \end{align*}

Mathematica [C]  time = 0.0360176, size = 61, normalized size = 0.25 \[ \frac{x^{13/2} \left (\frac{13 a^5 (A b-a B)}{(a+b x)^5}+(13 a B-3 A b) \, _2F_1\left (5,\frac{13}{2};\frac{15}{2};-\frac{b x}{a}\right )\right )}{65 a^6 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(11/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

(x^(13/2)*((13*a^5*(A*b - a*B))/(a + b*x)^5 + (-3*A*b + 13*a*B)*Hypergeometric2F1[5, 13/2, 15/2, -((b*x)/a)]))
/(65*a^6*b)

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Maple [A]  time = 0.027, size = 266, normalized size = 1.1 \begin{align*}{\frac{2\,B}{3\,{b}^{6}}{x}^{{\frac{3}{2}}}}+2\,{\frac{A\sqrt{x}}{{b}^{6}}}-12\,{\frac{aB\sqrt{x}}{{b}^{7}}}+{\frac{843\,aA}{128\,{b}^{2} \left ( bx+a \right ) ^{5}}{x}^{{\frac{9}{2}}}}-{\frac{2373\,B{a}^{2}}{128\,{b}^{3} \left ( bx+a \right ) ^{5}}{x}^{{\frac{9}{2}}}}+{\frac{1327\,A{a}^{2}}{64\,{b}^{3} \left ( bx+a \right ) ^{5}}{x}^{{\frac{7}{2}}}}-{\frac{12131\,B{a}^{3}}{192\,{b}^{4} \left ( bx+a \right ) ^{5}}{x}^{{\frac{7}{2}}}}+{\frac{131\,A{a}^{3}}{5\,{b}^{4} \left ( bx+a \right ) ^{5}}{x}^{{\frac{5}{2}}}}-{\frac{1253\,B{a}^{4}}{15\,{b}^{5} \left ( bx+a \right ) ^{5}}{x}^{{\frac{5}{2}}}}+{\frac{977\,A{a}^{4}}{64\,{b}^{5} \left ( bx+a \right ) ^{5}}{x}^{{\frac{3}{2}}}}-{\frac{9629\,B{a}^{5}}{192\,{b}^{6} \left ( bx+a \right ) ^{5}}{x}^{{\frac{3}{2}}}}+{\frac{437\,A{a}^{5}}{128\,{b}^{6} \left ( bx+a \right ) ^{5}}\sqrt{x}}-{\frac{1467\,B{a}^{6}}{128\,{b}^{7} \left ( bx+a \right ) ^{5}}\sqrt{x}}-{\frac{693\,aA}{128\,{b}^{6}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3003\,B{a}^{2}}{128\,{b}^{7}}\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(x^(11/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x)

[Out]

2/3/b^6*B*x^(3/2)+2/b^6*A*x^(1/2)-12/b^7*a*B*x^(1/2)+843/128*a/b^2/(b*x+a)^5*x^(9/2)*A-2373/128*a^2/b^3/(b*x+a
)^5*x^(9/2)*B+1327/64*a^2/b^3/(b*x+a)^5*x^(7/2)*A-12131/192*a^3/b^4/(b*x+a)^5*x^(7/2)*B+131/5*a^3/b^4/(b*x+a)^
5*x^(5/2)*A-1253/15*a^4/b^5/(b*x+a)^5*x^(5/2)*B+977/64*a^4/b^5/(b*x+a)^5*A*x^(3/2)-9629/192*a^5/b^6/(b*x+a)^5*
B*x^(3/2)+437/128*a^5/b^6/(b*x+a)^5*x^(1/2)*A-1467/128*a^6/b^7/(b*x+a)^5*x^(1/2)*B-693/128*a/b^6/(a*b)^(1/2)*a
rctan(x^(1/2)*b/(a*b)^(1/2))*A+3003/128*a^2/b^7/(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(x^(11/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.02932, size = 1617, normalized size = 6.74 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")

[Out]

[-1/3840*(3465*(13*B*a^6 - 3*A*a^5*b + (13*B*a*b^5 - 3*A*b^6)*x^5 + 5*(13*B*a^2*b^4 - 3*A*a*b^5)*x^4 + 10*(13*
B*a^3*b^3 - 3*A*a^2*b^4)*x^3 + 10*(13*B*a^4*b^2 - 3*A*a^3*b^3)*x^2 + 5*(13*B*a^5*b - 3*A*a^4*b^2)*x)*sqrt(-a/b
)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(1280*B*b^6*x^6 - 45045*B*a^6 + 10395*A*a^5*b - 1280*(
13*B*a*b^5 - 3*A*b^6)*x^5 - 10615*(13*B*a^2*b^4 - 3*A*a*b^5)*x^4 - 26070*(13*B*a^3*b^3 - 3*A*a^2*b^4)*x^3 - 29
568*(13*B*a^4*b^2 - 3*A*a^3*b^3)*x^2 - 16170*(13*B*a^5*b - 3*A*a^4*b^2)*x)*sqrt(x))/(b^12*x^5 + 5*a*b^11*x^4 +
10*a^2*b^10*x^3 + 10*a^3*b^9*x^2 + 5*a^4*b^8*x + a^5*b^7), 1/1920*(3465*(13*B*a^6 - 3*A*a^5*b + (13*B*a*b^5 -
3*A*b^6)*x^5 + 5*(13*B*a^2*b^4 - 3*A*a*b^5)*x^4 + 10*(13*B*a^3*b^3 - 3*A*a^2*b^4)*x^3 + 10*(13*B*a^4*b^2 - 3*
A*a^3*b^3)*x^2 + 5*(13*B*a^5*b - 3*A*a^4*b^2)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (1280*B*b^6*x^6 - 4
5045*B*a^6 + 10395*A*a^5*b - 1280*(13*B*a*b^5 - 3*A*b^6)*x^5 - 10615*(13*B*a^2*b^4 - 3*A*a*b^5)*x^4 - 26070*(1
3*B*a^3*b^3 - 3*A*a^2*b^4)*x^3 - 29568*(13*B*a^4*b^2 - 3*A*a^3*b^3)*x^2 - 16170*(13*B*a^5*b - 3*A*a^4*b^2)*x)*
sqrt(x))/(b^12*x^5 + 5*a*b^11*x^4 + 10*a^2*b^10*x^3 + 10*a^3*b^9*x^2 + 5*a^4*b^8*x + a^5*b^7)]

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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(x**(11/2)*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.25288, size = 258, normalized size = 1.08 \begin{align*} \frac{231 \,{\left (13 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{128 \, \sqrt{a b} b^{7}} - \frac{35595 \, B a^{2} b^{4} x^{\frac{9}{2}} - 12645 \, A a b^{5} x^{\frac{9}{2}} + 121310 \, B a^{3} b^{3} x^{\frac{7}{2}} - 39810 \, A a^{2} b^{4} x^{\frac{7}{2}} + 160384 \, B a^{4} b^{2} x^{\frac{5}{2}} - 50304 \, A a^{3} b^{3} x^{\frac{5}{2}} + 96290 \, B a^{5} b x^{\frac{3}{2}} - 29310 \, A a^{4} b^{2} x^{\frac{3}{2}} + 22005 \, B a^{6} \sqrt{x} - 6555 \, A a^{5} b \sqrt{x}}{1920 \,{\left (b x + a\right )}^{5} b^{7}} + \frac{2 \,{\left (B b^{12} x^{\frac{3}{2}} - 18 \, B a b^{11} \sqrt{x} + 3 \, A b^{12} \sqrt{x}\right )}}{3 \, b^{18}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(11/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")

[Out]

231/128*(13*B*a^2 - 3*A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^7) - 1/1920*(35595*B*a^2*b^4*x^(9/2) - 1
2645*A*a*b^5*x^(9/2) + 121310*B*a^3*b^3*x^(7/2) - 39810*A*a^2*b^4*x^(7/2) + 160384*B*a^4*b^2*x^(5/2) - 50304*A
*a^3*b^3*x^(5/2) + 96290*B*a^5*b*x^(3/2) - 29310*A*a^4*b^2*x^(3/2) + 22005*B*a^6*sqrt(x) - 6555*A*a^5*b*sqrt(x
))/((b*x + a)^5*b^7) + 2/3*(B*b^12*x^(3/2) - 18*B*a*b^11*sqrt(x) + 3*A*b^12*sqrt(x))/b^18