3.776 \(\int \frac{x^{5/2} (A+B x)}{(a^2+2 a b x+b^2 x^2)^3} \, dx\)

Optimal. Leaf size=195 \[ \frac{\sqrt{x} (7 a B+3 A b)}{128 a^2 b^4 (a+b x)}+\frac{(7 a B+3 A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{128 a^{5/2} b^{9/2}}-\frac{x^{5/2} (7 a B+3 A b)}{40 a b^2 (a+b x)^4}-\frac{x^{3/2} (7 a B+3 A b)}{48 a b^3 (a+b x)^3}-\frac{\sqrt{x} (7 a B+3 A b)}{64 a b^4 (a+b x)^2}+\frac{x^{7/2} (A b-a B)}{5 a b (a+b x)^5} \]

[Out]

((A*b - a*B)*x^(7/2))/(5*a*b*(a + b*x)^5) - ((3*A*b + 7*a*B)*x^(5/2))/(40*a*b^2*(a + b*x)^4) - ((3*A*b + 7*a*B
)*x^(3/2))/(48*a*b^3*(a + b*x)^3) - ((3*A*b + 7*a*B)*Sqrt[x])/(64*a*b^4*(a + b*x)^2) + ((3*A*b + 7*a*B)*Sqrt[x
])/(128*a^2*b^4*(a + b*x)) + ((3*A*b + 7*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(128*a^(5/2)*b^(9/2))

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Rubi [A]  time = 0.0897173, antiderivative size = 195, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.207, Rules used = {27, 78, 47, 51, 63, 205} \[ \frac{\sqrt{x} (7 a B+3 A b)}{128 a^2 b^4 (a+b x)}+\frac{(7 a B+3 A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{128 a^{5/2} b^{9/2}}-\frac{x^{5/2} (7 a B+3 A b)}{40 a b^2 (a+b x)^4}-\frac{x^{3/2} (7 a B+3 A b)}{48 a b^3 (a+b x)^3}-\frac{\sqrt{x} (7 a B+3 A b)}{64 a b^4 (a+b x)^2}+\frac{x^{7/2} (A b-a B)}{5 a b (a+b x)^5} \]

Antiderivative was successfully verified.

[In]

Int[(x^(5/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

((A*b - a*B)*x^(7/2))/(5*a*b*(a + b*x)^5) - ((3*A*b + 7*a*B)*x^(5/2))/(40*a*b^2*(a + b*x)^4) - ((3*A*b + 7*a*B
)*x^(3/2))/(48*a*b^3*(a + b*x)^3) - ((3*A*b + 7*a*B)*Sqrt[x])/(64*a*b^4*(a + b*x)^2) + ((3*A*b + 7*a*B)*Sqrt[x
])/(128*a^2*b^4*(a + b*x)) + ((3*A*b + 7*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(128*a^(5/2)*b^(9/2))

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 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{x^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac{x^{5/2} (A+B x)}{(a+b x)^6} \, dx\\ &=\frac{(A b-a B) x^{7/2}}{5 a b (a+b x)^5}+\frac{(3 A b+7 a B) \int \frac{x^{5/2}}{(a+b x)^5} \, dx}{10 a b}\\ &=\frac{(A b-a B) x^{7/2}}{5 a b (a+b x)^5}-\frac{(3 A b+7 a B) x^{5/2}}{40 a b^2 (a+b x)^4}+\frac{(3 A b+7 a B) \int \frac{x^{3/2}}{(a+b x)^4} \, dx}{16 a b^2}\\ &=\frac{(A b-a B) x^{7/2}}{5 a b (a+b x)^5}-\frac{(3 A b+7 a B) x^{5/2}}{40 a b^2 (a+b x)^4}-\frac{(3 A b+7 a B) x^{3/2}}{48 a b^3 (a+b x)^3}+\frac{(3 A b+7 a B) \int \frac{\sqrt{x}}{(a+b x)^3} \, dx}{32 a b^3}\\ &=\frac{(A b-a B) x^{7/2}}{5 a b (a+b x)^5}-\frac{(3 A b+7 a B) x^{5/2}}{40 a b^2 (a+b x)^4}-\frac{(3 A b+7 a B) x^{3/2}}{48 a b^3 (a+b x)^3}-\frac{(3 A b+7 a B) \sqrt{x}}{64 a b^4 (a+b x)^2}+\frac{(3 A b+7 a B) \int \frac{1}{\sqrt{x} (a+b x)^2} \, dx}{128 a b^4}\\ &=\frac{(A b-a B) x^{7/2}}{5 a b (a+b x)^5}-\frac{(3 A b+7 a B) x^{5/2}}{40 a b^2 (a+b x)^4}-\frac{(3 A b+7 a B) x^{3/2}}{48 a b^3 (a+b x)^3}-\frac{(3 A b+7 a B) \sqrt{x}}{64 a b^4 (a+b x)^2}+\frac{(3 A b+7 a B) \sqrt{x}}{128 a^2 b^4 (a+b x)}+\frac{(3 A b+7 a B) \int \frac{1}{\sqrt{x} (a+b x)} \, dx}{256 a^2 b^4}\\ &=\frac{(A b-a B) x^{7/2}}{5 a b (a+b x)^5}-\frac{(3 A b+7 a B) x^{5/2}}{40 a b^2 (a+b x)^4}-\frac{(3 A b+7 a B) x^{3/2}}{48 a b^3 (a+b x)^3}-\frac{(3 A b+7 a B) \sqrt{x}}{64 a b^4 (a+b x)^2}+\frac{(3 A b+7 a B) \sqrt{x}}{128 a^2 b^4 (a+b x)}+\frac{(3 A b+7 a B) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{x}\right )}{128 a^2 b^4}\\ &=\frac{(A b-a B) x^{7/2}}{5 a b (a+b x)^5}-\frac{(3 A b+7 a B) x^{5/2}}{40 a b^2 (a+b x)^4}-\frac{(3 A b+7 a B) x^{3/2}}{48 a b^3 (a+b x)^3}-\frac{(3 A b+7 a B) \sqrt{x}}{64 a b^4 (a+b x)^2}+\frac{(3 A b+7 a B) \sqrt{x}}{128 a^2 b^4 (a+b x)}+\frac{(3 A b+7 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{128 a^{5/2} b^{9/2}}\\ \end{align*}

Mathematica [C]  time = 0.0316064, size = 61, normalized size = 0.31 \[ \frac{x^{7/2} \left (\frac{7 a^5 (A b-a B)}{(a+b x)^5}+(7 a B+3 A b) \, _2F_1\left (\frac{7}{2},5;\frac{9}{2};-\frac{b x}{a}\right )\right )}{35 a^6 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^(5/2)*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^3,x]

[Out]

(x^(7/2)*((7*a^5*(A*b - a*B))/(a + b*x)^5 + (3*A*b + 7*a*B)*Hypergeometric2F1[7/2, 5, 9/2, -((b*x)/a)]))/(35*a
^6*b)

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Maple [A]  time = 0.019, size = 154, normalized size = 0.8 \begin{align*} 2\,{\frac{1}{ \left ( bx+a \right ) ^{5}} \left ({\frac{ \left ( 3\,Ab+7\,aB \right ){x}^{9/2}}{256\,{a}^{2}}}+{\frac{ \left ( 21\,Ab-79\,aB \right ){x}^{7/2}}{384\,ab}}-1/30\,{\frac{ \left ( 3\,Ab+7\,aB \right ){x}^{5/2}}{{b}^{2}}}-{\frac{7\,a \left ( 3\,Ab+7\,aB \right ){x}^{3/2}}{384\,{b}^{3}}}-{\frac{ \left ( 3\,Ab+7\,aB \right ){a}^{2}\sqrt{x}}{256\,{b}^{4}}} \right ) }+{\frac{3\,A}{128\,{a}^{2}{b}^{3}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{7\,B}{128\,a{b}^{4}}\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^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x)

[Out]

2*(1/256*(3*A*b+7*B*a)/a^2*x^(9/2)+1/384*(21*A*b-79*B*a)/a/b*x^(7/2)-1/30/b^2*(3*A*b+7*B*a)*x^(5/2)-7/384*a/b^
3*(3*A*b+7*B*a)*x^(3/2)-1/256*(3*A*b+7*B*a)*a^2/b^4*x^(1/2))/(b*x+a)^5+3/128/a^2/b^3/(a*b)^(1/2)*arctan(x^(1/2
)*b/(a*b)^(1/2))*A+7/128/a/b^4/(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^(5/2)*(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 = 1.60831, size = 1435, normalized size = 7.36 \begin{align*} \left [-\frac{15 \,{\left (7 \, B a^{6} + 3 \, A a^{5} b +{\left (7 \, B a b^{5} + 3 \, A b^{6}\right )} x^{5} + 5 \,{\left (7 \, B a^{2} b^{4} + 3 \, A a b^{5}\right )} x^{4} + 10 \,{\left (7 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{3} + 10 \,{\left (7 \, B a^{4} b^{2} + 3 \, A a^{3} b^{3}\right )} x^{2} + 5 \,{\left (7 \, B a^{5} b + 3 \, A a^{4} b^{2}\right )} x\right )} \sqrt{-a b} \log \left (\frac{b x - a - 2 \, \sqrt{-a b} \sqrt{x}}{b x + a}\right ) + 2 \,{\left (105 \, B a^{6} b + 45 \, A a^{5} b^{2} - 15 \,{\left (7 \, B a^{2} b^{5} + 3 \, A a b^{6}\right )} x^{4} + 10 \,{\left (79 \, B a^{3} b^{4} - 21 \, A a^{2} b^{5}\right )} x^{3} + 128 \,{\left (7 \, B a^{4} b^{3} + 3 \, A a^{3} b^{4}\right )} x^{2} + 70 \,{\left (7 \, B a^{5} b^{2} + 3 \, A a^{4} b^{3}\right )} x\right )} \sqrt{x}}{3840 \,{\left (a^{3} b^{10} x^{5} + 5 \, a^{4} b^{9} x^{4} + 10 \, a^{5} b^{8} x^{3} + 10 \, a^{6} b^{7} x^{2} + 5 \, a^{7} b^{6} x + a^{8} b^{5}\right )}}, -\frac{15 \,{\left (7 \, B a^{6} + 3 \, A a^{5} b +{\left (7 \, B a b^{5} + 3 \, A b^{6}\right )} x^{5} + 5 \,{\left (7 \, B a^{2} b^{4} + 3 \, A a b^{5}\right )} x^{4} + 10 \,{\left (7 \, B a^{3} b^{3} + 3 \, A a^{2} b^{4}\right )} x^{3} + 10 \,{\left (7 \, B a^{4} b^{2} + 3 \, A a^{3} b^{3}\right )} x^{2} + 5 \,{\left (7 \, B a^{5} b + 3 \, A a^{4} b^{2}\right )} x\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{b \sqrt{x}}\right ) +{\left (105 \, B a^{6} b + 45 \, A a^{5} b^{2} - 15 \,{\left (7 \, B a^{2} b^{5} + 3 \, A a b^{6}\right )} x^{4} + 10 \,{\left (79 \, B a^{3} b^{4} - 21 \, A a^{2} b^{5}\right )} x^{3} + 128 \,{\left (7 \, B a^{4} b^{3} + 3 \, A a^{3} b^{4}\right )} x^{2} + 70 \,{\left (7 \, B a^{5} b^{2} + 3 \, A a^{4} b^{3}\right )} x\right )} \sqrt{x}}{1920 \,{\left (a^{3} b^{10} x^{5} + 5 \, a^{4} b^{9} x^{4} + 10 \, a^{5} b^{8} x^{3} + 10 \, a^{6} b^{7} x^{2} + 5 \, a^{7} b^{6} x + a^{8} b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="fricas")

[Out]

[-1/3840*(15*(7*B*a^6 + 3*A*a^5*b + (7*B*a*b^5 + 3*A*b^6)*x^5 + 5*(7*B*a^2*b^4 + 3*A*a*b^5)*x^4 + 10*(7*B*a^3*
b^3 + 3*A*a^2*b^4)*x^3 + 10*(7*B*a^4*b^2 + 3*A*a^3*b^3)*x^2 + 5*(7*B*a^5*b + 3*A*a^4*b^2)*x)*sqrt(-a*b)*log((b
*x - a - 2*sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(105*B*a^6*b + 45*A*a^5*b^2 - 15*(7*B*a^2*b^5 + 3*A*a*b^6)*x^4 +
 10*(79*B*a^3*b^4 - 21*A*a^2*b^5)*x^3 + 128*(7*B*a^4*b^3 + 3*A*a^3*b^4)*x^2 + 70*(7*B*a^5*b^2 + 3*A*a^4*b^3)*x
)*sqrt(x))/(a^3*b^10*x^5 + 5*a^4*b^9*x^4 + 10*a^5*b^8*x^3 + 10*a^6*b^7*x^2 + 5*a^7*b^6*x + a^8*b^5), -1/1920*(
15*(7*B*a^6 + 3*A*a^5*b + (7*B*a*b^5 + 3*A*b^6)*x^5 + 5*(7*B*a^2*b^4 + 3*A*a*b^5)*x^4 + 10*(7*B*a^3*b^3 + 3*A*
a^2*b^4)*x^3 + 10*(7*B*a^4*b^2 + 3*A*a^3*b^3)*x^2 + 5*(7*B*a^5*b + 3*A*a^4*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/
(b*sqrt(x))) + (105*B*a^6*b + 45*A*a^5*b^2 - 15*(7*B*a^2*b^5 + 3*A*a*b^6)*x^4 + 10*(79*B*a^3*b^4 - 21*A*a^2*b^
5)*x^3 + 128*(7*B*a^4*b^3 + 3*A*a^3*b^4)*x^2 + 70*(7*B*a^5*b^2 + 3*A*a^4*b^3)*x)*sqrt(x))/(a^3*b^10*x^5 + 5*a^
4*b^9*x^4 + 10*a^5*b^8*x^3 + 10*a^6*b^7*x^2 + 5*a^7*b^6*x + a^8*b^5)]

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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**(5/2)*(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 = 1.32255, size = 211, normalized size = 1.08 \begin{align*} \frac{{\left (7 \, B a + 3 \, A b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{128 \, \sqrt{a b} a^{2} b^{4}} + \frac{105 \, B a b^{4} x^{\frac{9}{2}} + 45 \, A b^{5} x^{\frac{9}{2}} - 790 \, B a^{2} b^{3} x^{\frac{7}{2}} + 210 \, A a b^{4} x^{\frac{7}{2}} - 896 \, B a^{3} b^{2} x^{\frac{5}{2}} - 384 \, A a^{2} b^{3} x^{\frac{5}{2}} - 490 \, B a^{4} b x^{\frac{3}{2}} - 210 \, A a^{3} b^{2} x^{\frac{3}{2}} - 105 \, B a^{5} \sqrt{x} - 45 \, A a^{4} b \sqrt{x}}{1920 \,{\left (b x + a\right )}^{5} a^{2} b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x, algorithm="giac")

[Out]

1/128*(7*B*a + 3*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^2*b^4) + 1/1920*(105*B*a*b^4*x^(9/2) + 45*A*b^5
*x^(9/2) - 790*B*a^2*b^3*x^(7/2) + 210*A*a*b^4*x^(7/2) - 896*B*a^3*b^2*x^(5/2) - 384*A*a^2*b^3*x^(5/2) - 490*B
*a^4*b*x^(3/2) - 210*A*a^3*b^2*x^(3/2) - 105*B*a^5*sqrt(x) - 45*A*a^4*b*sqrt(x))/((b*x + a)^5*a^2*b^4)