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

Optimal. Leaf size=185 \[ \frac{3 \sqrt{x} (a B+A b)}{128 a^3 b^3 (a+b x)}+\frac{\sqrt{x} (a B+A b)}{64 a^2 b^3 (a+b x)^2}+\frac{3 (a B+A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{128 a^{7/2} b^{7/2}}-\frac{x^{3/2} (a B+A b)}{8 a b^2 (a+b x)^4}-\frac{\sqrt{x} (a B+A b)}{16 a b^3 (a+b x)^3}+\frac{x^{5/2} (A b-a B)}{5 a b (a+b x)^5} \]

[Out]

((A*b - a*B)*x^(5/2))/(5*a*b*(a + b*x)^5) - ((A*b + a*B)*x^(3/2))/(8*a*b^2*(a + b*x)^4) - ((A*b + a*B)*Sqrt[x]
)/(16*a*b^3*(a + b*x)^3) + ((A*b + a*B)*Sqrt[x])/(64*a^2*b^3*(a + b*x)^2) + (3*(A*b + a*B)*Sqrt[x])/(128*a^3*b
^3*(a + b*x)) + (3*(A*b + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(128*a^(7/2)*b^(7/2))

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Rubi [A]  time = 0.0925746, antiderivative size = 185, 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} \[ \frac{3 \sqrt{x} (a B+A b)}{128 a^3 b^3 (a+b x)}+\frac{\sqrt{x} (a B+A b)}{64 a^2 b^3 (a+b x)^2}+\frac{3 (a B+A b) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{128 a^{7/2} b^{7/2}}-\frac{x^{3/2} (a B+A b)}{8 a b^2 (a+b x)^4}-\frac{\sqrt{x} (a B+A b)}{16 a b^3 (a+b x)^3}+\frac{x^{5/2} (A b-a B)}{5 a b (a+b x)^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b - a*B)*x^(5/2))/(5*a*b*(a + b*x)^5) - ((A*b + a*B)*x^(3/2))/(8*a*b^2*(a + b*x)^4) - ((A*b + a*B)*Sqrt[x]
)/(16*a*b^3*(a + b*x)^3) + ((A*b + a*B)*Sqrt[x])/(64*a^2*b^3*(a + b*x)^2) + (3*(A*b + a*B)*Sqrt[x])/(128*a^3*b
^3*(a + b*x)) + (3*(A*b + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(128*a^(7/2)*b^(7/2))

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^{3/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^3} \, dx &=\int \frac{x^{3/2} (A+B x)}{(a+b x)^6} \, dx\\ &=\frac{(A b-a B) x^{5/2}}{5 a b (a+b x)^5}+\frac{(A b+a B) \int \frac{x^{3/2}}{(a+b x)^5} \, dx}{2 a b}\\ &=\frac{(A b-a B) x^{5/2}}{5 a b (a+b x)^5}-\frac{(A b+a B) x^{3/2}}{8 a b^2 (a+b x)^4}+\frac{(3 (A b+a B)) \int \frac{\sqrt{x}}{(a+b x)^4} \, dx}{16 a b^2}\\ &=\frac{(A b-a B) x^{5/2}}{5 a b (a+b x)^5}-\frac{(A b+a B) x^{3/2}}{8 a b^2 (a+b x)^4}-\frac{(A b+a B) \sqrt{x}}{16 a b^3 (a+b x)^3}+\frac{(A b+a B) \int \frac{1}{\sqrt{x} (a+b x)^3} \, dx}{32 a b^3}\\ &=\frac{(A b-a B) x^{5/2}}{5 a b (a+b x)^5}-\frac{(A b+a B) x^{3/2}}{8 a b^2 (a+b x)^4}-\frac{(A b+a B) \sqrt{x}}{16 a b^3 (a+b x)^3}+\frac{(A b+a B) \sqrt{x}}{64 a^2 b^3 (a+b x)^2}+\frac{(3 (A b+a B)) \int \frac{1}{\sqrt{x} (a+b x)^2} \, dx}{128 a^2 b^3}\\ &=\frac{(A b-a B) x^{5/2}}{5 a b (a+b x)^5}-\frac{(A b+a B) x^{3/2}}{8 a b^2 (a+b x)^4}-\frac{(A b+a B) \sqrt{x}}{16 a b^3 (a+b x)^3}+\frac{(A b+a B) \sqrt{x}}{64 a^2 b^3 (a+b x)^2}+\frac{3 (A b+a B) \sqrt{x}}{128 a^3 b^3 (a+b x)}+\frac{(3 (A b+a B)) \int \frac{1}{\sqrt{x} (a+b x)} \, dx}{256 a^3 b^3}\\ &=\frac{(A b-a B) x^{5/2}}{5 a b (a+b x)^5}-\frac{(A b+a B) x^{3/2}}{8 a b^2 (a+b x)^4}-\frac{(A b+a B) \sqrt{x}}{16 a b^3 (a+b x)^3}+\frac{(A b+a B) \sqrt{x}}{64 a^2 b^3 (a+b x)^2}+\frac{3 (A b+a B) \sqrt{x}}{128 a^3 b^3 (a+b x)}+\frac{(3 (A b+a B)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{x}\right )}{128 a^3 b^3}\\ &=\frac{(A b-a B) x^{5/2}}{5 a b (a+b x)^5}-\frac{(A b+a B) x^{3/2}}{8 a b^2 (a+b x)^4}-\frac{(A b+a B) \sqrt{x}}{16 a b^3 (a+b x)^3}+\frac{(A b+a B) \sqrt{x}}{64 a^2 b^3 (a+b x)^2}+\frac{3 (A b+a B) \sqrt{x}}{128 a^3 b^3 (a+b x)}+\frac{3 (A b+a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{128 a^{7/2} b^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0330053, size = 60, normalized size = 0.32 \[ \frac{x^{5/2} \left (\frac{5 a^5 (A b-a B)}{(a+b x)^5}+5 (a B+A b) \, _2F_1\left (\frac{5}{2},5;\frac{7}{2};-\frac{b x}{a}\right )\right )}{25 a^6 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(5/2)*((5*a^5*(A*b - a*B))/(a + b*x)^5 + 5*(A*b + a*B)*Hypergeometric2F1[5/2, 5, 7/2, -((b*x)/a)]))/(25*a^6
*b)

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Maple [A]  time = 0.017, size = 143, normalized size = 0.8 \begin{align*} 2\,{\frac{1}{ \left ( bx+a \right ) ^{5}} \left ({\frac{ \left ( 3\,Ab+3\,aB \right ) b{x}^{9/2}}{256\,{a}^{3}}}+{\frac{ \left ( 7\,Ab+7\,aB \right ){x}^{7/2}}{128\,{a}^{2}}}+1/10\,{\frac{ \left ( Ab-aB \right ){x}^{5/2}}{ab}}-{\frac{ \left ( 7\,Ab+7\,aB \right ){x}^{3/2}}{128\,{b}^{2}}}-{\frac{ \left ( 3\,Ab+3\,aB \right ) a\sqrt{x}}{256\,{b}^{3}}} \right ) }+{\frac{3\,A}{128\,{a}^{3}{b}^{2}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{3\,B}{128\,{a}^{2}{b}^{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(x^(3/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^3,x)

[Out]

2*(3/256*(A*b+B*a)/a^3*b*x^(9/2)+7/128*(A*b+B*a)/a^2*x^(7/2)+1/10*(A*b-B*a)/a/b*x^(5/2)-7/128*(A*b+B*a)/b^2*x^
(3/2)-3/256*(A*b+B*a)*a/b^3*x^(1/2))/(b*x+a)^5+3/128/a^3/b^2/(a*b)^(1/2)*arctan(x^(1/2)*b/(a*b)^(1/2))*A+3/128
/a^2/b^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(x^(3/2)*(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 = 1.7149, size = 1315, normalized size = 7.11 \begin{align*} \left [-\frac{15 \,{\left (B a^{6} + A a^{5} b +{\left (B a b^{5} + A b^{6}\right )} x^{5} + 5 \,{\left (B a^{2} b^{4} + A a b^{5}\right )} x^{4} + 10 \,{\left (B a^{3} b^{3} + A a^{2} b^{4}\right )} x^{3} + 10 \,{\left (B a^{4} b^{2} + A a^{3} b^{3}\right )} x^{2} + 5 \,{\left (B a^{5} b + 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 (15 \, B a^{6} b + 15 \, A a^{5} b^{2} - 15 \,{\left (B a^{2} b^{5} + A a b^{6}\right )} x^{4} - 70 \,{\left (B a^{3} b^{4} + A a^{2} b^{5}\right )} x^{3} + 128 \,{\left (B a^{4} b^{3} - A a^{3} b^{4}\right )} x^{2} + 70 \,{\left (B a^{5} b^{2} + A a^{4} b^{3}\right )} x\right )} \sqrt{x}}{1280 \,{\left (a^{4} b^{9} x^{5} + 5 \, a^{5} b^{8} x^{4} + 10 \, a^{6} b^{7} x^{3} + 10 \, a^{7} b^{6} x^{2} + 5 \, a^{8} b^{5} x + a^{9} b^{4}\right )}}, -\frac{15 \,{\left (B a^{6} + A a^{5} b +{\left (B a b^{5} + A b^{6}\right )} x^{5} + 5 \,{\left (B a^{2} b^{4} + A a b^{5}\right )} x^{4} + 10 \,{\left (B a^{3} b^{3} + A a^{2} b^{4}\right )} x^{3} + 10 \,{\left (B a^{4} b^{2} + A a^{3} b^{3}\right )} x^{2} + 5 \,{\left (B a^{5} b + A a^{4} b^{2}\right )} x\right )} \sqrt{a b} \arctan \left (\frac{\sqrt{a b}}{b \sqrt{x}}\right ) +{\left (15 \, B a^{6} b + 15 \, A a^{5} b^{2} - 15 \,{\left (B a^{2} b^{5} + A a b^{6}\right )} x^{4} - 70 \,{\left (B a^{3} b^{4} + A a^{2} b^{5}\right )} x^{3} + 128 \,{\left (B a^{4} b^{3} - A a^{3} b^{4}\right )} x^{2} + 70 \,{\left (B a^{5} b^{2} + A a^{4} b^{3}\right )} x\right )} \sqrt{x}}{640 \,{\left (a^{4} b^{9} x^{5} + 5 \, a^{5} b^{8} x^{4} + 10 \, a^{6} b^{7} x^{3} + 10 \, a^{7} b^{6} x^{2} + 5 \, a^{8} b^{5} x + a^{9} b^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/1280*(15*(B*a^6 + A*a^5*b + (B*a*b^5 + A*b^6)*x^5 + 5*(B*a^2*b^4 + A*a*b^5)*x^4 + 10*(B*a^3*b^3 + A*a^2*b^
4)*x^3 + 10*(B*a^4*b^2 + A*a^3*b^3)*x^2 + 5*(B*a^5*b + A*a^4*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*sqrt(-a*b)*sq
rt(x))/(b*x + a)) + 2*(15*B*a^6*b + 15*A*a^5*b^2 - 15*(B*a^2*b^5 + A*a*b^6)*x^4 - 70*(B*a^3*b^4 + A*a^2*b^5)*x
^3 + 128*(B*a^4*b^3 - A*a^3*b^4)*x^2 + 70*(B*a^5*b^2 + A*a^4*b^3)*x)*sqrt(x))/(a^4*b^9*x^5 + 5*a^5*b^8*x^4 + 1
0*a^6*b^7*x^3 + 10*a^7*b^6*x^2 + 5*a^8*b^5*x + a^9*b^4), -1/640*(15*(B*a^6 + A*a^5*b + (B*a*b^5 + A*b^6)*x^5 +
5*(B*a^2*b^4 + A*a*b^5)*x^4 + 10*(B*a^3*b^3 + A*a^2*b^4)*x^3 + 10*(B*a^4*b^2 + A*a^3*b^3)*x^2 + 5*(B*a^5*b +
A*a^4*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (15*B*a^6*b + 15*A*a^5*b^2 - 15*(B*a^2*b^5 + A*a*b^6)*
x^4 - 70*(B*a^3*b^4 + A*a^2*b^5)*x^3 + 128*(B*a^4*b^3 - A*a^3*b^4)*x^2 + 70*(B*a^5*b^2 + A*a^4*b^3)*x)*sqrt(x)
)/(a^4*b^9*x^5 + 5*a^5*b^8*x^4 + 10*a^6*b^7*x^3 + 10*a^7*b^6*x^2 + 5*a^8*b^5*x + a^9*b^4)]

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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**(3/2)*(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 = 1.21849, size = 208, normalized size = 1.12 \begin{align*} \frac{3 \,{\left (B a + A b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{128 \, \sqrt{a b} a^{3} b^{3}} + \frac{15 \, B a b^{4} x^{\frac{9}{2}} + 15 \, A b^{5} x^{\frac{9}{2}} + 70 \, B a^{2} b^{3} x^{\frac{7}{2}} + 70 \, A a b^{4} x^{\frac{7}{2}} - 128 \, B a^{3} b^{2} x^{\frac{5}{2}} + 128 \, A a^{2} b^{3} x^{\frac{5}{2}} - 70 \, B a^{4} b x^{\frac{3}{2}} - 70 \, A a^{3} b^{2} x^{\frac{3}{2}} - 15 \, B a^{5} \sqrt{x} - 15 \, A a^{4} b \sqrt{x}}{640 \,{\left (b x + a\right )}^{5} a^{3} b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/128*(B*a + A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3*b^3) + 1/640*(15*B*a*b^4*x^(9/2) + 15*A*b^5*x^(9/
2) + 70*B*a^2*b^3*x^(7/2) + 70*A*a*b^4*x^(7/2) - 128*B*a^3*b^2*x^(5/2) + 128*A*a^2*b^3*x^(5/2) - 70*B*a^4*b*x^
(3/2) - 70*A*a^3*b^2*x^(3/2) - 15*B*a^5*sqrt(x) - 15*A*a^4*b*sqrt(x))/((b*x + a)^5*a^3*b^3)