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

Optimal. Leaf size=173 \[ \frac{x^{7/2} (A b-3 a B)}{4 a b^2 (a+b x)^2}+\frac{7 x^{5/2} (A b-3 a B)}{8 a b^3 (a+b x)}-\frac{35 x^{3/2} (A b-3 a B)}{24 a b^4}+\frac{35 \sqrt{x} (A b-3 a B)}{8 b^5}-\frac{35 \sqrt{a} (A b-3 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{8 b^{11/2}}+\frac{x^{9/2} (A b-a B)}{3 a b (a+b x)^3} \]

[Out]

(35*(A*b - 3*a*B)*Sqrt[x])/(8*b^5) - (35*(A*b - 3*a*B)*x^(3/2))/(24*a*b^4) + ((A*b - a*B)*x^(9/2))/(3*a*b*(a +
b*x)^3) + ((A*b - 3*a*B)*x^(7/2))/(4*a*b^2*(a + b*x)^2) + (7*(A*b - 3*a*B)*x^(5/2))/(8*a*b^3*(a + b*x)) - (35
*Sqrt[a]*(A*b - 3*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8*b^(11/2))

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Rubi [A]  time = 0.0783583, antiderivative size = 173, 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, 50, 63, 205} \[ \frac{x^{7/2} (A b-3 a B)}{4 a b^2 (a+b x)^2}+\frac{7 x^{5/2} (A b-3 a B)}{8 a b^3 (a+b x)}-\frac{35 x^{3/2} (A b-3 a B)}{24 a b^4}+\frac{35 \sqrt{x} (A b-3 a B)}{8 b^5}-\frac{35 \sqrt{a} (A b-3 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{8 b^{11/2}}+\frac{x^{9/2} (A b-a B)}{3 a b (a+b x)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(35*(A*b - 3*a*B)*Sqrt[x])/(8*b^5) - (35*(A*b - 3*a*B)*x^(3/2))/(24*a*b^4) + ((A*b - a*B)*x^(9/2))/(3*a*b*(a +
b*x)^3) + ((A*b - 3*a*B)*x^(7/2))/(4*a*b^2*(a + b*x)^2) + (7*(A*b - 3*a*B)*x^(5/2))/(8*a*b^3*(a + b*x)) - (35
*Sqrt[a]*(A*b - 3*a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(8*b^(11/2))

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^{7/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{x^{7/2} (A+B x)}{(a+b x)^4} \, dx\\ &=\frac{(A b-a B) x^{9/2}}{3 a b (a+b x)^3}-\frac{\left (\frac{3 A b}{2}-\frac{9 a B}{2}\right ) \int \frac{x^{7/2}}{(a+b x)^3} \, dx}{3 a b}\\ &=\frac{(A b-a B) x^{9/2}}{3 a b (a+b x)^3}+\frac{(A b-3 a B) x^{7/2}}{4 a b^2 (a+b x)^2}-\frac{(7 (A b-3 a B)) \int \frac{x^{5/2}}{(a+b x)^2} \, dx}{8 a b^2}\\ &=\frac{(A b-a B) x^{9/2}}{3 a b (a+b x)^3}+\frac{(A b-3 a B) x^{7/2}}{4 a b^2 (a+b x)^2}+\frac{7 (A b-3 a B) x^{5/2}}{8 a b^3 (a+b x)}-\frac{(35 (A b-3 a B)) \int \frac{x^{3/2}}{a+b x} \, dx}{16 a b^3}\\ &=-\frac{35 (A b-3 a B) x^{3/2}}{24 a b^4}+\frac{(A b-a B) x^{9/2}}{3 a b (a+b x)^3}+\frac{(A b-3 a B) x^{7/2}}{4 a b^2 (a+b x)^2}+\frac{7 (A b-3 a B) x^{5/2}}{8 a b^3 (a+b x)}+\frac{(35 (A b-3 a B)) \int \frac{\sqrt{x}}{a+b x} \, dx}{16 b^4}\\ &=\frac{35 (A b-3 a B) \sqrt{x}}{8 b^5}-\frac{35 (A b-3 a B) x^{3/2}}{24 a b^4}+\frac{(A b-a B) x^{9/2}}{3 a b (a+b x)^3}+\frac{(A b-3 a B) x^{7/2}}{4 a b^2 (a+b x)^2}+\frac{7 (A b-3 a B) x^{5/2}}{8 a b^3 (a+b x)}-\frac{(35 a (A b-3 a B)) \int \frac{1}{\sqrt{x} (a+b x)} \, dx}{16 b^5}\\ &=\frac{35 (A b-3 a B) \sqrt{x}}{8 b^5}-\frac{35 (A b-3 a B) x^{3/2}}{24 a b^4}+\frac{(A b-a B) x^{9/2}}{3 a b (a+b x)^3}+\frac{(A b-3 a B) x^{7/2}}{4 a b^2 (a+b x)^2}+\frac{7 (A b-3 a B) x^{5/2}}{8 a b^3 (a+b x)}-\frac{(35 a (A b-3 a B)) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\sqrt{x}\right )}{8 b^5}\\ &=\frac{35 (A b-3 a B) \sqrt{x}}{8 b^5}-\frac{35 (A b-3 a B) x^{3/2}}{24 a b^4}+\frac{(A b-a B) x^{9/2}}{3 a b (a+b x)^3}+\frac{(A b-3 a B) x^{7/2}}{4 a b^2 (a+b x)^2}+\frac{7 (A b-3 a B) x^{5/2}}{8 a b^3 (a+b x)}-\frac{35 \sqrt{a} (A b-3 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{8 b^{11/2}}\\ \end{align*}

Mathematica [C]  time = 0.0319197, size = 61, normalized size = 0.35 \[ \frac{x^{9/2} \left (\frac{9 a^3 (A b-a B)}{(a+b x)^3}+(9 a B-3 A b) \, _2F_1\left (3,\frac{9}{2};\frac{11}{2};-\frac{b x}{a}\right )\right )}{27 a^4 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(9/2)*((9*a^3*(A*b - a*B))/(a + b*x)^3 + (-3*A*b + 9*a*B)*Hypergeometric2F1[3, 9/2, 11/2, -((b*x)/a)]))/(27
*a^4*b)

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Maple [A]  time = 0.02, size = 190, normalized size = 1.1 \begin{align*}{\frac{2\,B}{3\,{b}^{4}}{x}^{{\frac{3}{2}}}}+2\,{\frac{A\sqrt{x}}{{b}^{4}}}-8\,{\frac{aB\sqrt{x}}{{b}^{5}}}+{\frac{29\,aA}{8\,{b}^{2} \left ( bx+a \right ) ^{3}}{x}^{{\frac{5}{2}}}}-{\frac{55\,B{a}^{2}}{8\,{b}^{3} \left ( bx+a \right ) ^{3}}{x}^{{\frac{5}{2}}}}+{\frac{17\,A{a}^{2}}{3\,{b}^{3} \left ( bx+a \right ) ^{3}}{x}^{{\frac{3}{2}}}}-{\frac{35\,B{a}^{3}}{3\,{b}^{4} \left ( bx+a \right ) ^{3}}{x}^{{\frac{3}{2}}}}+{\frac{19\,A{a}^{3}}{8\,{b}^{4} \left ( bx+a \right ) ^{3}}\sqrt{x}}-{\frac{41\,B{a}^{4}}{8\,{b}^{5} \left ( bx+a \right ) ^{3}}\sqrt{x}}-{\frac{35\,aA}{8\,{b}^{4}}\arctan \left ({b\sqrt{x}{\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}+{\frac{105\,B{a}^{2}}{8\,{b}^{5}}\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^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x)

[Out]

2/3/b^4*B*x^(3/2)+2/b^4*A*x^(1/2)-8/b^5*a*B*x^(1/2)+29/8*a/b^2/(b*x+a)^3*x^(5/2)*A-55/8*a^2/b^3/(b*x+a)^3*x^(5
/2)*B+17/3*a^2/b^3/(b*x+a)^3*A*x^(3/2)-35/3*a^3/b^4/(b*x+a)^3*B*x^(3/2)+19/8*a^3/b^4/(b*x+a)^3*x^(1/2)*A-41/8*
a^4/b^5/(b*x+a)^3*x^(1/2)*B-35/8*a/b^4/(a*b)^(1/2)*arctan(x^(1/2)*b/(a*b)^(1/2))*A+105/8*a^2/b^5/(a*b)^(1/2)*a
rctan(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^(7/2)*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.61833, size = 994, normalized size = 5.75 \begin{align*} \left [-\frac{105 \,{\left (3 \, B a^{4} - A a^{3} b +{\left (3 \, B a b^{3} - A b^{4}\right )} x^{3} + 3 \,{\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 3 \,{\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt{-\frac{a}{b}} \log \left (\frac{b x - 2 \, b \sqrt{x} \sqrt{-\frac{a}{b}} - a}{b x + a}\right ) - 2 \,{\left (16 \, B b^{4} x^{4} - 315 \, B a^{4} + 105 \, A a^{3} b - 48 \,{\left (3 \, B a b^{3} - A b^{4}\right )} x^{3} - 231 \,{\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 280 \,{\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt{x}}{48 \,{\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}, \frac{105 \,{\left (3 \, B a^{4} - A a^{3} b +{\left (3 \, B a b^{3} - A b^{4}\right )} x^{3} + 3 \,{\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 3 \,{\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) +{\left (16 \, B b^{4} x^{4} - 315 \, B a^{4} + 105 \, A a^{3} b - 48 \,{\left (3 \, B a b^{3} - A b^{4}\right )} x^{3} - 231 \,{\left (3 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} - 280 \,{\left (3 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \sqrt{x}}{24 \,{\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/48*(105*(3*B*a^4 - A*a^3*b + (3*B*a*b^3 - A*b^4)*x^3 + 3*(3*B*a^2*b^2 - A*a*b^3)*x^2 + 3*(3*B*a^3*b - A*a^
2*b^2)*x)*sqrt(-a/b)*log((b*x - 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(16*B*b^4*x^4 - 315*B*a^4 + 105*A*a
^3*b - 48*(3*B*a*b^3 - A*b^4)*x^3 - 231*(3*B*a^2*b^2 - A*a*b^3)*x^2 - 280*(3*B*a^3*b - A*a^2*b^2)*x)*sqrt(x))/
(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5), 1/24*(105*(3*B*a^4 - A*a^3*b + (3*B*a*b^3 - A*b^4)*x^3 + 3*(3
*B*a^2*b^2 - A*a*b^3)*x^2 + 3*(3*B*a^3*b - A*a^2*b^2)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (16*B*b^4*x
^4 - 315*B*a^4 + 105*A*a^3*b - 48*(3*B*a*b^3 - A*b^4)*x^3 - 231*(3*B*a^2*b^2 - A*a*b^3)*x^2 - 280*(3*B*a^3*b -
A*a^2*b^2)*x)*sqrt(x))/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*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**(7/2)*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**2,x)

[Out]

Timed out

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Giac [A]  time = 1.16413, size = 193, normalized size = 1.12 \begin{align*} \frac{35 \,{\left (3 \, B a^{2} - A a b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{8 \, \sqrt{a b} b^{5}} - \frac{165 \, B a^{2} b^{2} x^{\frac{5}{2}} - 87 \, A a b^{3} x^{\frac{5}{2}} + 280 \, B a^{3} b x^{\frac{3}{2}} - 136 \, A a^{2} b^{2} x^{\frac{3}{2}} + 123 \, B a^{4} \sqrt{x} - 57 \, A a^{3} b \sqrt{x}}{24 \,{\left (b x + a\right )}^{3} b^{5}} + \frac{2 \,{\left (B b^{8} x^{\frac{3}{2}} - 12 \, B a b^{7} \sqrt{x} + 3 \, A b^{8} \sqrt{x}\right )}}{3 \, b^{12}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

35/8*(3*B*a^2 - A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^5) - 1/24*(165*B*a^2*b^2*x^(5/2) - 87*A*a*b^3*
x^(5/2) + 280*B*a^3*b*x^(3/2) - 136*A*a^2*b^2*x^(3/2) + 123*B*a^4*sqrt(x) - 57*A*a^3*b*sqrt(x))/((b*x + a)^3*b
^5) + 2/3*(B*b^8*x^(3/2) - 12*B*a*b^7*sqrt(x) + 3*A*b^8*sqrt(x))/b^12