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

Optimal. Leaf size=255 \[ \frac{x^{7/2} (A b-a B)}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^{5/2} (3 A b-7 a B)}{4 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 x^{3/2} (a+b x) (3 A b-7 a B)}{12 a b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 \sqrt{x} (a+b x) (3 A b-7 a B)}{4 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 \sqrt{a} (a+b x) (3 A b-7 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

((3*A*b - 7*a*B)*x^(5/2))/(4*a*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - a*B)*x^(7/2))/(2*a*b*(a + b*x)*Sqr
t[a^2 + 2*a*b*x + b^2*x^2]) + (5*(3*A*b - 7*a*B)*Sqrt[x]*(a + b*x))/(4*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5
*(3*A*b - 7*a*B)*x^(3/2)*(a + b*x))/(12*a*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*Sqrt[a]*(3*A*b - 7*a*B)*(a +
b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*b^(9/2)*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.121023, antiderivative size = 255, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {770, 78, 47, 50, 63, 205} \[ \frac{x^{7/2} (A b-a B)}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{x^{5/2} (3 A b-7 a B)}{4 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 x^{3/2} (a+b x) (3 A b-7 a B)}{12 a b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 \sqrt{x} (a+b x) (3 A b-7 a B)}{4 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 \sqrt{a} (a+b x) (3 A b-7 a B) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A*b - 7*a*B)*x^(5/2))/(4*a*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - a*B)*x^(7/2))/(2*a*b*(a + b*x)*Sqr
t[a^2 + 2*a*b*x + b^2*x^2]) + (5*(3*A*b - 7*a*B)*Sqrt[x]*(a + b*x))/(4*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5
*(3*A*b - 7*a*B)*x^(3/2)*(a + b*x))/(12*a*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (5*Sqrt[a]*(3*A*b - 7*a*B)*(a +
b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*b^(9/2)*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 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^{5/2} (A+B x)}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{x^{5/2} (A+B x)}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{(A b-a B) x^{7/2}}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left ((3 A b-7 a B) \left (a b+b^2 x\right )\right ) \int \frac{x^{5/2}}{\left (a b+b^2 x\right )^2} \, dx}{4 a \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{(3 A b-7 a B) x^{5/2}}{4 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) x^{7/2}}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (5 (3 A b-7 a B) \left (a b+b^2 x\right )\right ) \int \frac{x^{3/2}}{a b+b^2 x} \, dx}{8 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{(3 A b-7 a B) x^{5/2}}{4 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) x^{7/2}}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 (3 A b-7 a B) x^{3/2} (a+b x)}{12 a b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (5 (3 A b-7 a B) \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{x}}{a b+b^2 x} \, dx}{8 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{(3 A b-7 a B) x^{5/2}}{4 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) x^{7/2}}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 (3 A b-7 a B) \sqrt{x} (a+b x)}{4 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 (3 A b-7 a B) x^{3/2} (a+b x)}{12 a b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (5 a (3 A b-7 a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{\sqrt{x} \left (a b+b^2 x\right )} \, dx}{8 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{(3 A b-7 a B) x^{5/2}}{4 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) x^{7/2}}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 (3 A b-7 a B) \sqrt{x} (a+b x)}{4 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 (3 A b-7 a B) x^{3/2} (a+b x)}{12 a b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (5 a (3 A b-7 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 )}{4 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{(3 A b-7 a B) x^{5/2}}{4 a b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(A b-a B) x^{7/2}}{2 a b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{5 (3 A b-7 a B) \sqrt{x} (a+b x)}{4 b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 (3 A b-7 a B) x^{3/2} (a+b x)}{12 a b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{5 \sqrt{a} (3 A b-7 a B) (a+b x) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0321729, size = 79, normalized size = 0.31 \[ \frac{x^{7/2} \left (7 a^2 (A b-a B)+(a+b x)^2 (7 a B-3 A b) \, _2F_1\left (2,\frac{7}{2};\frac{9}{2};-\frac{b x}{a}\right )\right )}{14 a^3 b (a+b x) \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(7/2)*(7*a^2*(A*b - a*B) + (-3*A*b + 7*a*B)*(a + b*x)^2*Hypergeometric2F1[2, 7/2, 9/2, -((b*x)/a)]))/(14*a^
3*b*(a + b*x)*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.017, size = 247, normalized size = 1. \begin{align*}{\frac{bx+a}{12\,{b}^{4}} \left ( 8\,B\sqrt{ab}{x}^{7/2}{b}^{3}-56\,B\sqrt{ab}{x}^{5/2}a{b}^{2}+75\,A\sqrt{ab}{x}^{3/2}a{b}^{2}+24\,A\sqrt{ab}{x}^{5/2}{b}^{3}-45\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{2}a{b}^{3}-175\,B\sqrt{ab}{x}^{3/2}{a}^{2}b+105\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){x}^{2}{a}^{2}{b}^{2}-90\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ) x{a}^{2}{b}^{2}+210\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ) x{a}^{3}b+45\,A\sqrt{ab}\sqrt{x}{a}^{2}b-45\,A\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){a}^{3}b-105\,B\sqrt{ab}\sqrt{x}{a}^{3}+105\,B\arctan \left ({\frac{\sqrt{x}b}{\sqrt{ab}}} \right ){a}^{4} \right ){\frac{1}{\sqrt{ab}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \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)^(3/2),x)

[Out]

1/12*(8*B*(a*b)^(1/2)*x^(7/2)*b^3-56*B*(a*b)^(1/2)*x^(5/2)*a*b^2+75*A*(a*b)^(1/2)*x^(3/2)*a*b^2+24*A*(a*b)^(1/
2)*x^(5/2)*b^3-45*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x^2*a*b^3-175*B*(a*b)^(1/2)*x^(3/2)*a^2*b+105*B*arctan(x^(1/
2)*b/(a*b)^(1/2))*x^2*a^2*b^2-90*A*arctan(x^(1/2)*b/(a*b)^(1/2))*x*a^2*b^2+210*B*arctan(x^(1/2)*b/(a*b)^(1/2))
*x*a^3*b+45*A*(a*b)^(1/2)*x^(1/2)*a^2*b-45*A*arctan(x^(1/2)*b/(a*b)^(1/2))*a^3*b-105*B*(a*b)^(1/2)*x^(1/2)*a^3
+105*B*arctan(x^(1/2)*b/(a*b)^(1/2))*a^4)*(b*x+a)/(a*b)^(1/2)/b^4/((b*x+a)^2)^(3/2)

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.29158, size = 775, normalized size = 3.04 \begin{align*} \left [-\frac{15 \,{\left (7 \, B a^{3} - 3 \, A a^{2} b +{\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} + 2 \,{\left (7 \, B a^{2} b - 3 \, A a 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 (8 \, B b^{3} x^{3} - 105 \, B a^{3} + 45 \, A a^{2} b - 8 \,{\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} - 25 \,{\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt{x}}{24 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\right )}}, \frac{15 \,{\left (7 \, B a^{3} - 3 \, A a^{2} b +{\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} + 2 \,{\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt{\frac{a}{b}} \arctan \left (\frac{b \sqrt{x} \sqrt{\frac{a}{b}}}{a}\right ) +{\left (8 \, B b^{3} x^{3} - 105 \, B a^{3} + 45 \, A a^{2} b - 8 \,{\left (7 \, B a b^{2} - 3 \, A b^{3}\right )} x^{2} - 25 \,{\left (7 \, B a^{2} b - 3 \, A a b^{2}\right )} x\right )} \sqrt{x}}{12 \,{\left (b^{6} x^{2} + 2 \, a b^{5} x + a^{2} b^{4}\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)^(3/2),x, algorithm="fricas")

[Out]

[-1/24*(15*(7*B*a^3 - 3*A*a^2*b + (7*B*a*b^2 - 3*A*b^3)*x^2 + 2*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(-a/b)*log((b*x
- 2*b*sqrt(x)*sqrt(-a/b) - a)/(b*x + a)) - 2*(8*B*b^3*x^3 - 105*B*a^3 + 45*A*a^2*b - 8*(7*B*a*b^2 - 3*A*b^3)*
x^2 - 25*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(x))/(b^6*x^2 + 2*a*b^5*x + a^2*b^4), 1/12*(15*(7*B*a^3 - 3*A*a^2*b +
(7*B*a*b^2 - 3*A*b^3)*x^2 + 2*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(a/b)*arctan(b*sqrt(x)*sqrt(a/b)/a) + (8*B*b^3*x^
3 - 105*B*a^3 + 45*A*a^2*b - 8*(7*B*a*b^2 - 3*A*b^3)*x^2 - 25*(7*B*a^2*b - 3*A*a*b^2)*x)*sqrt(x))/(b^6*x^2 + 2
*a*b^5*x + a^2*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**(5/2)*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.17924, size = 193, normalized size = 0.76 \begin{align*} \frac{5 \,{\left (7 \, B a^{2} - 3 \, A a b\right )} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right )}{4 \, \sqrt{a b} b^{4} \mathrm{sgn}\left (b x + a\right )} - \frac{13 \, B a^{2} b x^{\frac{3}{2}} - 9 \, A a b^{2} x^{\frac{3}{2}} + 11 \, B a^{3} \sqrt{x} - 7 \, A a^{2} b \sqrt{x}}{4 \,{\left (b x + a\right )}^{2} b^{4} \mathrm{sgn}\left (b x + a\right )} + \frac{2 \,{\left (B b^{6} x^{\frac{3}{2}} - 9 \, B a b^{5} \sqrt{x} + 3 \, A b^{6} \sqrt{x}\right )}}{3 \, b^{9} \mathrm{sgn}\left (b x + a\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)^(3/2),x, algorithm="giac")

[Out]

5/4*(7*B*a^2 - 3*A*a*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^4*sgn(b*x + a)) - 1/4*(13*B*a^2*b*x^(3/2) - 9
*A*a*b^2*x^(3/2) + 11*B*a^3*sqrt(x) - 7*A*a^2*b*sqrt(x))/((b*x + a)^2*b^4*sgn(b*x + a)) + 2/3*(B*b^6*x^(3/2) -
9*B*a*b^5*sqrt(x) + 3*A*b^6*sqrt(x))/(b^9*sgn(b*x + a))