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

Optimal. Leaf size=206 \[ \frac {x^{5/2} (A b-a B)}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^{3/2} (A b-5 a B)}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 (a+b x) (A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 \sqrt {x} (a+b x) (A b-5 a B)}{4 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

[Out]

1/4*(A*b-5*B*a)*x^(3/2)/a/b^2/((b*x+a)^2)^(1/2)+1/2*(A*b-B*a)*x^(5/2)/a/b/(b*x+a)/((b*x+a)^2)^(1/2)+3/4*(A*b-5
*B*a)*(b*x+a)*arctan(b^(1/2)*x^(1/2)/a^(1/2))/b^(7/2)/a^(1/2)/((b*x+a)^2)^(1/2)-3/4*(A*b-5*B*a)*(b*x+a)*x^(1/2
)/a/b^3/((b*x+a)^2)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {x^{5/2} (A b-a B)}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^{3/2} (A b-5 a B)}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 \sqrt {x} (a+b x) (A b-5 a B)}{4 a b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 (a+b x) (A b-5 a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 \sqrt {a} b^{7/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 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]

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]

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 79, normalized size = 0.38 \[ \frac {x^{5/2} \left (5 a^2 (A b-a B)+(a+b x)^2 (5 a B-A b) \, _2F_1\left (2,\frac {5}{2};\frac {7}{2};-\frac {b x}{a}\right )\right )}{10 a^3 b (a+b x) \sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(x^(5/2)*(5*a^2*(A*b - a*B) + (-(A*b) + 5*a*B)*(a + b*x)^2*Hypergeometric2F1[2, 5/2, 7/2, -((b*x)/a)]))/(10*a^
3*b*(a + b*x)*Sqrt[(a + b*x)^2])

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fricas [A]  time = 1.07, size = 319, normalized size = 1.55 \[ \left [\frac {3 \, {\left (5 \, B a^{3} - A a^{2} b + {\left (5 \, B a b^{2} - A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {-a b} \log \left (\frac {b x - a - 2 \, \sqrt {-a b} \sqrt {x}}{b x + a}\right ) + 2 \, {\left (8 \, B a b^{3} x^{2} + 15 \, B a^{3} b - 3 \, A a^{2} b^{2} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{8 \, {\left (a b^{6} x^{2} + 2 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}, \frac {3 \, {\left (5 \, B a^{3} - A a^{2} b + {\left (5 \, B a b^{2} - A b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{2} b - A a b^{2}\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (8 \, B a b^{3} x^{2} + 15 \, B a^{3} b - 3 \, A a^{2} b^{2} + 5 \, {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{4 \, {\left (a b^{6} x^{2} + 2 \, a^{2} b^{5} x + a^{3} b^{4}\right )}}\right ] \]

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)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(3*(5*B*a^3 - A*a^2*b + (5*B*a*b^2 - A*b^3)*x^2 + 2*(5*B*a^2*b - A*a*b^2)*x)*sqrt(-a*b)*log((b*x - a - 2*
sqrt(-a*b)*sqrt(x))/(b*x + a)) + 2*(8*B*a*b^3*x^2 + 15*B*a^3*b - 3*A*a^2*b^2 + 5*(5*B*a^2*b^2 - A*a*b^3)*x)*sq
rt(x))/(a*b^6*x^2 + 2*a^2*b^5*x + a^3*b^4), 1/4*(3*(5*B*a^3 - A*a^2*b + (5*B*a*b^2 - A*b^3)*x^2 + 2*(5*B*a^2*b
- A*a*b^2)*x)*sqrt(a*b)*arctan(sqrt(a*b)/(b*sqrt(x))) + (8*B*a*b^3*x^2 + 15*B*a^3*b - 3*A*a^2*b^2 + 5*(5*B*a^
2*b^2 - A*a*b^3)*x)*sqrt(x))/(a*b^6*x^2 + 2*a^2*b^5*x + a^3*b^4)]

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giac [A]  time = 0.21, size = 111, normalized size = 0.54 \[ \frac {2 \, B \sqrt {x}}{b^{3} \mathrm {sgn}\left (b x + a\right )} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3} \mathrm {sgn}\left (b x + a\right )} + \frac {9 \, B a b x^{\frac {3}{2}} - 5 \, A b^{2} x^{\frac {3}{2}} + 7 \, B a^{2} \sqrt {x} - 3 \, A a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} b^{3} \mathrm {sgn}\left (b x + a\right )} \]

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)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 208, normalized size = 1.01 \[ -\frac {\left (-3 A \,b^{3} x^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+15 B a \,b^{2} x^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-6 A a \,b^{2} x \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+30 B \,a^{2} b x \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-8 \sqrt {a b}\, B \,b^{2} x^{\frac {5}{2}}-3 A \,a^{2} b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+15 B \,a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+5 \sqrt {a b}\, A \,b^{2} x^{\frac {3}{2}}-25 \sqrt {a b}\, B a b \,x^{\frac {3}{2}}+3 \sqrt {a b}\, A a b \sqrt {x}-15 \sqrt {a b}\, B \,a^{2} \sqrt {x}\right ) \left (b x +a \right )}{4 \sqrt {a b}\, \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{3}} \]

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)^(3/2),x)

[Out]

-1/4*(5*(a*b)^(1/2)*A*b^2*x^(3/2)-3*A*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x^2*b^3-25*(a*b)^(1/2)*B*a*b*x^(3/2)-8*(
a*b)^(1/2)*B*b^2*x^(5/2)+15*B*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x^2*a*b^2-6*A*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x*
a*b^2+30*B*arctan(1/(a*b)^(1/2)*b*x^(1/2))*x*a^2*b+3*(a*b)^(1/2)*A*a*b*x^(1/2)-3*A*a^2*b*arctan(1/(a*b)^(1/2)*
b*x^(1/2))-15*(a*b)^(1/2)*B*a^2*x^(1/2)+15*B*a^3*arctan(1/(a*b)^(1/2)*b*x^(1/2)))*(b*x+a)/(a*b)^(1/2)/b^3/((b*
x+a)^2)^(3/2)

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maxima [A]  time = 1.58, size = 237, normalized size = 1.15 \[ \frac {{\left (5 \, {\left (7 \, B a b^{3} - A b^{4}\right )} x^{2} + 3 \, {\left (5 \, B a^{2} b^{2} + A a b^{3}\right )} x\right )} x^{\frac {5}{2}} + 12 \, {\left (4 \, B a^{2} b^{2} x^{2} + {\left (B a^{3} b + A a^{2} b^{2}\right )} x\right )} x^{\frac {3}{2}} + {\left (3 \, {\left (7 \, B a^{3} b - A a^{2} b^{2}\right )} x^{2} + {\left (5 \, B a^{4} + A a^{3} b\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{2} b^{5} x^{3} + 3 \, a^{3} b^{4} x^{2} + 3 \, a^{4} b^{3} x + a^{5} b^{2}\right )}} - \frac {3 \, {\left (5 \, B a - A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} b^{3}} - \frac {5 \, {\left (7 \, B a b - A b^{2}\right )} x^{\frac {3}{2}} - 18 \, {\left (5 \, B a^{2} - A a b\right )} \sqrt {x}}{24 \, a^{2} b^{3}} \]

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)^(3/2),x, algorithm="maxima")

[Out]

1/24*((5*(7*B*a*b^3 - A*b^4)*x^2 + 3*(5*B*a^2*b^2 + A*a*b^3)*x)*x^(5/2) + 12*(4*B*a^2*b^2*x^2 + (B*a^3*b + A*a
^2*b^2)*x)*x^(3/2) + (3*(7*B*a^3*b - A*a^2*b^2)*x^2 + (5*B*a^4 + A*a^3*b)*x)*sqrt(x))/(a^2*b^5*x^3 + 3*a^3*b^4
*x^2 + 3*a^4*b^3*x + a^5*b^2) - 3/4*(5*B*a - A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*b^3) - 1/24*(5*(7*B*a
*b - A*b^2)*x^(3/2) - 18*(5*B*a^2 - A*a*b)*sqrt(x))/(a^2*b^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^{3/2}\,\left (A+B\,x\right )}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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)**(3/2),x)

[Out]

Timed out

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