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

Optimal. Leaf size=209 \[ \frac {A b-a B}{2 a b \sqrt {x} (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 A b-a B}{4 a^2 b \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 (a+b x) (5 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2} \sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 (a+b x) (5 A b-a B)}{4 a^3 b \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.11, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {770, 78, 51, 63, 205} \begin {gather*} \frac {A b-a B}{2 a b \sqrt {x} (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 (a+b x) (5 A b-a B)}{4 a^3 b \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {5 A b-a B}{4 a^2 b \sqrt {x} \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 (a+b x) (5 A b-a B) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2} \sqrt {b} \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

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Mathematica [C]  time = 0.03, size = 77, normalized size = 0.37 \begin {gather*} \frac {a^2 (A b-a B)+(a+b x)^2 (a B-5 A b) \, _2F_1\left (-\frac {1}{2},2;\frac {1}{2};-\frac {b x}{a}\right )}{2 a^3 b \sqrt {x} (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(A*b - a*B) + (-5*A*b + a*B)*(a + b*x)^2*Hypergeometric2F1[-1/2, 2, 1/2, -((b*x)/a)])/(2*a^3*b*Sqrt[x]*(a
+ b*x)*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [A]  time = 17.13, size = 113, normalized size = 0.54 \begin {gather*} \frac {(a+b x) \left (\frac {3 (a B-5 A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{7/2} \sqrt {b}}+\frac {-8 a^2 A+5 a^2 B x-25 a A b x+3 a b B x^2-15 A b^2 x^2}{4 a^3 \sqrt {x} (a+b x)^2}\right )}{\sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*((-8*a^2*A - 25*a*A*b*x + 5*a^2*B*x - 15*A*b^2*x^2 + 3*a*b*B*x^2)/(4*a^3*Sqrt[x]*(a + b*x)^2) + (3*
(-5*A*b + a*B)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(7/2)*Sqrt[b])))/Sqrt[(a + b*x)^2]

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fricas [A]  time = 0.43, size = 331, normalized size = 1.58 \begin {gather*} \left [\frac {3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (B a^{3} - 5 \, A a^{2} b\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 \, A a^{3} b - 3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} - 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{8 \, {\left (a^{4} b^{3} x^{3} + 2 \, a^{5} b^{2} x^{2} + a^{6} b x\right )}}, -\frac {3 \, {\left ({\left (B a b^{2} - 5 \, A b^{3}\right )} x^{3} + 2 \, {\left (B a^{2} b - 5 \, A a b^{2}\right )} x^{2} + {\left (B a^{3} - 5 \, A a^{2} b\right )} x\right )} \sqrt {a b} \arctan \left (\frac {\sqrt {a b}}{b \sqrt {x}}\right ) + {\left (8 \, A a^{3} b - 3 \, {\left (B a^{2} b^{2} - 5 \, A a b^{3}\right )} x^{2} - 5 \, {\left (B a^{3} b - 5 \, A a^{2} b^{2}\right )} x\right )} \sqrt {x}}{4 \, {\left (a^{4} b^{3} x^{3} + 2 \, a^{5} b^{2} x^{2} + a^{6} b x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.21, size = 110, normalized size = 0.53 \begin {gather*} \frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3} \mathrm {sgn}\left (b x + a\right )} - \frac {2 \, A}{a^{3} \sqrt {x} \mathrm {sgn}\left (b x + a\right )} + \frac {3 \, B a b x^{\frac {3}{2}} - 7 \, A b^{2} x^{\frac {3}{2}} + 5 \, B a^{2} \sqrt {x} - 9 \, A a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a^{3} \mathrm {sgn}\left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.07, size = 214, normalized size = 1.02 \begin {gather*} -\frac {\left (15 A \,b^{3} x^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-3 B a \,b^{2} x^{\frac {5}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+30 A a \,b^{2} x^{\frac {3}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-6 B \,a^{2} b \,x^{\frac {3}{2}} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+15 A \,a^{2} b \sqrt {x}\, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )-3 B \,a^{3} \sqrt {x}\, \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+15 \sqrt {a b}\, A \,b^{2} x^{2}-3 \sqrt {a b}\, B a b \,x^{2}+25 \sqrt {a b}\, A a b x -5 \sqrt {a b}\, B \,a^{2} x +8 \sqrt {a b}\, A \,a^{2}\right ) \left (b x +a \right )}{4 \sqrt {a b}\, \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} a^{3} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.65, size = 280, normalized size = 1.34 \begin {gather*} \frac {60 \, {\left (B a^{3} b^{2} - 7 \, A a^{2} b^{3}\right )} x^{\frac {5}{2}} - {\left ({\left (B a b^{4} + 5 \, A b^{5}\right )} x^{2} - 15 \, {\left (B a^{2} b^{3} - 7 \, A a b^{4}\right )} x\right )} x^{\frac {5}{2}} + {\left (9 \, {\left (B a^{3} b^{2} + 5 \, A a^{2} b^{3}\right )} x^{2} + 85 \, {\left (B a^{4} b - 7 \, A a^{3} b^{2}\right )} x\right )} \sqrt {x} + \frac {16 \, {\left ({\left (B a^{4} b + 5 \, A a^{3} b^{2}\right )} x^{2} + 3 \, {\left (B a^{5} - 7 \, A a^{4} b\right )} x\right )}}{\sqrt {x}} + \frac {48 \, {\left (A a^{4} b x^{2} - A a^{5} x\right )}}{x^{\frac {3}{2}}}}{24 \, {\left (a^{5} b^{3} x^{3} + 3 \, a^{6} b^{2} x^{2} + 3 \, a^{7} b x + a^{8}\right )}} + \frac {3 \, {\left (B a - 5 \, A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a^{3}} + \frac {{\left (B a b + 5 \, A b^{2}\right )} x^{\frac {3}{2}} - 18 \, {\left (B a^{2} - 5 \, A a b\right )} \sqrt {x}}{24 \, a^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(60*(B*a^3*b^2 - 7*A*a^2*b^3)*x^(5/2) - ((B*a*b^4 + 5*A*b^5)*x^2 - 15*(B*a^2*b^3 - 7*A*a*b^4)*x)*x^(5/2)
+ (9*(B*a^3*b^2 + 5*A*a^2*b^3)*x^2 + 85*(B*a^4*b - 7*A*a^3*b^2)*x)*sqrt(x) + 16*((B*a^4*b + 5*A*a^3*b^2)*x^2 +
3*(B*a^5 - 7*A*a^4*b)*x)/sqrt(x) + 48*(A*a^4*b*x^2 - A*a^5*x)/x^(3/2))/(a^5*b^3*x^3 + 3*a^6*b^2*x^2 + 3*a^7*b
*x + a^8) + 3/4*(B*a - 5*A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3) + 1/24*((B*a*b + 5*A*b^2)*x^(3/2) -
18*(B*a^2 - 5*A*a*b)*sqrt(x))/a^5

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {A + B x}{x^{\frac {3}{2}} \left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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