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

Optimal. Leaf size=158 \[ -\frac {\sqrt {x} (3 a B+A b)}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^{3/2} (A b-a B)}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 158, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {770, 78, 47, 63, 205} \begin {gather*} \frac {x^{3/2} (A b-a B)}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\sqrt {x} (3 a B+A b)}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(a+b x) (3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((A*b + 3*a*B)*Sqrt[x])/(4*a*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - a*B)*x^(3/2))/(2*a*b*(a + b*x)*Sqrt
[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b + 3*a*B)*(a + b*x)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(3/2)*b^(5/2)*Sq
rt[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 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 {\sqrt {x} (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 {\sqrt {x} (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^{3/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((A b+3 a B) \left (a b+b^2 x\right )\right ) \int \frac {\sqrt {x}}{\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+3 a B) \sqrt {x}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{3/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((A b+3 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 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(A b+3 a B) \sqrt {x}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{3/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left ((A b+3 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 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(A b+3 a B) \sqrt {x}}{4 a b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) x^{3/2}}{2 a b (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b+3 a B) (a+b x) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{5/2} \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 105, normalized size = 0.66 \begin {gather*} \frac {\sqrt {a} \sqrt {b} \sqrt {x} \left (-3 a^2 B-a b (A+5 B x)+A b^2 x\right )+(a+b x)^2 (3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{5/2} (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 10.32, size = 103, normalized size = 0.65 \begin {gather*} \frac {(a+b x) \left (\frac {(3 a B+A b) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{5/2}}-\frac {\sqrt {x} \left (3 a^2 B+a A b+5 a b B x-A b^2 x\right )}{4 a b^2 (a+b x)^2}\right )}{\sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*(-1/4*(Sqrt[x]*(a*A*b + 3*a^2*B - A*b^2*x + 5*a*b*B*x))/(a*b^2*(a + b*x)^2) + ((A*b + 3*a*B)*ArcTan
[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(4*a^(3/2)*b^(5/2))))/Sqrt[(a + b*x)^2]

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fricas [A]  time = 0.45, size = 291, normalized size = 1.84 \begin {gather*} \left [-\frac {{\left (3 \, B a^{3} + A a^{2} b + {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 2 \, {\left (3 \, 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 (3 \, B a^{3} b + A a^{2} b^{2} + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{8 \, {\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\right )}}, -\frac {{\left (3 \, B a^{3} + A a^{2} b + {\left (3 \, B a b^{2} + A b^{3}\right )} x^{2} + 2 \, {\left (3 \, 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 (3 \, B a^{3} b + A a^{2} b^{2} + {\left (5 \, B a^{2} b^{2} - A a b^{3}\right )} x\right )} \sqrt {x}}{4 \, {\left (a^{2} b^{5} x^{2} + 2 \, a^{3} b^{4} x + a^{4} b^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.30, size = 98, normalized size = 0.62 \begin {gather*} \frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b^{2} \mathrm {sgn}\left (b x + a\right )} - \frac {5 \, B a b x^{\frac {3}{2}} - A b^{2} x^{\frac {3}{2}} + 3 \, B a^{2} \sqrt {x} + A a b \sqrt {x}}{4 \, {\left (b x + a\right )}^{2} a b^{2} \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^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.07, size = 194, normalized size = 1.23 \begin {gather*} \frac {\left (A \,b^{3} x^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+3 B a \,b^{2} x^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+2 A a \,b^{2} x \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+6 B \,a^{2} b x \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+A \,a^{2} b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+3 B \,a^{3} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )+\sqrt {a b}\, A \,b^{2} x^{\frac {3}{2}}-5 \sqrt {a b}\, B a b \,x^{\frac {3}{2}}-\sqrt {a b}\, A a b \sqrt {x}-3 \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}} a \,b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*x^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

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

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maxima [B]  time = 1.68, size = 218, normalized size = 1.38 \begin {gather*} \frac {12 \, {\left (B a^{3} b + A a^{2} b^{2}\right )} x^{\frac {5}{2}} - {\left ({\left (5 \, B a b^{3} + A b^{4}\right )} x^{2} - 3 \, {\left (B a^{2} b^{2} + A a b^{3}\right )} x\right )} x^{\frac {5}{2}} - {\left (3 \, {\left (B a^{3} b - 3 \, A a^{2} b^{2}\right )} x^{2} - {\left (B a^{4} + 17 \, A a^{3} b\right )} x\right )} \sqrt {x}}{24 \, {\left (a^{3} b^{4} x^{3} + 3 \, a^{4} b^{3} x^{2} + 3 \, a^{5} b^{2} x + a^{6} b\right )}} + \frac {{\left (3 \, B a + A b\right )} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \, \sqrt {a b} a b^{2}} + \frac {{\left (5 \, B a b + A b^{2}\right )} x^{\frac {3}{2}} - 6 \, {\left (3 \, B a^{2} + A a b\right )} \sqrt {x}}{24 \, a^{3} b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*x^(1/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

1/24*(12*(B*a^3*b + A*a^2*b^2)*x^(5/2) - ((5*B*a*b^3 + A*b^4)*x^2 - 3*(B*a^2*b^2 + A*a*b^3)*x)*x^(5/2) - (3*(B
*a^3*b - 3*A*a^2*b^2)*x^2 - (B*a^4 + 17*A*a^3*b)*x)*sqrt(x))/(a^3*b^4*x^3 + 3*a^4*b^3*x^2 + 3*a^5*b^2*x + a^6*
b) + 1/4*(3*B*a + A*b)*arctan(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a*b^2) + 1/24*((5*B*a*b + A*b^2)*x^(3/2) - 6*(3*
B*a^2 + A*a*b)*sqrt(x))/(a^3*b^2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {x}\,\left (A+B\,x\right )}{{\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((x^(1/2)*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)

[Out]

int((x^(1/2)*(A + B*x))/(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 {\sqrt {x} \left (A + B x\right )}{\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**(1/2)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)

[Out]

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

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