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

Optimal. Leaf size=113 \[ -\frac {A b-2 a B}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a (A b-a B)}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.07, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {770, 77} \begin {gather*} -\frac {A b-2 a B}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a (A b-a B)}{2 b^3 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B (a+b x) \log (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

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Mathematica [A]  time = 0.03, size = 65, normalized size = 0.58 \begin {gather*} \frac {3 a^2 B-a b (A-4 B x)+2 B (a+b x)^2 \log (a+b x)-2 A b^2 x}{2 b^3 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 1.32, size = 1415, normalized size = 12.52 \begin {gather*} \frac {\left (\frac {4 A x}{\sqrt {b^2}}-\frac {4 A \sqrt {a^2+2 b x a+b^2 x^2}}{b^2}\right ) \left (a^2+b x a+b^2 x^2-\sqrt {b^2} x \sqrt {a^2+2 b x a+b^2 x^2}\right )}{\left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )^2 \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )^2}+\frac {8 b B \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) x^4-\frac {16 a b B x^3}{\sqrt {b^2}}+16 a B \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) x^3-\frac {8 b B \sqrt {a^2+2 b x a+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) x^3}{\sqrt {b^2}}-\frac {24 a^2 B x^2}{\sqrt {b^2}}+\frac {8 a^2 B \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) x^2}{b}-\frac {8 a B \sqrt {a^2+2 b x a+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) x^2}{\sqrt {b^2}}+\frac {16 a B \sqrt {a^2+2 b x a+b^2 x^2} x^2}{b}-\frac {8 a A b x^2}{\sqrt {b^2}}-\frac {12 a^3 B x}{b \sqrt {b^2}}+\frac {8 a^2 B \sqrt {a^2+2 b x a+b^2 x^2} x}{b^2}+\frac {8 a A \sqrt {a^2+2 b x a+b^2 x^2} x}{b}-\frac {8 a^2 A x}{\sqrt {b^2}}+\frac {4 a^3 B \sqrt {a^2+2 b x a+b^2 x^2}}{b^3}-\frac {4 a^3 A}{b \sqrt {b^2}}}{\left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )^2 \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )^2}+\frac {\frac {4 \sqrt {b^2} B a^4}{b^4}+\frac {12 \sqrt {b^2} B x a^3}{b^3}+\frac {12 \left (b^2\right )^{3/2} B x^2 a^2}{b^4}-\frac {4 \left (b^2\right )^{3/2} B x^2 \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^2}{b^4}-\frac {4 \left (b^2\right )^{3/2} B x^2 \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a^2}{b^4}-\frac {12 B x \sqrt {a^2+2 b x a+b^2 x^2} a^2}{b^2}-\frac {8 \sqrt {b^2} B x^3 \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a}{b}+\frac {4 B x^2 \sqrt {a^2+2 b x a+b^2 x^2} \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a}{b}-\frac {8 \sqrt {b^2} B x^3 \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a}{b}+\frac {4 B x^2 \sqrt {a^2+2 b x a+b^2 x^2} \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) a}{b}-4 \sqrt {b^2} B x^4 \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )+4 B x^3 \sqrt {a^2+2 b x a+b^2 x^2} \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )-4 \sqrt {b^2} B x^4 \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )+4 B x^3 \sqrt {a^2+2 b x a+b^2 x^2} \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )}{\left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )^2 \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((4*A*x)/Sqrt[b^2] - (4*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2)*(a^2 + a*b*x + b^2*x^2 - Sqrt[b^2]*x*Sqrt[a^2 +
2*a*b*x + b^2*x^2]))/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*
b*x + b^2*x^2])^2) + ((-4*a^3*A)/(b*Sqrt[b^2]) - (8*a^2*A*x)/Sqrt[b^2] - (12*a^3*B*x)/(b*Sqrt[b^2]) - (8*a*A*b
*x^2)/Sqrt[b^2] - (24*a^2*B*x^2)/Sqrt[b^2] - (16*a*b*B*x^3)/Sqrt[b^2] + (4*a^3*B*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
)/b^3 + (8*a*A*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + (8*a^2*B*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + (16*a*B*x^
2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + (8*a^2*B*x^2*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])
/b + 16*a*B*x^3*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] + 8*b*B*x^4*ArcTanh[(-(Sqrt[b^2]*x
) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] - (8*a*B*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqr
t[a^2 + 2*a*b*x + b^2*x^2])/a])/Sqrt[b^2] - (8*b*B*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) +
Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/Sqrt[b^2])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*(a - Sqrt
[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2) + ((4*a^4*Sqrt[b^2]*B)/b^4 + (12*a^3*Sqrt[b^2]*B*x)/b^3 + (12*a^2*
(b^2)^(3/2)*B*x^2)/b^4 - (12*a^2*B*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 - (4*a^2*(b^2)^(3/2)*B*x^2*Log[-a - Sq
rt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b^4 - (8*a*Sqrt[b^2]*B*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b
*x + b^2*x^2]])/b - 4*Sqrt[b^2]*B*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] + (4*a*B*x^2*Sqrt[
a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b + 4*B*x^3*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - (4*a^2*(b^2)^(3/2)*B*x^2*Log[a - Sqrt[b^2]
*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b^4 - (8*a*Sqrt[b^2]*B*x^3*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2
*x^2]])/b - 4*Sqrt[b^2]*B*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] + (4*a*B*x^2*Sqrt[a^2 + 2*a
*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b + 4*B*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2
]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2*
(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^2)

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fricas [A]  time = 0.43, size = 81, normalized size = 0.72 \begin {gather*} \frac {3 \, B a^{2} - A a b + 2 \, {\left (2 \, B a b - A b^{2}\right )} x + 2 \, {\left (B b^{2} x^{2} + 2 \, B a b x + B a^{2}\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x^{2} + 2 \, a b^{4} x + a^{2} b^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [A]  time = 0.06, size = 83, normalized size = 0.73 \begin {gather*} -\frac {\left (-2 B \,b^{2} x^{2} \ln \left (b x +a \right )-4 B a b x \ln \left (b x +a \right )+2 A \,b^{2} x -2 B \,a^{2} \ln \left (b x +a \right )-4 B a b x +A a b -3 B \,a^{2}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.50, size = 89, normalized size = 0.79 \begin {gather*} \frac {B \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {A}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {2 \, B a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {3 \, B a^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {A a}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

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