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

Optimal. Leaf size=202 \[ -\frac {a^2 (3 A b-4 a B)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a (a+b x) (A b-2 a B) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x (a+b x) (A b-3 a B)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 (A b-a B)}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.14, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {770, 77} \begin {gather*} \frac {a^3 (A b-a B)}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^2 (3 A b-4 a B)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x (a+b x) (A b-3 a B)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a (a+b x) (A b-2 a B) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((a^2*(3*A*b - 4*a*B))/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + (a^3*(A*b - a*B))/(2*b^5*(a + b*x)*Sqrt[a^2 + 2
*a*b*x + b^2*x^2]) + ((A*b - 3*a*B)*x*(a + b*x))/(b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*x^2*(a + b*x))/(2*b^
3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (3*a*(A*b - 2*a*B)*(a + b*x)*Log[a + b*x])/(b^5*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^3 (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 (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 b-3 a B}{b^7}+\frac {B x}{b^6}+\frac {a^3 (-A b+a B)}{b^7 (a+b x)^3}-\frac {a^2 (-3 A b+4 a B)}{b^7 (a+b x)^2}+\frac {3 a (-A b+2 a B)}{b^7 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {a^2 (3 A b-4 a B)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 (A b-a B)}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-3 a B) x (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^2 (a+b x)}{2 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a (A b-2 a B) (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 117, normalized size = 0.58 \begin {gather*} \frac {7 a^4 B+a^3 (2 b B x-5 A b)-a^2 b^2 x (4 A+11 B x)+4 a b^3 x^2 (A-B x)+6 a (a+b x)^2 (2 a B-A b) \log (a+b x)+b^4 x^3 (2 A+B x)}{2 b^5 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*a^4*B + 4*a*b^3*x^2*(A - B*x) + b^4*x^3*(2*A + B*x) - a^2*b^2*x*(4*A + 11*B*x) + a^3*(-5*A*b + 2*b*B*x) + 6
*a*(-(A*b) + 2*a*B)*(a + b*x)^2*Log[a + b*x])/(2*b^5*(a + b*x)*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [B]  time = 2.19, size = 2737, normalized size = 13.55 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((16*a^4*A*Sqrt[b^2]*x)/b^4 + (32*a^3*A*Sqrt[b^2]*x^2)/b^3 + (8*a^2*A*(b^2)^(3/2)*x^3)/b^4 - (20*a*A*Sqrt[b^2]
*x^4)/b - 8*A*Sqrt[b^2]*x^5 - (4*a^4*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^4 - (12*a^3*A*x*Sqrt[a^2 + 2*a*b*x + b
^2*x^2])/b^3 - (20*a^2*A*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + (12*a*A*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b
+ 8*A*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - (24*a^3*A*x^2*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^
2])/a])/b^2 - (48*a^2*A*x^3*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b - 24*a*A*x^4*ArcTan
h[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] + (24*a^2*A*Sqrt[b^2]*x^2*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])/b^3 + (24*a*A*(b^2)^(3/2)*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])/b^4)/((-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^5*A)/(b^3*Sqrt[b^2]) - (16*
a^4*A*x)/(b^2)^(3/2) - (20*a^5*B*x)/(b^3*Sqrt[b^2]) - (16*a^3*A*x^2)/(b*Sqrt[b^2]) - (34*a^4*B*x^2)/(b^2)^(3/2
) + (24*a^3*B*x^3)/(b*Sqrt[b^2]) + (70*a^2*B*x^4)/Sqrt[b^2] + (28*a*b*B*x^5)/Sqrt[b^2] + (4*a^5*B*Sqrt[a^2 + 2
*a*b*x + b^2*x^2])/b^5 + (16*a^3*A*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 + (16*a^4*B*x*Sqrt[a^2 + 2*a*b*x + b^2
*x^2])/b^4 + (18*a^3*B*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 - (42*a^2*B*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b
^2 - (28*a*B*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + (48*a^4*B*x^2*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2])/a])/b^3 + (96*a^3*B*x^3*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^2 + (48*a^
2*B*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b - (48*a^3*B*x^2*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])/(b^2)^(3/2) - (48*a^2*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])/(b*Sqrt[b^2]) + (12*a^3*A*x^2*Log
[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*Sqrt[b^2]) + (24*a^2*A*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[
a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (12*a*A*b*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sq
rt[b^2] - (12*a^2*A*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
^2 - (12*a*A*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]])/b + (12*
a^3*A*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*Sqrt[b^2]) + (24*a^2*A*x^3*Log[a - Sqrt[b^2
]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (12*a*A*b*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2
*x^2]])/Sqrt[b^2] - (12*a^2*A*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^2 - (12*a*A*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]])/
b)/((-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^6*B)/(b^4*Sqrt[b^2]) + (20*a^5*B*x)/(b^3*Sqrt[b^2]) + (19*a^4*B*x^2)/(b^2)^(3/2) - (6*a^3*B*x^3)/(b*Sq
rt[b^2]) - (13*a^2*B*x^4)/Sqrt[b^2] - (12*a*b*B*x^5)/Sqrt[b^2] - 4*Sqrt[b^2]*B*x^6 - (20*a^4*B*x*Sqrt[a^2 + 2*
a*b*x + b^2*x^2])/b^4 + (a^3*B*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 + (5*a^2*B*x^3*Sqrt[a^2 + 2*a*b*x + b^2*
x^2])/b^2 + (8*a*B*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + 4*B*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - (24*a^4*B*x^
2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) - (48*a^3*B*x^3*Log[-a - Sqrt[b^2]*x + Sq
rt[a^2 + 2*a*b*x + b^2*x^2]])/(b*Sqrt[b^2]) - (24*a^2*B*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^
2]])/Sqrt[b^2] + (24*a^3*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^3 + (24*a^2*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]])
/b^2 - (24*a^4*B*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) - (48*a^3*B*x^3*Log[a -
Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*Sqrt[b^2]) - (24*a^2*B*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 +
2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (24*a^3*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^3 + (24*a^2*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]])/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)

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fricas [A]  time = 0.41, size = 171, normalized size = 0.85 \begin {gather*} \frac {B b^{4} x^{4} + 7 \, B a^{4} - 5 \, A a^{3} b - 2 \, {\left (2 \, B a b^{3} - A b^{4}\right )} x^{3} - {\left (11 \, B a^{2} b^{2} - 4 \, A a b^{3}\right )} x^{2} + 2 \, {\left (B a^{3} b - 2 \, A a^{2} b^{2}\right )} x + 6 \, {\left (2 \, B a^{4} - A a^{3} b + {\left (2 \, B a^{2} b^{2} - A a b^{3}\right )} x^{2} + 2 \, {\left (2 \, B a^{3} b - A a^{2} b^{2}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{7} x^{2} + 2 \, a b^{6} x + a^{2} b^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*b^4*x^4 + 7*B*a^4 - 5*A*a^3*b - 2*(2*B*a*b^3 - A*b^4)*x^3 - (11*B*a^2*b^2 - 4*A*a*b^3)*x^2 + 2*(B*a^3*b
- 2*A*a^2*b^2)*x + 6*(2*B*a^4 - A*a^3*b + (2*B*a^2*b^2 - A*a*b^3)*x^2 + 2*(2*B*a^3*b - A*a^2*b^2)*x)*log(b*x
+ a))/(b^7*x^2 + 2*a*b^6*x + a^2*b^5)

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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^3*(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 = 0.10, size = 191, normalized size = 0.95 \begin {gather*} -\frac {\left (-B \,b^{4} x^{4}+6 A a \,b^{3} x^{2} \ln \left (b x +a \right )-2 A \,b^{4} x^{3}-12 B \,a^{2} b^{2} x^{2} \ln \left (b x +a \right )+4 B a \,b^{3} x^{3}+12 A \,a^{2} b^{2} x \ln \left (b x +a \right )-4 A a \,b^{3} x^{2}-24 B \,a^{3} b x \ln \left (b x +a \right )+11 B \,a^{2} b^{2} x^{2}+6 A \,a^{3} b \ln \left (b x +a \right )+4 A \,a^{2} b^{2} x -12 B \,a^{4} \ln \left (b x +a \right )-2 B \,a^{3} b x +5 A \,a^{3} b -7 B \,a^{4}\right ) \left (b x +a \right )}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(-B*b^4*x^4+6*A*ln(b*x+a)*x^2*a*b^3-2*A*b^4*x^3-12*B*ln(b*x+a)*x^2*a^2*b^2+4*B*x^3*a*b^3+12*A*ln(b*x+a)*x
*a^2*b^2-4*A*x^2*a*b^3-24*B*ln(b*x+a)*x*a^3*b+11*B*x^2*a^2*b^2+6*A*a^3*b*ln(b*x+a)+4*A*x*a^2*b^2-12*B*a^4*ln(b
*x+a)-2*B*a^3*b*x+5*A*a^3*b-7*B*a^4)*(b*x+a)/b^5/((b*x+a)^2)^(3/2)

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maxima [A]  time = 0.51, size = 242, normalized size = 1.20 \begin {gather*} \frac {B x^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {5 \, B a x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac {A x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {6 \, B a^{2} \log \left (x + \frac {a}{b}\right )}{b^{5}} - \frac {3 \, A a \log \left (x + \frac {a}{b}\right )}{b^{4}} - \frac {5 \, B a^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} + \frac {2 \, A a^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} + \frac {12 \, B a^{3} x}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {6 \, A a^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, B a^{4}}{2 \, b^{7} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, A a^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*B*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 5/2*B*a*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3) + A*x^2/(sqrt(
b^2*x^2 + 2*a*b*x + a^2)*b^2) + 6*B*a^2*log(x + a/b)/b^5 - 3*A*a*log(x + a/b)/b^4 - 5*B*a^3/(sqrt(b^2*x^2 + 2*
a*b*x + a^2)*b^5) + 2*A*a^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) + 12*B*a^3*x/(b^6*(x + a/b)^2) - 6*A*a^2*x/(b^
5*(x + a/b)^2) + 23/2*B*a^4/(b^7*(x + a/b)^2) - 11/2*A*a^3/(b^6*(x + a/b)^2)

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

[Out]

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

[Out]

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

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