3.7.41 \(\int \frac {x^4 (A+B x)}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)
Optimal. Leaf size=249 \[ \frac {2 a^2 (a+b x) (3 A b-5 a B) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a x (a+b x) (A b-2 a B)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^2 (a+b x) (A b-3 a B)}{2 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^3 (a+b x)}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^4 (A b-a B)}{2 b^6 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 (4 A b-5 a B)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}} \]
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Rubi [A] time = 0.18, antiderivative size = 249, 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^4 (A b-a B)}{2 b^6 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {a^3 (4 A b-5 a B)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a x (a+b x) (A b-2 a B)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {x^2 (a+b x) (A b-3 a B)}{2 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^2 (a+b x) (3 A b-5 a B) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^3 (a+b x)}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(x^4*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
[Out]
(a^3*(4*A*b - 5*a*B))/(b^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (a^4*(A*b - a*B))/(2*b^6*(a + b*x)*Sqrt[a^2 + 2*a*
b*x + b^2*x^2]) - (3*a*(A*b - 2*a*B)*x*(a + b*x))/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((A*b - 3*a*B)*x^2*(a
+ b*x))/(2*b^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*x^3*(a + b*x))/(3*b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (2*a
^2*(3*A*b - 5*a*B)*(a + b*x)*Log[a + b*x])/(b^6*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^4 (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^4 (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 {3 a (-A b+2 a B)}{b^8}+\frac {(A b-3 a B) x}{b^7}+\frac {B x^2}{b^6}-\frac {a^4 (-A b+a B)}{b^8 (a+b x)^3}+\frac {a^3 (-4 A b+5 a B)}{b^8 (a+b x)^2}-\frac {2 a^2 (-3 A b+5 a B)}{b^8 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {a^3 (4 A b-5 a B)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {a^4 (A b-a B)}{2 b^6 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 a (A b-2 a B) x (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-3 a B) x^2 (a+b x)}{2 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B x^3 (a+b x)}{3 b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {2 a^2 (3 A b-5 a B) (a+b x) \log (a+b x)}{b^6 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}
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Mathematica [A] time = 0.06, size = 140, normalized size = 0.56 \begin {gather*} \frac {-27 a^5 B+3 a^4 b (7 A+2 B x)+3 a^3 b^2 x (2 A+21 B x)+a^2 b^3 x^2 (20 B x-33 A)-12 a^2 (a+b x)^2 (5 a B-3 A b) \log (a+b x)-a b^4 x^3 (12 A+5 B x)+b^5 x^4 (3 A+2 B x)}{6 b^6 (a+b x) \sqrt {(a+b x)^2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(x^4*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
[Out]
(-27*a^5*B + b^5*x^4*(3*A + 2*B*x) + 3*a^4*b*(7*A + 2*B*x) - a*b^4*x^3*(12*A + 5*B*x) + a^2*b^3*x^2*(-33*A + 2
0*B*x) + 3*a^3*b^2*x*(2*A + 21*B*x) - 12*a^2*(-3*A*b + 5*a*B)*(a + b*x)^2*Log[a + b*x])/(6*b^6*(a + b*x)*Sqrt[
(a + b*x)^2])
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IntegrateAlgebraic [B] time = 2.46, size = 3062, normalized size = 12.30 \begin {gather*} \text {Result too large to show} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[(x^4*(A + B*x))/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]
[Out]
((-20*a^5*A*x)/(b^3*Sqrt[b^2]) - (34*a^4*A*x^2)/(b^2)^(3/2) + (24*a^3*A*x^3)/(b*Sqrt[b^2]) + (70*a^2*A*x^4)/Sq
rt[b^2] + (28*a*A*b*x^5)/Sqrt[b^2] + (4*a^5*A*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^5 + (16*a^4*A*x*Sqrt[a^2 + 2*a*
b*x + b^2*x^2])/b^4 + (18*a^3*A*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 - (42*a^2*A*x^3*Sqrt[a^2 + 2*a*b*x + b^
2*x^2])/b^2 - (28*a*A*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + (48*a^4*A*x^2*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2
+ 2*a*b*x + b^2*x^2])/a])/b^3 + (96*a^3*A*x^3*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^2
+ (48*a^2*A*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b - (48*a^3*A*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*A*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]))/((-a - Sq
rt[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*A)
/(b^4*Sqrt[b^2]) + (20*a^5*A*x)/(b^3*Sqrt[b^2]) + (24*a^6*B*x)/(b^4*Sqrt[b^2]) + (19*a^4*A*x^2)/(b^2)^(3/2) +
(86*a^5*B*x^2)/(3*b^3*Sqrt[b^2]) - (6*a^3*A*x^3)/(b*Sqrt[b^2]) - (260*a^4*B*x^3)/(3*(b^2)^(3/2)) - (13*a^2*A*x
^4)/Sqrt[b^2] - (490*a^3*B*x^4)/(3*b*Sqrt[b^2]) - (12*a*A*b*x^5)/Sqrt[b^2] - (224*a^2*B*x^5)/(3*Sqrt[b^2]) - 4
*A*Sqrt[b^2]*x^6 - (28*a*b*B*x^6)/(3*Sqrt[b^2]) - (8*Sqrt[b^2]*B*x^7)/3 - (4*a^6*B*Sqrt[a^2 + 2*a*b*x + b^2*x^
2])/b^6 - (20*a^4*A*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^4 - (20*a^5*B*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^5 + (a
^3*A*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 - (26*a^4*B*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*b^4) + (5*a^2*A*
x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + (286*a^3*B*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*b^3) + (8*a*A*x^4*Sq
rt[a^2 + 2*a*b*x + b^2*x^2])/b + (68*a^2*B*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + 4*A*x^5*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2] + (20*a*B*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*b) + (8*B*x^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/3 - (
80*a^5*B*x^2*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^4 - (160*a^4*B*x^3*ArcTanh[(-(Sqrt
[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^3 - (80*a^3*B*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2])/a])/b^2 + (80*a^4*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^3*Sqrt[b^2]) + (80*a^3*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^2)^(3/2) - (24*a^4*A*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*A*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
*A*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*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^3 + (24*a^2*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^2 - (24*a^4*A*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*A*x^3*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*Sqr
t[b^2]) - (24*a^2*A*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*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^3 + (24*a^2*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^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^7*B)/(b^5*Sqrt[b^2]) - (24*
a^6*B*x)/(b^4*Sqrt[b^2]) - (20*a^5*B*x^2)/(b^3*Sqrt[b^2]) + (24*a^4*B*x^3)/(b^2)^(3/2) + (52*a^3*B*x^4)/(b*Sqr
t[b^2]) + (48*a^2*B*x^5)/Sqrt[b^2] + (16*a*b*B*x^6)/Sqrt[b^2] + (24*a^5*B*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^5
- (4*a^4*B*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^4 - (20*a^3*B*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 - (32*a^
2*B*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 - (16*a*B*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + (40*a^5*B*x^2*Log[
-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^3*Sqrt[b^2]) + (80*a^4*B*x^3*Log[-a - Sqrt[b^2]*x + Sqrt
[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) + (40*a^3*B*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])
/(b*Sqrt[b^2]) - (40*a^4*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 - (40*a^3*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^3 + (40*a^5*B*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^3*Sqrt[b^2]) + (80*a^4*B*x^3*Log
[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) + (40*a^3*B*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2
+ 2*a*b*x + b^2*x^2]])/(b*Sqrt[b^2]) - (40*a^4*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 - (40*a^3*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^3)/((-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 = 0.42, size = 197, normalized size = 0.79 \begin {gather*} \frac {2 \, B b^{5} x^{5} - 27 \, B a^{5} + 21 \, A a^{4} b - {\left (5 \, B a b^{4} - 3 \, A b^{5}\right )} x^{4} + 4 \, {\left (5 \, B a^{2} b^{3} - 3 \, A a b^{4}\right )} x^{3} + 3 \, {\left (21 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3}\right )} x^{2} + 6 \, {\left (B a^{4} b + A a^{3} b^{2}\right )} x - 12 \, {\left (5 \, B a^{5} - 3 \, A a^{4} b + {\left (5 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3}\right )} x^{2} + 2 \, {\left (5 \, B a^{4} b - 3 \, A a^{3} b^{2}\right )} x\right )} \log \left (b x + a\right )}{6 \, {\left (b^{8} x^{2} + 2 \, a b^{7} x + a^{2} b^{6}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="fricas")
[Out]
1/6*(2*B*b^5*x^5 - 27*B*a^5 + 21*A*a^4*b - (5*B*a*b^4 - 3*A*b^5)*x^4 + 4*(5*B*a^2*b^3 - 3*A*a*b^4)*x^3 + 3*(21
*B*a^3*b^2 - 11*A*a^2*b^3)*x^2 + 6*(B*a^4*b + A*a^3*b^2)*x - 12*(5*B*a^5 - 3*A*a^4*b + (5*B*a^3*b^2 - 3*A*a^2*
b^3)*x^2 + 2*(5*B*a^4*b - 3*A*a^3*b^2)*x)*log(b*x + a))/(b^8*x^2 + 2*a*b^7*x + a^2*b^6)
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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^4*(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 = 0.07, size = 217, normalized size = 0.87 \begin {gather*} \frac {\left (2 B \,b^{5} x^{5}+3 A \,b^{5} x^{4}-5 B a \,b^{4} x^{4}+36 A \,a^{2} b^{3} x^{2} \ln \left (b x +a \right )-12 A a \,b^{4} x^{3}-60 B \,a^{3} b^{2} x^{2} \ln \left (b x +a \right )+20 B \,a^{2} b^{3} x^{3}+72 A \,a^{3} b^{2} x \ln \left (b x +a \right )-33 A \,a^{2} b^{3} x^{2}-120 B \,a^{4} b x \ln \left (b x +a \right )+63 B \,a^{3} b^{2} x^{2}+36 A \,a^{4} b \ln \left (b x +a \right )+6 A \,a^{3} b^{2} x -60 B \,a^{5} \ln \left (b x +a \right )+6 B \,a^{4} b x +21 A \,a^{4} b -27 B \,a^{5}\right ) \left (b x +a \right )}{6 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} b^{6}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)
[Out]
1/6*(2*B*b^5*x^5+3*A*b^5*x^4-5*B*a*b^4*x^4+36*A*ln(b*x+a)*x^2*a^2*b^3-12*A*a*b^4*x^3-60*B*ln(b*x+a)*x^2*a^3*b^
2+20*B*a^2*b^3*x^3+72*A*ln(b*x+a)*x*a^3*b^2-33*A*a^2*b^3*x^2-120*B*ln(b*x+a)*x*a^4*b+63*B*a^3*b^2*x^2+36*A*a^4
*b*ln(b*x+a)+6*A*a^3*b^2*x-60*B*a^5*ln(b*x+a)+6*B*a^4*b*x+21*A*a^4*b-27*B*a^5)*(b*x+a)/b^6/((b*x+a)^2)^(3/2)
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maxima [A] time = 0.59, size = 303, normalized size = 1.22 \begin {gather*} \frac {B x^{4}}{3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {7 \, B a x^{3}}{6 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} + \frac {A x^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {9 \, B a^{2} x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {5 \, A a x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{3}} - \frac {10 \, B a^{3} \log \left (x + \frac {a}{b}\right )}{b^{6}} + \frac {6 \, A a^{2} \log \left (x + \frac {a}{b}\right )}{b^{5}} + \frac {9 \, B a^{4}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{6}} - \frac {5 \, A a^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{5}} - \frac {20 \, B a^{4} x}{b^{7} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {12 \, A a^{3} x}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {39 \, B a^{5}}{2 \, b^{8} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, A a^{4}}{2 \, b^{7} {\left (x + \frac {a}{b}\right )}^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(x^4*(B*x+A)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")
[Out]
1/3*B*x^4/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 7/6*B*a*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^3) + 1/2*A*x^3/(s
qrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 9/2*B*a^2*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - 5/2*A*a*x^2/(sqrt(b^2*
x^2 + 2*a*b*x + a^2)*b^3) - 10*B*a^3*log(x + a/b)/b^6 + 6*A*a^2*log(x + a/b)/b^5 + 9*B*a^4/(sqrt(b^2*x^2 + 2*a
*b*x + a^2)*b^6) - 5*A*a^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^5) - 20*B*a^4*x/(b^7*(x + a/b)^2) + 12*A*a^3*x/(b^
6*(x + a/b)^2) - 39/2*B*a^5/(b^8*(x + a/b)^2) + 23/2*A*a^4/(b^7*(x + a/b)^2)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {x^4\,\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^4*(A + B*x))/(a^2 + b^2*x^2 + 2*a*b*x)^(3/2),x)
[Out]
int((x^4*(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^{4} \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**4*(B*x+A)/(b**2*x**2+2*a*b*x+a**2)**(3/2),x)
[Out]
Integral(x**4*(A + B*x)/((a + b*x)**2)**(3/2), x)
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