3.19.9 \(\int \frac {(a+b x) (d+e x)^4}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)
Optimal. Leaf size=201 \[ -\frac {(d+e x)^4}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {4 e^3 (a+b x) (b d-a e) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {4 e^2 (b d-a e)^2}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 e (b d-a e)^3}{3 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 e^4 x (a+b x)}{3 b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]
________________________________________________________________________________________
Rubi [A] time = 0.14, antiderivative size = 201, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used =
{768, 646, 43} \begin {gather*} -\frac {4 e^2 (b d-a e)^2}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 e^3 (a+b x) (b d-a e) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 e (b d-a e)^3}{3 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^4}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {4 e^4 x (a+b x)}{3 b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((a + b*x)*(d + e*x)^4)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
[Out]
-(d + e*x)^4/(3*b*(a^2 + 2*a*b*x + b^2*x^2)^(3/2)) - (4*e^2*(b*d - a*e)^2)/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
- (2*e*(b*d - a*e)^3)/(3*b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (4*e^4*x*(a + b*x))/(3*b^4*Sqrt[a^2 +
2*a*b*x + b^2*x^2]) + (4*e^3*(b*d - a*e)*(a + b*x)*Log[a + b*x])/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
Rule 43
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Rule 646
Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p] && NeQ[2*c*d - b*e, 0]
Rule 768
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]
Rubi steps
\begin {align*} \int \frac {(a+b x) (d+e x)^4}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=-\frac {(d+e x)^4}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {(4 e) \int \frac {(d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx}{3 b}\\ &=-\frac {(d+e x)^4}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {\left (4 b e \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^3}{\left (a b+b^2 x\right )^3} \, dx}{3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^4}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {\left (4 b e \left (a b+b^2 x\right )\right ) \int \left (\frac {e^3}{b^6}+\frac {(b d-a e)^3}{b^6 (a+b x)^3}+\frac {3 e (b d-a e)^2}{b^6 (a+b x)^2}+\frac {3 e^2 (b d-a e)}{b^6 (a+b x)}\right ) \, dx}{3 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^4}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {4 e^2 (b d-a e)^2}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {2 e (b d-a e)^3}{3 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 e^4 x (a+b x)}{3 b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {4 e^3 (b d-a e) (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}
________________________________________________________________________________________
Mathematica [A] time = 0.08, size = 170, normalized size = 0.85 \begin {gather*} \frac {-13 a^4 e^4+a^3 b e^3 (22 d-27 e x)-3 a^2 b^2 e^2 \left (2 d^2-18 d e x+3 e^2 x^2\right )+a b^3 e \left (-2 d^3-18 d^2 e x+36 d e^2 x^2+9 e^3 x^3\right )-12 e^3 (a+b x)^3 (a e-b d) \log (a+b x)-\left (b^4 \left (d^4+6 d^3 e x+18 d^2 e^2 x^2-3 e^4 x^4\right )\right )}{3 b^5 \left ((a+b x)^2\right )^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((a + b*x)*(d + e*x)^4)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
[Out]
(-13*a^4*e^4 + a^3*b*e^3*(22*d - 27*e*x) - 3*a^2*b^2*e^2*(2*d^2 - 18*d*e*x + 3*e^2*x^2) + a*b^3*e*(-2*d^3 - 18
*d^2*e*x + 36*d*e^2*x^2 + 9*e^3*x^3) - b^4*(d^4 + 6*d^3*e*x + 18*d^2*e^2*x^2 - 3*e^4*x^4) - 12*e^3*(-(b*d) + a
*e)*(a + b*x)^3*Log[a + b*x])/(3*b^5*((a + b*x)^2)^(3/2))
________________________________________________________________________________________
IntegrateAlgebraic [B] time = 4.77, size = 4330, normalized size = 21.54 \begin {gather*} \text {Result too large to show} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[((a + b*x)*(d + e*x)^4)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
[Out]
(4*Sqrt[b^2]*d^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(a^2*b*d^2 - 4*a^3*d*e - a*b^2*d^2*x + 4*a^2*b*d*e*x - 18*a^3*e
^2*x + b^3*d^2*x^2 - 4*a*b^2*d*e*x^2 - 27*a^2*b*e^2*x^2 + 6*b^3*d*e*x^3 - 27*a*b^2*e^2*x^3) + 4*d^2*(a^3*b^2*d
^2 - 4*a^4*b*d*e + 6*a^5*e^2 + 18*a^4*b*e^2*x + 45*a^3*b^2*e^2*x^2 - b^5*d^2*x^3 - 2*a*b^4*d*e*x^3 + 54*a^2*b^
3*e^2*x^3 - 6*b^5*d*e*x^4 + 27*a*b^4*e^2*x^4))/(3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(4*a^2*b^6 + 8*a*b^7*x + 4
*b^8*x^2) + 3*Sqrt[b^2]*x^3*(-4*a^3*b^5 - 12*a^2*b^6*x - 12*a*b^7*x^2 - 4*b^8*x^3)) + ((192*a^4*d^2*e^2*x)/(b*
Sqrt[b^2]) + (48*a^6*e^4*x)/(b^3*Sqrt[b^2]) + (480*a^3*d^2*e^2*x^2)/Sqrt[b^2] + (208*a^5*e^4*x^2)/(b^2)^(3/2)
+ (640*a^2*b*d^2*e^2*x^3)/Sqrt[b^2] + (1328*a^4*e^4*x^3)/(3*b*Sqrt[b^2]) + 480*a*Sqrt[b^2]*d^2*e^2*x^4 + (400*
a^3*e^4*x^4)/Sqrt[b^2] + (192*b^3*d^2*e^2*x^5)/Sqrt[b^2] + (48*a^2*b*e^4*x^5)/Sqrt[b^2] - 112*a*Sqrt[b^2]*e^4*
x^6 - (32*b^3*e^4*x^7)/Sqrt[b^2] - (64*a^4*d^2*e^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 - (32*a^6*e^4*Sqrt[a^2 +
2*a*b*x + b^2*x^2])/(3*b^5) - (128*a^3*d^2*e^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 - (112*a^5*e^4*x*Sqrt[a^2
+ 2*a*b*x + b^2*x^2])/(3*b^4) - (352*a^2*d^2*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b - (512*a^4*e^4*x^2*Sqrt
[a^2 + 2*a*b*x + b^2*x^2])/(3*b^3) - 288*a*d^2*e^2*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - (272*a^3*e^4*x^3*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/b^2 - 192*b*d^2*e^2*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - (128*a^2*e^4*x^4*Sqrt[a^2 + 2
*a*b*x + b^2*x^2])/b + 80*a*e^4*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2] + 32*b*e^4*x^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]
- (128*a^4*e^4*x^3*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^2 - (384*a^3*e^4*x^4*ArcTan
h[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b - 384*a^2*e^4*x^5*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2
+ 2*a*b*x + b^2*x^2])/a] - 128*a*b*e^4*x^6*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] + (128*
a^3*e^4*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]) + (256*a^2*e^4*x^4*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] + (128*a*b*e^4*x^5*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])^3*(a - Sqrt[b^2]*x + Sqrt[a^2
+ 2*a*b*x + b^2*x^2])^3) + ((128*a^6*d*e^3)/(3*b^3*Sqrt[b^2]) + (160*a^5*d*e^3*x)/(b^2)^(3/2) + (544*a^4*d*e^3
*x^2)/(b*Sqrt[b^2]) + (768*a^3*d*e^3*x^3)/Sqrt[b^2] + (384*a^2*b*d*e^3*x^4)/Sqrt[b^2] - (160*a^4*d*e^3*x*Sqrt[
a^2 + 2*a*b*x + b^2*x^2])/b^3 - (384*a^3*d*e^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 - (384*a^2*d*e^3*x^3*Sqr
t[a^2 + 2*a*b*x + b^2*x^2])/b - (64*a^3*d*e^3*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[
b^2] - (192*a^2*b*d*e^3*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - 192*a*Sqrt[b^2]
*d*e^3*x^5*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - (64*b^3*d*e^3*x^6*Log[-a - Sqrt[b^2]*x + Sq
rt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (64*a^2*d*e^3*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 + 128*a*d*e^3*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sq
rt[a^2 + 2*a*b*x + b^2*x^2]] + 64*b*d*e^3*x^5*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]] - (64*a^3*d*e^3*x^3*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (192
*a^2*b*d*e^3*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - 192*a*Sqrt[b^2]*d*e^3*x^5*L
og[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - (64*b^3*d*e^3*x^6*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b
*x + b^2*x^2]])/Sqrt[b^2] + (64*a^2*d*e^3*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 + 128*a*d*e^3*x^4*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]] + 64*b*d*e^3*x^5*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])^3*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^3)
+ ((-32*a^7*e^4)/(3*b^4*Sqrt[b^2]) - (160*a^5*d*e^3*x)/(b^2)^(3/2) - (48*a^6*e^4*x)/(b^3*Sqrt[b^2]) - (544*a^
4*d*e^3*x^2)/(b*Sqrt[b^2]) - (208*a^5*e^4*x^2)/(b^2)^(3/2) - (3008*a^3*d*e^3*x^3)/(3*Sqrt[b^2]) - (320*a^4*e^4
*x^3)/(b*Sqrt[b^2]) - (960*a^2*b*d*e^3*x^4)/Sqrt[b^2] - (160*a^3*e^4*x^4)/Sqrt[b^2] - 384*a*Sqrt[b^2]*d*e^3*x^
5 + (128*a^5*d*e^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*b^4) + (352*a^4*d*e^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3
*b^3) + (48*a^5*e^4*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^4 + (1280*a^3*d*e^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/
(3*b^2) + (160*a^4*e^4*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 + (576*a^2*d*e^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^
2])/b + (160*a^3*e^4*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + 384*a*d*e^3*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2] +
(128*a^3*d*e^3*x^3*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b + 384*a^2*d*e^3*x^4*ArcTanh[
(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] + 384*a*b*d*e^3*x^5*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2
*a*b*x + b^2*x^2])/a] + 128*b^2*d*e^3*x^6*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] - (128*a
^2*d*e^3*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] - (256*a*b*d*e^3*x^4*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] - 128*Sqrt[b^2]*d*e^3*x^5*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] + (64*a^4*e^4*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*Sqrt[b^2]) +
(192*a^3*e^4*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (192*a^2*b*e^4*x^5*Log[-a
- Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + 64*a*Sqrt[b^2]*e^4*x^6*Log[-a - Sqrt[b^2]*x + Sqr
t[a^2 + 2*a*b*x + b^2*x^2]] - (64*a^3*e^4*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 - (128*a^2*e^4*x^4*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 - 64*a*e^4*x^5*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]] + (64*a^4*e^4*x^3*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*Sqrt[b^2]) + (192*a^3*e^4
*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (192*a^2*b*e^4*x^5*Log[a - Sqrt[b^2]*x
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + 64*a*Sqrt[b^2]*e^4*x^6*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]] - (64*a^3*e^4*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 - (128*a^2*e^4*x^4*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 - 64*a*e^4*x^5*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 - S
qrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^3*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^3)
________________________________________________________________________________________
fricas [A] time = 0.49, size = 292, normalized size = 1.45 \begin {gather*} \frac {3 \, b^{4} e^{4} x^{4} + 9 \, a b^{3} e^{4} x^{3} - b^{4} d^{4} - 2 \, a b^{3} d^{3} e - 6 \, a^{2} b^{2} d^{2} e^{2} + 22 \, a^{3} b d e^{3} - 13 \, a^{4} e^{4} - 9 \, {\left (2 \, b^{4} d^{2} e^{2} - 4 \, a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} - 3 \, {\left (2 \, b^{4} d^{3} e + 6 \, a b^{3} d^{2} e^{2} - 18 \, a^{2} b^{2} d e^{3} + 9 \, a^{3} b e^{4}\right )} x + 12 \, {\left (a^{3} b d e^{3} - a^{4} e^{4} + {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 3 \, {\left (a b^{3} d e^{3} - a^{2} b^{2} e^{4}\right )} x^{2} + 3 \, {\left (a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x\right )} \log \left (b x + a\right )}{3 \, {\left (b^{8} x^{3} + 3 \, a b^{7} x^{2} + 3 \, a^{2} b^{6} x + a^{3} b^{5}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")
[Out]
1/3*(3*b^4*e^4*x^4 + 9*a*b^3*e^4*x^3 - b^4*d^4 - 2*a*b^3*d^3*e - 6*a^2*b^2*d^2*e^2 + 22*a^3*b*d*e^3 - 13*a^4*e
^4 - 9*(2*b^4*d^2*e^2 - 4*a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 - 3*(2*b^4*d^3*e + 6*a*b^3*d^2*e^2 - 18*a^2*b^2*d*e^3
+ 9*a^3*b*e^4)*x + 12*(a^3*b*d*e^3 - a^4*e^4 + (b^4*d*e^3 - a*b^3*e^4)*x^3 + 3*(a*b^3*d*e^3 - a^2*b^2*e^4)*x^
2 + 3*(a^2*b^2*d*e^3 - a^3*b*e^4)*x)*log(b*x + a))/(b^8*x^3 + 3*a*b^7*x^2 + 3*a^2*b^6*x + a^3*b^5)
________________________________________________________________________________________
giac [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (b x + a\right )} {\left (e x + d\right )}^{4}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}}}\,{d x} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")
[Out]
integrate((b*x + a)*(e*x + d)^4/(b^2*x^2 + 2*a*b*x + a^2)^(5/2), x)
________________________________________________________________________________________
maple [B] time = 0.07, size = 322, normalized size = 1.60 \begin {gather*} -\frac {\left (12 a \,b^{3} e^{4} x^{3} \ln \left (b x +a \right )-12 b^{4} d \,e^{3} x^{3} \ln \left (b x +a \right )-3 b^{4} e^{4} x^{4}+36 a^{2} b^{2} e^{4} x^{2} \ln \left (b x +a \right )-36 a \,b^{3} d \,e^{3} x^{2} \ln \left (b x +a \right )-9 a \,b^{3} e^{4} x^{3}+36 a^{3} b \,e^{4} x \ln \left (b x +a \right )-36 a^{2} b^{2} d \,e^{3} x \ln \left (b x +a \right )+9 a^{2} b^{2} e^{4} x^{2}-36 a \,b^{3} d \,e^{3} x^{2}+18 b^{4} d^{2} e^{2} x^{2}+12 a^{4} e^{4} \ln \left (b x +a \right )-12 a^{3} b d \,e^{3} \ln \left (b x +a \right )+27 a^{3} b \,e^{4} x -54 a^{2} b^{2} d \,e^{3} x +18 a \,b^{3} d^{2} e^{2} x +6 b^{4} d^{3} e x +13 a^{4} e^{4}-22 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}+2 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \left (b x +a \right )^{2}}{3 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{5}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
[Out]
-1/3*(12*ln(b*x+a)*x^3*a*b^3*e^4-12*ln(b*x+a)*x^3*b^4*d*e^3-3*b^4*e^4*x^4+36*ln(b*x+a)*x^2*a^2*b^2*e^4-36*ln(b
*x+a)*x^2*a*b^3*d*e^3-9*a*b^3*e^4*x^3+36*a^3*b*e^4*x*ln(b*x+a)-36*a^2*b^2*d*e^3*x*ln(b*x+a)+9*a^2*b^2*e^4*x^2-
36*a*b^3*d*e^3*x^2+18*b^4*d^2*e^2*x^2+12*a^4*e^4*ln(b*x+a)-12*a^3*b*d*e^3*ln(b*x+a)+27*a^3*b*e^4*x-54*a^2*b^2*
d*e^3*x+18*a*b^3*d^2*e^2*x+6*b^4*d^3*e*x+13*a^4*e^4-22*a^3*b*d*e^3+6*a^2*b^2*d^2*e^2+2*a*b^3*d^3*e+b^4*d^4)*(b
*x+a)^2/b^5/((b*x+a)^2)^(5/2)
________________________________________________________________________________________
maxima [B] time = 0.99, size = 755, normalized size = 3.76 \begin {gather*} \frac {1}{12} \, b e^{4} {\left (\frac {12 \, b^{5} x^{5} + 48 \, a b^{4} x^{4} - 48 \, a^{2} b^{3} x^{3} - 252 \, a^{3} b^{2} x^{2} - 248 \, a^{4} b x - 77 \, a^{5}}{b^{10} x^{4} + 4 \, a b^{9} x^{3} + 6 \, a^{2} b^{8} x^{2} + 4 \, a^{3} b^{7} x + a^{4} b^{6}} - \frac {60 \, a \log \left (b x + a\right )}{b^{6}}\right )} + \frac {1}{3} \, b d e^{3} {\left (\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} + \frac {1}{12} \, a e^{4} {\left (\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} - \frac {1}{2} \, b d^{2} e^{2} {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{3} \, a d e^{3} {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{12} \, b d^{4} {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{3} \, a d^{3} e {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{3} \, b d^{3} e {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{2} \, a d^{2} e^{2} {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {a d^{4}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)*(e*x+d)^4/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")
[Out]
1/12*b*e^4*((12*b^5*x^5 + 48*a*b^4*x^4 - 48*a^2*b^3*x^3 - 252*a^3*b^2*x^2 - 248*a^4*b*x - 77*a^5)/(b^10*x^4 +
4*a*b^9*x^3 + 6*a^2*b^8*x^2 + 4*a^3*b^7*x + a^4*b^6) - 60*a*log(b*x + a)/b^6) + 1/3*b*d*e^3*((48*a*b^3*x^3 + 1
08*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5) + 12*log
(b*x + a)/b^5) + 1/12*a*e^4*((48*a*b^3*x^3 + 108*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^4 + 4*a*b^8*x^3 + 6
*a^2*b^7*x^2 + 4*a^3*b^6*x + a^4*b^5) + 12*log(b*x + a)/b^5) - 1/2*b*d^2*e^2*(12*x^2/((b^2*x^2 + 2*a*b*x + a^2
)^(3/2)*b^2) + 8*a^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) + 6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7*(x + a/b)^3) -
3*a^3/(b^8*(x + a/b)^4)) - 1/3*a*d*e^3*(12*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 8*a^2/((b^2*x^2 + 2*a*
b*x + a^2)^(3/2)*b^4) + 6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7*(x + a/b)^3) - 3*a^3/(b^8*(x + a/b)^4)) - 1/12*b*d^
4*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/(b^6*(x + a/b)^4)) - 1/3*a*d^3*e*(4/((b^2*x^2 + 2*a*b*x + a^2
)^(3/2)*b^2) - 3*a/(b^6*(x + a/b)^4)) - 1/3*b*d^3*e*(6/(b^5*(x + a/b)^2) - 8*a/(b^6*(x + a/b)^3) + 3*a^2/(b^7*
(x + a/b)^4)) - 1/2*a*d^2*e^2*(6/(b^5*(x + a/b)^2) - 8*a/(b^6*(x + a/b)^3) + 3*a^2/(b^7*(x + a/b)^4)) - 1/4*a*
d^4/(b^5*(x + a/b)^4)
________________________________________________________________________________________
mupad [F] time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a+b\,x\right )\,{\left (d+e\,x\right )}^4}{{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((a + b*x)*(d + e*x)^4)/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
[Out]
int(((a + b*x)*(d + e*x)^4)/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)
________________________________________________________________________________________
sympy [F] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right ) \left (d + e x\right )^{4}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)*(e*x+d)**4/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
[Out]
Integral((a + b*x)*(d + e*x)**4/((a + b*x)**2)**(5/2), x)
________________________________________________________________________________________