3.16.57 \(\int \frac {(A+B x) (d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)
Optimal. Leaf size=227 \[ -\frac {(d+e x)^4 (A b-a B)}{4 b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {3 B e^2 (b d-a e)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 B e (b d-a e)^2}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {B (b d-a e)^3}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B e^3 (a+b x) \log (a+b x)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}} \]
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Rubi [A] time = 0.17, antiderivative size = 227, 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 =
{770, 78, 43} \begin {gather*} -\frac {(d+e x)^4 (A b-a B)}{4 b (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {3 B e^2 (b d-a e)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 B e (b d-a e)^2}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {B (b d-a e)^3}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B e^3 (a+b x) \log (a+b x)}{b^5 \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)^3)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
[Out]
(-3*B*e^2*(b*d - a*e))/(b^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (B*(b*d - a*e)^3)/(3*b^5*(a + b*x)^2*Sqrt[a^2 + 2
*a*b*x + b^2*x^2]) - (3*B*e*(b*d - a*e)^2)/(2*b^5*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - ((A*b - a*B)*(d +
e*x)^4)/(4*b*(b*d - a*e)*(a + b*x)^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (B*e^3*(a + b*x)*Log[a + b*x])/(b^5*Sqr
t[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 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 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 {(A+B x) (d+e x)^3}{\left (a^2+2 a b x+b^2 x^2\right )^{5/2}} \, dx &=\frac {\left (b^4 \left (a b+b^2 x\right )\right ) \int \frac {(A+B x) (d+e x)^3}{\left (a b+b^2 x\right )^5} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(A b-a B) (d+e x)^4}{4 b (b d-a e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b^2 B \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^3}{\left (a b+b^2 x\right )^4} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(A b-a B) (d+e x)^4}{4 b (b d-a e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (b^2 B \left (a b+b^2 x\right )\right ) \int \left (\frac {(b d-a e)^3}{b^7 (a+b x)^4}+\frac {3 e (b d-a e)^2}{b^7 (a+b x)^3}+\frac {3 e^2 (b d-a e)}{b^7 (a+b x)^2}+\frac {e^3}{b^7 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {3 B e^2 (b d-a e)}{b^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {B (b d-a e)^3}{3 b^5 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {3 B e (b d-a e)^2}{2 b^5 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (d+e x)^4}{4 b (b d-a e) (a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {B e^3 (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 = 0.13, size = 239, normalized size = 1.05 \begin {gather*} \frac {-3 A b \left (a^3 e^3+a^2 b e^2 (d+4 e x)+a b^2 e \left (d^2+4 d e x+6 e^2 x^2\right )+b^3 \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )\right )+B \left (25 a^4 e^3+a^3 b e^2 (88 e x-9 d)-3 a^2 b^2 e \left (d^2+12 d e x-36 e^2 x^2\right )-a b^3 \left (d^3+12 d^2 e x+54 d e^2 x^2-48 e^3 x^3\right )-2 b^4 d x \left (2 d^2+9 d e x+18 e^2 x^2\right )\right )+12 B e^3 (a+b x)^4 \log (a+b x)}{12 b^5 (a+b x)^3 \sqrt {(a+b x)^2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((A + B*x)*(d + e*x)^3)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
[Out]
(B*(25*a^4*e^3 + a^3*b*e^2*(-9*d + 88*e*x) - 3*a^2*b^2*e*(d^2 + 12*d*e*x - 36*e^2*x^2) - 2*b^4*d*x*(2*d^2 + 9*
d*e*x + 18*e^2*x^2) - a*b^3*(d^3 + 12*d^2*e*x + 54*d*e^2*x^2 - 48*e^3*x^3)) - 3*A*b*(a^3*e^3 + a^2*b*e^2*(d +
4*e*x) + a*b^2*e*(d^2 + 4*d*e*x + 6*e^2*x^2) + b^3*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3)) + 12*B*e^3*(a
+ b*x)^4*Log[a + b*x])/(12*b^5*(a + b*x)^3*Sqrt[(a + b*x)^2])
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IntegrateAlgebraic [B] time = 9.66, size = 4832, normalized size = 21.29 \begin {gather*} \text {Result too large to show} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[((A + B*x)*(d + e*x)^3)/(a^2 + 2*a*b*x + b^2*x^2)^(5/2),x]
[Out]
(24*a^4*A*b^3*d^3 - 24*a^5*b^2*B*d^3 - 72*a^5*A*b^2*d^2*e + 72*a^6*b*B*d^2*e + 72*a^6*A*b*d*e^2 - 72*a^7*B*d*e
^2 - 24*a^7*A*e^3 - 64*a^4*b^3*B*d^3*x + 231*a^5*A*b^2*d*e^2*x - 288*a^6*b*B*d*e^2*x - 96*a^6*A*b*e^3*x - 136*
a^3*b^4*B*d^3*x^2 + 672*a^4*A*b^3*d*e^2*x^2 - 1008*a^5*b^2*B*d*e^2*x^2 - 336*a^5*A*b^2*e^3*x^2 - 144*a^2*b^5*B
*d^3*x^3 + 1008*a^3*A*b^4*d*e^2*x^3 - 2016*a^4*b^3*B*d*e^2*x^3 - 672*a^4*A*b^3*e^3*x^3 + 24*A*b^7*d^3*x^4 - 72
*a*b^6*B*d^3*x^4 + 24*a*A*b^6*d^2*e*x^4 + 24*a^2*b^5*B*d^2*e*x^4 + 864*a^2*A*b^5*d*e^2*x^4 - 2448*a^3*b^4*B*d*
e^2*x^4 - 816*a^3*A*b^4*e^3*x^4 + 96*A*b^7*d^2*e*x^5 + 96*a*b^6*B*d^2*e*x^5 + 432*a*A*b^6*d*e^2*x^5 - 1728*a^2
*b^5*B*d*e^2*x^5 - 576*a^2*A*b^5*e^3*x^5 + 144*b^7*B*d^2*e*x^6 + 144*A*b^7*d*e^2*x^6 - 576*a*b^6*B*d*e^2*x^6 -
192*a*A*b^6*e^3*x^6 + Sqrt[b^2]*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(24*a^3*A*b^2*d^3 - 72*a^4*A*b*d^2*e + 72*a^5*B
*d^2*e + 9*a^5*A*d*e^2 - 24*a^2*A*b^3*d^3*x + 64*a^3*b^2*B*d^3*x + 72*a^3*A*b^2*d^2*e*x - 72*a^4*b*B*d^2*e*x -
240*a^4*A*b*d*e^2*x + 288*a^5*B*d*e^2*x + 96*a^5*A*e^3*x + 24*a*A*b^4*d^3*x^2 + 72*a^2*b^3*B*d^3*x^2 - 72*a^2
*A*b^3*d^2*e*x^2 + 72*a^3*b^2*B*d^2*e*x^2 - 432*a^3*A*b^2*d*e^2*x^2 + 720*a^4*b*B*d*e^2*x^2 + 240*a^4*A*b*e^3*
x^2 - 24*A*b^5*d^3*x^3 + 72*a*b^4*B*d^3*x^3 + 72*a*A*b^4*d^2*e*x^3 - 72*a^2*b^3*B*d^2*e*x^3 - 576*a^2*A*b^3*d*
e^2*x^3 + 1296*a^3*b^2*B*d*e^2*x^3 + 432*a^3*A*b^2*e^3*x^3 - 96*A*b^5*d^2*e*x^4 + 48*a*b^4*B*d^2*e*x^4 - 288*a
*A*b^4*d*e^2*x^4 + 1152*a^2*b^3*B*d*e^2*x^4 + 384*a^2*A*b^3*e^3*x^4 - 144*b^5*B*d^2*e*x^5 - 144*A*b^5*d*e^2*x^
5 + 576*a*b^4*B*d*e^2*x^5 + 192*a*A*b^4*e^3*x^5))/(12*b^3*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-8*a^3*b^5 - 24*a
^2*b^6*x - 24*a*b^7*x^2 - 8*b^8*x^3) + 12*b^3*Sqrt[b^2]*x^4*(8*a^4*b^4 + 32*a^3*b^5*x + 48*a^2*b^6*x^2 + 32*a*
b^7*x^3 + 8*b^8*x^4)) + ((-308*a^5*A*d*e^2*x)/(b*Sqrt[b^2]) - (448*a^7*B*e^3*x)/(3*b^3*Sqrt[b^2]) - (896*a^4*A
*d*e^2*x^2)/Sqrt[b^2] - (1888*a^6*B*e^3*x^2)/(3*(b^2)^(3/2)) - (1344*a^3*A*b*d*e^2*x^3)/Sqrt[b^2] - (1600*a^5*
B*e^3*x^3)/(b*Sqrt[b^2]) - 1120*a^2*A*Sqrt[b^2]*d*e^2*x^4 - (8000*a^4*B*e^3*x^4)/(3*Sqrt[b^2]) - (448*a*A*b^3*
d*e^2*x^5)/Sqrt[b^2] - (8576*a^3*b*B*e^3*x^5)/(3*Sqrt[b^2]) - 1792*a^2*Sqrt[b^2]*B*e^3*x^6 - (512*a*b^3*B*e^3*
x^7)/Sqrt[b^2] + (84*a^5*A*d*e^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 + (32*a^7*B*e^3*Sqrt[a^2 + 2*a*b*x + b^2*x
^2])/b^5 + (224*a^4*A*d*e^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + (352*a^6*B*e^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x
^2])/(3*b^4) + (672*a^3*A*d*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b + (512*a^5*B*e^3*x^2*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/b^3 + 672*a^2*A*d*e^2*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2] + (1088*a^4*B*e^3*x^3*Sqrt[a^2 + 2*a*b*x +
b^2*x^2])/b^2 + 448*a*A*b*d*e^2*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2] + (4736*a^3*B*e^3*x^4*Sqrt[a^2 + 2*a*b*x + b
^2*x^2])/(3*b) + 1280*a^2*B*e^3*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2] + 512*a*b*B*e^3*x^6*Sqrt[a^2 + 2*a*b*x + b^2
*x^2] + (128*a^4*B*e^3*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b + 512*a^3*B*e^3*x^5*
ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] + 768*a^2*b*B*e^3*x^6*ArcTanh[(-(Sqrt[b^2]*x) + Sq
rt[a^2 + 2*a*b*x + b^2*x^2])/a] + 512*a*b^2*B*e^3*x^7*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])
/a] + 128*b^3*B*e^3*x^8*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] - (128*a^3*B*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] - (384*a^2*b*B
*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])/Sqrt[b^2]
- 384*a*Sqrt[b^2]*B*e^3*x^6*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] - (128*b^3*B*e^3*x^7*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])^4*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b
*x + b^2*x^2])^4) + ((32*a^8*B*e^3)/(b^4*Sqrt[b^2]) + (256*a^4*B*d^3*x)/(3*Sqrt[b^2]) + (384*a^6*B*d*e^2*x)/(b
^2)^(3/2) + (128*a^6*A*e^3*x)/(b^2)^(3/2) + (448*a^7*B*e^3*x)/(3*b^3*Sqrt[b^2]) + (544*a^3*b*B*d^3*x^2)/(3*Sqr
t[b^2]) + (1344*a^5*B*d*e^2*x^2)/(b*Sqrt[b^2]) + (448*a^5*A*e^3*x^2)/(b*Sqrt[b^2]) + (1888*a^6*B*e^3*x^2)/(3*(
b^2)^(3/2)) + 192*a^2*Sqrt[b^2]*B*d^3*x^3 + (2688*a^4*B*d*e^2*x^3)/Sqrt[b^2] + (896*a^4*A*e^3*x^3)/Sqrt[b^2] +
(1600*a^5*B*e^3*x^3)/(b*Sqrt[b^2]) + (320*a*b^3*B*d^3*x^4)/(3*Sqrt[b^2]) + (3360*a^3*b*B*d*e^2*x^4)/Sqrt[b^2]
+ (1120*a^3*A*b*e^3*x^4)/Sqrt[b^2] + (2400*a^4*B*e^3*x^4)/Sqrt[b^2] + (128*b^4*B*d^3*x^5)/(3*Sqrt[b^2]) + 268
8*a^2*Sqrt[b^2]*B*d*e^2*x^5 + 896*a^2*A*Sqrt[b^2]*e^3*x^5 + (1920*a^3*b*B*e^3*x^5)/Sqrt[b^2] + (1344*a*b^3*B*d
*e^2*x^6)/Sqrt[b^2] + (448*a*A*b^3*e^3*x^6)/Sqrt[b^2] + 640*a^2*Sqrt[b^2]*B*e^3*x^6 + (384*b^4*B*d*e^2*x^7)/Sq
rt[b^2] + (128*A*b^4*e^3*x^7)/Sqrt[b^2] - (32*a^4*B*d^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 - (96*a^6*B*d*e^2*S
qrt[a^2 + 2*a*b*x + b^2*x^2])/b^4 - (32*a^6*A*e^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^4 - (160*a^3*B*d^3*x*Sqrt[a
^2 + 2*a*b*x + b^2*x^2])/(3*b) - (288*a^5*B*d*e^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 - (96*a^5*A*e^3*x*Sqrt[
a^2 + 2*a*b*x + b^2*x^2])/b^3 - (448*a^6*B*e^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*b^4) - 128*a^2*B*d^3*x^2*Sq
rt[a^2 + 2*a*b*x + b^2*x^2] - (1056*a^4*B*d*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 - (352*a^4*A*e^3*x^2*Sq
rt[a^2 + 2*a*b*x + b^2*x^2])/b^2 - (480*a^5*B*e^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 - 64*a*b*B*d^3*x^3*Sq
rt[a^2 + 2*a*b*x + b^2*x^2] - (1632*a^3*B*d*e^2*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b - (544*a^3*A*e^3*x^3*Sqrt
[a^2 + 2*a*b*x + b^2*x^2])/b - (1120*a^4*B*e^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 - (128*b^2*B*d^3*x^4*Sqr
t[a^2 + 2*a*b*x + b^2*x^2])/3 - 1728*a^2*B*d*e^2*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - 576*a^2*A*e^3*x^4*Sqrt[a^
2 + 2*a*b*x + b^2*x^2] - (1280*a^3*B*e^3*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b - 960*a*b*B*d*e^2*x^5*Sqrt[a^2 +
2*a*b*x + b^2*x^2] - 320*a*A*b*e^3*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - 640*a^2*B*e^3*x^5*Sqrt[a^2 + 2*a*b*x +
b^2*x^2] - 384*b^2*B*d*e^2*x^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - 128*A*b^2*e^3*x^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2
] - (64*a^4*B*e^3*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (256*a^3*b*B*e^3*x^5*
Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - 384*a^2*Sqrt[b^2]*B*e^3*x^6*Log[-a - Sqrt[b
^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - (256*a*b^3*B*e^3*x^7*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*
x^2]])/Sqrt[b^2] - (64*b^4*B*e^3*x^8*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (64*a^
3*B*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]])/b + 192*a^2*B
*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]] + 192*a*b*B*e^3*x
^6*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^2*B*e^3*x^7*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*B*e^3*x^4*Log[a - Sq
rt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (256*a^3*b*B*e^3*x^5*Log[a - Sqrt[b^2]*x + Sqrt[a^2 +
2*a*b*x + b^2*x^2]])/Sqrt[b^2] - 384*a^2*Sqrt[b^2]*B*e^3*x^6*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^
2]] - (256*a*b^3*B*e^3*x^7*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (64*b^4*B*e^3*x^8
*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (64*a^3*B*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]])/b + 192*a^2*B*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]] + 192*a*b*B*e^3*x^6*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^2*B*e^3*x^7*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])^4*(a - Sqrt[b^2]*x
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2])^4)
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fricas [B] time = 0.42, size = 359, normalized size = 1.58 \begin {gather*} -\frac {{\left (B a b^{3} + 3 \, A b^{4}\right )} d^{3} + 3 \, {\left (B a^{2} b^{2} + A a b^{3}\right )} d^{2} e + 3 \, {\left (3 \, B a^{3} b + A a^{2} b^{2}\right )} d e^{2} - {\left (25 \, B a^{4} - 3 \, A a^{3} b\right )} e^{3} + 12 \, {\left (3 \, B b^{4} d e^{2} - {\left (4 \, B a b^{3} - A b^{4}\right )} e^{3}\right )} x^{3} + 18 \, {\left (B b^{4} d^{2} e + {\left (3 \, B a b^{3} + A b^{4}\right )} d e^{2} - {\left (6 \, B a^{2} b^{2} - A a b^{3}\right )} e^{3}\right )} x^{2} + 4 \, {\left (B b^{4} d^{3} + 3 \, {\left (B a b^{3} + A b^{4}\right )} d^{2} e + 3 \, {\left (3 \, B a^{2} b^{2} + A a b^{3}\right )} d e^{2} - {\left (22 \, B a^{3} b - 3 \, A a^{2} b^{2}\right )} e^{3}\right )} x - 12 \, {\left (B b^{4} e^{3} x^{4} + 4 \, B a b^{3} e^{3} x^{3} + 6 \, B a^{2} b^{2} e^{3} x^{2} + 4 \, B a^{3} b e^{3} x + B a^{4} e^{3}\right )} \log \left (b x + a\right )}{12 \, {\left (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}\right )}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")
[Out]
-1/12*((B*a*b^3 + 3*A*b^4)*d^3 + 3*(B*a^2*b^2 + A*a*b^3)*d^2*e + 3*(3*B*a^3*b + A*a^2*b^2)*d*e^2 - (25*B*a^4 -
3*A*a^3*b)*e^3 + 12*(3*B*b^4*d*e^2 - (4*B*a*b^3 - A*b^4)*e^3)*x^3 + 18*(B*b^4*d^2*e + (3*B*a*b^3 + A*b^4)*d*e
^2 - (6*B*a^2*b^2 - A*a*b^3)*e^3)*x^2 + 4*(B*b^4*d^3 + 3*(B*a*b^3 + A*b^4)*d^2*e + 3*(3*B*a^2*b^2 + A*a*b^3)*d
*e^2 - (22*B*a^3*b - 3*A*a^2*b^2)*e^3)*x - 12*(B*b^4*e^3*x^4 + 4*B*a*b^3*e^3*x^3 + 6*B*a^2*b^2*e^3*x^2 + 4*B*a
^3*b*e^3*x + B*a^4*e^3)*log(b*x + a))/(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)
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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((B*x+A)*(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")
[Out]
sage0*x
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maple [B] time = 0.06, size = 385, normalized size = 1.70 \begin {gather*} -\frac {\left (-12 B \,b^{4} e^{3} x^{4} \ln \left (b x +a \right )-48 B a \,b^{3} e^{3} x^{3} \ln \left (b x +a \right )+12 A \,b^{4} e^{3} x^{3}-72 B \,a^{2} b^{2} e^{3} x^{2} \ln \left (b x +a \right )-48 B a \,b^{3} e^{3} x^{3}+36 B \,b^{4} d \,e^{2} x^{3}+18 A a \,b^{3} e^{3} x^{2}+18 A \,b^{4} d \,e^{2} x^{2}-48 B \,a^{3} b \,e^{3} x \ln \left (b x +a \right )-108 B \,a^{2} b^{2} e^{3} x^{2}+54 B a \,b^{3} d \,e^{2} x^{2}+18 B \,b^{4} d^{2} e \,x^{2}+12 A \,a^{2} b^{2} e^{3} x +12 A a \,b^{3} d \,e^{2} x +12 A \,b^{4} d^{2} e x -12 B \,a^{4} e^{3} \ln \left (b x +a \right )-88 B \,a^{3} b \,e^{3} x +36 B \,a^{2} b^{2} d \,e^{2} x +12 B a \,b^{3} d^{2} e x +4 B \,b^{4} d^{3} x +3 A \,a^{3} b \,e^{3}+3 A \,a^{2} b^{2} d \,e^{2}+3 A a \,b^{3} d^{2} e +3 A \,b^{4} d^{3}-25 B \,a^{4} e^{3}+9 B \,a^{3} b d \,e^{2}+3 B \,a^{2} b^{2} d^{2} e +B a \,b^{3} d^{3}\right ) \left (b x +a \right )}{12 \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)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)
[Out]
-1/12*(-48*B*a^3*b*e^3*x*ln(b*x+a)-72*B*a^2*b^2*e^3*x^2*ln(b*x+a)-48*B*ln(b*x+a)*x^3*a*b^3*e^3+12*B*a*b^3*d^2*
e*x+3*A*a^3*b*e^3+36*B*a^2*b^2*d*e^2*x+12*A*a*b^3*d*e^2*x+54*B*a*b^3*d*e^2*x^2+3*A*b^4*d^3-25*B*a^4*e^3+9*B*a^
3*b*d*e^2+B*a*b^3*d^3+12*A*b^4*e^3*x^3-12*B*a^4*e^3*ln(b*x+a)+4*B*b^4*d^3*x+18*A*b^4*d*e^2*x^2+12*A*a^2*b^2*e^
3*x+3*B*a^2*b^2*d^2*e+3*A*a*b^3*d^2*e-88*B*a^3*b*e^3*x-108*B*a^2*b^2*e^3*x^2+18*B*b^4*d^2*e*x^2-48*B*a*b^3*e^3
*x^3+36*B*b^4*d*e^2*x^3+18*A*a*b^3*e^3*x^2+3*A*a^2*b^2*d*e^2+12*A*b^4*d^2*e*x-12*B*ln(b*x+a)*x^4*b^4*e^3)*(b*x
+a)/b^5/((b*x+a)^2)^(5/2)
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maxima [B] time = 0.67, size = 533, normalized size = 2.35 \begin {gather*} \frac {1}{12} \, B 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}{4} \, B d 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}{12} \, A 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^{3} {\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}{4} \, A d^{2} 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}{4} \, B d^{2} 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}{4} \, A d 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^{3}}{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)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="maxima")
[Out]
1/12*B*e^3*((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) - 1/4*B*d*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)) - 1/12*A*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^3*(4/((b^2*x^2 + 2*a
*b*x + a^2)^(3/2)*b^2) - 3*a/(b^6*(x + a/b)^4)) - 1/4*A*d^2*e*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/(
b^6*(x + a/b)^4)) - 1/4*B*d^2*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)) - 1/4*
A*d*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^3/(b^5*(x + a/b)^4)
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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 )}^3}{{\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)^3)/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2),x)
[Out]
int(((A + B*x)*(d + e*x)^3)/(a^2 + b^2*x^2 + 2*a*b*x)^(5/2), x)
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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 )^{3}}{\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)**3/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)
[Out]
Integral((A + B*x)*(d + e*x)**3/((a + b*x)**2)**(5/2), x)
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