3.23.32 \(\int \frac {(A+B x) (d+e x)^3}{(a+b x+c x^2)^{3/2}} \, dx\)
Optimal. Leaf size=325 \[ \frac {3 e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (B \left (-4 c e (a e+3 b d)+5 b^2 e^2+8 c^2 d^2\right )+4 A c e (2 c d-b e)\right )}{8 c^{7/2}}+\frac {2 (d+e x)^2 \left (-x \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )-b (a B e+A c d)+2 a c (A e+B d)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {e \sqrt {a+b x+c x^2} \left (-2 c e x \left (-4 c (3 a B e+A b e+b B d)+8 A c^2 d+5 b^2 B e\right )+4 b c \left (-13 a B e^2+6 A c d e+4 B c d^2\right )-32 c^2 \left (-a A e^2-3 a B d e+A c d^2\right )-12 b^2 c e (A e+3 B d)+15 b^3 B e^2\right )}{4 c^3 \left (b^2-4 a c\right )} \]
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Rubi [A] time = 0.34, antiderivative size = 325, normalized size of antiderivative = 1.00,
number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used =
{818, 779, 621, 206} \begin {gather*} -\frac {e \sqrt {a+b x+c x^2} \left (-2 c e x \left (-4 c (3 a B e+A b e+b B d)+8 A c^2 d+5 b^2 B e\right )+4 b c \left (-13 a B e^2+6 A c d e+4 B c d^2\right )-32 c^2 \left (-a A e^2-3 a B d e+A c d^2\right )-12 b^2 c e (A e+3 B d)+15 b^3 B e^2\right )}{4 c^3 \left (b^2-4 a c\right )}+\frac {3 e \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (B \left (-4 c e (a e+3 b d)+5 b^2 e^2+8 c^2 d^2\right )+4 A c e (2 c d-b e)\right )}{8 c^{7/2}}+\frac {2 (d+e x)^2 \left (-x \left (2 c (A c d-a B e)-b c (A e+B d)+b^2 B e\right )-b (a B e+A c d)+2 a c (A e+B d)\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((A + B*x)*(d + e*x)^3)/(a + b*x + c*x^2)^(3/2),x]
[Out]
(2*(d + e*x)^2*(2*a*c*(B*d + A*e) - b*(A*c*d + a*B*e) - (b^2*B*e - b*c*(B*d + A*e) + 2*c*(A*c*d - a*B*e))*x))/
(c*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) - (e*(15*b^3*B*e^2 - 12*b^2*c*e*(3*B*d + A*e) - 32*c^2*(A*c*d^2 - 3*a*
B*d*e - a*A*e^2) + 4*b*c*(4*B*c*d^2 + 6*A*c*d*e - 13*a*B*e^2) - 2*c*e*(8*A*c^2*d + 5*b^2*B*e - 4*c*(b*B*d + A*
b*e + 3*a*B*e))*x)*Sqrt[a + b*x + c*x^2])/(4*c^3*(b^2 - 4*a*c)) + (3*e*(4*A*c*e*(2*c*d - b*e) + B*(8*c^2*d^2 +
5*b^2*e^2 - 4*c*e*(3*b*d + a*e)))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c*x^2])])/(8*c^(7/2))
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 621
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 779
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((b
*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x)*(a + b*x + c*x^2)^(p + 1))/(2*c^2*(p + 1)*(2*p + 3
)), x] + Dist[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)), Int[(a
+ b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && !LeQ[p, -1]
Rule 818
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m - 1)*(a + b*x + c*x^2)^(p + 1)*(2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g
- c*(b*e*f + b*d*g + 2*a*e*g))*x))/(c*(p + 1)*(b^2 - 4*a*c)), x] - Dist[1/(c*(p + 1)*(b^2 - 4*a*c)), Int[(d +
e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Simp[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a
*e*(e*f*(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*(m + p + 1) + 2*c^2*d*f*(m +
2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2*p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && Ne
Q[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] &&
RationalQ[a, b, c, d, e, f, g]) || !ILtQ[m + 2*p + 3, 0])
Rubi steps
\begin {align*} \int \frac {(A+B x) (d+e x)^3}{\left (a+b x+c x^2\right )^{3/2}} \, dx &=\frac {2 (d+e x)^2 \left (2 a c (B d+A e)-b (A c d+a B e)-\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {2 \int \frac {(d+e x) \left (\frac {1}{2} e \left (b^2 B d+4 A b c d-12 a B c d+4 a b B e-8 a A c e\right )+\frac {1}{2} e \left (8 A c^2 d+5 b^2 B e-4 c (b B d+A b e+3 a B e)\right ) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{c \left (b^2-4 a c\right )}\\ &=\frac {2 (d+e x)^2 \left (2 a c (B d+A e)-b (A c d+a B e)-\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {e \left (15 b^3 B e^2-12 b^2 c e (3 B d+A e)-32 c^2 \left (A c d^2-3 a B d e-a A e^2\right )+4 b c \left (4 B c d^2+6 A c d e-13 a B e^2\right )-2 c e \left (8 A c^2 d+5 b^2 B e-4 c (b B d+A b e+3 a B e)\right ) x\right ) \sqrt {a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac {\left (3 e \left (4 A c e (2 c d-b e)+B \left (8 c^2 d^2+5 b^2 e^2-4 c e (3 b d+a e)\right )\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{8 c^3}\\ &=\frac {2 (d+e x)^2 \left (2 a c (B d+A e)-b (A c d+a B e)-\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {e \left (15 b^3 B e^2-12 b^2 c e (3 B d+A e)-32 c^2 \left (A c d^2-3 a B d e-a A e^2\right )+4 b c \left (4 B c d^2+6 A c d e-13 a B e^2\right )-2 c e \left (8 A c^2 d+5 b^2 B e-4 c (b B d+A b e+3 a B e)\right ) x\right ) \sqrt {a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac {\left (3 e \left (4 A c e (2 c d-b e)+B \left (8 c^2 d^2+5 b^2 e^2-4 c e (3 b d+a e)\right )\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{4 c^3}\\ &=\frac {2 (d+e x)^2 \left (2 a c (B d+A e)-b (A c d+a B e)-\left (b^2 B e-b c (B d+A e)+2 c (A c d-a B e)\right ) x\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {e \left (15 b^3 B e^2-12 b^2 c e (3 B d+A e)-32 c^2 \left (A c d^2-3 a B d e-a A e^2\right )+4 b c \left (4 B c d^2+6 A c d e-13 a B e^2\right )-2 c e \left (8 A c^2 d+5 b^2 B e-4 c (b B d+A b e+3 a B e)\right ) x\right ) \sqrt {a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac {3 e \left (4 A c e (2 c d-b e)+B \left (8 c^2 d^2+5 b^2 e^2-4 c e (3 b d+a e)\right )\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{7/2}}\\ \end {align*}
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Mathematica [A] time = 0.64, size = 426, normalized size = 1.31 \begin {gather*} \frac {-2 \sqrt {c} \left (4 A c \left (-4 c \left (2 a^2 e^3+a c e \left (-3 d^2-3 d e x+e^2 x^2\right )+c^2 d^3 x\right )+b^2 e^2 (3 a e+c x (e x-6 d))-2 b c \left (a e^2 (3 d+5 e x)+c d^2 (d-3 e x)\right )+3 b^3 e^3 x\right )+B \left (-4 a^2 c e^2 (6 c (4 d+e x)-13 b e)+a \left (-15 b^3 e^3+2 b^2 c e^2 (18 d+31 e x)+4 b c^2 e \left (-6 d^2-30 d e x+5 e^2 x^2\right )+8 c^3 \left (2 d^3+6 d^2 e x-6 d e^2 x^2-e^3 x^3\right )\right )+b x \left (-15 b^3 e^3+b^2 c e^2 (36 d-5 e x)+2 b c^2 e \left (-12 d^2+6 d e x+e^2 x^2\right )+8 c^3 d^3\right )\right )\right )-3 e \left (b^2-4 a c\right ) \sqrt {a+x (b+c x)} \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right ) \left (B \left (-4 c e (a e+3 b d)+5 b^2 e^2+8 c^2 d^2\right )+4 A c e (2 c d-b e)\right )}{8 c^{7/2} \left (4 a c-b^2\right ) \sqrt {a+x (b+c x)}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((A + B*x)*(d + e*x)^3)/(a + b*x + c*x^2)^(3/2),x]
[Out]
(-2*Sqrt[c]*(4*A*c*(3*b^3*e^3*x + b^2*e^2*(3*a*e + c*x*(-6*d + e*x)) - 2*b*c*(c*d^2*(d - 3*e*x) + a*e^2*(3*d +
5*e*x)) - 4*c*(2*a^2*e^3 + c^2*d^3*x + a*c*e*(-3*d^2 - 3*d*e*x + e^2*x^2))) + B*(-4*a^2*c*e^2*(-13*b*e + 6*c*
(4*d + e*x)) + b*x*(8*c^3*d^3 - 15*b^3*e^3 + b^2*c*e^2*(36*d - 5*e*x) + 2*b*c^2*e*(-12*d^2 + 6*d*e*x + e^2*x^2
)) + a*(-15*b^3*e^3 + 2*b^2*c*e^2*(18*d + 31*e*x) + 4*b*c^2*e*(-6*d^2 - 30*d*e*x + 5*e^2*x^2) + 8*c^3*(2*d^3 +
6*d^2*e*x - 6*d*e^2*x^2 - e^3*x^3)))) - 3*(b^2 - 4*a*c)*e*(4*A*c*e*(2*c*d - b*e) + B*(8*c^2*d^2 + 5*b^2*e^2 -
4*c*e*(3*b*d + a*e)))*Sqrt[a + x*(b + c*x)]*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(8*c^(7/2
)*(-b^2 + 4*a*c)*Sqrt[a + x*(b + c*x)])
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IntegrateAlgebraic [A] time = 3.78, size = 538, normalized size = 1.66 \begin {gather*} -\frac {-32 a^2 A c^2 e^3+52 a^2 b B c e^3-96 a^2 B c^2 d e^2-24 a^2 B c^2 e^3 x+12 a A b^2 c e^3-24 a A b c^2 d e^2-40 a A b c^2 e^3 x+48 a A c^3 d^2 e+48 a A c^3 d e^2 x-16 a A c^3 e^3 x^2-15 a b^3 B e^3+36 a b^2 B c d e^2+62 a b^2 B c e^3 x-24 a b B c^2 d^2 e-120 a b B c^2 d e^2 x+20 a b B c^2 e^3 x^2+16 a B c^3 d^3+48 a B c^3 d^2 e x-48 a B c^3 d e^2 x^2-8 a B c^3 e^3 x^3+12 A b^3 c e^3 x-24 A b^2 c^2 d e^2 x+4 A b^2 c^2 e^3 x^2-8 A b c^3 d^3+24 A b c^3 d^2 e x-16 A c^4 d^3 x-15 b^4 B e^3 x+36 b^3 B c d e^2 x-5 b^3 B c e^3 x^2-24 b^2 B c^2 d^2 e x+12 b^2 B c^2 d e^2 x^2+2 b^2 B c^2 e^3 x^3+8 b B c^3 d^3 x}{4 c^3 \left (4 a c-b^2\right ) \sqrt {a+b x+c x^2}}-\frac {3 \log \left (-2 c^{7/2} \sqrt {a+b x+c x^2}+b c^3+2 c^4 x\right ) \left (-4 a B c e^3-4 A b c e^3+8 A c^2 d e^2+5 b^2 B e^3-12 b B c d e^2+8 B c^2 d^2 e\right )}{8 c^{7/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
IntegrateAlgebraic[((A + B*x)*(d + e*x)^3)/(a + b*x + c*x^2)^(3/2),x]
[Out]
-1/4*(-8*A*b*c^3*d^3 + 16*a*B*c^3*d^3 - 24*a*b*B*c^2*d^2*e + 48*a*A*c^3*d^2*e + 36*a*b^2*B*c*d*e^2 - 24*a*A*b*
c^2*d*e^2 - 96*a^2*B*c^2*d*e^2 - 15*a*b^3*B*e^3 + 12*a*A*b^2*c*e^3 + 52*a^2*b*B*c*e^3 - 32*a^2*A*c^2*e^3 + 8*b
*B*c^3*d^3*x - 16*A*c^4*d^3*x - 24*b^2*B*c^2*d^2*e*x + 24*A*b*c^3*d^2*e*x + 48*a*B*c^3*d^2*e*x + 36*b^3*B*c*d*
e^2*x - 24*A*b^2*c^2*d*e^2*x - 120*a*b*B*c^2*d*e^2*x + 48*a*A*c^3*d*e^2*x - 15*b^4*B*e^3*x + 12*A*b^3*c*e^3*x
+ 62*a*b^2*B*c*e^3*x - 40*a*A*b*c^2*e^3*x - 24*a^2*B*c^2*e^3*x + 12*b^2*B*c^2*d*e^2*x^2 - 48*a*B*c^3*d*e^2*x^2
- 5*b^3*B*c*e^3*x^2 + 4*A*b^2*c^2*e^3*x^2 + 20*a*b*B*c^2*e^3*x^2 - 16*a*A*c^3*e^3*x^2 + 2*b^2*B*c^2*e^3*x^3 -
8*a*B*c^3*e^3*x^3)/(c^3*(-b^2 + 4*a*c)*Sqrt[a + b*x + c*x^2]) - (3*(8*B*c^2*d^2*e - 12*b*B*c*d*e^2 + 8*A*c^2*
d*e^2 + 5*b^2*B*e^3 - 4*A*b*c*e^3 - 4*a*B*c*e^3)*Log[b*c^3 + 2*c^4*x - 2*c^(7/2)*Sqrt[a + b*x + c*x^2]])/(8*c^
(7/2))
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fricas [B] time = 3.12, size = 1605, normalized size = 4.94
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(e*x+d)^3/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")
[Out]
[-1/16*(3*(8*(B*a*b^2*c^2 - 4*B*a^2*c^3)*d^2*e - 4*(3*B*a*b^3*c + 8*A*a^2*c^3 - 2*(6*B*a^2*b + A*a*b^2)*c^2)*d
*e^2 + (5*B*a*b^4 + 16*(B*a^3 + A*a^2*b)*c^2 - 4*(6*B*a^2*b^2 + A*a*b^3)*c)*e^3 + (8*(B*b^2*c^3 - 4*B*a*c^4)*d
^2*e - 4*(3*B*b^3*c^2 + 8*A*a*c^4 - 2*(6*B*a*b + A*b^2)*c^3)*d*e^2 + (5*B*b^4*c + 16*(B*a^2 + A*a*b)*c^3 - 4*(
6*B*a*b^2 + A*b^3)*c^2)*e^3)*x^2 + (8*(B*b^3*c^2 - 4*B*a*b*c^3)*d^2*e - 4*(3*B*b^4*c + 8*A*a*b*c^3 - 2*(6*B*a*
b^2 + A*b^3)*c^2)*d*e^2 + (5*B*b^5 + 16*(B*a^2*b + A*a*b^2)*c^2 - 4*(6*B*a*b^3 + A*b^4)*c)*e^3)*x)*sqrt(c)*log
(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(c) - 4*a*c) - 4*(8*(2*B*a - A*b)*c^4*d^
3 + 2*(B*b^2*c^3 - 4*B*a*c^4)*e^3*x^3 - 24*(B*a*b*c^3 - 2*A*a*c^4)*d^2*e + 12*(3*B*a*b^2*c^2 - 2*(4*B*a^2 + A*
a*b)*c^3)*d*e^2 - (15*B*a*b^3*c + 32*A*a^2*c^3 - 4*(13*B*a^2*b + 3*A*a*b^2)*c^2)*e^3 + (12*(B*b^2*c^3 - 4*B*a*
c^4)*d*e^2 - (5*B*b^3*c^2 + 16*A*a*c^4 - 4*(5*B*a*b + A*b^2)*c^3)*e^3)*x^2 + (8*(B*b*c^4 - 2*A*c^5)*d^3 - 24*(
B*b^2*c^3 - (2*B*a + A*b)*c^4)*d^2*e + 12*(3*B*b^3*c^2 + 4*A*a*c^4 - 2*(5*B*a*b + A*b^2)*c^3)*d*e^2 - (15*B*b^
4*c + 8*(3*B*a^2 + 5*A*a*b)*c^3 - 2*(31*B*a*b^2 + 6*A*b^3)*c^2)*e^3)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^4 - 4*
a^2*c^5 + (b^2*c^5 - 4*a*c^6)*x^2 + (b^3*c^4 - 4*a*b*c^5)*x), -1/8*(3*(8*(B*a*b^2*c^2 - 4*B*a^2*c^3)*d^2*e - 4
*(3*B*a*b^3*c + 8*A*a^2*c^3 - 2*(6*B*a^2*b + A*a*b^2)*c^2)*d*e^2 + (5*B*a*b^4 + 16*(B*a^3 + A*a^2*b)*c^2 - 4*(
6*B*a^2*b^2 + A*a*b^3)*c)*e^3 + (8*(B*b^2*c^3 - 4*B*a*c^4)*d^2*e - 4*(3*B*b^3*c^2 + 8*A*a*c^4 - 2*(6*B*a*b + A
*b^2)*c^3)*d*e^2 + (5*B*b^4*c + 16*(B*a^2 + A*a*b)*c^3 - 4*(6*B*a*b^2 + A*b^3)*c^2)*e^3)*x^2 + (8*(B*b^3*c^2 -
4*B*a*b*c^3)*d^2*e - 4*(3*B*b^4*c + 8*A*a*b*c^3 - 2*(6*B*a*b^2 + A*b^3)*c^2)*d*e^2 + (5*B*b^5 + 16*(B*a^2*b +
A*a*b^2)*c^2 - 4*(6*B*a*b^3 + A*b^4)*c)*e^3)*x)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c
)/(c^2*x^2 + b*c*x + a*c)) - 2*(8*(2*B*a - A*b)*c^4*d^3 + 2*(B*b^2*c^3 - 4*B*a*c^4)*e^3*x^3 - 24*(B*a*b*c^3 -
2*A*a*c^4)*d^2*e + 12*(3*B*a*b^2*c^2 - 2*(4*B*a^2 + A*a*b)*c^3)*d*e^2 - (15*B*a*b^3*c + 32*A*a^2*c^3 - 4*(13*B
*a^2*b + 3*A*a*b^2)*c^2)*e^3 + (12*(B*b^2*c^3 - 4*B*a*c^4)*d*e^2 - (5*B*b^3*c^2 + 16*A*a*c^4 - 4*(5*B*a*b + A*
b^2)*c^3)*e^3)*x^2 + (8*(B*b*c^4 - 2*A*c^5)*d^3 - 24*(B*b^2*c^3 - (2*B*a + A*b)*c^4)*d^2*e + 12*(3*B*b^3*c^2 +
4*A*a*c^4 - 2*(5*B*a*b + A*b^2)*c^3)*d*e^2 - (15*B*b^4*c + 8*(3*B*a^2 + 5*A*a*b)*c^3 - 2*(31*B*a*b^2 + 6*A*b^
3)*c^2)*e^3)*x)*sqrt(c*x^2 + b*x + a))/(a*b^2*c^4 - 4*a^2*c^5 + (b^2*c^5 - 4*a*c^6)*x^2 + (b^3*c^4 - 4*a*b*c^5
)*x)]
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giac [A] time = 0.33, size = 530, normalized size = 1.63 \begin {gather*} \frac {{\left ({\left (\frac {2 \, {\left (B b^{2} c^{2} e^{3} - 4 \, B a c^{3} e^{3}\right )} x}{b^{2} c^{3} - 4 \, a c^{4}} + \frac {12 \, B b^{2} c^{2} d e^{2} - 48 \, B a c^{3} d e^{2} - 5 \, B b^{3} c e^{3} + 20 \, B a b c^{2} e^{3} + 4 \, A b^{2} c^{2} e^{3} - 16 \, A a c^{3} e^{3}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x + \frac {8 \, B b c^{3} d^{3} - 16 \, A c^{4} d^{3} - 24 \, B b^{2} c^{2} d^{2} e + 48 \, B a c^{3} d^{2} e + 24 \, A b c^{3} d^{2} e + 36 \, B b^{3} c d e^{2} - 120 \, B a b c^{2} d e^{2} - 24 \, A b^{2} c^{2} d e^{2} + 48 \, A a c^{3} d e^{2} - 15 \, B b^{4} e^{3} + 62 \, B a b^{2} c e^{3} + 12 \, A b^{3} c e^{3} - 24 \, B a^{2} c^{2} e^{3} - 40 \, A a b c^{2} e^{3}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x + \frac {16 \, B a c^{3} d^{3} - 8 \, A b c^{3} d^{3} - 24 \, B a b c^{2} d^{2} e + 48 \, A a c^{3} d^{2} e + 36 \, B a b^{2} c d e^{2} - 96 \, B a^{2} c^{2} d e^{2} - 24 \, A a b c^{2} d e^{2} - 15 \, B a b^{3} e^{3} + 52 \, B a^{2} b c e^{3} + 12 \, A a b^{2} c e^{3} - 32 \, A a^{2} c^{2} e^{3}}{b^{2} c^{3} - 4 \, a c^{4}}}{4 \, \sqrt {c x^{2} + b x + a}} - \frac {3 \, {\left (8 \, B c^{2} d^{2} e - 12 \, B b c d e^{2} + 8 \, A c^{2} d e^{2} + 5 \, B b^{2} e^{3} - 4 \, B a c e^{3} - 4 \, A b c e^{3}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{8 \, c^{\frac {7}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(e*x+d)^3/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")
[Out]
1/4*(((2*(B*b^2*c^2*e^3 - 4*B*a*c^3*e^3)*x/(b^2*c^3 - 4*a*c^4) + (12*B*b^2*c^2*d*e^2 - 48*B*a*c^3*d*e^2 - 5*B*
b^3*c*e^3 + 20*B*a*b*c^2*e^3 + 4*A*b^2*c^2*e^3 - 16*A*a*c^3*e^3)/(b^2*c^3 - 4*a*c^4))*x + (8*B*b*c^3*d^3 - 16*
A*c^4*d^3 - 24*B*b^2*c^2*d^2*e + 48*B*a*c^3*d^2*e + 24*A*b*c^3*d^2*e + 36*B*b^3*c*d*e^2 - 120*B*a*b*c^2*d*e^2
- 24*A*b^2*c^2*d*e^2 + 48*A*a*c^3*d*e^2 - 15*B*b^4*e^3 + 62*B*a*b^2*c*e^3 + 12*A*b^3*c*e^3 - 24*B*a^2*c^2*e^3
- 40*A*a*b*c^2*e^3)/(b^2*c^3 - 4*a*c^4))*x + (16*B*a*c^3*d^3 - 8*A*b*c^3*d^3 - 24*B*a*b*c^2*d^2*e + 48*A*a*c^3
*d^2*e + 36*B*a*b^2*c*d*e^2 - 96*B*a^2*c^2*d*e^2 - 24*A*a*b*c^2*d*e^2 - 15*B*a*b^3*e^3 + 52*B*a^2*b*c*e^3 + 12
*A*a*b^2*c*e^3 - 32*A*a^2*c^2*e^3)/(b^2*c^3 - 4*a*c^4))/sqrt(c*x^2 + b*x + a) - 3/8*(8*B*c^2*d^2*e - 12*B*b*c*
d*e^2 + 8*A*c^2*d*e^2 + 5*B*b^2*e^3 - 4*B*a*c*e^3 - 4*A*b*c*e^3)*log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a)
)*sqrt(c) - b))/c^(7/2)
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maple [B] time = 0.07, size = 1451, normalized size = 4.46
result too large to display
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((B*x+A)*(e*x+d)^3/(c*x^2+b*x+a)^(3/2),x)
[Out]
-2*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*B*d^3-b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*B*d^3-9/2*b/c^(5/2)*ln((c*x
+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*B*d*e^2+6*a/c^2/(c*x^2+b*x+a)^(1/2)*B*d*e^2+3*x^2/c/(c*x^2+b*x+a)^(1/2)*B
*d*e^2-9/4*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*B*d*e^2+2*a/c^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*A*e^3-3
/2*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*A*e^3-3*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*A*d^2*e+3/2*b^3/c^2
/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*A*d*e^2+3/2*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*B*d^2*e-6*b/(4*a*c-b^2)/(
c*x^2+b*x+a)^(1/2)*x*A*d^2*e-9/2*b^3/c^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*B*d*e^2+4*a/c*b/(4*a*c-b^2)/(c*x^2+
b*x+a)^(1/2)*x*A*e^3+3*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*A*d*e^2+3*b^2/c/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)
*x*B*d^2*e+12*a/c*b/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*x*B*d*e^2+15/8*B*e^3*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/
2)*x+15/16*B*e^3*b^5/c^4/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-5/4*B*e^3*b/c^2*x^2/(c*x^2+b*x+a)^(1/2)-3*x/c/(c*x^2+
b*x+a)^(1/2)*A*d*e^2+3/2*b/c^2/(c*x^2+b*x+a)^(1/2)*B*d^2*e-1/c/(c*x^2+b*x+a)^(1/2)*B*d^3+1/2*B*e^3*x^3/c/(c*x^
2+b*x+a)^(1/2)+15/16*B*e^3*b^3/c^4/(c*x^2+b*x+a)^(1/2)+15/8*B*e^3*b^2/c^(7/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*
x+a)^(1/2))-3/2*B*e^3*a/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))+3/c^(3/2)*ln((c*x+1/2*b)/c^(1/2)+(
c*x^2+b*x+a)^(1/2))*A*d*e^2+2*a/c^2/(c*x^2+b*x+a)^(1/2)*A*e^3-3/2*b/c^(5/2)*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+
a)^(1/2))*A*e^3-3*x/c/(c*x^2+b*x+a)^(1/2)*B*d^2*e+3/2*b/c^2/(c*x^2+b*x+a)^(1/2)*A*d*e^2+3/c^(3/2)*ln((c*x+1/2*
b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))*B*d^2*e-3/c/(c*x^2+b*x+a)^(1/2)*A*d^2*e-13/2*B*e^3*b^2/c^2*a/(4*a*c-b^2)/(c*x^
2+b*x+a)^(1/2)*x+6*a/c^2*b^2/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*B*d*e^2-13/4*B*e^3*b^3/c^3*a/(4*a*c-b^2)/(c*x^2+b
*x+a)^(1/2)+9/2*b/c^2*x/(c*x^2+b*x+a)^(1/2)*B*d*e^2+2*A*d^3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+x^2/c/(c
*x^2+b*x+a)^(1/2)*A*e^3-3/4*b^2/c^3/(c*x^2+b*x+a)^(1/2)*A*e^3-3/4*b^4/c^3/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)*A*e^
3-13/4*B*e^3*b/c^3*a/(c*x^2+b*x+a)^(1/2)+3/2*B*e^3*a/c^2*x/(c*x^2+b*x+a)^(1/2)-15/8*B*e^3*b^2/c^3*x/(c*x^2+b*x
+a)^(1/2)+3/2*b/c^2*x/(c*x^2+b*x+a)^(1/2)*A*e^3-9/4*b^2/c^3/(c*x^2+b*x+a)^(1/2)*B*d*e^2
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(e*x+d)^3/(c*x^2+b*x+a)^(3/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?
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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 (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B*x)*(d + e*x)^3)/(a + b*x + c*x^2)^(3/2),x)
[Out]
int(((A + B*x)*(d + e*x)^3)/(a + b*x + c*x^2)^(3/2), 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 (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((B*x+A)*(e*x+d)**3/(c*x**2+b*x+a)**(3/2),x)
[Out]
Integral((A + B*x)*(d + e*x)**3/(a + b*x + c*x**2)**(3/2), x)
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