\(\int \frac {(d+e x)^4}{(a+b x+c x^2)^{3/2}} \, dx\) [2382]

Optimal result

Integrand size = 22, antiderivative size = 286 \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx=-\frac {2 (d+e x)^3 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {2 e (2 c d-b e) (d+e x)^2 \sqrt {a+b x+c x^2}}{c \left (b^2-4 a c\right )}+\frac {e \left (32 c^3 d^3-15 b^3 e^3+4 b c e^2 (12 b d+13 a e)-8 c^2 d e (5 b d+16 a e)+2 c e \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac {3 e^2 \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right ) \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{8 c^{7/2}} \]

[Out]

3/8*e^2*(16*c^2*d^2+5*b^2*e^2-4*c*e*(a*e+4*b*d))*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(7/2)-2*
(e*x+d)^3*(b*d-2*a*e+(-b*e+2*c*d)*x)/(-4*a*c+b^2)/(c*x^2+b*x+a)^(1/2)+2*e*(-b*e+2*c*d)*(e*x+d)^2*(c*x^2+b*x+a)
^(1/2)/c/(-4*a*c+b^2)+1/4*e*(32*c^3*d^3-15*b^3*e^3+4*b*c*e^2*(13*a*e+12*b*d)-8*c^2*d*e*(16*a*e+5*b*d)+2*c*e*(8
*c^2*d^2+5*b^2*e^2-4*c*e*(3*a*e+2*b*d))*x)*(c*x^2+b*x+a)^(1/2)/c^3/(-4*a*c+b^2)

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 286, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {752, 846, 793, 635, 212} \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {3 e^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right ) \left (-4 c e (a e+4 b d)+5 b^2 e^2+16 c^2 d^2\right )}{8 c^{7/2}}-\frac {2 (d+e x)^3 (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {2 e (d+e x)^2 \sqrt {a+b x+c x^2} (2 c d-b e)}{c \left (b^2-4 a c\right )}+\frac {e \sqrt {a+b x+c x^2} \left (2 c e x \left (-4 c e (3 a e+2 b d)+5 b^2 e^2+8 c^2 d^2\right )-8 c^2 d e (16 a e+5 b d)+4 b c e^2 (13 a e+12 b d)-15 b^3 e^3+32 c^3 d^3\right )}{4 c^3 \left (b^2-4 a c\right )} \]

[In]

Int[(d + e*x)^4/(a + b*x + c*x^2)^(3/2),x]

[Out]

(-2*(d + e*x)^3*(b*d - 2*a*e + (2*c*d - b*e)*x))/((b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (2*e*(2*c*d - b*e)*(d
+ e*x)^2*Sqrt[a + b*x + c*x^2])/(c*(b^2 - 4*a*c)) + (e*(32*c^3*d^3 - 15*b^3*e^3 + 4*b*c*e^2*(12*b*d + 13*a*e)
- 8*c^2*d*e*(5*b*d + 16*a*e) + 2*c*e*(8*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(2*b*d + 3*a*e))*x)*Sqrt[a + b*x + c*x^2]
)/(4*c^3*(b^2 - 4*a*c)) + (3*e^2*(16*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(4*b*d + a*e))*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]
*Sqrt[a + b*x + c*x^2])])/(8*c^(7/2))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 635

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 752

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(d
*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Dist[1/((p + 1)*(b^2 -
4*a*c)), Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2*c*d^2*(2*p + 3) + e*(b*e - 2*d*
c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] &
& NeQ[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, b, c, d,
e, m, p, x]

Rule 793

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 846

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)/(c*(m + 2*p + 2))), x] + Dist[1/(c*(m + 2*p + 2)), Int[(d + e*x)^(m
- 1)*(a + b*x + c*x^2)^p*Simp[m*(c*d*f - a*e*g) + d*(2*c*f - b*g)*(p + 1) + (m*(c*e*f + c*d*g - b*e*g) + e*(p
+ 1)*(2*c*f - b*g))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && GtQ[m, 0] && NeQ[m + 2*p + 2, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
&&  !(IGtQ[m, 0] && EqQ[f, 0])

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (d+e x)^3 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}-\frac {2 \int \frac {(d+e x)^2 (-3 e (b d-2 a e)-3 e (2 c d-b e) x)}{\sqrt {a+b x+c x^2}} \, dx}{b^2-4 a c} \\ & = -\frac {2 (d+e x)^3 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {2 e (2 c d-b e) (d+e x)^2 \sqrt {a+b x+c x^2}}{c \left (b^2-4 a c\right )}-\frac {2 \int \frac {(d+e x) \left (-\frac {3}{2} e \left (b^2 d e-20 a c d e+4 b \left (c d^2+a e^2\right )\right )-\frac {3}{2} e \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{3 c \left (b^2-4 a c\right )} \\ & = -\frac {2 (d+e x)^3 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {2 e (2 c d-b e) (d+e x)^2 \sqrt {a+b x+c x^2}}{c \left (b^2-4 a c\right )}+\frac {e \left (32 c^3 d^3-15 b^3 e^3+4 b c e^2 (12 b d+13 a e)-8 c^2 d e (5 b d+16 a e)+2 c e \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a 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^2 \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{8 c^3} \\ & = -\frac {2 (d+e x)^3 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {2 e (2 c d-b e) (d+e x)^2 \sqrt {a+b x+c x^2}}{c \left (b^2-4 a c\right )}+\frac {e \left (32 c^3 d^3-15 b^3 e^3+4 b c e^2 (12 b d+13 a e)-8 c^2 d e (5 b d+16 a e)+2 c e \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a 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^2 \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right )\right ) \text {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)^3 (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}+\frac {2 e (2 c d-b e) (d+e x)^2 \sqrt {a+b x+c x^2}}{c \left (b^2-4 a c\right )}+\frac {e \left (32 c^3 d^3-15 b^3 e^3+4 b c e^2 (12 b d+13 a e)-8 c^2 d e (5 b d+16 a e)+2 c e \left (8 c^2 d^2+5 b^2 e^2-4 c e (2 b d+3 a e)\right ) x\right ) \sqrt {a+b x+c x^2}}{4 c^3 \left (b^2-4 a c\right )}+\frac {3 e^2 \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\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*}

Mathematica [A] (verified)

Time = 1.51 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.11 \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c} \left (15 b^4 e^4 x+b^3 e^3 (15 a e+c x (-48 d+5 e x))+4 b c \left (-13 a^2 e^4+2 c^2 d^3 (d-4 e x)+a c e^2 \left (12 d^2+40 d e x-5 e^2 x^2\right )\right )-2 b^2 c e^2 \left (a e (24 d+31 e x)+c x \left (-24 d^2+8 d e x+e^2 x^2\right )\right )+8 c^2 \left (2 c^2 d^4 x+a^2 e^3 (16 d+3 e x)+a c e \left (-8 d^3-12 d^2 e x+8 d e^2 x^2+e^3 x^3\right )\right )\right )-3 \left (b^2-4 a c\right ) e^2 \left (16 c^2 d^2+5 b^2 e^2-4 c e (4 b d+a e)\right ) \sqrt {a+x (b+c x)} \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{4 c^{7/2} \left (-b^2+4 a c\right ) \sqrt {a+x (b+c x)}} \]

[In]

Integrate[(d + e*x)^4/(a + b*x + c*x^2)^(3/2),x]

[Out]

(Sqrt[c]*(15*b^4*e^4*x + b^3*e^3*(15*a*e + c*x*(-48*d + 5*e*x)) + 4*b*c*(-13*a^2*e^4 + 2*c^2*d^3*(d - 4*e*x) +
a*c*e^2*(12*d^2 + 40*d*e*x - 5*e^2*x^2)) - 2*b^2*c*e^2*(a*e*(24*d + 31*e*x) + c*x*(-24*d^2 + 8*d*e*x + e^2*x^
2)) + 8*c^2*(2*c^2*d^4*x + a^2*e^3*(16*d + 3*e*x) + a*c*e*(-8*d^3 - 12*d^2*e*x + 8*d*e^2*x^2 + e^3*x^3))) - 3*
(b^2 - 4*a*c)*e^2*(16*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(4*b*d + a*e))*Sqrt[a + x*(b + c*x)]*ArcTanh[(Sqrt[c]*x)/(-S
qrt[a] + Sqrt[a + x*(b + c*x)])])/(4*c^(7/2)*(-b^2 + 4*a*c)*Sqrt[a + x*(b + c*x)])

Maple [A] (verified)

Time = 0.50 (sec) , antiderivative size = 447, normalized size of antiderivative = 1.56

method	result	size
risch	\(-\frac {e^{3} \left (-2 c e x +7 b e -16 c d \right ) \sqrt {c \,x^{2}+b x +a}}{4 c^{3}}-\frac {-\frac {16 c^{3} d^{4} \left (2 c x +b \right )}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {8 a^{2} c \,e^{4} \left (2 c x +b \right )}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}-\frac {14 a \,b^{2} e^{4} \left (2 c x +b \right )}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\frac {32 a b c d \,e^{3} \left (2 c x +b \right )}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+\left (12 a \,c^{2} e^{4}-15 b^{2} c \,e^{4}+48 b \,c^{2} d \,e^{3}-48 c^{3} d^{2} e^{2}\right ) \left (-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}\right )+\left (-4 a b c \,e^{4}+32 a \,c^{2} d \,e^{3}-7 b^{3} e^{4}+16 b^{2} c d \,e^{3}-32 c^{3} d^{3} e \right ) \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{8 c^{3}}\)	\(447\)
default	\(\frac {2 d^{4} \left (2 c x +b \right )}{\left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}+e^{4} \left (\frac {x^{3}}{2 c \sqrt {c \,x^{2}+b x +a}}-\frac {5 b \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x +a}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}\right )}{2 c}-\frac {2 a \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{c}\right )}{4 c}-\frac {3 a \left (-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )+4 d \,e^{3} \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x +a}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}\right )}{2 c}-\frac {2 a \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{c}\right )+6 d^{2} e^{2} \left (-\frac {x}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{c^{\frac {3}{2}}}\right )+4 d^{3} e \left (-\frac {1}{c \sqrt {c \,x^{2}+b x +a}}-\frac {b \left (2 c x +b \right )}{c \left (4 a c -b^{2}\right ) \sqrt {c \,x^{2}+b x +a}}\right )\)	\(752\)

[In]

int((e*x+d)^4/(c*x^2+b*x+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*e^3*(-2*c*e*x+7*b*e-16*c*d)*(c*x^2+b*x+a)^(1/2)/c^3-1/8/c^3*(-16*c^3*d^4*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x
+a)^(1/2)+8*a^2*c*e^4*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)-14*a*b^2*e^4*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+
a)^(1/2)+32*a*b*c*d*e^3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)+(12*a*c^2*e^4-15*b^2*c*e^4+48*b*c^2*d*e^3-48
*c^3*d^2*e^2)*(-x/c/(c*x^2+b*x+a)^(1/2)-1/2*b/c*(-1/c/(c*x^2+b*x+a)^(1/2)-b/c*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x
+a)^(1/2))+1/c^(3/2)*ln((1/2*b+c*x)/c^(1/2)+(c*x^2+b*x+a)^(1/2)))+(-4*a*b*c*e^4+32*a*c^2*d*e^3-7*b^3*e^4+16*b^
2*c*d*e^3-32*c^3*d^3*e)*(-1/c/(c*x^2+b*x+a)^(1/2)-b/c*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(1/2)))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 585 vs. \(2 (268) = 536\).

Time = 0.48 (sec) , antiderivative size = 1173, normalized size of antiderivative = 4.10 \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate((e*x+d)^4/(c*x^2+b*x+a)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(3*(16*(a*b^2*c^2 - 4*a^2*c^3)*d^2*e^2 - 16*(a*b^3*c - 4*a^2*b*c^2)*d*e^3 + (5*a*b^4 - 24*a^2*b^2*c + 1
6*a^3*c^2)*e^4 + (16*(b^2*c^3 - 4*a*c^4)*d^2*e^2 - 16*(b^3*c^2 - 4*a*b*c^3)*d*e^3 + (5*b^4*c - 24*a*b^2*c^2 +
16*a^2*c^3)*e^4)*x^2 + (16*(b^3*c^2 - 4*a*b*c^3)*d^2*e^2 - 16*(b^4*c - 4*a*b^2*c^2)*d*e^3 + (5*b^5 - 24*a*b^3*
c + 16*a^2*b*c^2)*e^4)*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*b*c^4*d^4 - 64*a*c^4*d^3*e + 48*a*b*c^3*d^2*e^2 - 2*(b^2*c^3 - 4*a*c^4)*e^4*x^3 - 16*(3*a*b^2
*c^2 - 8*a^2*c^3)*d*e^3 + (15*a*b^3*c - 52*a^2*b*c^2)*e^4 - (16*(b^2*c^3 - 4*a*c^4)*d*e^3 - 5*(b^3*c^2 - 4*a*b
*c^3)*e^4)*x^2 + (16*c^5*d^4 - 32*b*c^4*d^3*e + 48*(b^2*c^3 - 2*a*c^4)*d^2*e^2 - 16*(3*b^3*c^2 - 10*a*b*c^3)*d
*e^3 + (15*b^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*e^4)*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*(16*(a*b^2*c^2 - 4*a^2*c^3)*d^2*e^2 - 16*(a*b^3*c - 4*a^2*
b*c^2)*d*e^3 + (5*a*b^4 - 24*a^2*b^2*c + 16*a^3*c^2)*e^4 + (16*(b^2*c^3 - 4*a*c^4)*d^2*e^2 - 16*(b^3*c^2 - 4*a
*b*c^3)*d*e^3 + (5*b^4*c - 24*a*b^2*c^2 + 16*a^2*c^3)*e^4)*x^2 + (16*(b^3*c^2 - 4*a*b*c^3)*d^2*e^2 - 16*(b^4*c
- 4*a*b^2*c^2)*d*e^3 + (5*b^5 - 24*a*b^3*c + 16*a^2*b*c^2)*e^4)*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*b*c^4*d^4 - 64*a*c^4*d^3*e + 48*a*b*c^3*d^2*e^2 - 2*(b^2*
c^3 - 4*a*c^4)*e^4*x^3 - 16*(3*a*b^2*c^2 - 8*a^2*c^3)*d*e^3 + (15*a*b^3*c - 52*a^2*b*c^2)*e^4 - (16*(b^2*c^3 -
4*a*c^4)*d*e^3 - 5*(b^3*c^2 - 4*a*b*c^3)*e^4)*x^2 + (16*c^5*d^4 - 32*b*c^4*d^3*e + 48*(b^2*c^3 - 2*a*c^4)*d^2
*e^2 - 16*(3*b^3*c^2 - 10*a*b*c^3)*d*e^3 + (15*b^4*c - 62*a*b^2*c^2 + 24*a^2*c^3)*e^4)*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)]

Sympy [F]

\[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{4}}{\left (a + b x + c x^{2}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate((e*x+d)**4/(c*x**2+b*x+a)**(3/2),x)

[Out]

Integral((d + e*x)**4/(a + b*x + c*x**2)**(3/2), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((e*x+d)^4/(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 deta

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.38 \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (\frac {2 \, {\left (b^{2} c^{2} e^{4} - 4 \, a c^{3} e^{4}\right )} x}{b^{2} c^{3} - 4 \, a c^{4}} + \frac {16 \, b^{2} c^{2} d e^{3} - 64 \, a c^{3} d e^{3} - 5 \, b^{3} c e^{4} + 20 \, a b c^{2} e^{4}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac {16 \, c^{4} d^{4} - 32 \, b c^{3} d^{3} e + 48 \, b^{2} c^{2} d^{2} e^{2} - 96 \, a c^{3} d^{2} e^{2} - 48 \, b^{3} c d e^{3} + 160 \, a b c^{2} d e^{3} + 15 \, b^{4} e^{4} - 62 \, a b^{2} c e^{4} + 24 \, a^{2} c^{2} e^{4}}{b^{2} c^{3} - 4 \, a c^{4}}\right )} x - \frac {8 \, b c^{3} d^{4} - 64 \, a c^{3} d^{3} e + 48 \, a b c^{2} d^{2} e^{2} - 48 \, a b^{2} c d e^{3} + 128 \, a^{2} c^{2} d e^{3} + 15 \, a b^{3} e^{4} - 52 \, a^{2} b c e^{4}}{b^{2} c^{3} - 4 \, a c^{4}}}{4 \, \sqrt {c x^{2} + b x + a}} - \frac {3 \, {\left (16 \, c^{2} d^{2} e^{2} - 16 \, b c d e^{3} + 5 \, b^{2} e^{4} - 4 \, a c e^{4}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {7}{2}}} \]

[In]

integrate((e*x+d)^4/(c*x^2+b*x+a)^(3/2),x, algorithm="giac")

[Out]

1/4*(((2*(b^2*c^2*e^4 - 4*a*c^3*e^4)*x/(b^2*c^3 - 4*a*c^4) + (16*b^2*c^2*d*e^3 - 64*a*c^3*d*e^3 - 5*b^3*c*e^4
+ 20*a*b*c^2*e^4)/(b^2*c^3 - 4*a*c^4))*x - (16*c^4*d^4 - 32*b*c^3*d^3*e + 48*b^2*c^2*d^2*e^2 - 96*a*c^3*d^2*e^
2 - 48*b^3*c*d*e^3 + 160*a*b*c^2*d*e^3 + 15*b^4*e^4 - 62*a*b^2*c*e^4 + 24*a^2*c^2*e^4)/(b^2*c^3 - 4*a*c^4))*x
- (8*b*c^3*d^4 - 64*a*c^3*d^3*e + 48*a*b*c^2*d^2*e^2 - 48*a*b^2*c*d*e^3 + 128*a^2*c^2*d*e^3 + 15*a*b^3*e^4 - 5
2*a^2*b*c*e^4)/(b^2*c^3 - 4*a*c^4))/sqrt(c*x^2 + b*x + a) - 3/8*(16*c^2*d^2*e^2 - 16*b*c*d*e^3 + 5*b^2*e^4 - 4
*a*c*e^4)*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(7/2)

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^4}{{\left (c\,x^2+b\,x+a\right )}^{3/2}} \,d x \]

[In]

int((d + e*x)^4/(a + b*x + c*x^2)^(3/2),x)

[Out]

int((d + e*x)^4/(a + b*x + c*x^2)^(3/2), x)