Integrand size = 22, antiderivative size = 295 \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {2 (d+e x)^3 (b d-2 a e+(2 c d-b e) x)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}+\frac {4 (d+e x) \left (4 b c d \left (c d^2+3 a e^2\right )-4 a c e \left (c d^2+3 a e^2\right )-b^2 \left (5 c d^2 e-a e^3\right )+(2 c d-b e) \left (4 c^2 d^2-b^2 e^2-4 c e (b d-2 a e)\right ) x\right )}{3 c \left (b^2-4 a c\right )^2 \sqrt {a+b x+c x^2}}-\frac {2 e (2 c d-b e) \left (8 c^2 d^2-3 b^2 e^2-4 c e (2 b d-5 a e)\right ) \sqrt {a+b x+c x^2}}{3 c^2 \left (b^2-4 a c\right )^2}+\frac {e^4 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{c^{5/2}} \]
-2/3*(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)^(3/2) +e^4*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(5/2)+4/3*(e*x+d )*(4*b*c*d*(3*a*e^2+c*d^2)-4*a*c*e*(3*a*e^2+c*d^2)-b^2*(-a*e^3+5*c*d^2*e)+ (-b*e+2*c*d)*(4*c^2*d^2-b^2*e^2-4*c*e*(-2*a*e+b*d))*x)/c/(-4*a*c+b^2)^2/(c *x^2+b*x+a)^(1/2)-2/3*e*(-b*e+2*c*d)*(8*c^2*d^2-3*b^2*e^2-4*c*e*(-5*a*e+2* b*d))*(c*x^2+b*x+a)^(1/2)/c^2/(-4*a*c+b^2)^2
Time = 2.13 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.36 \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\frac {2 \left (-3 b^5 e^4 x^2-2 b^4 e^4 x \left (3 a+2 c x^2\right )+b^3 \left (-3 a^2 e^4+18 a c e^4 x^2-c^2 d \left (d^3+12 d^2 e x-18 d e^2 x^2-4 e^3 x^3\right )\right )+4 b c \left (5 a^3 e^4+2 c^3 d^3 x^2 (3 d-4 e x)+12 a^2 c d e^2 (d-2 e x)+3 a c^2 d \left (d^3-4 d^2 e x+6 d e^2 x^2-4 e^3 x^3\right )\right )-8 c^2 \left (-2 c^3 d^4 x^3+a^3 e^3 (8 d+3 e x)-3 a c^2 d^2 x \left (d^2+2 e^2 x^2\right )+4 a^2 c e \left (d^3+3 d e^2 x^2+e^3 x^3\right )\right )+2 b^2 c \left (21 a^2 e^4 x+3 c^2 d^2 x \left (d^2-8 d e x+2 e^2 x^2\right )+2 a c e \left (-2 d^3+18 d^2 e x-6 d e^2 x^2+7 e^3 x^3\right )\right )\right )}{3 c^2 \left (b^2-4 a c\right )^2 (a+x (b+c x))^{3/2}}+\frac {2 e^4 \text {arctanh}\left (\frac {\sqrt {c} x}{-\sqrt {a}+\sqrt {a+x (b+c x)}}\right )}{c^{5/2}} \]
(2*(-3*b^5*e^4*x^2 - 2*b^4*e^4*x*(3*a + 2*c*x^2) + b^3*(-3*a^2*e^4 + 18*a* c*e^4*x^2 - c^2*d*(d^3 + 12*d^2*e*x - 18*d*e^2*x^2 - 4*e^3*x^3)) + 4*b*c*( 5*a^3*e^4 + 2*c^3*d^3*x^2*(3*d - 4*e*x) + 12*a^2*c*d*e^2*(d - 2*e*x) + 3*a *c^2*d*(d^3 - 4*d^2*e*x + 6*d*e^2*x^2 - 4*e^3*x^3)) - 8*c^2*(-2*c^3*d^4*x^ 3 + a^3*e^3*(8*d + 3*e*x) - 3*a*c^2*d^2*x*(d^2 + 2*e^2*x^2) + 4*a^2*c*e*(d ^3 + 3*d*e^2*x^2 + e^3*x^3)) + 2*b^2*c*(21*a^2*e^4*x + 3*c^2*d^2*x*(d^2 - 8*d*e*x + 2*e^2*x^2) + 2*a*c*e*(-2*d^3 + 18*d^2*e*x - 6*d*e^2*x^2 + 7*e^3* x^3))))/(3*c^2*(b^2 - 4*a*c)^2*(a + x*(b + c*x))^(3/2)) + (2*e^4*ArcTanh[( Sqrt[c]*x)/(-Sqrt[a] + Sqrt[a + x*(b + c*x)])])/c^(5/2)
Time = 0.62 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {1164, 1233, 27, 1160, 1092, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 1164 |
| \(\displaystyle -\frac {2 \int \frac {(d+e x)^2 \left (4 c d^2-e (5 b d-6 a e)-e (2 c d-b e) x\right )}{\left (c x^2+b x+a\right )^{3/2}}dx}{3 \left (b^2-4 a c\right )}-\frac {2 (d+e x)^3 (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1233 |
| \(\displaystyle -\frac {2 \left (\frac {2 \int \frac {e \left (d e^2 b^3-2 \left (6 c d^2 e-a e^3\right ) b^2+4 c d \left (2 c d^2+5 a e^2\right ) b-24 a^2 c e^3+(2 c d-b e) \left (8 c^2 d^2-3 b^2 e^2-4 c e (2 b d-5 a e)\right ) x\right )}{2 \sqrt {c x^2+b x+a}}dx}{c \left (b^2-4 a c\right )}-\frac {2 (d+e x) \left (x (2 c d-b e) \left (-4 c e (b d-2 a e)-b^2 e^2+4 c^2 d^2\right )-\left (b^2 \left (5 c d^2 e-a e^3\right )\right )+4 b c d \left (3 a e^2+c d^2\right )-4 a c e \left (3 a e^2+c d^2\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\right )}{3 \left (b^2-4 a c\right )}-\frac {2 (d+e x)^3 (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {2 \left (\frac {e \int \frac {d e^2 b^3-2 \left (6 c d^2 e-a e^3\right ) b^2+4 c d \left (2 c d^2+5 a e^2\right ) b-24 a^2 c e^3+(2 c d-b e) \left (8 c^2 d^2-3 b^2 e^2-4 c e (2 b d-5 a e)\right ) x}{\sqrt {c x^2+b x+a}}dx}{c \left (b^2-4 a c\right )}-\frac {2 (d+e x) \left (x (2 c d-b e) \left (-4 c e (b d-2 a e)-b^2 e^2+4 c^2 d^2\right )-\left (b^2 \left (5 c d^2 e-a e^3\right )\right )+4 b c d \left (3 a e^2+c d^2\right )-4 a c e \left (3 a e^2+c d^2\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\right )}{3 \left (b^2-4 a c\right )}-\frac {2 (d+e x)^3 (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle -\frac {2 \left (\frac {e \left (\frac {\sqrt {a+b x+c x^2} (2 c d-b e) \left (-4 c e (2 b d-5 a e)-3 b^2 e^2+8 c^2 d^2\right )}{c}-\frac {3 e^3 \left (b^2-4 a c\right )^2 \int \frac {1}{\sqrt {c x^2+b x+a}}dx}{2 c}\right )}{c \left (b^2-4 a c\right )}-\frac {2 (d+e x) \left (x (2 c d-b e) \left (-4 c e (b d-2 a e)-b^2 e^2+4 c^2 d^2\right )-\left (b^2 \left (5 c d^2 e-a e^3\right )\right )+4 b c d \left (3 a e^2+c d^2\right )-4 a c e \left (3 a e^2+c d^2\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\right )}{3 \left (b^2-4 a c\right )}-\frac {2 (d+e x)^3 (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle -\frac {2 \left (\frac {e \left (\frac {\sqrt {a+b x+c x^2} (2 c d-b e) \left (-4 c e (2 b d-5 a e)-3 b^2 e^2+8 c^2 d^2\right )}{c}-\frac {3 e^3 \left (b^2-4 a c\right )^2 \int \frac {1}{4 c-\frac {(b+2 c x)^2}{c x^2+b x+a}}d\frac {b+2 c x}{\sqrt {c x^2+b x+a}}}{c}\right )}{c \left (b^2-4 a c\right )}-\frac {2 (d+e x) \left (x (2 c d-b e) \left (-4 c e (b d-2 a e)-b^2 e^2+4 c^2 d^2\right )-\left (b^2 \left (5 c d^2 e-a e^3\right )\right )+4 b c d \left (3 a e^2+c d^2\right )-4 a c e \left (3 a e^2+c d^2\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\right )}{3 \left (b^2-4 a c\right )}-\frac {2 (d+e x)^3 (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle -\frac {2 \left (\frac {e \left (\frac {\sqrt {a+b x+c x^2} (2 c d-b e) \left (-4 c e (2 b d-5 a e)-3 b^2 e^2+8 c^2 d^2\right )}{c}-\frac {3 e^3 \left (b^2-4 a c\right )^2 \text {arctanh}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{2 c^{3/2}}\right )}{c \left (b^2-4 a c\right )}-\frac {2 (d+e x) \left (x (2 c d-b e) \left (-4 c e (b d-2 a e)-b^2 e^2+4 c^2 d^2\right )-\left (b^2 \left (5 c d^2 e-a e^3\right )\right )+4 b c d \left (3 a e^2+c d^2\right )-4 a c e \left (3 a e^2+c d^2\right )\right )}{c \left (b^2-4 a c\right ) \sqrt {a+b x+c x^2}}\right )}{3 \left (b^2-4 a c\right )}-\frac {2 (d+e x)^3 (-2 a e+x (2 c d-b e)+b d)}{3 \left (b^2-4 a c\right ) \left (a+b x+c x^2\right )^{3/2}}\) |
(-2*(d + e*x)^3*(b*d - 2*a*e + (2*c*d - b*e)*x))/(3*(b^2 - 4*a*c)*(a + b*x + c*x^2)^(3/2)) - (2*((-2*(d + e*x)*(4*b*c*d*(c*d^2 + 3*a*e^2) - 4*a*c*e* (c*d^2 + 3*a*e^2) - b^2*(5*c*d^2*e - a*e^3) + (2*c*d - b*e)*(4*c^2*d^2 - b ^2*e^2 - 4*c*e*(b*d - 2*a*e))*x))/(c*(b^2 - 4*a*c)*Sqrt[a + b*x + c*x^2]) + (e*(((2*c*d - b*e)*(8*c^2*d^2 - 3*b^2*e^2 - 4*c*e*(2*b*d - 5*a*e))*Sqrt[ a + b*x + c*x^2])/c - (3*(b^2 - 4*a*c)^2*e^3*ArcTanh[(b + 2*c*x)/(2*Sqrt[c ]*Sqrt[a + b*x + c*x^2])])/(2*c^(3/2))))/(c*(b^2 - 4*a*c))))/(3*(b^2 - 4*a *c))
3.24.92.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> 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] + Simp[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] && LtQ[p, -1] && GtQ[m, 1] && Int QuadraticQ[a, b, c, d, e, m, p, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(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] - Simp[1/(c*( p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim p[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] && 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])
Leaf count of result is larger than twice the leaf count of optimal. \(1131\) vs. \(2(275)=550\).
Time = 0.45 (sec) , antiderivative size = 1132, normalized size of antiderivative = 3.84
d^4*(2/3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2 *c*x+b)/(c*x^2+b*x+a)^(1/2))+e^4*(-1/3*x^3/c/(c*x^2+b*x+a)^(3/2)-1/2*b/c*( -x^2/c/(c*x^2+b*x+a)^(3/2)+1/2*b/c*(-1/2*x/c/(c*x^2+b*x+a)^(3/2)-1/4*b/c*( -1/3/c/(c*x^2+b*x+a)^(3/2)-1/2*b/c*(2/3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a )^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/(c*x^2+b*x+a)^(1/2)))+1/2*a/c*(2/3* (2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/( c*x^2+b*x+a)^(1/2)))+2*a/c*(-1/3/c/(c*x^2+b*x+a)^(3/2)-1/2*b/c*(2/3*(2*c*x +b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/(c*x^2+ b*x+a)^(1/2))))+1/c*(-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*d*e^3*(-x^2/c/(c*x^2+b*x+a)^(3/2)+1/2 *b/c*(-1/2*x/c/(c*x^2+b*x+a)^(3/2)-1/4*b/c*(-1/3/c/(c*x^2+b*x+a)^(3/2)-1/2 *b/c*(2/3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*( 2*c*x+b)/(c*x^2+b*x+a)^(1/2)))+1/2*a/c*(2/3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b *x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/(c*x^2+b*x+a)^(1/2)))+2*a/c*(-1 /3/c/(c*x^2+b*x+a)^(3/2)-1/2*b/c*(2/3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^ (3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/(c*x^2+b*x+a)^(1/2))))+6*d^2*e^2*(-1/ 2*x/c/(c*x^2+b*x+a)^(3/2)-1/4*b/c*(-1/3/c/(c*x^2+b*x+a)^(3/2)-1/2*b/c*(2/3 *(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^(3/2)+16/3*c/(4*a*c-b^2)^2*(2*c*x+b)/ (c*x^2+b*x+a)^(1/2)))+1/2*a/c*(2/3*(2*c*x+b)/(4*a*c-b^2)/(c*x^2+b*x+a)^...
Leaf count of result is larger than twice the leaf count of optimal. 756 vs. \(2 (275) = 550\).
Time = 1.45 (sec) , antiderivative size = 1515, normalized size of antiderivative = 5.14 \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Too large to display} \]
[1/6*(3*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*e^4*x^4 + 2*(b^5*c - 8*a*b^3 *c^2 + 16*a^2*b*c^3)*e^4*x^3 + (b^6 - 6*a*b^4*c + 32*a^3*c^3)*e^4*x^2 + 2* (a*b^5 - 8*a^2*b^3*c + 16*a^3*b*c^2)*e^4*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a ^4*c^2)*e^4)*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*(48*a^2*b*c^3*d^2*e^2 - 64*a^3*c^3*d* e^3 - (b^3*c^3 - 12*a*b*c^4)*d^4 - 8*(a*b^2*c^3 + 4*a^2*c^4)*d^3*e - (3*a^ 2*b^3*c - 20*a^3*b*c^2)*e^4 + 4*(4*c^6*d^4 - 8*b*c^5*d^3*e + 3*(b^2*c^4 + 4*a*c^5)*d^2*e^2 + (b^3*c^3 - 12*a*b*c^4)*d*e^3 - (b^4*c^2 - 7*a*b^2*c^3 + 8*a^2*c^4)*e^4)*x^3 + 3*(8*b*c^5*d^4 - 16*b^2*c^4*d^3*e + 6*(b^3*c^3 + 4* a*b*c^4)*d^2*e^2 - 8*(a*b^2*c^3 + 4*a^2*c^4)*d*e^3 - (b^5*c - 6*a*b^3*c^2) *e^4)*x^2 + 6*(12*a*b^2*c^3*d^2*e^2 - 16*a^2*b*c^3*d*e^3 + (b^2*c^4 + 4*a* c^5)*d^4 - 2*(b^3*c^3 + 4*a*b*c^4)*d^3*e - (a*b^4*c - 7*a^2*b^2*c^2 + 4*a^ 3*c^3)*e^4)*x)*sqrt(c*x^2 + b*x + a))/(a^2*b^4*c^3 - 8*a^3*b^2*c^4 + 16*a^ 4*c^5 + (b^4*c^5 - 8*a*b^2*c^6 + 16*a^2*c^7)*x^4 + 2*(b^5*c^4 - 8*a*b^3*c^ 5 + 16*a^2*b*c^6)*x^3 + (b^6*c^3 - 6*a*b^4*c^4 + 32*a^3*c^6)*x^2 + 2*(a*b^ 5*c^3 - 8*a^2*b^3*c^4 + 16*a^3*b*c^5)*x), -1/3*(3*((b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4)*e^4*x^4 + 2*(b^5*c - 8*a*b^3*c^2 + 16*a^2*b*c^3)*e^4*x^3 + ( b^6 - 6*a*b^4*c + 32*a^3*c^3)*e^4*x^2 + 2*(a*b^5 - 8*a^2*b^3*c + 16*a^3*b* c^2)*e^4*x + (a^2*b^4 - 8*a^3*b^2*c + 16*a^4*c^2)*e^4)*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)) - ...
Timed out. \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` for more deta
Leaf count of result is larger than twice the leaf count of optimal. 564 vs. \(2 (275) = 550\).
Time = 0.31 (sec) , antiderivative size = 564, normalized size of antiderivative = 1.91 \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=-\frac {e^{4} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} + b \right |}\right )}{c^{\frac {5}{2}}} + \frac {2 \, {\left ({\left ({\left (\frac {4 \, {\left (4 \, c^{5} d^{4} - 8 \, b c^{4} d^{3} e + 3 \, b^{2} c^{3} d^{2} e^{2} + 12 \, a c^{4} d^{2} e^{2} + b^{3} c^{2} d e^{3} - 12 \, a b c^{3} d e^{3} - b^{4} c e^{4} + 7 \, a b^{2} c^{2} e^{4} - 8 \, a^{2} c^{3} e^{4}\right )} x}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}} + \frac {3 \, {\left (8 \, b c^{4} d^{4} - 16 \, b^{2} c^{3} d^{3} e + 6 \, b^{3} c^{2} d^{2} e^{2} + 24 \, a b c^{3} d^{2} e^{2} - 8 \, a b^{2} c^{2} d e^{3} - 32 \, a^{2} c^{3} d e^{3} - b^{5} e^{4} + 6 \, a b^{3} c e^{4}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x + \frac {6 \, {\left (b^{2} c^{3} d^{4} + 4 \, a c^{4} d^{4} - 2 \, b^{3} c^{2} d^{3} e - 8 \, a b c^{3} d^{3} e + 12 \, a b^{2} c^{2} d^{2} e^{2} - 16 \, a^{2} b c^{2} d e^{3} - a b^{4} e^{4} + 7 \, a^{2} b^{2} c e^{4} - 4 \, a^{3} c^{2} e^{4}\right )}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )} x - \frac {b^{3} c^{2} d^{4} - 12 \, a b c^{3} d^{4} + 8 \, a b^{2} c^{2} d^{3} e + 32 \, a^{2} c^{3} d^{3} e - 48 \, a^{2} b c^{2} d^{2} e^{2} + 64 \, a^{3} c^{2} d e^{3} + 3 \, a^{2} b^{3} e^{4} - 20 \, a^{3} b c e^{4}}{b^{4} c^{2} - 8 \, a b^{2} c^{3} + 16 \, a^{2} c^{4}}\right )}}{3 \, {\left (c x^{2} + b x + a\right )}^{\frac {3}{2}}} \]
-e^4*log(abs(2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) + b))/c^(5/2) + 2/3*(((4*(4*c^5*d^4 - 8*b*c^4*d^3*e + 3*b^2*c^3*d^2*e^2 + 12*a*c^4*d^2*e^ 2 + b^3*c^2*d*e^3 - 12*a*b*c^3*d*e^3 - b^4*c*e^4 + 7*a*b^2*c^2*e^4 - 8*a^2 *c^3*e^4)*x/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4) + 3*(8*b*c^4*d^4 - 16*b^2 *c^3*d^3*e + 6*b^3*c^2*d^2*e^2 + 24*a*b*c^3*d^2*e^2 - 8*a*b^2*c^2*d*e^3 - 32*a^2*c^3*d*e^3 - b^5*e^4 + 6*a*b^3*c*e^4)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^ 2*c^4))*x + 6*(b^2*c^3*d^4 + 4*a*c^4*d^4 - 2*b^3*c^2*d^3*e - 8*a*b*c^3*d^3 *e + 12*a*b^2*c^2*d^2*e^2 - 16*a^2*b*c^2*d*e^3 - a*b^4*e^4 + 7*a^2*b^2*c*e ^4 - 4*a^3*c^2*e^4)/(b^4*c^2 - 8*a*b^2*c^3 + 16*a^2*c^4))*x - (b^3*c^2*d^4 - 12*a*b*c^3*d^4 + 8*a*b^2*c^2*d^3*e + 32*a^2*c^3*d^3*e - 48*a^2*b*c^2*d^ 2*e^2 + 64*a^3*c^2*d*e^3 + 3*a^2*b^3*e^4 - 20*a^3*b*c*e^4)/(b^4*c^2 - 8*a* b^2*c^3 + 16*a^2*c^4))/(c*x^2 + b*x + a)^(3/2)
Timed out. \[ \int \frac {(d+e x)^4}{\left (a+b x+c x^2\right )^{5/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^4}{{\left (c\,x^2+b\,x+a\right )}^{5/2}} \,d x \]