Integrand size = 28, antiderivative size = 307 \[ \int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{(d+e x)^5} \, dx=-\frac {5 \left (b^2-4 a c\right ) e (2 c d-b e) (b d-2 a e+(2 c d-b e) x) \sqrt {a+b x+c x^2}}{64 \left (c d^2-b d e+a e^2\right )^3 (d+e x)^2}+\frac {(2 c d-b e) \left (a+b x+c x^2\right )^{3/2}}{4 \left (c d^2-b d e+a e^2\right ) (d+e x)^4}+\frac {\left (4 c^2 d^2+5 b^2 e^2-4 c e (b d+4 a e)\right ) \left (a+b x+c x^2\right )^{3/2}}{24 \left (c d^2-b d e+a e^2\right )^2 (d+e x)^3}+\frac {5 \left (b^2-4 a c\right )^2 e (2 c d-b e) \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x+c x^2}}\right )}{128 \left (c d^2-b d e+a e^2\right )^{7/2}} \]
1/4*(-b*e+2*c*d)*(c*x^2+b*x+a)^(3/2)/(a*e^2-b*d*e+c*d^2)/(e*x+d)^4+1/24*(4 *c^2*d^2+5*b^2*e^2-4*c*e*(4*a*e+b*d))*(c*x^2+b*x+a)^(3/2)/(a*e^2-b*d*e+c*d ^2)^2/(e*x+d)^3+5/128*(-4*a*c+b^2)^2*e*(-b*e+2*c*d)*arctanh(1/2*(b*d-2*a*e +(-b*e+2*c*d)*x)/(a*e^2-b*d*e+c*d^2)^(1/2)/(c*x^2+b*x+a)^(1/2))/(a*e^2-b*d *e+c*d^2)^(7/2)-5/64*(-4*a*c+b^2)*e*(-b*e+2*c*d)*(b*d-2*a*e+(-b*e+2*c*d)*x )*(c*x^2+b*x+a)^(1/2)/(a*e^2-b*d*e+c*d^2)^3/(e*x+d)^2
Time = 10.51 (sec) , antiderivative size = 290, normalized size of antiderivative = 0.94 \[ \int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{(d+e x)^5} \, dx=\frac {\frac {6 (2 c d-b e) \left (c d^2+e (-b d+a e)\right ) (a+x (b+c x))^{3/2}}{(d+e x)^4}+\frac {\left (4 c^2 d^2+5 b^2 e^2-4 c e (b d+4 a e)\right ) (a+x (b+c x))^{3/2}}{(d+e x)^3}+\frac {15}{2} \left (b^2-4 a c\right ) e (-2 c d+b e) \left (\frac {\sqrt {a+x (b+c x)} (-2 a e+2 c d x+b (d-e x))}{4 \left (c d^2+e (-b d+a e)\right ) (d+e x)^2}+\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {-b d+2 a e-2 c d x+b e x}{2 \sqrt {c d^2+e (-b d+a e)} \sqrt {a+x (b+c x)}}\right )}{8 \left (c d^2+e (-b d+a e)\right )^{3/2}}\right )}{24 \left (c d^2+e (-b d+a e)\right )^2} \]
((6*(2*c*d - b*e)*(c*d^2 + e*(-(b*d) + a*e))*(a + x*(b + c*x))^(3/2))/(d + e*x)^4 + ((4*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(b*d + 4*a*e))*(a + x*(b + c*x)) ^(3/2))/(d + e*x)^3 + (15*(b^2 - 4*a*c)*e*(-2*c*d + b*e)*((Sqrt[a + x*(b + c*x)]*(-2*a*e + 2*c*d*x + b*(d - e*x)))/(4*(c*d^2 + e*(-(b*d) + a*e))*(d + e*x)^2) + ((b^2 - 4*a*c)*ArcTanh[(-(b*d) + 2*a*e - 2*c*d*x + b*e*x)/(2*S qrt[c*d^2 + e*(-(b*d) + a*e)]*Sqrt[a + x*(b + c*x)])])/(8*(c*d^2 + e*(-(b* d) + a*e))^(3/2))))/2)/(24*(c*d^2 + e*(-(b*d) + a*e))^2)
Time = 0.57 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {1237, 27, 1228, 1152, 1154, 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.
| \(\displaystyle \int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{(d+e x)^5} \, dx\) |
| \(\Big \downarrow \) 1237 |
| \(\displaystyle \frac {\left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{4 (d+e x)^4 \left (a e^2-b d e+c d^2\right )}-\frac {\int -\frac {\left (-5 e b^2+2 c d b+16 a c e+2 c (2 c d-b e) x\right ) \sqrt {c x^2+b x+a}}{2 (d+e x)^4}dx}{4 \left (a e^2-b d e+c d^2\right )}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {\left (-5 e b^2+2 c d b+16 a c e+2 c (2 c d-b e) x\right ) \sqrt {c x^2+b x+a}}{(d+e x)^4}dx}{8 \left (a e^2-b d e+c d^2\right )}+\frac {\left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{4 (d+e x)^4 \left (a e^2-b d e+c d^2\right )}\) |
| \(\Big \downarrow \) 1228 |
| \(\displaystyle \frac {\frac {\left (a+b x+c x^2\right )^{3/2} \left (-4 c e (4 a e+b d)+5 b^2 e^2+4 c^2 d^2\right )}{3 (d+e x)^3 \left (a e^2-b d e+c d^2\right )}-\frac {5 e \left (b^2-4 a c\right ) (2 c d-b e) \int \frac {\sqrt {c x^2+b x+a}}{(d+e x)^3}dx}{2 \left (a e^2-b d e+c d^2\right )}}{8 \left (a e^2-b d e+c d^2\right )}+\frac {\left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{4 (d+e x)^4 \left (a e^2-b d e+c d^2\right )}\) |
| \(\Big \downarrow \) 1152 |
| \(\displaystyle \frac {\frac {\left (a+b x+c x^2\right )^{3/2} \left (-4 c e (4 a e+b d)+5 b^2 e^2+4 c^2 d^2\right )}{3 (d+e x)^3 \left (a e^2-b d e+c d^2\right )}-\frac {5 e \left (b^2-4 a c\right ) (2 c d-b e) \left (\frac {\sqrt {a+b x+c x^2} (-2 a e+x (2 c d-b e)+b d)}{4 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{(d+e x) \sqrt {c x^2+b x+a}}dx}{8 \left (a e^2-b d e+c d^2\right )}\right )}{2 \left (a e^2-b d e+c d^2\right )}}{8 \left (a e^2-b d e+c d^2\right )}+\frac {\left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{4 (d+e x)^4 \left (a e^2-b d e+c d^2\right )}\) |
| \(\Big \downarrow \) 1154 |
| \(\displaystyle \frac {\frac {\left (a+b x+c x^2\right )^{3/2} \left (-4 c e (4 a e+b d)+5 b^2 e^2+4 c^2 d^2\right )}{3 (d+e x)^3 \left (a e^2-b d e+c d^2\right )}-\frac {5 e \left (b^2-4 a c\right ) (2 c d-b e) \left (\frac {\left (b^2-4 a c\right ) \int \frac {1}{4 \left (c d^2-b e d+a e^2\right )-\frac {(b d-2 a e+(2 c d-b e) x)^2}{c x^2+b x+a}}d\left (-\frac {b d-2 a e+(2 c d-b e) x}{\sqrt {c x^2+b x+a}}\right )}{4 \left (a e^2-b d e+c d^2\right )}+\frac {\sqrt {a+b x+c x^2} (-2 a e+x (2 c d-b e)+b d)}{4 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}\right )}{2 \left (a e^2-b d e+c d^2\right )}}{8 \left (a e^2-b d e+c d^2\right )}+\frac {\left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{4 (d+e x)^4 \left (a e^2-b d e+c d^2\right )}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\left (a+b x+c x^2\right )^{3/2} \left (-4 c e (4 a e+b d)+5 b^2 e^2+4 c^2 d^2\right )}{3 (d+e x)^3 \left (a e^2-b d e+c d^2\right )}-\frac {5 e \left (b^2-4 a c\right ) (2 c d-b e) \left (\frac {\sqrt {a+b x+c x^2} (-2 a e+x (2 c d-b e)+b d)}{4 (d+e x)^2 \left (a e^2-b d e+c d^2\right )}-\frac {\left (b^2-4 a c\right ) \text {arctanh}\left (\frac {-2 a e+x (2 c d-b e)+b d}{2 \sqrt {a+b x+c x^2} \sqrt {a e^2-b d e+c d^2}}\right )}{8 \left (a e^2-b d e+c d^2\right )^{3/2}}\right )}{2 \left (a e^2-b d e+c d^2\right )}}{8 \left (a e^2-b d e+c d^2\right )}+\frac {\left (a+b x+c x^2\right )^{3/2} (2 c d-b e)}{4 (d+e x)^4 \left (a e^2-b d e+c d^2\right )}\) |
((2*c*d - b*e)*(a + b*x + c*x^2)^(3/2))/(4*(c*d^2 - b*d*e + a*e^2)*(d + e* x)^4) + (((4*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(b*d + 4*a*e))*(a + b*x + c*x^2)^ (3/2))/(3*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^3) - (5*(b^2 - 4*a*c)*e*(2*c*d - b*e)*(((b*d - 2*a*e + (2*c*d - b*e)*x)*Sqrt[a + b*x + c*x^2])/(4*(c*d^2 - b*d*e + a*e^2)*(d + e*x)^2) - ((b^2 - 4*a*c)*ArcTanh[(b*d - 2*a*e + (2* c*d - b*e)*x)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x + c*x^2])])/(8*( c*d^2 - b*d*e + a*e^2)^(3/2))))/(2*(c*d^2 - b*d*e + a*e^2)))/(8*(c*d^2 - b *d*e + a*e^2))
3.16.55.3.1 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[((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/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[p*((b^2 - 4*a *c)/(2*(m + 1)*(c*d^2 - b*d*e + a*e^2))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[m + 2*p + 2, 0] && GtQ[p, 0]
Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Sym bol] :> Simp[-2 Subst[Int[1/(4*c*d^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, ( 2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c , d, e}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + b*x + c*x^2)^(p + 1)/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2))), x] - Simp[(b*(e *f + d*g) - 2*(c*d*f + a*e*g))/(2*(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^ (m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x ] && EqQ[Simplify[m + 2*p + 3], 0]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*f - d*g)*(d + e*x)^(m + 1)*((a + b* x + c*x^2)^(p + 1)/((m + 1)*(c*d^2 - b*d*e + a*e^2))), x] + Simp[1/((m + 1) *(c*d^2 - b*d*e + a*e^2)) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[ (c*d*f - f*b*e + a*e*g)*(m + 1) + b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && LtQ[m, -1 ] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Leaf count of result is larger than twice the leaf count of optimal. \(3787\) vs. \(2(285)=570\).
Time = 1.15 (sec) , antiderivative size = 3788, normalized size of antiderivative = 12.34
2*c/e^5*(-1/3/(a*e^2-b*d*e+c*d^2)*e^2/(x+d/e)^3*((x+d/e)^2*c+(b*e-2*c*d)/e *(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)-1/2*(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d ^2)*(-1/2/(a*e^2-b*d*e+c*d^2)*e^2/(x+d/e)^2*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+ d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(3/2)-1/4*(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d^2)* (-1/(a*e^2-b*d*e+c*d^2)*e^2/(x+d/e)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a* e^2-b*d*e+c*d^2)/e^2)^(3/2)+1/2*(b*e-2*c*d)*e/(a*e^2-b*d*e+c*d^2)*(((x+d/e )^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/2*(b*e-2*c*d) /e*ln((1/2*(b*e-2*c*d)/e+c*(x+d/e))/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d)/e*(x+ d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/c^(1/2)-(a*e^2-b*d*e+c*d^2)/e^2/((a*e ^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-2*c*d)/e*(x+ d/e)+2*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+ (a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/(x+d/e)))+2*c/(a*e^2-b*d*e+c*d^2)*e^2*(1/4 *(2*c*(x+d/e)+(b*e-2*c*d)/e)/c*((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b *d*e+c*d^2)/e^2)^(1/2)+1/8*(4*c*(a*e^2-b*d*e+c*d^2)/e^2-(b*e-2*c*d)^2/e^2) /c^(3/2)*ln((1/2*(b*e-2*c*d)/e+c*(x+d/e))/c^(1/2)+((x+d/e)^2*c+(b*e-2*c*d) /e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))))+1/2*c/(a*e^2-b*d*e+c*d^2)*e^2 *(((x+d/e)^2*c+(b*e-2*c*d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2)+1/2*(b *e-2*c*d)/e*ln((1/2*(b*e-2*c*d)/e+c*(x+d/e))/c^(1/2)+((x+d/e)^2*c+(b*e-2*c *d)/e*(x+d/e)+(a*e^2-b*d*e+c*d^2)/e^2)^(1/2))/c^(1/2)-(a*e^2-b*d*e+c*d^2)/ e^2/((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)*ln((2*(a*e^2-b*d*e+c*d^2)/e^2+(b*e-...
Leaf count of result is larger than twice the leaf count of optimal. 1977 vs. \(2 (285) = 570\).
Time = 12.18 (sec) , antiderivative size = 3996, normalized size of antiderivative = 13.02 \[ \int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{(d+e x)^5} \, dx=\text {Too large to display} \]
[-1/768*(15*(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d^5*e - (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*d^4*e^2 + (2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d*e^5 - ( b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*e^6)*x^4 + 4*(2*(b^4*c - 8*a*b^2*c^2 + 16* a^2*c^3)*d^2*e^4 - (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)*d*e^5)*x^3 + 6*(2*(b^4 *c - 8*a*b^2*c^2 + 16*a^2*c^3)*d^3*e^3 - (b^5 - 8*a*b^3*c + 16*a^2*b*c^2)* d^2*e^4)*x^2 + 4*(2*(b^4*c - 8*a*b^2*c^2 + 16*a^2*c^3)*d^4*e^2 - (b^5 - 8* a*b^3*c + 16*a^2*b*c^2)*d^3*e^3)*x)*sqrt(c*d^2 - b*d*e + a*e^2)*log((8*a*b *d*e - 8*a^2*e^2 - (b^2 + 4*a*c)*d^2 - (8*c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a *c)*e^2)*x^2 + 4*sqrt(c*d^2 - b*d*e + a*e^2)*sqrt(c*x^2 + b*x + a)*(b*d - 2*a*e + (2*c*d - b*e)*x) - 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e) *x)/(e^2*x^2 + 2*d*e*x + d^2)) - 4*(128*a*c^4*d^7 - 48*a^4*b*e^7 - 6*(5*b^ 3*c^2 + 52*a*b*c^3)*d^6*e + (45*b^4*c + 448*a*b^2*c^2 - 16*a^2*c^3)*d^5*e^ 2 - (15*b^5 + 412*a*b^3*c + 112*a^2*b*c^2)*d^4*e^3 + (133*a*b^4 + 472*a^2* b^2*c - 176*a^3*c^2)*d^3*e^4 - 2*(127*a^2*b^3 + 44*a^3*b*c)*d^2*e^5 + 8*(2 3*a^3*b^2 - 4*a^4*c)*d*e^6 + (32*c^5*d^6*e - 96*b*c^4*d^5*e^2 + 4*(19*b^2* c^3 + 44*a*c^4)*d^4*e^3 + 8*(b^3*c^2 - 44*a*b*c^3)*d^3*e^4 - (35*b^4*c - 2 56*a*b^2*c^2 - 16*a^2*c^3)*d^2*e^5 + (15*b^5 - 80*a*b^3*c - 16*a^2*b*c^2)* d*e^6 - (15*a*b^4 - 100*a^2*b^2*c + 128*a^3*c^2)*e^7)*x^3 + (128*c^5*d^7 - 400*b*c^4*d^6*e + 352*(b^2*c^3 + 2*a*c^4)*d^5*e^2 - 2*(3*b^3*c^2 + 748*a* b*c^3)*d^4*e^3 - (129*b^4*c - 1080*a*b^2*c^2 - 304*a^2*c^3)*d^3*e^4 + (...
\[ \int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{(d+e x)^5} \, dx=\int \frac {\left (b + 2 c x\right ) \sqrt {a + b x + c x^{2}}}{\left (d + e x\right )^{5}}\, dx \]
Exception generated. \[ \int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{(d+e x)^5} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*e^2-b*d*e>0)', see `assume?` f or more de
\[ \int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{(d+e x)^5} \, dx=\int { \frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )}}{{\left (e x + d\right )}^{5}} \,d x } \]
Timed out. \[ \int \frac {(b+2 c x) \sqrt {a+b x+c x^2}}{(d+e x)^5} \, dx=\int \frac {\left (b+2\,c\,x\right )\,\sqrt {c\,x^2+b\,x+a}}{{\left (d+e\,x\right )}^5} \,d x \]