Integrand size = 22, antiderivative size = 262 \[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\frac {5 e (2 c d-b e) \sqrt [4]{a+b x+c x^2}}{3 c^2}+\frac {2 e (d+e x) \sqrt [4]{a+b x+c x^2}}{3 c}+\frac {\sqrt [4]{b^2-4 a c} \left (12 c^2 d^2+5 b^2 e^2-4 c e (3 b d+2 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} c^{9/4} (b+2 c x)} \]
5/3*e*(-b*e+2*c*d)*(c*x^2+b*x+a)^(1/4)/c^2+2/3*e*(e*x+d)*(c*x^2+b*x+a)^(1/ 4)/c+1/12*(-4*a*c+b^2)^(1/4)*(12*c^2*d^2+5*b^2*e^2-4*c*e*(2*a*e+3*b*d))*(c os(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(1/4)))^2)^(1 /2)/cos(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(1/4)))* EllipticF(sin(2*arctan(c^(1/4)*(c*x^2+b*x+a)^(1/4)*2^(1/2)/(-4*a*c+b^2)^(1 /4))),1/2*2^(1/2))*(1+2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))*(( 2*c*x+b)^2/(-4*a*c+b^2)/(1+2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2 ))^2)^(1/2)/c^(9/4)/(2*c*x+b)*2^(1/2)
Time = 10.17 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.57 \[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\frac {2 c e (a+x (b+c x)) (-5 b e+2 c (6 d+e x))+\sqrt {2} \sqrt {b^2-4 a c} \left (12 c^2 d^2+5 b^2 e^2-4 c e (3 b d+2 a e)\right ) \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ),2\right )}{6 c^3 (a+x (b+c x))^{3/4}} \]
(2*c*e*(a + x*(b + c*x))*(-5*b*e + 2*c*(6*d + e*x)) + Sqrt[2]*Sqrt[b^2 - 4 *a*c]*(12*c^2*d^2 + 5*b^2*e^2 - 4*c*e*(3*b*d + 2*a*e))*((c*(a + x*(b + c*x )))/(-b^2 + 4*a*c))^(3/4)*EllipticF[ArcSin[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]]/ 2, 2])/(6*c^3*(a + x*(b + c*x))^(3/4))
Time = 0.37 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.20, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {1166, 27, 1160, 1094, 761}
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.
| \(\displaystyle \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{3/4}} \, dx\) |
| \(\Big \downarrow \) 1166 |
| \(\displaystyle \frac {2 \int \frac {6 c d^2-e (b d+4 a e)+5 e (2 c d-b e) x}{4 \left (c x^2+b x+a\right )^{3/4}}dx}{3 c}+\frac {2 e (d+e x) \sqrt [4]{a+b x+c x^2}}{3 c}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \frac {6 c d^2-e (b d+4 a e)+5 e (2 c d-b e) x}{\left (c x^2+b x+a\right )^{3/4}}dx}{6 c}+\frac {2 e (d+e x) \sqrt [4]{a+b x+c x^2}}{3 c}\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {\frac {\left (-4 c e (2 a e+3 b d)+5 b^2 e^2+12 c^2 d^2\right ) \int \frac {1}{\left (c x^2+b x+a\right )^{3/4}}dx}{2 c}+\frac {10 e \sqrt [4]{a+b x+c x^2} (2 c d-b e)}{c}}{6 c}+\frac {2 e (d+e x) \sqrt [4]{a+b x+c x^2}}{3 c}\) |
| \(\Big \downarrow \) 1094 |
| \(\displaystyle \frac {\frac {2 \sqrt {(b+2 c x)^2} \left (-4 c e (2 a e+3 b d)+5 b^2 e^2+12 c^2 d^2\right ) \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{c (b+2 c x)}+\frac {10 e \sqrt [4]{a+b x+c x^2} (2 c d-b e)}{c}}{6 c}+\frac {2 e (d+e x) \sqrt [4]{a+b x+c x^2}}{3 c}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {\frac {\sqrt [4]{b^2-4 a c} \sqrt {(b+2 c x)^2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (-4 c e (2 a e+3 b d)+5 b^2 e^2+12 c^2 d^2\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{\sqrt {2} c^{5/4} (b+2 c x) \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}+\frac {10 e \sqrt [4]{a+b x+c x^2} (2 c d-b e)}{c}}{6 c}+\frac {2 e (d+e x) \sqrt [4]{a+b x+c x^2}}{3 c}\) |
(2*e*(d + e*x)*(a + b*x + c*x^2)^(1/4))/(3*c) + ((10*e*(2*c*d - b*e)*(a + b*x + c*x^2)^(1/4))/c + ((b^2 - 4*a*c)^(1/4)*(12*c^2*d^2 + 5*b^2*e^2 - 4*c *e*(3*b*d + 2*a*e))*Sqrt[(b + 2*c*x)^2]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x ^2])/Sqrt[b^2 - 4*a*c])*Sqrt[(b^2 - 4*a*c + 4*c*(a + b*x + c*x^2))/((b^2 - 4*a*c)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])^2)]*Elli pticF[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/ 4)], 1/2])/(Sqrt[2]*c^(5/4)*(b + 2*c*x)*Sqrt[b^2 - 4*a*c + 4*c*(a + b*x + c*x^2)]))/(6*c)
3.26.36.3.1 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[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[4*(Sqrt[(b + 2*c*x)^2]/(b + 2*c*x)) Subst[Int[x^(4*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4 *c*x^4], x], x, (a + b*x + c*x^2)^(1/4)], x] /; FreeQ[{a, b, c}, x] && Inte gerQ[4*p]
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[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[1/(c*(m + 2*p + 1)) Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* (a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat icQ[a, b, c, d, e, m, p, x]
\[\int \frac {\left (e x +d \right )^{2}}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}d x\]
\[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\int \frac {\left (d + e x\right )^{2}}{\left (a + b x + c x^{2}\right )^{\frac {3}{4}}}\, dx \]
\[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{4}}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{{\left (c\,x^2+b\,x+a\right )}^{3/4}} \,d x \]