Integrand size = 20, antiderivative size = 200 \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\frac {2 e \sqrt [4]{a+b x+c x^2}}{c}+\frac {\sqrt [4]{b^2-4 a c} (2 c d-b e) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{\sqrt {2} c^{5/4} (b+2 c x)} \]
2*e*(c*x^2+b*x+a)^(1/4)/c+1/2*(-4*a*c+b^2)^(1/4)*(-b*e+2*c*d)*(cos(2*arcta n(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^(5/4)/(2*c*x+b)*2^(1/2)
Time = 10.12 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.56 \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\frac {2 c e (a+x (b+c x))-\sqrt {2} \sqrt {b^2-4 a c} (-2 c d+b e) \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ),2\right )}{c^2 (a+x (b+c x))^{3/4}} \]
(2*c*e*(a + x*(b + c*x)) - Sqrt[2]*Sqrt[b^2 - 4*a*c]*(-2*c*d + b*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])/(c^2*(a + x*(b + c*x))^(3/4))
Time = 0.28 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.25, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {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.
\(\displaystyle \int \frac {d+e x}{\left (a+b x+c x^2\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {(2 c d-b e) \int \frac {1}{\left (c x^2+b x+a\right )^{3/4}}dx}{2 c}+\frac {2 e \sqrt [4]{a+b x+c x^2}}{c}\) |
\(\Big \downarrow \) 1094 |
\(\displaystyle \frac {2 \sqrt {(b+2 c x)^2} (2 c d-b e) \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{c (b+2 c x)}+\frac {2 e \sqrt [4]{a+b x+c x^2}}{c}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {\sqrt [4]{b^2-4 a c} \sqrt {(b+2 c x)^2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} (2 c d-b e) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{\sqrt {2} c^{5/4} (b+2 c x) \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}+\frac {2 e \sqrt [4]{a+b x+c x^2}}{c}\) |
(2*e*(a + b*x + c*x^2)^(1/4))/c + ((b^2 - 4*a*c)^(1/4)*(2*c*d - b*e)*Sqrt[ (b + 2*c*x)^2]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])*S qrt[(b^2 - 4*a*c + 4*c*(a + b*x + c*x^2))/((b^2 - 4*a*c)*(1 + (2*Sqrt[c]*S qrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])^2)]*EllipticF[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)])
3.26.37.3.1 Defintions of rubi rules used
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]
\[\int \frac {e x +d}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}d x\]
\[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\int { \frac {e x + d}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\int \frac {d + e x}{\left (a + b x + c x^{2}\right )^{\frac {3}{4}}}\, dx \]
\[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\int { \frac {e x + d}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{4}}} \,d x } \]
\[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\int { \frac {e x + d}{{\left (c x^{2} + b x + a\right )}^{\frac {3}{4}}} \,d x } \]
Timed out. \[ \int \frac {d+e x}{\left (a+b x+c x^2\right )^{3/4}} \, dx=\int \frac {d+e\,x}{{\left (c\,x^2+b\,x+a\right )}^{3/4}} \,d x \]