Integrand size = 20, antiderivative size = 241 \[ \int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx=\frac {(2 c d-b e) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{6 c^2}+\frac {2 e \left (a+b x+c x^2\right )^{5/4}}{5 c}-\frac {\left (b^2-4 a c\right )^{5/4} (2 c d-b e) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{12 \sqrt {2} c^{9/4} (b+2 c x)} \]
1/6*(-b*e+2*c*d)*(2*c*x+b)*(c*x^2+b*x+a)^(1/4)/c^2+2/5*e*(c*x^2+b*x+a)^(5/ 4)/c-1/24*(-4*a*c+b^2)^(5/4)*(-b*e+2*c*d)*(cos(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 = 8.28 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.58 \[ \int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx=\frac {24 c^2 e (a+x (b+c x))^2+5 (2 c d-b e) \left (2 c (b+2 c x) (a+x (b+c x))-\sqrt {2} \left (b^2-4 a c\right )^{3/2} \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ),2\right )\right )}{60 c^3 (a+x (b+c x))^{3/4}} \]
(24*c^2*e*(a + x*(b + c*x))^2 + 5*(2*c*d - b*e)*(2*c*(b + 2*c*x)*(a + x*(b + c*x)) - Sqrt[2]*(b^2 - 4*a*c)^(3/2)*((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]))/(60*c^3* (a + x*(b + c*x))^(3/4))
Time = 0.31 (sec) , antiderivative size = 290, normalized size of antiderivative = 1.20, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {1160, 1087, 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.
| \(\displaystyle \int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx\) |
| \(\Big \downarrow \) 1160 |
| \(\displaystyle \frac {(2 c d-b e) \int \sqrt [4]{c x^2+b x+a}dx}{2 c}+\frac {2 e \left (a+b x+c x^2\right )^{5/4}}{5 c}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\left (c x^2+b x+a\right )^{3/4}}dx}{12 c}\right )}{2 c}+\frac {2 e \left (a+b x+c x^2\right )^{5/4}}{5 c}\) |
| \(\Big \downarrow \) 1094 |
| \(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{3 c (b+2 c x)}\right )}{2 c}+\frac {2 e \left (a+b x+c x^2\right )^{5/4}}{5 c}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {(2 c d-b e) \left (\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right )^{5/4} \sqrt {(b+2 c x)^2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} c^{5/4} (b+2 c x) \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}\right )}{2 c}+\frac {2 e \left (a+b x+c x^2\right )^{5/4}}{5 c}\) |
(2*e*(a + b*x + c*x^2)^(5/4))/(5*c) + ((2*c*d - b*e)*(((b + 2*c*x)*(a + b* x + c*x^2)^(1/4))/(3*c) - ((b^2 - 4*a*c)^(5/4)*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)]*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])/(6*Sqrt[2]*c^(5/4)*(b + 2*c*x)*Sqrt[ b^2 - 4*a*c + 4*c*(a + b*x + c*x^2)])))/(2*c)
3.26.12.3.1 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[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
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]
\[\int \left (e x +d \right ) \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}d x\]
\[ \int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )} \,d x } \]
\[ \int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx=\int \left (d + e x\right ) \sqrt [4]{a + b x + c x^{2}}\, dx \]
\[ \int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )} \,d x } \]
\[ \int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} {\left (e x + d\right )} \,d x } \]
Timed out. \[ \int (d+e x) \sqrt [4]{a+b x+c x^2} \, dx=\int \left (d+e\,x\right )\,{\left (c\,x^2+b\,x+a\right )}^{1/4} \,d x \]