Integrand size = 14, antiderivative size = 123 \[ \int \sqrt [4]{a+b x+c x^2} \, dx=\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}+\frac {\left (-b^2+4 a c\right )^{3/2} \left (-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right ),2\right )}{3 \sqrt {2} c^2 \left (a+b x+c x^2\right )^{3/4}} \] Output:
1/3*(2*c*x+b)*(c*x^2+b*x+a)^(1/4)/c+1/6*(4*a*c-b^2)^(3/2)*(-c*(c*x^2+b*x+a )/(-4*a*c+b^2))^(3/4)*InverseJacobiAM(1/2*arctan((2*c*x+b)/(4*a*c-b^2)^(1/ 2)),2^(1/2))*2^(1/2)/c^2/(c*x^2+b*x+a)^(3/4)
Time = 10.18 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.81 \[ \int \sqrt [4]{a+b x+c x^2} \, dx=\frac {\sqrt [4]{a+x (b+c x)} \left (2 (b+2 c x)+\frac {\sqrt {2} \sqrt {b^2-4 a c} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ),2\right )}{\sqrt [4]{\frac {c (a+x (b+c x))}{-b^2+4 a c}}}\right )}{6 c} \] Input:
Integrate[(a + b*x + c*x^2)^(1/4),x]
Output:
((a + x*(b + c*x))^(1/4)*(2*(b + 2*c*x) + (Sqrt[2]*Sqrt[b^2 - 4*a*c]*Ellip ticF[ArcSin[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]]/2, 2])/((c*(a + x*(b + c*x)))/( -b^2 + 4*a*c))^(1/4)))/(6*c)
Leaf count is larger than twice the leaf count of optimal. \(251\) vs. \(2(123)=246\).
Time = 0.32 (sec) , antiderivative size = 251, normalized size of antiderivative = 2.04, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {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.
\(\displaystyle \int \sqrt [4]{a+b x+c x^2} \, dx\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\left (c x^2+b x+a\right )^{3/4}}dx}{12 c}\) |
\(\Big \downarrow \) 1094 |
\(\displaystyle \frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{3 c (b+2 c x)}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right )^{5/4} \sqrt {(b+2 c x)^2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} c^{5/4} (b+2 c x) \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}\) |
Input:
Int[(a + b*x + c*x^2)^(1/4),x]
Output:
((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])*Sqr t[(b^2 - 4*a*c + 4*c*(a + b*x + c*x^2))/((b^2 - 4*a*c)*(1 + (2*Sqrt[c]*Sqr t[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)])
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]
\[\int \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}}d x\]
Input:
int((c*x^2+b*x+a)^(1/4),x)
Output:
int((c*x^2+b*x+a)^(1/4),x)
\[ \int \sqrt [4]{a+b x+c x^2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} \,d x } \] Input:
integrate((c*x^2+b*x+a)^(1/4),x, algorithm="fricas")
Output:
integral((c*x^2 + b*x + a)^(1/4), x)
\[ \int \sqrt [4]{a+b x+c x^2} \, dx=\int \sqrt [4]{a + b x + c x^{2}}\, dx \] Input:
integrate((c*x**2+b*x+a)**(1/4),x)
Output:
Integral((a + b*x + c*x**2)**(1/4), x)
\[ \int \sqrt [4]{a+b x+c x^2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} \,d x } \] Input:
integrate((c*x^2+b*x+a)^(1/4),x, algorithm="maxima")
Output:
integrate((c*x^2 + b*x + a)^(1/4), x)
\[ \int \sqrt [4]{a+b x+c x^2} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {1}{4}} \,d x } \] Input:
integrate((c*x^2+b*x+a)^(1/4),x, algorithm="giac")
Output:
integrate((c*x^2 + b*x + a)^(1/4), x)
Timed out. \[ \int \sqrt [4]{a+b x+c x^2} \, dx=\int {\left (c\,x^2+b\,x+a\right )}^{1/4} \,d x \] Input:
int((a + b*x + c*x^2)^(1/4),x)
Output:
int((a + b*x + c*x^2)^(1/4), x)
\[ \int \sqrt [4]{a+b x+c x^2} \, dx=\frac {8 \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} a +4 \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} b x -4 \left (\int \frac {x}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}d x \right ) a c +\left (\int \frac {x}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}d x \right ) b^{2}}{6 b} \] Input:
int((c*x^2+b*x+a)^(1/4),x)
Output:
(8*(a + b*x + c*x**2)**(1/4)*a + 4*(a + b*x + c*x**2)**(1/4)*b*x - 4*int(( (a + b*x + c*x**2)**(1/4)*x)/(a + b*x + c*x**2),x)*a*c + int(((a + b*x + c *x**2)**(1/4)*x)/(a + b*x + c*x**2),x)*b**2)/(6*b)