Integrand size = 14, antiderivative size = 452 \[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {3 \sqrt {b^2-4 a c} (b+2 c x) \sqrt [4]{a+b x+c x^2}}{10 c^{3/2} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}+\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{10 \sqrt {2} c^{7/4} (b+2 c x)}-\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{20 \sqrt {2} c^{7/4} (b+2 c x)} \]
1/5*(2*c*x+b)*(c*x^2+b*x+a)^(3/4)/c-3/10*(2*c*x+b)*(c*x^2+b*x+a)^(1/4)*(-4 *a*c+b^2)^(1/2)/c^(3/2)/(1+2*c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2 ))+3/20*(-4*a*c+b^2)^(7/4)*(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)))*EllipticE(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^(7/4)/(2*c*x+b)*2^(1/2)-3/40*(- 4*a*c+b^2)^(7/4)*(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^(7/4)/(2*c*x+b)*2^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.07 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.26 \[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\frac {(b+2 c x) (a+x (b+c x))^{3/4} \left (8 \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{3/4}+3 \sqrt {2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{40 c \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{3/4}} \]
((b + 2*c*x)*(a + x*(b + c*x))^(3/4)*(8*((c*(a + x*(b + c*x)))/(-b^2 + 4*a *c))^(3/4) + 3*Sqrt[2]*Hypergeometric2F1[1/4, 1/2, 3/2, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/(40*c*((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^(3/4))
Time = 0.46 (sec) , antiderivative size = 577, normalized size of antiderivative = 1.28, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1087, 1094, 834, 761, 1510}
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.
| \(\displaystyle \int \left (a+b x+c x^2\right )^{3/4} \, dx\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {3 \left (b^2-4 a c\right ) \int \frac {1}{\sqrt [4]{c x^2+b x+a}}dx}{20 c}\) |
| \(\Big \downarrow \) 1094 |
| \(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {3 \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \int \frac {\sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{5 c (b+2 c x)}\) |
| \(\Big \downarrow \) 834 |
| \(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {3 \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \left (\frac {\sqrt {b^2-4 a c} \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}-\frac {\sqrt {b^2-4 a c} \int \frac {1-\frac {2 \sqrt {c} \sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}\right )}{5 c (b+2 c x)}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {3 \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \left (\frac {\left (b^2-4 a c\right )^{3/4} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{4 \sqrt {2} c^{3/4} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}-\frac {\sqrt {b^2-4 a c} \int \frac {1-\frac {2 \sqrt {c} \sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}\right )}{5 c (b+2 c x)}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {3 \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \left (\frac {\left (b^2-4 a c\right )^{3/4} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{4 \sqrt {2} c^{3/4} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}-\frac {\sqrt {b^2-4 a c} \left (\frac {\sqrt [4]{b^2-4 a c} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt [4]{c} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}-\frac {\sqrt [4]{a+b x+c x^2} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )}\right )}{2 \sqrt {c}}\right )}{5 c (b+2 c x)}\) |
((b + 2*c*x)*(a + b*x + c*x^2)^(3/4))/(5*c) - (3*(b^2 - 4*a*c)*Sqrt[(b + 2 *c*x)^2]*(-1/2*(Sqrt[b^2 - 4*a*c]*(-(((a + b*x + c*x^2)^(1/4)*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]))) + ((b^2 - 4*a*c)^(1/4)*(1 + (2*Sqrt[c]*Sqr t[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)]*EllipticE[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^(1/4)*Sqrt[b^2 - 4*a*c + 4*c*(a + b*x + c*x^2)])))/Sqrt[c] + ((b^2 - 4*a*c)^(3/4)*(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])/(4*Sqrt[2]*c^(3/4)*Sqrt[b^2 - 4*a*c + 4*c*(a + b*x + c*x^2 )])))/(5*c*(b + 2*c*x))
3.26.19.3.1 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[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], 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_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
\[\int \left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}d x\]
\[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {3}{4}} \,d x } \]
\[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\int \left (a + b x + c x^{2}\right )^{\frac {3}{4}}\, dx \]
\[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {3}{4}} \,d x } \]
\[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {3}{4}} \,d x } \]
Timed out. \[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\int {\left (c\,x^2+b\,x+a\right )}^{3/4} \,d x \]