\(\int (a+b x+c x^2)^{3/4} \, dx\) [2519]

Optimal result

Integrand size = 14, antiderivative size = 452 \[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {3 \sqrt {b^2-4 a c} (b+2 c x) \sqrt [4]{a+b x+c x^2}}{10 c^{3/2} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}+\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{10 \sqrt {2} c^{7/4} (b+2 c x)}-\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{20 \sqrt {2} c^{7/4} (b+2 c x)} \]

[Out]

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)))*Ellip
ticE(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(s
in(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)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 452, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {626, 637, 311, 226, 1210} \[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=-\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{20 \sqrt {2} c^{7/4} (b+2 c x)}+\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{10 \sqrt {2} c^{7/4} (b+2 c x)}-\frac {3 \sqrt {b^2-4 a c} (b+2 c x) \sqrt [4]{a+b x+c x^2}}{10 c^{3/2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c} \]

[In]

Int[(a + b*x + c*x^2)^(3/4),x]

[Out]

((b + 2*c*x)*(a + b*x + c*x^2)^(3/4))/(5*c) - (3*Sqrt[b^2 - 4*a*c]*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(10*c^
(3/2)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])) + (3*(b^2 - 4*a*c)^(7/4)*Sqrt[(b + 2*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)]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x
^2])/Sqrt[b^2 - 4*a*c])*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
])/(10*Sqrt[2]*c^(7/4)*(b + 2*c*x)) - (3*(b^2 - 4*a*c)^(7/4)*Sqrt[(b + 2*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)]*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])*Ellip
ticF[2*ArcTan[(Sqrt[2]*c^(1/4)*(a + b*x + c*x^2)^(1/4))/(b^2 - 4*a*c)^(1/4)], 1/2])/(20*Sqrt[2]*c^(7/4)*(b + 2
*c*x))

Rule 226

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]

Rule 311

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 626

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] - Dist[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]
&& NeQ[b^2 - 4*a*c, 0] && GtQ[p, 0] && IntegerQ[4*p]

Rule 637

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{d = Denominator[p]}, Dist[d*(Sqrt[(b + 2*c*x)
^2]/(b + 2*c*x)), Subst[Int[x^(d*(p + 1) - 1)/Sqrt[b^2 - 4*a*c + 4*c*x^d], x], x, (a + b*x + c*x^2)^(1/d)], x]
/; 3 <= d <= 4] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && RationalQ[p]

Rule 1210

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
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps \begin{align*} \text {integral}& = \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {\left (3 \left (b^2-4 a c\right )\right ) \int \frac {1}{\sqrt [4]{a+b x+c x^2}} \, dx}{20 c} \\ & = \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {\left (3 \left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {x^2}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{5 c (b+2 c x)} \\ & = \frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c}-\frac {\left (3 \left (b^2-4 a c\right )^{3/2} \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{10 c^{3/2} (b+2 c x)}+\frac {\left (3 \left (b^2-4 a c\right )^{3/2} \sqrt {(b+2 c x)^2}\right ) \text {Subst}\left (\int \frac {1-\frac {2 \sqrt {c} x^2}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c x^4}} \, dx,x,\sqrt [4]{a+b x+c x^2}\right )}{10 c^{3/2} (b+2 c x)} \\ & = \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 \tan ^{-1}\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 ) F\left (2 \tan ^{-1}\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)} \\ \end{align*}

Mathematica [C] (verified)

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 \[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\frac {(b+2 c x) (a+x (b+c x))^{3/4} \left (8 \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{3/4}+3 \sqrt {2} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{40 c \left (\frac {c (a+x (b+c x))}{-b^2+4 a c}\right )^{3/4}} \]

[In]

Integrate[(a + b*x + c*x^2)^(3/4),x]

[Out]

((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]*Hypergeometri
c2F1[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))

Maple [F]

\[\int \left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}d x\]

[In]

int((c*x^2+b*x+a)^(3/4),x)

[Out]

int((c*x^2+b*x+a)^(3/4),x)

Fricas [F]

\[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {3}{4}} \,d x } \]

[In]

integrate((c*x^2+b*x+a)^(3/4),x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^(3/4), x)

Sympy [F]

\[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\int \left (a + b x + c x^{2}\right )^{\frac {3}{4}}\, dx \]

[In]

integrate((c*x**2+b*x+a)**(3/4),x)

[Out]

Integral((a + b*x + c*x**2)**(3/4), x)

Maxima [F]

\[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {3}{4}} \,d x } \]

[In]

integrate((c*x^2+b*x+a)^(3/4),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(3/4), x)

Giac [F]

\[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {3}{4}} \,d x } \]

[In]

integrate((c*x^2+b*x+a)^(3/4),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(3/4), x)

Mupad [F(-1)]

Timed out. \[ \int \left (a+b x+c x^2\right )^{3/4} \, dx=\int {\left (c\,x^2+b\,x+a\right )}^{3/4} \,d x \]

[In]

int((a + b*x + c*x^2)^(3/4),x)

[Out]

int((a + b*x + c*x^2)^(3/4), x)