3.2519 \(\int (a+b x+c x^2)^{3/4} \, dx\)

Optimal. Leaf size=452 \[ -\frac {3 \sqrt {b^2-4 a c} (b+2 c x) \sqrt [4]{a+b x+c x^2}}{10 c^{3/2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )}-\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{20 \sqrt {2} c^{7/4} (b+2 c x)}+\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{10 \sqrt {2} c^{7/4} (b+2 c x)}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c} \]

[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)

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Rubi [A]  time = 0.36, antiderivative size = 452, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {612, 623, 305, 220, 1196} \[ -\frac {3 \sqrt {b^2-4 a c} (b+2 c x) \sqrt [4]{a+b x+c x^2}}{10 c^{3/2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )}-\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{20 \sqrt {2} c^{7/4} (b+2 c x)}+\frac {3 \left (b^2-4 a c\right )^{7/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) E\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{10 \sqrt {2} c^{7/4} (b+2 c x)}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{3/4}}{5 c} \]

Antiderivative was successfully verified.

[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 220

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)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 305

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 612

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 623

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 1196

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)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin {align*} \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 {\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 ) \operatorname {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 ) \operatorname {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 ) \operatorname {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*}

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Mathematica [C]  time = 0.09, size = 119, normalized size = 0.26 \[ \frac {(b+2 c x) (a+x (b+c x))^{3/4} \left (3 \sqrt {2} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {3}{2};\frac {(b+2 c x)^2}{b^2-4 a c}\right )+8 \left (\frac {c (a+x (b+c x))}{4 a c-b^2}\right )^{3/4}\right )}{40 c \left (\frac {c (a+x (b+c x))}{4 a c-b^2}\right )^{3/4}} \]

Antiderivative was successfully verified.

[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))

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fricas [F]  time = 1.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c x^{2} + b x + a\right )}^{\frac {3}{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((c*x^2 + b*x + a)^(3/4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + b x + a\right )}^{\frac {3}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x + a)^(3/4), x)

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maple [F]  time = 2.48, size = 0, normalized size = 0.00 \[ \int \left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + b x + a\right )}^{\frac {3}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*x^2 + b*x + a)^(3/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (c\,x^2+b\,x+a\right )}^{3/4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(3/4),x)

[Out]

int((a + b*x + c*x^2)^(3/4), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b x + c x^{2}\right )^{\frac {3}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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