3.2525 \(\int (a+b x+c x^2)^{5/4} \, dx\)

Optimal. Leaf size=236 \[ \frac {5 \left (b^2-4 a c\right )^{9/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{168 \sqrt {2} c^{9/4} (b+2 c x)}-\frac {5 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{7 c} \]

[Out]

-5/84*(-4*a*c+b^2)*(2*c*x+b)*(c*x^2+b*x+a)^(1/4)/c^2+1/7*(2*c*x+b)*(c*x^2+b*x+a)^(5/4)/c+5/336*(-4*a*c+b^2)^(9
/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^(9/4)/(2*c*x+b)*2^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.16, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {612, 623, 220} \[ -\frac {5 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^2}+\frac {5 \left (b^2-4 a c\right )^{9/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) F\left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{168 \sqrt {2} c^{9/4} (b+2 c x)}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{7 c} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)^(5/4),x]

[Out]

(-5*(b^2 - 4*a*c)*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/(84*c^2) + ((b + 2*c*x)*(a + b*x + c*x^2)^(5/4))/(7*c)
+ (5*(b^2 - 4*a*c)^(9/4)*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])*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])/(168*Sqrt[2]*c^(9/4)*(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 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]

Rubi steps

\begin {align*} \int \left (a+b x+c x^2\right )^{5/4} \, dx &=\frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{7 c}-\frac {\left (5 \left (b^2-4 a c\right )\right ) \int \sqrt [4]{a+b x+c x^2} \, dx}{28 c}\\ &=-\frac {5 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{7 c}+\frac {\left (5 \left (b^2-4 a c\right )^2\right ) \int \frac {1}{\left (a+b x+c x^2\right )^{3/4}} \, dx}{336 c^2}\\ &=-\frac {5 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{7 c}+\frac {\left (5 \left (b^2-4 a c\right )^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 )}{84 c^2 (b+2 c x)}\\ &=-\frac {5 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{7 c}+\frac {5 \left (b^2-4 a c\right )^{9/4} \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 )}{168 \sqrt {2} c^{9/4} (b+2 c x)}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.36, size = 125, normalized size = 0.53 \[ \frac {\sqrt [4]{a+x (b+c x)} \left (2 (b+2 c x) \left (4 c \left (8 a+3 c x^2\right )-5 b^2+12 b c x\right )-\frac {5 \sqrt {2} \left (b^2-4 a c\right )^{3/2} F\left (\left .\frac {1}{2} \sin ^{-1}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )\right |2\right )}{\sqrt [4]{\frac {c (a+x (b+c x))}{4 a c-b^2}}}\right )}{168 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)^(5/4),x]

[Out]

((a + x*(b + c*x))^(1/4)*(2*(b + 2*c*x)*(-5*b^2 + 12*b*c*x + 4*c*(8*a + 3*c*x^2)) - (5*Sqrt[2]*(b^2 - 4*a*c)^(
3/2)*EllipticF[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)))/(16
8*c^2)

________________________________________________________________________________________

fricas [F]  time = 1.37, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/4),x, algorithm="fricas")

[Out]

integral((c*x^2 + b*x + a)^(5/4), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/4),x, algorithm="giac")

[Out]

integrate((c*x^2 + b*x + a)^(5/4), x)

________________________________________________________________________________________

maple [F]  time = 2.32, size = 0, normalized size = 0.00 \[ \int \left (c \,x^{2}+b x +a \right )^{\frac {5}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)^(5/4),x)

[Out]

int((c*x^2+b*x+a)^(5/4),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((c*x^2 + b*x + a)^(5/4), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (c\,x^2+b\,x+a\right )}^{5/4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x + c*x^2)^(5/4),x)

[Out]

int((a + b*x + c*x^2)^(5/4), x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b x + c x^{2}\right )^{\frac {5}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)**(5/4),x)

[Out]

Integral((a + b*x + c*x**2)**(5/4), x)

________________________________________________________________________________________