\(\int (a+b x+c x^2)^{5/4} \, dx\) [737]

Optimal result

Integrand size = 14, antiderivative size = 158 \[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=-\frac {5 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{7 c}+\frac {5 \left (-b^2+4 a c\right )^{5/2} \left (-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right ),2\right )}{84 \sqrt {2} c^3 \left (a+b x+c x^2\right )^{3/4}} \] Output:

-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+5/168*(4*a*c-b^2)^(5/2)*(-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^3/(c*x^2+b*x+a)^(3/4)
 

Mathematica [A] (warning: unable to verify)

Time = 10.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.79 \[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=\frac {\sqrt [4]{a+x (b+c x)} \left (2 (b+2 c x) \left (-5 b^2+12 b c x+4 c \left (8 a+3 c x^2\right )\right )-\frac {5 \sqrt {2} \left (b^2-4 a c\right )^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ),2\right )}{\sqrt [4]{\frac {c (a+x (b+c x))}{-b^2+4 a c}}}\right )}{168 c^2} \] Input:

Integrate[(a + b*x + c*x^2)^(5/4),x]
 

Output:

((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)))/(168* 
c^2)
 

Rubi [A] (warning: unable to verify)

Time = 0.35 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.86, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1087, 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.

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

\(\Big \downarrow \) 1087

\(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{7 c}-\frac {5 \left (b^2-4 a c\right ) \int \sqrt [4]{c x^2+b x+a}dx}{28 c}\)

\(\Big \downarrow \) 1087

\(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{7 c}-\frac {5 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right ) \int \frac {1}{\left (c x^2+b x+a\right )^{3/4}}dx}{12 c}\right )}{28 c}\)

\(\Big \downarrow \) 1094

\(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{7 c}-\frac {5 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right ) \sqrt {(b+2 c x)^2} \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{3 c (b+2 c x)}\right )}{28 c}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{7 c}-\frac {5 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right )^{5/4} \sqrt {(b+2 c x)^2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} c^{5/4} (b+2 c x) \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}\right )}{28 c}\)

Input:

Int[(a + b*x + c*x^2)^(5/4),x]
 

Output:

((b + 2*c*x)*(a + b*x + c*x^2)^(5/4))/(7*c) - (5*(b^2 - 4*a*c)*(((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])*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])/(6*Sqrt[2]*c^(5/4)*(b + 2* 
c*x)*Sqrt[b^2 - 4*a*c + 4*c*(a + b*x + c*x^2)])))/(28*c)
 

Defintions of rubi rules used

rule 761

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 1087

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])
 

rule 1094

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]
 

Maple [F]

\[\int \left (c \,x^{2}+b x +a \right )^{\frac {5}{4}}d x\]

Input:

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

Output:

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

Fricas [F]

\[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{4}} \,d x } \] Input:

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

Output:

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

Sympy [F]

\[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=\int \left (a + b x + c x^{2}\right )^{\frac {5}{4}}\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{4}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{4}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=\int {\left (c\,x^2+b\,x+a\right )}^{5/4} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=\frac {160 \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} a^{2} c -16 \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} a \,b^{2}+128 \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} a b c x +4 \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} b^{3} x +72 \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} b^{2} c \,x^{2}+48 \left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} b \,c^{2} x^{3}-80 \left (\int \frac {x}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}d x \right ) a^{2} c^{2}+40 \left (\int \frac {x}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}d x \right ) a \,b^{2} c -5 \left (\int \frac {x}{\left (c \,x^{2}+b x +a \right )^{\frac {3}{4}}}d x \right ) b^{4}}{168 b c} \] Input:

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

Output:

(160*(a + b*x + c*x**2)**(1/4)*a**2*c - 16*(a + b*x + c*x**2)**(1/4)*a*b** 
2 + 128*(a + b*x + c*x**2)**(1/4)*a*b*c*x + 4*(a + b*x + c*x**2)**(1/4)*b* 
*3*x + 72*(a + b*x + c*x**2)**(1/4)*b**2*c*x**2 + 48*(a + b*x + c*x**2)**( 
1/4)*b*c**2*x**3 - 80*int(((a + b*x + c*x**2)**(1/4)*x)/(a + b*x + c*x**2) 
,x)*a**2*c**2 + 40*int(((a + b*x + c*x**2)**(1/4)*x)/(a + b*x + c*x**2),x) 
*a*b**2*c - 5*int(((a + b*x + c*x**2)**(1/4)*x)/(a + b*x + c*x**2),x)*b**4 
)/(168*b*c)