\(\int \frac {1}{(a+b x+c x^2)^{5/4}} \, dx\) [757]

Optimal result

Integrand size = 14, antiderivative size = 119 \[ \int \frac {1}{\left (a+b x+c x^2\right )^{5/4}} \, dx=-\frac {4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {4 \sqrt {2} \sqrt [4]{-\frac {c \left (a+b x+c x^2\right )}{b^2-4 a c}} E\left (\left .\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )\right |2\right )}{\sqrt {b^2-4 a c} \sqrt [4]{a+b x+c x^2}} \] Output:

(-8*c*x-4*b)/(-4*a*c+b^2)/(c*x^2+b*x+a)^(1/4)+4*2^(1/2)*(-c*(c*x^2+b*x+a)/ 
(-4*a*c+b^2))^(1/4)*EllipticE(sin(1/2*arcsin((2*c*x+b)/(-4*a*c+b^2)^(1/2)) 
),2^(1/2))/(-4*a*c+b^2)^(1/2)/(c*x^2+b*x+a)^(1/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.07 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.84 \[ \int \frac {1}{\left (a+b x+c x^2\right )^{5/4}} \, dx=-\frac {2 \sqrt {2} (b+2 c x) \left (\sqrt {2}-\sqrt [4]{\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{\left (b^2-4 a c\right ) \sqrt [4]{a+x (b+c x)}} \] Input:

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

Output:

(-2*Sqrt[2]*(b + 2*c*x)*(Sqrt[2] - ((c*(a + x*(b + c*x)))/(-b^2 + 4*a*c))^ 
(1/4)*Hypergeometric2F1[1/4, 1/2, 3/2, (b + 2*c*x)^2/(b^2 - 4*a*c)]))/((b^ 
2 - 4*a*c)*(a + x*(b + c*x))^(1/4))
 

Rubi [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(580\) vs. \(2(119)=238\).

Time = 0.49 (sec) , antiderivative size = 580, normalized size of antiderivative = 4.87, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {1089, 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.

\(\displaystyle \int \frac {1}{\left (a+b x+c x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 1089

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

\(\Big \downarrow \) 1094

\(\displaystyle \frac {16 c \sqrt {(b+2 c x)^2} \int \frac {\sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{\left (b^2-4 a c\right ) (b+2 c x)}-\frac {4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}\)

\(\Big \downarrow \) 834

\(\displaystyle \frac {16 c \sqrt {(b+2 c x)^2} \left (\frac {\sqrt {b^2-4 a c} \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}-\frac {\sqrt {b^2-4 a c} \int \frac {1-\frac {2 \sqrt {c} \sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}\right )}{\left (b^2-4 a c\right ) (b+2 c x)}-\frac {4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {16 c \sqrt {(b+2 c x)^2} \left (\frac {\left (b^2-4 a c\right )^{3/4} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{4 \sqrt {2} c^{3/4} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}-\frac {\sqrt {b^2-4 a c} \int \frac {1-\frac {2 \sqrt {c} \sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}\right )}{\left (b^2-4 a c\right ) (b+2 c x)}-\frac {4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}\)

\(\Big \downarrow \) 1510

\(\displaystyle \frac {16 c \sqrt {(b+2 c x)^2} \left (\frac {\left (b^2-4 a c\right )^{3/4} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{4 \sqrt {2} c^{3/4} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}-\frac {\sqrt {b^2-4 a c} \left (\frac {\sqrt [4]{b^2-4 a c} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt [4]{c} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}-\frac {\sqrt [4]{a+b x+c x^2} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )}\right )}{2 \sqrt {c}}\right )}{\left (b^2-4 a c\right ) (b+2 c x)}-\frac {4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}\)

Input:

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

Output:

(-4*(b + 2*c*x))/((b^2 - 4*a*c)*(a + b*x + c*x^2)^(1/4)) + (16*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]*S 
qrt[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)])))/((b^2 - 4*a*c)*(b + 2*c*x))
 

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 834

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]
 

rule 1089

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) 
*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] - Simp[2*c*((2*p + 
 3)/((p + 1)*(b^2 - 4*a*c)))   Int[(a + b*x + c*x^2)^(p + 1), x], x] /; Fre 
eQ[{a, b, c}, x] && LtQ[p, -1] && (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]
 

rule 1510

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]
 

Maple [F]

\[\int \frac {1}{\left (c \,x^{2}+b x +a \right )^{\frac {5}{4}}}d x\]

Input:

int(1/(c*x^2+b*x+a)^(5/4),x)
 

Output:

int(1/(c*x^2+b*x+a)^(5/4),x)
 

Fricas [F]

\[ \int \frac {1}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(1/(c*x^2+b*x+a)^(5/4),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {1}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int \frac {1}{\left (a + b x + c x^{2}\right )^{\frac {5}{4}}}\, dx \] Input:

integrate(1/(c*x**2+b*x+a)**(5/4),x)
 

Output:

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

Maxima [F]

\[ \int \frac {1}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(1/(c*x^2+b*x+a)^(5/4),x, algorithm="maxima")
 

Output:

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

Giac [F]

\[ \int \frac {1}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate(1/(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. \[ \int \frac {1}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int \frac {1}{{\left (c\,x^2+b\,x+a\right )}^{5/4}} \,d x \] Input:

int(1/(a + b*x + c*x^2)^(5/4),x)
 

Output:

int(1/(a + b*x + c*x^2)^(5/4), x)
 

Reduce [F]

\[ \int \frac {1}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int \frac {1}{\left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} a +\left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} b x +\left (c \,x^{2}+b x +a \right )^{\frac {1}{4}} c \,x^{2}}d x \] Input:

int(1/(c*x^2+b*x+a)^(5/4),x)
 

Output:

int(1/((a + b*x + c*x**2)**(1/4)*a + (a + b*x + c*x**2)**(1/4)*b*x + (a + 
b*x + c*x**2)**(1/4)*c*x**2),x)