\(\int \frac {1}{(a+b x+c x^2)^{5/4}} \, dx\) [2545]

Optimal result

Integrand size = 14, antiderivative size = 451 \[ \int \frac {1}{\left (a+b x+c x^2\right )^{5/4}} \, dx=-\frac {4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {8 \sqrt {c} (b+2 c x) \sqrt [4]{a+b x+c x^2}}{\left (b^2-4 a c\right )^{3/2} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}-\frac {4 \sqrt {2} \sqrt [4]{c} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac {2 \sqrt {2} \sqrt [4]{c} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)} \]

[Out]

-4*(2*c*x+b)/(-4*a*c+b^2)/(c*x^2+b*x+a)^(1/4)+8*(2*c*x+b)*(c*x^2+b*x+a)^(1/4)*c^(1/2)/(-4*a*c+b^2)^(3/2)/(1+2*
c^(1/2)*(c*x^2+b*x+a)^(1/2)/(-4*a*c+b^2)^(1/2))-4*c^(1/4)*(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)))*EllipticE(si
n(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))*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)/(-4*a*c+b^2)^(1/4)/(2*c*x+b)+2*c^(1/4)*(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*arcta
n(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))*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)
/(-4*a*c+b^2)^(1/4)/(2*c*x+b)

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 451, 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 = {628, 637, 311, 226, 1210} \[ \int \frac {1}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\frac {2 \sqrt {2} \sqrt [4]{c} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)}-\frac {4 \sqrt {2} \sqrt [4]{c} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b^2-4 a c} (b+2 c x)}-\frac {4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {8 \sqrt {c} (b+2 c x) \sqrt [4]{a+b x+c x^2}}{\left (b^2-4 a c\right )^{3/2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )} \]

[In]

Int[(a + b*x + c*x^2)^(-5/4),x]

[Out]

(-4*(b + 2*c*x))/((b^2 - 4*a*c)*(a + b*x + c*x^2)^(1/4)) + (8*Sqrt[c]*(b + 2*c*x)*(a + b*x + c*x^2)^(1/4))/((b
^2 - 4*a*c)^(3/2)*(1 + (2*Sqrt[c]*Sqrt[a + b*x + c*x^2])/Sqrt[b^2 - 4*a*c])) - (4*Sqrt[2]*c^(1/4)*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])/((b^2 - 4*a*c)^(1/4)*(b + 2*c*x)) + (2*Sqrt[2]*c^(1/4)*Sqrt[(b + 2*c*x)^2/((b^2 - 4*a*c)*(1 + (2*Sq
rt[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])/((b^2 - 4*a*c)^(1/4)*
(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 628

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] - Dist[2*c*((2*p + 3)/((p + 1)*(b^2 - 4*a*c))), Int[(a + b*x + c*x^2)^(p + 1), x], x] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && LtQ[p, -1] && NeQ[p, -3/2] && 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 {4 (b+2 c x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {(4 c) \int \frac {1}{\sqrt [4]{a+b x+c x^2}} \, 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}}+\frac {\left (16 c \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 )}{\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}}+\frac {\left (8 \sqrt {c} \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 )}{\sqrt {b^2-4 a c} (b+2 c x)}-\frac {\left (8 \sqrt {c} \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 )}{\sqrt {b^2-4 a c} (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}}+\frac {8 \sqrt {c} (b+2 c x) \sqrt [4]{a+b x+c x^2}}{\left (b^2-4 a c\right )^{3/2} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}-\frac {4 \sqrt {2} \sqrt [4]{c} \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 )}{\sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac {2 \sqrt {2} \sqrt [4]{c} \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 )}{\sqrt [4]{b^2-4 a c} (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 = 10.06 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.22 \[ \int \frac {1}{\left (a+b x+c x^2\right )^{5/4}} \, dx=-\frac {2 \sqrt {2} (b+2 c x) \left (\sqrt {2}-\sqrt [4]{\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )\right )}{\left (b^2-4 a c\right ) \sqrt [4]{a+x (b+c x)}} \]

[In]

Integrate[(a + b*x + c*x^2)^(-5/4),x]

[Out]

(-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))

Maple [F]

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

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(1/(a + b*x + c*x^2)^(5/4),x)

[Out]

int(1/(a + b*x + c*x^2)^(5/4), x)