Integrand size = 14, antiderivative size = 236 \[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=-\frac {5 \left (b^2-4 a c\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{84 c^2}+\frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{7 c}+\frac {5 \left (b^2-4 a c\right )^{9/4} \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{168 \sqrt {2} c^{9/4} (b+2 c x)} \]
-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)
Time = 10.24 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.53 \[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=\frac {\sqrt [4]{a+x (b+c x)} \left (2 (b+2 c x) \left (-5 b^2+12 b c x+4 c \left (8 a+3 c x^2\right )\right )-\frac {5 \sqrt {2} \left (b^2-4 a c\right )^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right ),2\right )}{\sqrt [4]{\frac {c (a+x (b+c x))}{-b^2+4 a c}}}\right )}{168 c^2} \]
((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)
Time = 0.32 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.25, 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 |
\(\displaystyle \frac {(b+2 c x) \left (a+b x+c x^2\right )^{5/4}}{7 c}-\frac {5 \left (b^2-4 a c\right ) \left (\frac {(b+2 c x) \sqrt [4]{a+b x+c x^2}}{3 c}-\frac {\left (b^2-4 a c\right )^{5/4} \sqrt {(b+2 c x)^2} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} c^{5/4} (b+2 c x) \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}\right )}{28 c}\) |
((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)
3.26.25.3.1 Defintions of rubi rules used
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]
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])
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]
\[\int \left (c \,x^{2}+b x +a \right )^{\frac {5}{4}}d x\]
\[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{4}} \,d x } \]
\[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=\int \left (a + b x + c x^{2}\right )^{\frac {5}{4}}\, dx \]
\[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{4}} \,d x } \]
\[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=\int { {\left (c x^{2} + b x + a\right )}^{\frac {5}{4}} \,d x } \]
Timed out. \[ \int \left (a+b x+c x^2\right )^{5/4} \, dx=\int {\left (c\,x^2+b\,x+a\right )}^{5/4} \,d x \]