Integrand size = 22, antiderivative size = 594 \[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=-\frac {4 (d+e x) (b d-2 a e+(2 c d-b e) x)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}+\frac {4 e (2 c d-b e) \left (a+b x+c x^2\right )^{3/4}}{c \left (b^2-4 a c\right )}+\frac {2 \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) (b+2 c x) \sqrt [4]{a+b x+c x^2}}{c^{3/2} \left (b^2-4 a c\right )^{3/2} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )}-\frac {\sqrt {2} \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{c^{7/4} \sqrt [4]{b^2-4 a c} (b+2 c x)}+\frac {\left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) \sqrt {\frac {(b+2 c x)^2}{\left (b^2-4 a c\right ) \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right )^2}} \left (1+\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{a+b x+c x^2}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{\sqrt {2} c^{7/4} \sqrt [4]{b^2-4 a c} (b+2 c x)} \]
-4*(e*x+d)*(b*d-2*a*e+(-b*e+2*c*d)*x)/(-4*a*c+b^2)/(c*x^2+b*x+a)^(1/4)+4*e *(-b*e+2*c*d)*(c*x^2+b*x+a)^(3/4)/c/(-4*a*c+b^2)+2*(4*c^2*d^2+3*b^2*e^2-4* c*e*(2*a*e+b*d))*(2*c*x+b)*(c*x^2+b*x+a)^(1/4)/c^(3/2)/(-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))+1/2*(4*c^2*d^2+3*b^2* e^2-4*c*e*(2*a*e+b*d))*(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^(7/4)/(-4*a*c+b^2)^(1/4)/(2*c*x+b)* 2^(1/2)-(4*c^2*d^2+3*b^2*e^2-4*c*e*(2*a*e+b*d))*(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(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))*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^(7 /4)/(-4*a*c+b^2)^(1/4)/(2*c*x+b)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.29 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.30 \[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\frac {-8 c \left (a b e^2+2 c^2 d^2 x+b^2 e^2 x+b c d (d-2 e x)-2 a c e (2 d+e x)\right )+\sqrt {2} \left (4 c^2 d^2+3 b^2 e^2-4 c e (b d+2 a e)\right ) (b+2 c x) \sqrt [4]{\frac {c (a+x (b+c x))}{-b^2+4 a c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {(b+2 c x)^2}{b^2-4 a c}\right )}{2 c^2 \left (b^2-4 a c\right ) \sqrt [4]{a+x (b+c x)}} \]
(-8*c*(a*b*e^2 + 2*c^2*d^2*x + b^2*e^2*x + b*c*d*(d - 2*e*x) - 2*a*c*e*(2* d + e*x)) + Sqrt[2]*(4*c^2*d^2 + 3*b^2*e^2 - 4*c*e*(b*d + 2*a*e))*(b + 2*c *x)*((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)])/(2*c^2*(b^2 - 4*a*c)*(a + x*(b + c*x ))^(1/4))
Time = 0.59 (sec) , antiderivative size = 661, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {1164, 27, 1160, 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.
\(\displaystyle \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 1164 |
\(\displaystyle -\frac {4 \int -\frac {2 c d^2+b e d-4 a e^2+3 e (2 c d-b e) x}{2 \sqrt [4]{c x^2+b x+a}}dx}{b^2-4 a c}-\frac {4 (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 \int \frac {2 c d^2+e (b d-4 a e)+3 e (2 c d-b e) x}{\sqrt [4]{c x^2+b x+a}}dx}{b^2-4 a c}-\frac {4 (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {2 \left (\frac {\left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) \int \frac {1}{\sqrt [4]{c x^2+b x+a}}dx}{2 c}+\frac {2 e \left (a+b x+c x^2\right )^{3/4} (2 c d-b e)}{c}\right )}{b^2-4 a c}-\frac {4 (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}\) |
\(\Big \downarrow \) 1094 |
\(\displaystyle \frac {2 \left (\frac {2 \sqrt {(b+2 c x)^2} \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) \int \frac {\sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{c (b+2 c x)}+\frac {2 e \left (a+b x+c x^2\right )^{3/4} (2 c d-b e)}{c}\right )}{b^2-4 a c}-\frac {4 (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {2 \left (\frac {2 \sqrt {(b+2 c x)^2} \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) \left (\frac {\sqrt {b^2-4 a c} \int \frac {1}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}-\frac {\sqrt {b^2-4 a c} \int \frac {1-\frac {2 \sqrt {c} \sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}\right )}{c (b+2 c x)}+\frac {2 e \left (a+b x+c x^2\right )^{3/4} (2 c d-b e)}{c}\right )}{b^2-4 a c}-\frac {4 (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2 \left (\frac {2 \sqrt {(b+2 c x)^2} \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) \left (\frac {\left (b^2-4 a c\right )^{3/4} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{4 \sqrt {2} c^{3/4} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}-\frac {\sqrt {b^2-4 a c} \int \frac {1-\frac {2 \sqrt {c} \sqrt {c x^2+b x+a}}{\sqrt {b^2-4 a c}}}{\sqrt {b^2-4 a c+4 c \left (c x^2+b x+a\right )}}d\sqrt [4]{c x^2+b x+a}}{2 \sqrt {c}}\right )}{c (b+2 c x)}+\frac {2 e \left (a+b x+c x^2\right )^{3/4} (2 c d-b e)}{c}\right )}{b^2-4 a c}-\frac {4 (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {2 \left (\frac {2 \sqrt {(b+2 c x)^2} \left (-4 c e (2 a e+b d)+3 b^2 e^2+4 c^2 d^2\right ) \left (\frac {\left (b^2-4 a c\right )^{3/4} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right ),\frac {1}{2}\right )}{4 \sqrt {2} c^{3/4} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}-\frac {\sqrt {b^2-4 a c} \left (\frac {\sqrt [4]{b^2-4 a c} \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right ) \sqrt {\frac {4 c \left (a+b x+c x^2\right )-4 a c+b^2}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )^2}} E\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{c x^2+b x+a}}{\sqrt [4]{b^2-4 a c}}\right )|\frac {1}{2}\right )}{\sqrt {2} \sqrt [4]{c} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}-\frac {\sqrt [4]{a+b x+c x^2} \sqrt {4 c \left (a+b x+c x^2\right )-4 a c+b^2}}{\left (b^2-4 a c\right ) \left (\frac {2 \sqrt {c} \sqrt {a+b x+c x^2}}{\sqrt {b^2-4 a c}}+1\right )}\right )}{2 \sqrt {c}}\right )}{c (b+2 c x)}+\frac {2 e \left (a+b x+c x^2\right )^{3/4} (2 c d-b e)}{c}\right )}{b^2-4 a c}-\frac {4 (d+e x) (-2 a e+x (2 c d-b e)+b d)}{\left (b^2-4 a c\right ) \sqrt [4]{a+b x+c x^2}}\) |
(-4*(d + e*x)*(b*d - 2*a*e + (2*c*d - b*e)*x))/((b^2 - 4*a*c)*(a + b*x + c *x^2)^(1/4)) + (2*((2*e*(2*c*d - b*e)*(a + b*x + c*x^2)^(3/4))/c + (2*(4*c ^2*d^2 + 3*b^2*e^2 - 4*c*e*(b*d + 2*a*e))*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]*Sqrt[a + b*x + c*x^2])/S qrt[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*Sq rt[2]*c^(3/4)*Sqrt[b^2 - 4*a*c + 4*c*(a + b*x + c*x^2)])))/(c*(b + 2*c*x)) ))/(b^2 - 4*a*c)
3.26.43.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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[(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]
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[((d_.) + (e_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol ] :> Simp[e*((a + b*x + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + Simp[(2*c*d - b *e)/(2*c) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m - 1)*(d*b - 2*a*e + (2*c*d - b*e)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c))), x] + Simp[1/((p + 1)*(b^2 - 4*a* c)) Int[(d + e*x)^(m - 2)*Simp[e*(2*a*e*(m - 1) + b*d*(2*p - m + 4)) - 2* c*d^2*(2*p + 3) + e*(b*e - 2*d*c)*(m + 2*p + 2)*x, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && LtQ[p, -1] && GtQ[m, 1] && Int QuadraticQ[a, b, c, d, e, m, p, x]
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]
\[\int \frac {\left (e x +d \right )^{2}}{\left (c \,x^{2}+b x +a \right )^{\frac {5}{4}}}d x\]
\[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}} \,d x } \]
integral((e^2*x^2 + 2*d*e*x + d^2)*(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)
\[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int \frac {\left (d + e x\right )^{2}}{\left (a + b x + c x^{2}\right )^{\frac {5}{4}}}\, dx \]
\[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}} \,d x } \]
\[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{2} + b x + a\right )}^{\frac {5}{4}}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^2}{\left (a+b x+c x^2\right )^{5/4}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{{\left (c\,x^2+b\,x+a\right )}^{5/4}} \,d x \]