Integrand size = 19, antiderivative size = 270 \[ \int \frac {(d+e x)^2}{\left (a+c x^4\right )^{3/2}} \, dx=\frac {x (d+e x)^2}{2 a \sqrt {a+c x^4}}-\frac {e^2 x \sqrt {a+c x^4}}{2 a \sqrt {c} \left (\sqrt {a}+\sqrt {c} x^2\right )}+\frac {e^2 \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{2 a^{3/4} c^{3/4} \sqrt {a+c x^4}}+\frac {\left (\sqrt {c} d^2-\sqrt {a} e^2\right ) \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 a^{5/4} c^{3/4} \sqrt {a+c x^4}} \]
1/2*x*(e*x+d)^2/a/(c*x^4+a)^(1/2)-1/2*e^2*x*(c*x^4+a)^(1/2)/a/c^(1/2)/(a^( 1/2)+x^2*c^(1/2))+1/2*e^2*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^(1/2)/cos(2 *arctan(c^(1/4)*x/a^(1/4)))*EllipticE(sin(2*arctan(c^(1/4)*x/a^(1/4))),1/2 *2^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/ a^(3/4)/c^(3/4)/(c*x^4+a)^(1/2)+1/4*(cos(2*arctan(c^(1/4)*x/a^(1/4)))^2)^( 1/2)/cos(2*arctan(c^(1/4)*x/a^(1/4)))*EllipticF(sin(2*arctan(c^(1/4)*x/a^( 1/4))),1/2*2^(1/2))*(-e^2*a^(1/2)+d^2*c^(1/2))*(a^(1/2)+x^2*c^(1/2))*((c*x ^4+a)/(a^(1/2)+x^2*c^(1/2))^2)^(1/2)/a^(5/4)/c^(3/4)/(c*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.07 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.40 \[ \int \frac {(d+e x)^2}{\left (a+c x^4\right )^{3/2}} \, dx=\frac {x \left (3 d (d+2 e x)+3 d^2 \sqrt {1+\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^4}{a}\right )+2 e^2 x^2 \sqrt {1+\frac {c x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {c x^4}{a}\right )\right )}{6 a \sqrt {a+c x^4}} \]
(x*(3*d*(d + 2*e*x) + 3*d^2*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[1/4, 1/2 , 5/4, -((c*x^4)/a)] + 2*e^2*x^2*Sqrt[1 + (c*x^4)/a]*Hypergeometric2F1[3/4 , 3/2, 7/4, -((c*x^4)/a)]))/(6*a*Sqrt[a + c*x^4])
Time = 0.38 (sec) , antiderivative size = 268, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {2394, 25, 1512, 27, 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.
| \(\displaystyle \int \frac {(d+e x)^2}{\left (a+c x^4\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 2394 |
| \(\displaystyle \frac {x (d+e x)^2}{2 a \sqrt {a+c x^4}}-\frac {\int -\frac {d^2-e^2 x^2}{\sqrt {c x^4+a}}dx}{2 a}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {d^2-e^2 x^2}{\sqrt {c x^4+a}}dx}{2 a}+\frac {x (d+e x)^2}{2 a \sqrt {a+c x^4}}\) |
| \(\Big \downarrow \) 1512 |
| \(\displaystyle \frac {\left (d^2-\frac {\sqrt {a} e^2}{\sqrt {c}}\right ) \int \frac {1}{\sqrt {c x^4+a}}dx+\frac {\sqrt {a} e^2 \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {a} \sqrt {c x^4+a}}dx}{\sqrt {c}}}{2 a}+\frac {x (d+e x)^2}{2 a \sqrt {a+c x^4}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (d^2-\frac {\sqrt {a} e^2}{\sqrt {c}}\right ) \int \frac {1}{\sqrt {c x^4+a}}dx+\frac {e^2 \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+a}}dx}{\sqrt {c}}}{2 a}+\frac {x (d+e x)^2}{2 a \sqrt {a+c x^4}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {\frac {e^2 \int \frac {\sqrt {a}-\sqrt {c} x^2}{\sqrt {c x^4+a}}dx}{\sqrt {c}}+\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (d^2-\frac {\sqrt {a} e^2}{\sqrt {c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}}}{2 a}+\frac {x (d+e x)^2}{2 a \sqrt {a+c x^4}}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {\frac {\left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} \left (d^2-\frac {\sqrt {a} e^2}{\sqrt {c}}\right ) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 \sqrt [4]{a} \sqrt [4]{c} \sqrt {a+c x^4}}+\frac {e^2 \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {c} x^2\right ) \sqrt {\frac {a+c x^4}{\left (\sqrt {a}+\sqrt {c} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{c} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{c} \sqrt {a+c x^4}}-\frac {x \sqrt {a+c x^4}}{\sqrt {a}+\sqrt {c} x^2}\right )}{\sqrt {c}}}{2 a}+\frac {x (d+e x)^2}{2 a \sqrt {a+c x^4}}\) |
(x*(d + e*x)^2)/(2*a*Sqrt[a + c*x^4]) + ((e^2*(-((x*Sqrt[a + c*x^4])/(Sqrt [a] + Sqrt[c]*x^2)) + (a^(1/4)*(Sqrt[a] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(S qrt[a] + Sqrt[c]*x^2)^2]*EllipticE[2*ArcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(c ^(1/4)*Sqrt[a + c*x^4])))/Sqrt[c] + ((d^2 - (Sqrt[a]*e^2)/Sqrt[c])*(Sqrt[a ] + Sqrt[c]*x^2)*Sqrt[(a + c*x^4)/(Sqrt[a] + Sqrt[c]*x^2)^2]*EllipticF[2*A rcTan[(c^(1/4)*x)/a^(1/4)], 1/2])/(2*a^(1/4)*c^(1/4)*Sqrt[a + c*x^4]))/(2* a)
3.3.20.3.1 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[((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[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q Int[1/Sqrt[a + c*x^4], x], x] - Simp[e/q Int[(1 - q*x^2)/Sqrt[a + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, c , d, e}, x] && PosQ[c/a]
Int[(Pq_)*((a_) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(-x)*Pq*((a + b *x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[1/(a*n*(p + 1)) Int[ExpandToSum[n *(p + 1)*Pq + D[x*Pq, x], x]*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x ] && PolyQ[Pq, x] && IGtQ[n, 0] && LtQ[p, -1] && LtQ[Expon[Pq, x], n - 1]
Result contains complex when optimal does not.
Time = 1.11 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.86
| method | result | size |
| elliptic | \(-\frac {2 c \left (-\frac {e^{2} x^{3}}{4 a c}-\frac {e d \,x^{2}}{2 c a}-\frac {d^{2} x}{4 a c}\right )}{\sqrt {\left (x^{4}+\frac {a}{c}\right ) c}}+\frac {d^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}-\frac {i e^{2} \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\) | \(233\) |
| default | \(d^{2} \left (\frac {x}{2 a \sqrt {\left (x^{4}+\frac {a}{c}\right ) c}}+\frac {\sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )}{2 a \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}}\right )+e^{2} \left (\frac {x^{3}}{2 a \sqrt {\left (x^{4}+\frac {a}{c}\right ) c}}-\frac {i \sqrt {1-\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {c}\, x^{2}}{\sqrt {a}}}\, \left (F\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )-E\left (x \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {c}}{\sqrt {a}}}\, \sqrt {c \,x^{4}+a}\, \sqrt {c}}\right )+\frac {e d \,x^{2}}{\sqrt {c \,x^{4}+a}\, a}\) | \(239\) |
-2*c*(-1/4/a*e^2/c*x^3-1/2/c*e*d/a*x^2-1/4/a*d^2/c*x)/((x^4+a/c)*c)^(1/2)+ 1/2/a*d^2/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)*x^2)^(1/2)*(1+I/a ^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)*EllipticF(x*(I/a^(1/2)*c^(1/2))^ (1/2),I)-1/2*I/a^(1/2)*e^2/(I/a^(1/2)*c^(1/2))^(1/2)*(1-I/a^(1/2)*c^(1/2)* x^2)^(1/2)*(1+I/a^(1/2)*c^(1/2)*x^2)^(1/2)/(c*x^4+a)^(1/2)/c^(1/2)*(Ellipt icF(x*(I/a^(1/2)*c^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*c^(1/2))^(1/2),I ))
Time = 0.12 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.53 \[ \int \frac {(d+e x)^2}{\left (a+c x^4\right )^{3/2}} \, dx=\frac {{\left (c e^{2} x^{4} + a e^{2}\right )} \sqrt {a} \left (-\frac {c}{a}\right )^{\frac {3}{4}} E(\arcsin \left (x \left (-\frac {c}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left ({\left (c d^{2} + c e^{2}\right )} x^{4} + a d^{2} + a e^{2}\right )} \sqrt {a} \left (-\frac {c}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {c}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) + {\left (c e^{2} x^{3} + 2 \, c d e x^{2} + c d^{2} x\right )} \sqrt {c x^{4} + a}}{2 \, {\left (a c^{2} x^{4} + a^{2} c\right )}} \]
1/2*((c*e^2*x^4 + a*e^2)*sqrt(a)*(-c/a)^(3/4)*elliptic_e(arcsin(x*(-c/a)^( 1/4)), -1) - ((c*d^2 + c*e^2)*x^4 + a*d^2 + a*e^2)*sqrt(a)*(-c/a)^(3/4)*el liptic_f(arcsin(x*(-c/a)^(1/4)), -1) + (c*e^2*x^3 + 2*c*d*e*x^2 + c*d^2*x) *sqrt(c*x^4 + a))/(a*c^2*x^4 + a^2*c)
\[ \int \frac {(d+e x)^2}{\left (a+c x^4\right )^{3/2}} \, dx=\int \frac {\left (d + e x\right )^{2}}{\left (a + c x^{4}\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(d+e x)^2}{\left (a+c x^4\right )^{3/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(d+e x)^2}{\left (a+c x^4\right )^{3/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(d+e x)^2}{\left (a+c x^4\right )^{3/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{{\left (c\,x^4+a\right )}^{3/2}} \,d x \]