Integrand size = 37, antiderivative size = 234 \[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/4}} \, dx=\frac {5 \left (c d^2-a e^2\right ) \sqrt [4]{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c^2 d^2}+\frac {2 (d+e x) \sqrt [4]{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{3 c d}-\frac {5 \sqrt {e} \left (c d^2-a e^2\right )^{3/2} (a e+c d x)^{3/2} \left (\frac {c d (d+e x)}{e (a e+c d x)}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {c d^2-a e^2}}{\sqrt {e} \sqrt {a e+c d x}}\right ),2\right )}{3 c^3 d^3 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/4}} \] Output:
5/3*(-a*e^2+c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/4)/c^2/d^2+2/3*(e* x+d)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/4)/c/d-5/3*e^(1/2)*(-a*e^2+c*d^2 )^(3/2)*(c*d*x+a*e)^(3/2)*(c*d*(e*x+d)/e/(c*d*x+a*e))^(3/4)*InverseJacobiA M(1/2*arctan((-a*e^2+c*d^2)^(1/2)/e^(1/2)/(c*d*x+a*e)^(1/2)),2^(1/2))/c^3/ d^3/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.05 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.39 \[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/4}} \, dx=\frac {4 (d+e x) \sqrt [4]{(a e+c d x) (d+e x)} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {1}{4},\frac {5}{4},\frac {e (a e+c d x)}{-c d^2+a e^2}\right )}{c d \left (\frac {c d (d+e x)}{c d^2-a e^2}\right )^{5/4}} \] Input:
Integrate[(d + e*x)^2/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/4),x]
Output:
(4*(d + e*x)*((a*e + c*d*x)*(d + e*x))^(1/4)*Hypergeometric2F1[-5/4, 1/4, 5/4, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/(c*d*((c*d*(d + e*x))/(c*d^2 - a*e^2))^(5/4))
Leaf count is larger than twice the leaf count of optimal. \(509\) vs. \(2(234)=468\).
Time = 0.51 (sec) , antiderivative size = 509, normalized size of antiderivative = 2.18, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {1166, 27, 1160, 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.
\(\displaystyle \int \frac {(d+e x)^2}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 1166 |
\(\displaystyle \frac {2 \int \frac {5 e \left (c d^2-a e^2\right ) (d+e x)}{4 \left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )^{3/4}}dx}{3 c d e}+\frac {2 (d+e x) \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {5 \left (c d^2-a e^2\right ) \int \frac {d+e x}{\left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )^{3/4}}dx}{6 c d}+\frac {2 (d+e x) \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d}\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {5 \left (c d^2-a e^2\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \int \frac {1}{\left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )^{3/4}}dx}{2 d}+\frac {2 \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d}\right )}{6 c d}+\frac {2 (d+e x) \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d}\) |
\(\Big \downarrow \) 1094 |
\(\displaystyle \frac {5 \left (c d^2-a e^2\right ) \left (\frac {2 \left (d^2-\frac {a e^2}{c}\right ) \sqrt {\left (a e^2+c d^2+2 c d e x\right )^2} \int \frac {1}{\sqrt {\left (c d^2-a e^2\right )^2+4 c d e \left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )}}d\sqrt [4]{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}{d \left (a e^2+c d^2+2 c d e x\right )}+\frac {2 \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d}\right )}{6 c d}+\frac {2 (d+e x) \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {5 \left (c d^2-a e^2\right ) \left (\frac {\sqrt {c d^2-a e^2} \left (d^2-\frac {a e^2}{c}\right ) \sqrt {\left (a e^2+c d^2+2 c d e x\right )^2} \left (\frac {2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d^2-a e^2}+1\right ) \sqrt {\frac {4 c d e \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )+\left (c d^2-a e^2\right )^2}{\left (c d^2-a e^2\right )^2 \left (\frac {2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d^2-a e^2}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt [4]{e} \sqrt [4]{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}{\sqrt {c d^2-a e^2}}\right ),\frac {1}{2}\right )}{\sqrt {2} \sqrt [4]{c} d^{5/4} \sqrt [4]{e} \left (a e^2+c d^2+2 c d e x\right ) \sqrt {4 c d e \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )+\left (c d^2-a e^2\right )^2}}+\frac {2 \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d}\right )}{6 c d}+\frac {2 (d+e x) \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d}\) |
Input:
Int[(d + e*x)^2/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/4),x]
Output:
(2*(d + e*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(1/4))/(3*c*d) + (5*( c*d^2 - a*e^2)*((2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(1/4))/(c*d) + (Sqrt[c*d^2 - a*e^2]*(d^2 - (a*e^2)/c)*Sqrt[(c*d^2 + a*e^2 + 2*c*d*e*x)^2] *(1 + (2*Sqrt[c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^ 2])/(c*d^2 - a*e^2))*Sqrt[((c*d^2 - a*e^2)^2 + 4*c*d*e*(a*d*e + (c*d^2 + a *e^2)*x + c*d*e*x^2))/((c*d^2 - a*e^2)^2*(1 + (2*Sqrt[c]*Sqrt[d]*Sqrt[e]*S qrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/(c*d^2 - a*e^2))^2)]*EllipticF [2*ArcTan[(Sqrt[2]*c^(1/4)*d^(1/4)*e^(1/4)*(a*d*e + (c*d^2 + a*e^2)*x + c* d*e*x^2)^(1/4))/Sqrt[c*d^2 - a*e^2]], 1/2])/(Sqrt[2]*c^(1/4)*d^(5/4)*e^(1/ 4)*(c*d^2 + a*e^2 + 2*c*d*e*x)*Sqrt[(c*d^2 - a*e^2)^2 + 4*c*d*e*(a*d*e + ( c*d^2 + a*e^2)*x + c*d*e*x^2)])))/(6*c*d)
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[((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[e*(d + e*x)^(m - 1)*((a + b*x + c*x^2)^(p + 1)/(c*(m + 2*p + 1))), x] + Simp[1/(c*(m + 2*p + 1)) Int[(d + e*x)^(m - 2)*Simp[c*d^2*(m + 2*p + 1) - e*(a*e*(m - 1) + b*d*(p + 1)) + e*(2*c*d - b*e)*(m + p)*x, x]* (a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && If[Ration alQ[m], GtQ[m, 1], SumSimplerQ[m, -2]] && NeQ[m + 2*p + 1, 0] && IntQuadrat icQ[a, b, c, d, e, m, p, x]
\[\int \frac {\left (e x +d \right )^{2}}{{\left (a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d \,x^{2} e \right )}^{\frac {3}{4}}}d x\]
Input:
int((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(3/4),x)
Output:
int((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(3/4),x)
\[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/4),x, algorithm=" fricas")
Output:
integral((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(1/4)*(e*x + d)/(c*d*x + a*e), x)
\[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/4}} \, dx=\int \frac {\left (d + e x\right )^{2}}{\left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {3}{4}}}\, dx \] Input:
integrate((e*x+d)**2/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(3/4),x)
Output:
Integral((d + e*x)**2/((d + e*x)*(a*e + c*d*x))**(3/4), x)
\[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/4),x, algorithm=" maxima")
Output:
integrate((e*x + d)^2/(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(3/4), x)
\[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/4}} \, dx=\int { \frac {{\left (e x + d\right )}^{2}}{{\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {3}{4}}} \,d x } \] Input:
integrate((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/4),x, algorithm=" giac")
Output:
integrate((e*x + d)^2/(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(3/4), x)
Timed out. \[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/4}} \, dx=\int \frac {{\left (d+e\,x\right )}^2}{{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{3/4}} \,d x \] Input:
int((d + e*x)^2/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(3/4),x)
Output:
int((d + e*x)^2/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(3/4), x)
\[ \int \frac {(d+e x)^2}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/4}} \, dx=\left (\int \frac {x^{2}}{\left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{4}}}d x \right ) e^{2}+2 \left (\int \frac {x}{\left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{4}}}d x \right ) d e +\left (\int \frac {1}{\left (c d e \,x^{2}+a \,e^{2} x +c \,d^{2} x +a d e \right )^{\frac {3}{4}}}d x \right ) d^{2} \] Input:
int((e*x+d)^2/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/4),x)
Output:
int(x**2/(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(3/4),x)*e**2 + 2*int (x/(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(3/4),x)*d*e + int(1/(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(3/4),x)*d**2