Integrand size = 35, antiderivative size = 415 \[ \int (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4} \, dx=-\frac {5 \left (c d^2-a e^2\right )^4 \sqrt [4]{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{168 c^3 d^3 e^2}+\frac {\left (c d^2-a e^2\right )^3 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4}}{84 c^3 d^3 e (d+e x)}+\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{9/4}}{9 c d}+\frac {\left (c d^2-a e^2\right )^2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{9/4}}{14 c^3 d^3 (d+e x)^2}+\frac {\left (c d^2-a e^2\right ) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{9/4}}{7 c^2 d^2 (d+e x)}-\frac {5 \left (c d^2-a e^2\right )^{9/2} (a e+c d x)^{3/2} \left (\frac {c d (d+e x)}{e (a e+c d x)}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {c d^2-a e^2}}{\sqrt {e} \sqrt {a e+c d x}}\right ),2\right )}{168 c^4 d^4 e^{3/2} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/4}} \] Output:
-5/168*(-a*e^2+c*d^2)^4*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/4)/c^3/d^3/e^ 2+1/84*(-a*e^2+c*d^2)^3*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4)/c^3/d^3/e/ (e*x+d)+2/9*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(9/4)/c/d+1/14*(-a*e^2+c*d^2 )^2*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(9/4)/c^3/d^3/(e*x+d)^2+1/7*(-a*e^2+ c*d^2)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(9/4)/c^2/d^2/(e*x+d)-5/168*(-a*e ^2+c*d^2)^(9/2)*(c*d*x+a*e)^(3/2)*(c*d*(e*x+d)/e/(c*d*x+a*e))^(3/4)*Invers eJacobiAM(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^4/d^4/e^(3/2)/(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 = 112, normalized size of antiderivative = 0.27 \[ \int (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4} \, dx=\frac {4 \left (c d^2-a e^2\right )^2 (a e+c d x)^2 \sqrt [4]{(a e+c d x) (d+e x)} \operatorname {Hypergeometric2F1}\left (-\frac {9}{4},\frac {9}{4},\frac {13}{4},\frac {e (a e+c d x)}{-c d^2+a e^2}\right )}{9 c^3 d^3 \sqrt [4]{\frac {c d (d+e x)}{c d^2-a e^2}}} \] Input:
Integrate[(d + e*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/4),x]
Output:
(4*(c*d^2 - a*e^2)^2*(a*e + c*d*x)^2*((a*e + c*d*x)*(d + e*x))^(1/4)*Hyper geometric2F1[-9/4, 9/4, 13/4, (e*(a*e + c*d*x))/(-(c*d^2) + a*e^2)])/(9*c^ 3*d^3*((c*d*(d + e*x))/(c*d^2 - a*e^2))^(1/4))
Time = 0.92 (sec) , antiderivative size = 601, normalized size of antiderivative = 1.45, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {1160, 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.
\(\displaystyle \int (d+e x) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4} \, dx\) |
\(\Big \downarrow \) 1160 |
\(\displaystyle \frac {\left (d^2-\frac {a e^2}{c}\right ) \int \left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )^{5/4}dx}{2 d}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{9/4}}{9 c d}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (a e^2+c d^2+2 c d e x\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{7 c d e}-\frac {5 \left (c d^2-a e^2\right )^2 \int \sqrt [4]{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}dx}{28 c d e}\right )}{2 d}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{9/4}}{9 c d}\) |
\(\Big \downarrow \) 1087 |
\(\displaystyle \frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (a e^2+c d^2+2 c d e x\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{7 c d e}-\frac {5 \left (c d^2-a e^2\right )^2 \left (\frac {\left (a e^2+c d^2+2 c d e x\right ) \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}-\frac {\left (c d^2-a e^2\right )^2 \int \frac {1}{\left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )^{3/4}}dx}{12 c d e}\right )}{28 c d e}\right )}{2 d}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{9/4}}{9 c d}\) |
\(\Big \downarrow \) 1094 |
\(\displaystyle \frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (a e^2+c d^2+2 c d e x\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{7 c d e}-\frac {5 \left (c d^2-a e^2\right )^2 \left (\frac {\left (a e^2+c d^2+2 c d e x\right ) \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}-\frac {\left (c d^2-a e^2\right )^2 \sqrt {\left (a e^2+c d^2+2 c d e x\right )^2} \int \frac {1}{\sqrt {\left (c d^2-a e^2\right )^2+4 c d e \left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )}}d\sqrt [4]{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}{3 c d e \left (a e^2+c d^2+2 c d e x\right )}\right )}{28 c d e}\right )}{2 d}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{9/4}}{9 c d}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (a e^2+c d^2+2 c d e x\right ) \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/4}}{7 c d e}-\frac {5 \left (c d^2-a e^2\right )^2 \left (\frac {\left (a e^2+c d^2+2 c d e x\right ) \sqrt [4]{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{3 c d e}-\frac {\left (c d^2-a e^2\right )^{5/2} \sqrt {\left (a e^2+c d^2+2 c d e x\right )^2} \left (\frac {2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d^2-a e^2}+1\right ) \sqrt {\frac {4 c d e \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )+\left (c d^2-a e^2\right )^2}{\left (c d^2-a e^2\right )^2 \left (\frac {2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{c d^2-a e^2}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt {2} \sqrt [4]{c} \sqrt [4]{d} \sqrt [4]{e} \sqrt [4]{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}{\sqrt {c d^2-a e^2}}\right ),\frac {1}{2}\right )}{6 \sqrt {2} c^{5/4} d^{5/4} e^{5/4} \left (a e^2+c d^2+2 c d e x\right ) \sqrt {4 c d e \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )+\left (c d^2-a e^2\right )^2}}\right )}{28 c d e}\right )}{2 d}+\frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{9/4}}{9 c d}\) |
Input:
Int[(d + e*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/4),x]
Output:
(2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(9/4))/(9*c*d) + ((d^2 - (a*e^2 )/c)*(((c*d^2 + a*e^2 + 2*c*d*e*x)*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2) ^(5/4))/(7*c*d*e) - (5*(c*d^2 - a*e^2)^2*(((c*d^2 + a*e^2 + 2*c*d*e*x)*(a* d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(1/4))/(3*c*d*e) - ((c*d^2 - a*e^2)^( 5/2)*Sqrt[(c*d^2 + a*e^2 + 2*c*d*e*x)^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))*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]*Sqrt[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])/(6*Sqrt[2]*c^(5/4)*d^(5/4)*e^(5/4)*(c*d^2 + a*e^2 + 2*c*d*e*x)*S qrt[(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)])) )/(28*c*d*e)))/(2*d)
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[((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]
\[\int \left (e x +d \right ) {\left (a d e +\left (a \,e^{2}+c \,d^{2}\right ) x +c d \,x^{2} e \right )}^{\frac {5}{4}}d x\]
Input:
int((e*x+d)*(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(5/4),x)
Output:
int((e*x+d)*(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^(5/4),x)
\[ \int (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4} \, dx=\int { {\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{4}} {\left (e x + d\right )} \,d x } \] Input:
integrate((e*x+d)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4),x, algorithm="fr icas")
Output:
integral((c*d*e^2*x^3 + a*d^2*e + (2*c*d^2*e + a*e^3)*x^2 + (c*d^3 + 2*a*d *e^2)*x)*(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(1/4), x)
\[ \int (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4} \, dx=\int \left (\left (d + e x\right ) \left (a e + c d x\right )\right )^{\frac {5}{4}} \left (d + e x\right )\, dx \] Input:
integrate((e*x+d)*(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**(5/4),x)
Output:
Integral(((d + e*x)*(a*e + c*d*x))**(5/4)*(d + e*x), x)
\[ \int (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4} \, dx=\int { {\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{4}} {\left (e x + d\right )} \,d x } \] Input:
integrate((e*x+d)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4),x, algorithm="ma xima")
Output:
integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/4)*(e*x + d), x)
\[ \int (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4} \, dx=\int { {\left (c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x\right )}^{\frac {5}{4}} {\left (e x + d\right )} \,d x } \] Input:
integrate((e*x+d)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4),x, algorithm="gi ac")
Output:
integrate((c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)^(5/4)*(e*x + d), x)
Timed out. \[ \int (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4} \, dx=\int \left (d+e\,x\right )\,{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/4} \,d x \] Input:
int((d + e*x)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/4),x)
Output:
int((d + e*x)*(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^(5/4), x)
\[ \int (d+e x) \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/4} \, dx =\text {Too large to display} \] Input:
int((e*x+d)*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(5/4),x)
Output:
(48*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**4*d*e**7 - 12*(a* d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**4*e**8*x - 208*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**3*c*d**3*e**5 + 40*(a*d*e + a* e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**3*c*d**2*e**6*x + 8*(a*d*e + a*e **2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**3*c*d*e**7*x**2 + 656*(a*d*e + a* e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**2*c**2*d**5*e**3 + 896*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**2*c**2*d**4*e**4*x + 904*(a*d* e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**2*c**2*d**3*e**5*x**2 + 30 4*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a**2*c**2*d**2*e**6*x* *3 - 48*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a*c**3*d**7*e + 856*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a*c**3*d**6*e**2*x + 1336*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a*c**3*d**5*e**3*x **2 + 896*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a*c**3*d**4*e* *4*x**3 + 224*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*a*c**3*d** 3*e**5*x**4 + 12*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*c**4*d* *8*x + 440*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*c**4*d**7*e*x **2 + 592*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*c**4*d**6*e**2 *x**3 + 224*(a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*c**4*d**5*e* *3*x**4 + 15*int(((a*d*e + a*e**2*x + c*d**2*x + c*d*e*x**2)**(1/4)*x)/(a* *2*d*e**3 + a**2*e**4*x + a*c*d**3*e + 2*a*c*d**2*e**2*x + a*c*d*e**3*x...