Integrand size = 34, antiderivative size = 1213 \[ \int \frac {1}{\left (a d+(b d+a e) x+(c d+b e) x^2+c e x^3\right )^{5/2}} \, dx =\text {Too large to display} \] Output:
-2/3*(e*x+d)*(b*c*d-b^2*e+2*a*c*e+c*(-b*e+2*c*d)*x)*(c*x^2+b*x+a)/(-4*a*c+ b^2)/(a*e^2-b*d*e+c*d^2)/(a*d+(a*e+b*d)*x+(b*e+c*d)*x^2+c*e*x^3)^(5/2)-2/3 *(e*x+d)*(7*a*c*e*(-b*e+2*c*d)^2-(2*a*c*e-b^2*e+b*c*d)*(8*c^2*d^2-6*b^2*e^ 2-c*e*(-18*a*e+b*d))-2*c*(-b*e+2*c*d)*(4*c^2*d^2-3*b^2*e^2-4*c*e*(-4*a*e+b *d))*x)*(c*x^2+b*x+a)^2/(-4*a*c+b^2)^2/(a*e^2-b*d*e+c*d^2)^2/(a*d+(a*e+b*d )*x+(b*e+c*d)*x^2+c*e*x^3)^(5/2)+2/3*e*(16*c^4*d^4-8*b^4*e^4-4*c^3*d^2*e*( -21*a*e+8*b*d)+b^2*c*e^3*(51*a*e+13*b*d)+3*c^2*e^2*(-20*a^2*e^2-28*a*b*d*e +b^2*d^2))*(e*x+d)*(c*x^2+b*x+a)^3/(-4*a*c+b^2)^2/(a*e^2-b*d*e+c*d^2)^3/(a *d+(a*e+b*d)*x+(b*e+c*d)*x^2+c*e*x^3)^(5/2)+16/3*e*(-b*e+2*c*d)*(c^2*d^2-2 *b^2*e^2-c*e*(-9*a*e+b*d))*(c^2*d^2+b^2*e^2-c*e*(3*a*e+b*d))*(e*x+d)^2*(c* x^2+b*x+a)^3/(-4*a*c+b^2)^2/(a*e^2-b*d*e+c*d^2)^4/(a*d+(a*e+b*d)*x+(b*e+c* d)*x^2+c*e*x^3)^(5/2)-8/3*2^(1/2)*(-b*e+2*c*d)*(c^2*d^2-2*b^2*e^2-c*e*(-9* a*e+b*d))*(c^2*d^2+b^2*e^2-c*e*(3*a*e+b*d))*(e*x+d)^3*(c*x^2+b*x+a)^2*(-c* (c*x^2+b*x+a)/(-4*a*c+b^2))^(1/2)*EllipticE(1/2*(1+(2*c*x+b)/(-4*a*c+b^2)^ (1/2))^(1/2)*2^(1/2),(-2*(-4*a*c+b^2)^(1/2)*e/(2*c*d-(b+(-4*a*c+b^2)^(1/2) )*e))^(1/2))/(-4*a*c+b^2)^(3/2)/(a*e^2-b*d*e+c*d^2)^4/(c*(e*x+d)/(2*c*d-(b +(-4*a*c+b^2)^(1/2))*e))^(1/2)/(a*d+(a*e+b*d)*x+(b*e+c*d)*x^2+c*e*x^3)^(5/ 2)+2/3*2^(1/2)*(16*c^4*d^4-8*b^4*e^4-4*c^3*d^2*e*(-21*a*e+8*b*d)+b^2*c*e^3 *(51*a*e+13*b*d)+3*c^2*e^2*(-20*a^2*e^2-28*a*b*d*e+b^2*d^2))*(e*x+d)^2*(c* (e*x+d)/(2*c*d-(b+(-4*a*c+b^2)^(1/2))*e))^(1/2)*(c*x^2+b*x+a)^2*(-c*(c*...
Result contains complex when optimal does not.
Time = 24.37 (sec) , antiderivative size = 11303, normalized size of antiderivative = 9.32 \[ \int \frac {1}{\left (a d+(b d+a e) x+(c d+b e) x^2+c e x^3\right )^{5/2}} \, dx=\text {Result too large to show} \] Input:
Integrate[(a*d + (b*d + a*e)*x + (c*d + b*e)*x^2 + c*e*x^3)^(-5/2),x]
Output:
Result too large to show
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.
| \(\displaystyle \int \frac {1}{\left (x (a e+b d)+a d+x^2 (b e+c d)+c e x^3\right )^{5/2}} \, dx\) |
| \(\Big \downarrow \) 2481 |
| \(\displaystyle \int \frac {1}{\left (\frac {(2 c d-b e) \left (-c e (b d-9 a e)-2 b^2 e^2+c^2 d^2\right )}{27 c^2 e^2}+\frac {\left (3 c e (a e+b d)-(b e+c d)^2\right ) \left (\frac {b e+c d}{3 c e}+x\right )}{3 c e}+c e \left (\frac {b e+c d}{3 c e}+x\right )^3\right )^{5/2}}d\left (\frac {b e+c d}{3 c e}+x\right )\) |
Input:
Int[(a*d + (b*d + a*e)*x + (c*d + b*e)*x^2 + c*e*x^3)^(-5/2),x]
Output:
$Aborted
Int[(Px_)^(p_), x_Symbol] :> With[{a = Coeff[Px, x, 0], b = Coeff[Px, x, 1] , c = Coeff[Px, x, 2], d = Coeff[Px, x, 3]}, Subst[Int[Simp[(2*c^3 - 9*b*c* d + 27*a*d^2)/(27*d^2) - (c^2 - 3*b*d)*(x/(3*d)) + d*x^3, x]^p, x], x, c/(3 *d) + x]] /; FreeQ[p, x] && PolyQ[Px, x, 3]
Leaf count of result is larger than twice the leaf count of optimal. \(3394\) vs. \(2(1153)=2306\).
Time = 0.44 (sec) , antiderivative size = 3395, normalized size of antiderivative = 2.80
| method | result | size |
| default | \(\text {Expression too large to display}\) | \(3395\) |
| elliptic | \(\text {Expression too large to display}\) | \(3395\) |
Input:
int(1/(a*d+(a*e+b*d)*x+(b*e+c*d)*x^2+c*e*x^3)^(5/2),x,method=_RETURNVERBOS E)
Output:
(-4/3/c/e*(3*a*c*e^2-b^2*e^2+b*c*d*e-c^2*d^2)/(4*a^3*c*e^4-a^2*b^2*e^4-8*a ^2*b*c*d*e^3+8*a^2*c^2*d^2*e^2+2*a*b^3*d*e^3+2*a*b^2*c*d^2*e^2-8*a*b*c^2*d ^3*e+4*a*c^3*d^4-b^4*d^2*e^2+2*b^3*c*d^3*e-b^2*c^2*d^4)*x^2-2/3*(7*a*b*c*e ^3-2*a*c^2*d*e^2-2*b^3*e^3+b^2*c*d*e^2+b*c^2*d^2*e-2*c^3*d^3)/(4*a^3*c*e^4 -a^2*b^2*e^4-8*a^2*b*c*d*e^3+8*a^2*c^2*d^2*e^2+2*a*b^3*d*e^3+2*a*b^2*c*d^2 *e^2-8*a*b*c^2*d^3*e+4*a*c^3*d^4-b^4*d^2*e^2+2*b^3*c*d^3*e-b^2*c^2*d^4)/c^ 2/e^2*x-2/3*(4*a^2*c*e^3-a*b^2*e^3+3*a*b*c*d*e^2-4*a*c^2*d^2*e-b^3*d*e^2+2 *b^2*c*d^2*e-b*c^2*d^3)/(4*a^3*c*e^4-a^2*b^2*e^4-8*a^2*b*c*d*e^3+8*a^2*c^2 *d^2*e^2+2*a*b^3*d*e^3+2*a*b^2*c*d^2*e^2-8*a*b*c^2*d^3*e+4*a*c^3*d^4-b^4*d ^2*e^2+2*b^3*c*d^3*e-b^2*c^2*d^4)/c^2/e^2)*(c*e*x^3+b*e*x^2+c*d*x^2+a*e*x+ b*d*x+a*d)^(1/2)/(x^3+(b*e+c*d)/c/e*x^2+(a*e+b*d)/c/e*x+a*d/c/e)^2-2*c*e*( -8/3*(3*a*c*e^2-b^2*e^2+b*c*d*e-c^2*d^2)*(9*a*b*c*e^3-18*a*c^2*d*e^2-2*b^3 *e^3+3*b^2*c*d*e^2+3*b*c^2*d^2*e-2*c^3*d^3)/(4*a^3*c*e^4-a^2*b^2*e^4-8*a^2 *b*c*d*e^3+8*a^2*c^2*d^2*e^2+2*a*b^3*d*e^3+2*a*b^2*c*d^2*e^2-8*a*b*c^2*d^3 *e+4*a*c^3*d^4-b^4*d^2*e^2+2*b^3*c*d^3*e-b^2*c^2*d^4)^2*x^2+1/3*(60*a^3*c^ 3*e^6-267*a^2*b^2*c^2*e^6+456*a^2*b*c^3*d*e^5-24*a^2*c^4*d^2*e^4+128*a*b^4 *c*e^6-250*a*b^3*c^2*d*e^5+6*a*b^2*c^3*d^2*e^4+104*a*b*c^4*d^3*e^3-100*a*c ^5*d^4*e^2-16*b^6*e^6+32*b^5*c*d*e^5+5*b^4*c^2*d^2*e^4-26*b^3*c^3*d^3*e^3+ 5*b^2*c^4*d^4*e^2+32*b*c^5*d^5*e-16*c^6*d^6)/(4*a^3*c*e^4-a^2*b^2*e^4-8*a^ 2*b*c*d*e^3+8*a^2*c^2*d^2*e^2+2*a*b^3*d*e^3+2*a*b^2*c*d^2*e^2-8*a*b*c^2...
Leaf count of result is larger than twice the leaf count of optimal. 8140 vs. \(2 (1161) = 2322\).
Time = 1.28 (sec) , antiderivative size = 8140, normalized size of antiderivative = 6.71 \[ \int \frac {1}{\left (a d+(b d+a e) x+(c d+b e) x^2+c e x^3\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate(1/(a*d+(a*e+b*d)*x+(b*e+c*d)*x^2+c*e*x^3)^(5/2),x, algorithm="fr icas")
Output:
Too large to include
\[ \int \frac {1}{\left (a d+(b d+a e) x+(c d+b e) x^2+c e x^3\right )^{5/2}} \, dx=\int \frac {1}{\left (a d + c e x^{3} + x^{2} \left (b e + c d\right ) + x \left (a e + b d\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(1/(a*d+(a*e+b*d)*x+(b*e+c*d)*x**2+c*e*x**3)**(5/2),x)
Output:
Integral((a*d + c*e*x**3 + x**2*(b*e + c*d) + x*(a*e + b*d))**(-5/2), x)
\[ \int \frac {1}{\left (a d+(b d+a e) x+(c d+b e) x^2+c e x^3\right )^{5/2}} \, dx=\int { \frac {1}{{\left (c e x^{3} + {\left (c d + b e\right )} x^{2} + a d + {\left (b d + a e\right )} x\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(a*d+(a*e+b*d)*x+(b*e+c*d)*x^2+c*e*x^3)^(5/2),x, algorithm="ma xima")
Output:
integrate((c*e*x^3 + (c*d + b*e)*x^2 + a*d + (b*d + a*e)*x)^(-5/2), x)
\[ \int \frac {1}{\left (a d+(b d+a e) x+(c d+b e) x^2+c e x^3\right )^{5/2}} \, dx=\int { \frac {1}{{\left (c e x^{3} + {\left (c d + b e\right )} x^{2} + a d + {\left (b d + a e\right )} x\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate(1/(a*d+(a*e+b*d)*x+(b*e+c*d)*x^2+c*e*x^3)^(5/2),x, algorithm="gi ac")
Output:
integrate((c*e*x^3 + (c*d + b*e)*x^2 + a*d + (b*d + a*e)*x)^(-5/2), x)
Timed out. \[ \int \frac {1}{\left (a d+(b d+a e) x+(c d+b e) x^2+c e x^3\right )^{5/2}} \, dx=\int \frac {1}{{\left (c\,e\,x^3+\left (b\,e+c\,d\right )\,x^2+\left (a\,e+b\,d\right )\,x+a\,d\right )}^{5/2}} \,d x \] Input:
int(1/(a*d + x*(a*e + b*d) + x^2*(b*e + c*d) + c*e*x^3)^(5/2),x)
Output:
int(1/(a*d + x*(a*e + b*d) + x^2*(b*e + c*d) + c*e*x^3)^(5/2), x)
\[ \int \frac {1}{\left (a d+(b d+a e) x+(c d+b e) x^2+c e x^3\right )^{5/2}} \, dx=\int \frac {\sqrt {e x +d}\, \sqrt {c \,x^{2}+b x +a}}{c^{3} e^{3} x^{9}+3 b \,c^{2} e^{3} x^{8}+3 c^{3} d \,e^{2} x^{8}+3 a \,c^{2} e^{3} x^{7}+3 b^{2} c \,e^{3} x^{7}+9 b \,c^{2} d \,e^{2} x^{7}+3 c^{3} d^{2} e \,x^{7}+6 a b c \,e^{3} x^{6}+9 a \,c^{2} d \,e^{2} x^{6}+b^{3} e^{3} x^{6}+9 b^{2} c d \,e^{2} x^{6}+9 b \,c^{2} d^{2} e \,x^{6}+c^{3} d^{3} x^{6}+3 a^{2} c \,e^{3} x^{5}+3 a \,b^{2} e^{3} x^{5}+18 a b c d \,e^{2} x^{5}+9 a \,c^{2} d^{2} e \,x^{5}+3 b^{3} d \,e^{2} x^{5}+9 b^{2} c \,d^{2} e \,x^{5}+3 b \,c^{2} d^{3} x^{5}+3 a^{2} b \,e^{3} x^{4}+9 a^{2} c d \,e^{2} x^{4}+9 a \,b^{2} d \,e^{2} x^{4}+18 a b c \,d^{2} e \,x^{4}+3 a \,c^{2} d^{3} x^{4}+3 b^{3} d^{2} e \,x^{4}+3 b^{2} c \,d^{3} x^{4}+a^{3} e^{3} x^{3}+9 a^{2} b d \,e^{2} x^{3}+9 a^{2} c \,d^{2} e \,x^{3}+9 a \,b^{2} d^{2} e \,x^{3}+6 a b c \,d^{3} x^{3}+b^{3} d^{3} x^{3}+3 a^{3} d \,e^{2} x^{2}+9 a^{2} b \,d^{2} e \,x^{2}+3 a^{2} c \,d^{3} x^{2}+3 a \,b^{2} d^{3} x^{2}+3 a^{3} d^{2} e x +3 a^{2} b \,d^{3} x +a^{3} d^{3}}d x \] Input:
int(1/(a*d+(a*e+b*d)*x+(b*e+c*d)*x^2+c*e*x^3)^(5/2),x)
Output:
int((sqrt(d + e*x)*sqrt(a + b*x + c*x**2))/(a**3*d**3 + 3*a**3*d**2*e*x + 3*a**3*d*e**2*x**2 + a**3*e**3*x**3 + 3*a**2*b*d**3*x + 9*a**2*b*d**2*e*x* *2 + 9*a**2*b*d*e**2*x**3 + 3*a**2*b*e**3*x**4 + 3*a**2*c*d**3*x**2 + 9*a* *2*c*d**2*e*x**3 + 9*a**2*c*d*e**2*x**4 + 3*a**2*c*e**3*x**5 + 3*a*b**2*d* *3*x**2 + 9*a*b**2*d**2*e*x**3 + 9*a*b**2*d*e**2*x**4 + 3*a*b**2*e**3*x**5 + 6*a*b*c*d**3*x**3 + 18*a*b*c*d**2*e*x**4 + 18*a*b*c*d*e**2*x**5 + 6*a*b *c*e**3*x**6 + 3*a*c**2*d**3*x**4 + 9*a*c**2*d**2*e*x**5 + 9*a*c**2*d*e**2 *x**6 + 3*a*c**2*e**3*x**7 + b**3*d**3*x**3 + 3*b**3*d**2*e*x**4 + 3*b**3* d*e**2*x**5 + b**3*e**3*x**6 + 3*b**2*c*d**3*x**4 + 9*b**2*c*d**2*e*x**5 + 9*b**2*c*d*e**2*x**6 + 3*b**2*c*e**3*x**7 + 3*b*c**2*d**3*x**5 + 9*b*c**2 *d**2*e*x**6 + 9*b*c**2*d*e**2*x**7 + 3*b*c**2*e**3*x**8 + c**3*d**3*x**6 + 3*c**3*d**2*e*x**7 + 3*c**3*d*e**2*x**8 + c**3*e**3*x**9),x)