Integrand size = 24, antiderivative size = 141 \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {\left (4-e^2 x^2\right )^{3/4}}{15 \sqrt [4]{3} e (2+e x)^{9/2}}-\frac {\left (4-e^2 x^2\right )^{3/4}}{55 \sqrt [4]{3} e (2+e x)^{7/2}}-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{385 \sqrt [4]{3} e (2+e x)^{5/2}}-\frac {2 \left (4-e^2 x^2\right )^{3/4}}{1155 \sqrt [4]{3} e (2+e x)^{3/2}} \] Output:
-1/45*(-e^2*x^2+4)^(3/4)*3^(3/4)/e/(e*x+2)^(9/2)-1/165*(-e^2*x^2+4)^(3/4)* 3^(3/4)/e/(e*x+2)^(7/2)-2/1155*(-e^2*x^2+4)^(3/4)*3^(3/4)/e/(e*x+2)^(5/2)- 2/3465*(-e^2*x^2+4)^(3/4)*3^(3/4)/e/(e*x+2)^(3/2)
Time = 0.57 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.40 \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {\left (4-e^2 x^2\right )^{3/4} \left (159+69 e x+18 e^2 x^2+2 e^3 x^3\right )}{1155 \sqrt [4]{3} e (2+e x)^{9/2}} \] Input:
Integrate[1/((2 + e*x)^(9/2)*(12 - 3*e^2*x^2)^(1/4)),x]
Output:
-1/1155*((4 - e^2*x^2)^(3/4)*(159 + 69*e*x + 18*e^2*x^2 + 2*e^3*x^3))/(3^( 1/4)*e*(2 + e*x)^(9/2))
Time = 0.34 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {456, 55, 27, 55, 55, 48}
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.
\(\displaystyle \int \frac {1}{(e x+2)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx\) |
\(\Big \downarrow \) 456 |
\(\displaystyle \int \frac {1}{\sqrt [4]{6-3 e x} (e x+2)^{19/4}}dx\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {1}{5} \int \frac {1}{\sqrt [4]{3} \sqrt [4]{2-e x} (e x+2)^{15/4}}dx-\frac {(2-e x)^{3/4}}{15 \sqrt [4]{3} e (e x+2)^{15/4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {1}{\sqrt [4]{2-e x} (e x+2)^{15/4}}dx}{5 \sqrt [4]{3}}-\frac {(2-e x)^{3/4}}{15 \sqrt [4]{3} e (e x+2)^{15/4}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\frac {2}{11} \int \frac {1}{\sqrt [4]{2-e x} (e x+2)^{11/4}}dx-\frac {(2-e x)^{3/4}}{11 e (e x+2)^{11/4}}}{5 \sqrt [4]{3}}-\frac {(2-e x)^{3/4}}{15 \sqrt [4]{3} e (e x+2)^{15/4}}\) |
\(\Big \downarrow \) 55 |
\(\displaystyle \frac {\frac {2}{11} \left (\frac {1}{7} \int \frac {1}{\sqrt [4]{2-e x} (e x+2)^{7/4}}dx-\frac {(2-e x)^{3/4}}{7 e (e x+2)^{7/4}}\right )-\frac {(2-e x)^{3/4}}{11 e (e x+2)^{11/4}}}{5 \sqrt [4]{3}}-\frac {(2-e x)^{3/4}}{15 \sqrt [4]{3} e (e x+2)^{15/4}}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {\frac {2}{11} \left (-\frac {(2-e x)^{3/4}}{21 e (e x+2)^{3/4}}-\frac {(2-e x)^{3/4}}{7 e (e x+2)^{7/4}}\right )-\frac {(2-e x)^{3/4}}{11 e (e x+2)^{11/4}}}{5 \sqrt [4]{3}}-\frac {(2-e x)^{3/4}}{15 \sqrt [4]{3} e (e x+2)^{15/4}}\) |
Input:
Int[1/((2 + e*x)^(9/2)*(12 - 3*e^2*x^2)^(1/4)),x]
Output:
-1/15*(2 - e*x)^(3/4)/(3^(1/4)*e*(2 + e*x)^(15/4)) + (-1/11*(2 - e*x)^(3/4 )/(e*(2 + e*x)^(11/4)) + (2*(-1/7*(2 - e*x)^(3/4)/(e*(2 + e*x)^(7/4)) - (2 - e*x)^(3/4)/(21*e*(2 + e*x)^(3/4))))/11)/(5*3^(1/4))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S implify[m + n + 2]/((b*c - a*d)*(m + 1))) Int[(a + b*x)^Simplify[m + 1]*( c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 2], 0] && NeQ[m, -1] && !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] || !SumSimp lerQ[n, 1])
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ (c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !Integ erQ[n]))
Time = 0.17 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.37
method | result | size |
gosper | \(\frac {\left (e x -2\right ) \left (2 e^{3} x^{3}+18 e^{2} x^{2}+69 e x +159\right )}{1155 \left (e x +2\right )^{\frac {7}{2}} e \left (-3 e^{2} x^{2}+12\right )^{\frac {1}{4}}}\) | \(52\) |
orering | \(\frac {\left (e x -2\right ) \left (2 e^{3} x^{3}+18 e^{2} x^{2}+69 e x +159\right )}{1155 \left (e x +2\right )^{\frac {7}{2}} e \left (-3 e^{2} x^{2}+12\right )^{\frac {1}{4}}}\) | \(52\) |
Input:
int(1/(e*x+2)^(9/2)/(-3*e^2*x^2+12)^(1/4),x,method=_RETURNVERBOSE)
Output:
1/1155*(e*x-2)*(2*e^3*x^3+18*e^2*x^2+69*e*x+159)/(e*x+2)^(7/2)/e/(-3*e^2*x ^2+12)^(1/4)
Time = 0.09 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.61 \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {{\left (2 \, e^{3} x^{3} + 18 \, e^{2} x^{2} + 69 \, e x + 159\right )} {\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {3}{4}} \sqrt {e x + 2}}{3465 \, {\left (e^{6} x^{5} + 10 \, e^{5} x^{4} + 40 \, e^{4} x^{3} + 80 \, e^{3} x^{2} + 80 \, e^{2} x + 32 \, e\right )}} \] Input:
integrate(1/(e*x+2)^(9/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="fricas")
Output:
-1/3465*(2*e^3*x^3 + 18*e^2*x^2 + 69*e*x + 159)*(-3*e^2*x^2 + 12)^(3/4)*sq rt(e*x + 2)/(e^6*x^5 + 10*e^5*x^4 + 40*e^4*x^3 + 80*e^3*x^2 + 80*e^2*x + 3 2*e)
Timed out. \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=\text {Timed out} \] Input:
integrate(1/(e*x+2)**(9/2)/(-3*e**2*x**2+12)**(1/4),x)
Output:
Timed out
\[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=\int { \frac {1}{{\left (-3 \, e^{2} x^{2} + 12\right )}^{\frac {1}{4}} {\left (e x + 2\right )}^{\frac {9}{2}}} \,d x } \] Input:
integrate(1/(e*x+2)^(9/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((-3*e^2*x^2 + 12)^(1/4)*(e*x + 2)^(9/2)), x)
Time = 0.15 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.49 \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {3^{\frac {3}{4}} {\left (77 \, {\left (\frac {4}{e x + 2} - 1\right )}^{\frac {15}{4}} + 315 \, {\left (\frac {4}{e x + 2} - 1\right )}^{\frac {11}{4}} + 495 \, {\left (\frac {4}{e x + 2} - 1\right )}^{\frac {7}{4}} + 385 \, {\left (\frac {4}{e x + 2} - 1\right )}^{\frac {3}{4}}\right )}}{221760 \, e} \] Input:
integrate(1/(e*x+2)^(9/2)/(-3*e^2*x^2+12)^(1/4),x, algorithm="giac")
Output:
-1/221760*3^(3/4)*(77*(4/(e*x + 2) - 1)^(15/4) + 315*(4/(e*x + 2) - 1)^(11 /4) + 495*(4/(e*x + 2) - 1)^(7/4) + 385*(4/(e*x + 2) - 1)^(3/4))/e
Time = 6.94 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.79 \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=-\frac {{\left (12-3\,e^2\,x^2\right )}^{3/4}\,\left (\frac {23\,x}{1155\,e^4}+\frac {53}{1155\,e^5}+\frac {2\,x^3}{3465\,e^2}+\frac {2\,x^2}{385\,e^3}\right )}{\frac {16\,\sqrt {e\,x+2}}{e^4}+x^4\,\sqrt {e\,x+2}+\frac {32\,x\,\sqrt {e\,x+2}}{e^3}+\frac {8\,x^3\,\sqrt {e\,x+2}}{e}+\frac {24\,x^2\,\sqrt {e\,x+2}}{e^2}} \] Input:
int(1/((12 - 3*e^2*x^2)^(1/4)*(e*x + 2)^(9/2)),x)
Output:
-((12 - 3*e^2*x^2)^(3/4)*((23*x)/(1155*e^4) + 53/(1155*e^5) + (2*x^3)/(346 5*e^2) + (2*x^2)/(385*e^3)))/((16*(e*x + 2)^(1/2))/e^4 + x^4*(e*x + 2)^(1/ 2) + (32*x*(e*x + 2)^(1/2))/e^3 + (8*x^3*(e*x + 2)^(1/2))/e + (24*x^2*(e*x + 2)^(1/2))/e^2)
Time = 0.30 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(2+e x)^{9/2} \sqrt [4]{12-3 e^2 x^2}} \, dx=\frac {\sqrt {e x +2}\, \left (-e^{2} x^{2}+4\right )^{\frac {1}{4}} 3^{\frac {1}{4}} \left (2 e^{3} x^{3}+14 e^{2} x^{2}+37 e x -146\right )}{585 \sqrt {-e^{2} x^{2}+4}\, \sqrt {3}\, e \left (e^{4} x^{4}+8 e^{3} x^{3}+24 e^{2} x^{2}+32 e x +16\right )} \] Input:
int(1/(e*x+2)^(9/2)/(-3*e^2*x^2+12)^(1/4),x)
Output:
(sqrt(e*x + 2)*( - e**2*x**2 + 4)**(1/4)*3**(1/4)*(2*e**3*x**3 + 14*e**2*x **2 + 37*e*x - 146))/(585*sqrt( - e**2*x**2 + 4)*sqrt(3)*e*(e**4*x**4 + 8* e**3*x**3 + 24*e**2*x**2 + 32*e*x + 16))