Integrand size = 24, antiderivative size = 131 \[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\frac {2 (d+e x)^3}{9 e \left (d^2-e^2 x^2\right )^{9/2}}+\frac {2 (d+e x)}{21 e \left (d^2-e^2 x^2\right )^{7/2}}+\frac {x}{21 d^2 \left (d^2-e^2 x^2\right )^{5/2}}+\frac {4 x}{63 d^4 \left (d^2-e^2 x^2\right )^{3/2}}+\frac {8 x}{63 d^6 \sqrt {d^2-e^2 x^2}} \] Output:
2/9*(e*x+d)^3/e/(-e^2*x^2+d^2)^(9/2)+2/21*(e*x+d)/e/(-e^2*x^2+d^2)^(7/2)+1 /21*x/d^2/(-e^2*x^2+d^2)^(5/2)+4/63*x/d^4/(-e^2*x^2+d^2)^(3/2)+8/63*x/d^6/ (-e^2*x^2+d^2)^(1/2)
Time = 0.50 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.71 \[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (20 d^5-17 d^4 e x-16 d^3 e^2 x^2+44 d^2 e^3 x^3-32 d e^4 x^4+8 e^5 x^5\right )}{63 d^6 e (d-e x)^5 (d+e x)} \] Input:
Integrate[(d + e*x)^4/(d^2 - e^2*x^2)^(11/2),x]
Output:
(Sqrt[d^2 - e^2*x^2]*(20*d^5 - 17*d^4*e*x - 16*d^3*e^2*x^2 + 44*d^2*e^3*x^ 3 - 32*d*e^4*x^4 + 8*e^5*x^5))/(63*d^6*e*(d - e*x)^5*(d + e*x))
Time = 0.44 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.41, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {464, 461, 461, 461, 470, 208}
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.
\(\displaystyle \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx\) |
\(\Big \downarrow \) 464 |
\(\displaystyle \int \frac {1}{(d-e x)^4 \left (d^2-e^2 x^2\right )^{3/2}}dx\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {5 \int \frac {1}{(d-e x)^3 \left (d^2-e^2 x^2\right )^{3/2}}dx}{9 d}+\frac {1}{9 d e (d-e x)^4 \sqrt {d^2-e^2 x^2}}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {5 \left (\frac {4 \int \frac {1}{(d-e x)^2 \left (d^2-e^2 x^2\right )^{3/2}}dx}{7 d}+\frac {1}{7 d e (d-e x)^3 \sqrt {d^2-e^2 x^2}}\right )}{9 d}+\frac {1}{9 d e (d-e x)^4 \sqrt {d^2-e^2 x^2}}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {5 \left (\frac {4 \left (\frac {3 \int \frac {1}{(d-e x) \left (d^2-e^2 x^2\right )^{3/2}}dx}{5 d}+\frac {1}{5 d e (d-e x)^2 \sqrt {d^2-e^2 x^2}}\right )}{7 d}+\frac {1}{7 d e (d-e x)^3 \sqrt {d^2-e^2 x^2}}\right )}{9 d}+\frac {1}{9 d e (d-e x)^4 \sqrt {d^2-e^2 x^2}}\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {5 \left (\frac {4 \left (\frac {3 \left (\frac {2 \int \frac {1}{\left (d^2-e^2 x^2\right )^{3/2}}dx}{3 d}+\frac {1}{3 d e (d-e x) \sqrt {d^2-e^2 x^2}}\right )}{5 d}+\frac {1}{5 d e (d-e x)^2 \sqrt {d^2-e^2 x^2}}\right )}{7 d}+\frac {1}{7 d e (d-e x)^3 \sqrt {d^2-e^2 x^2}}\right )}{9 d}+\frac {1}{9 d e (d-e x)^4 \sqrt {d^2-e^2 x^2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {1}{9 d e (d-e x)^4 \sqrt {d^2-e^2 x^2}}+\frac {5 \left (\frac {1}{7 d e (d-e x)^3 \sqrt {d^2-e^2 x^2}}+\frac {4 \left (\frac {1}{5 d e (d-e x)^2 \sqrt {d^2-e^2 x^2}}+\frac {3 \left (\frac {1}{3 d e (d-e x) \sqrt {d^2-e^2 x^2}}+\frac {2 x}{3 d^3 \sqrt {d^2-e^2 x^2}}\right )}{5 d}\right )}{7 d}\right )}{9 d}\) |
Input:
Int[(d + e*x)^4/(d^2 - e^2*x^2)^(11/2),x]
Output:
1/(9*d*e*(d - e*x)^4*Sqrt[d^2 - e^2*x^2]) + (5*(1/(7*d*e*(d - e*x)^3*Sqrt[ d^2 - e^2*x^2]) + (4*(1/(5*d*e*(d - e*x)^2*Sqrt[d^2 - e^2*x^2]) + (3*((2*x )/(3*d^3*Sqrt[d^2 - e^2*x^2]) + 1/(3*d*e*(d - e*x)*Sqrt[d^2 - e^2*x^2])))/ (5*d)))/(7*d)))/(9*d)
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl ify[n + 2*p + 2]/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[Simp lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[( a + b*x^2)^(n + p)/(a/c + b*(x/d))^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b *c^2 + a*d^2, 0] && IntegerQ[n] && RationalQ[p] && (LtQ[0, -n, p] || LtQ[p, -n, 0]) && NeQ[n, 2] && NeQ[n, -1]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[(n + 2*p + 2)/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + p + 1, 0] && IntegerQ[2*p]
Time = 0.28 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.67
method | result | size |
gosper | \(\frac {\left (e x +d \right )^{5} \left (-e x +d \right ) \left (8 e^{5} x^{5}-32 d \,e^{4} x^{4}+44 d^{2} e^{3} x^{3}-16 d^{3} e^{2} x^{2}-17 d^{4} e x +20 d^{5}\right )}{63 d^{6} e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {11}{2}}}\) | \(88\) |
orering | \(\frac {\left (e x +d \right )^{5} \left (-e x +d \right ) \left (8 e^{5} x^{5}-32 d \,e^{4} x^{4}+44 d^{2} e^{3} x^{3}-16 d^{3} e^{2} x^{2}-17 d^{4} e x +20 d^{5}\right )}{63 d^{6} e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {11}{2}}}\) | \(88\) |
trager | \(\frac {\left (8 e^{5} x^{5}-32 d \,e^{4} x^{4}+44 d^{2} e^{3} x^{3}-16 d^{3} e^{2} x^{2}-17 d^{4} e x +20 d^{5}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{63 d^{6} \left (-e x +d \right )^{5} e \left (e x +d \right )}\) | \(90\) |
default | \(d^{4} \left (\frac {x}{9 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {9}{2}}}+\frac {\frac {8 x}{63 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}+\frac {8 \left (\frac {6 x}{35 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{7 d^{2}}\right )}{9 d^{2}}}{d^{2}}\right )+e^{4} \left (\frac {x^{3}}{6 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {9}{2}}}-\frac {d^{2} \left (\frac {x}{8 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {9}{2}}}-\frac {d^{2} \left (\frac {x}{9 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {9}{2}}}+\frac {\frac {8 x}{63 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}+\frac {8 \left (\frac {6 x}{35 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{7 d^{2}}\right )}{9 d^{2}}}{d^{2}}\right )}{8 e^{2}}\right )}{2 e^{2}}\right )+4 d \,e^{3} \left (\frac {x^{2}}{7 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {9}{2}}}-\frac {2 d^{2}}{63 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {9}{2}}}\right )+6 d^{2} e^{2} \left (\frac {x}{8 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {9}{2}}}-\frac {d^{2} \left (\frac {x}{9 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {9}{2}}}+\frac {\frac {8 x}{63 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}+\frac {8 \left (\frac {6 x}{35 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {6 \left (\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{7 d^{2}}\right )}{9 d^{2}}}{d^{2}}\right )}{8 e^{2}}\right )+\frac {4 d^{3}}{9 e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {9}{2}}}\) | \(537\) |
Input:
int((e*x+d)^4/(-e^2*x^2+d^2)^(11/2),x,method=_RETURNVERBOSE)
Output:
1/63*(e*x+d)^5*(-e*x+d)*(8*e^5*x^5-32*d*e^4*x^4+44*d^2*e^3*x^3-16*d^3*e^2* x^2-17*d^4*e*x+20*d^5)/d^6/e/(-e^2*x^2+d^2)^(11/2)
Time = 0.11 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.40 \[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\frac {20 \, e^{6} x^{6} - 80 \, d e^{5} x^{5} + 100 \, d^{2} e^{4} x^{4} - 100 \, d^{4} e^{2} x^{2} + 80 \, d^{5} e x - 20 \, d^{6} - {\left (8 \, e^{5} x^{5} - 32 \, d e^{4} x^{4} + 44 \, d^{2} e^{3} x^{3} - 16 \, d^{3} e^{2} x^{2} - 17 \, d^{4} e x + 20 \, d^{5}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{63 \, {\left (d^{6} e^{7} x^{6} - 4 \, d^{7} e^{6} x^{5} + 5 \, d^{8} e^{5} x^{4} - 5 \, d^{10} e^{3} x^{2} + 4 \, d^{11} e^{2} x - d^{12} e\right )}} \] Input:
integrate((e*x+d)^4/(-e^2*x^2+d^2)^(11/2),x, algorithm="fricas")
Output:
1/63*(20*e^6*x^6 - 80*d*e^5*x^5 + 100*d^2*e^4*x^4 - 100*d^4*e^2*x^2 + 80*d ^5*e*x - 20*d^6 - (8*e^5*x^5 - 32*d*e^4*x^4 + 44*d^2*e^3*x^3 - 16*d^3*e^2* x^2 - 17*d^4*e*x + 20*d^5)*sqrt(-e^2*x^2 + d^2))/(d^6*e^7*x^6 - 4*d^7*e^6* x^5 + 5*d^8*e^5*x^4 - 5*d^10*e^3*x^2 + 4*d^11*e^2*x - d^12*e)
\[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\int \frac {\left (d + e x\right )^{4}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {11}{2}}}\, dx \] Input:
integrate((e*x+d)**4/(-e**2*x**2+d**2)**(11/2),x)
Output:
Integral((d + e*x)**4/(-(-d + e*x)*(d + e*x))**(11/2), x)
Time = 0.04 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.24 \[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\frac {e^{2} x^{3}}{6 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {9}{2}}} + \frac {4 \, d e x^{2}}{7 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {9}{2}}} + \frac {13 \, d^{2} x}{18 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {9}{2}}} + \frac {20 \, d^{3}}{63 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {9}{2}} e} + \frac {5 \, x}{126 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}}} + \frac {x}{21 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2}} + \frac {4 \, x}{63 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4}} + \frac {8 \, x}{63 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{6}} \] Input:
integrate((e*x+d)^4/(-e^2*x^2+d^2)^(11/2),x, algorithm="maxima")
Output:
1/6*e^2*x^3/(-e^2*x^2 + d^2)^(9/2) + 4/7*d*e*x^2/(-e^2*x^2 + d^2)^(9/2) + 13/18*d^2*x/(-e^2*x^2 + d^2)^(9/2) + 20/63*d^3/((-e^2*x^2 + d^2)^(9/2)*e) + 5/126*x/(-e^2*x^2 + d^2)^(7/2) + 1/21*x/((-e^2*x^2 + d^2)^(5/2)*d^2) + 4 /63*x/((-e^2*x^2 + d^2)^(3/2)*d^4) + 8/63*x/(sqrt(-e^2*x^2 + d^2)*d^6)
\[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\int { \frac {{\left (e x + d\right )}^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {11}{2}}} \,d x } \] Input:
integrate((e*x+d)^4/(-e^2*x^2+d^2)^(11/2),x, algorithm="giac")
Output:
integrate((e*x + d)^4/(-e^2*x^2 + d^2)^(11/2), x)
Time = 6.92 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.27 \[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (\frac {8\,x}{63\,d^6}+\frac {65}{1008\,d^5\,e}\right )}{\left (d+e\,x\right )\,\left (d-e\,x\right )}+\frac {\sqrt {d^2-e^2\,x^2}}{18\,d^2\,e\,{\left (d-e\,x\right )}^5}+\frac {17\,\sqrt {d^2-e^2\,x^2}}{252\,d^3\,e\,{\left (d-e\,x\right )}^4}+\frac {11\,\sqrt {d^2-e^2\,x^2}}{168\,d^4\,e\,{\left (d-e\,x\right )}^3}+\frac {65\,\sqrt {d^2-e^2\,x^2}}{1008\,d^5\,e\,{\left (d-e\,x\right )}^2} \] Input:
int((d + e*x)^4/(d^2 - e^2*x^2)^(11/2),x)
Output:
((d^2 - e^2*x^2)^(1/2)*((8*x)/(63*d^6) + 65/(1008*d^5*e)))/((d + e*x)*(d - e*x)) + (d^2 - e^2*x^2)^(1/2)/(18*d^2*e*(d - e*x)^5) + (17*(d^2 - e^2*x^2 )^(1/2))/(252*d^3*e*(d - e*x)^4) + (11*(d^2 - e^2*x^2)^(1/2))/(168*d^4*e*( d - e*x)^3) + (65*(d^2 - e^2*x^2)^(1/2))/(1008*d^5*e*(d - e*x)^2)
Time = 0.25 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95 \[ \int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\frac {\sqrt {-e^{2} x^{2}+d^{2}}\, \left (8 e^{5} x^{5}-32 d \,e^{4} x^{4}+44 d^{2} e^{3} x^{3}-16 d^{3} e^{2} x^{2}-17 d^{4} e x +20 d^{5}\right )}{63 d^{6} e \left (-e^{6} x^{6}+4 d \,e^{5} x^{5}-5 d^{2} e^{4} x^{4}+5 d^{4} e^{2} x^{2}-4 d^{5} e x +d^{6}\right )} \] Input:
int((e*x+d)^4/(-e^2*x^2+d^2)^(11/2),x)
Output:
(sqrt(d**2 - e**2*x**2)*(20*d**5 - 17*d**4*e*x - 16*d**3*e**2*x**2 + 44*d* *2*e**3*x**3 - 32*d*e**4*x**4 + 8*e**5*x**5))/(63*d**6*e*(d**6 - 4*d**5*e* x + 5*d**4*e**2*x**2 - 5*d**2*e**4*x**4 + 4*d*e**5*x**5 - e**6*x**6))