Integrand size = 24, antiderivative size = 86 \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {2 (d+e x)^2}{5 e \left (d^2-e^2 x^2\right )^{5/2}}+\frac {d+e x}{15 d e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 x}{15 d^3 \sqrt {d^2-e^2 x^2}} \] Output:
2/5*(e*x+d)^2/e/(-e^2*x^2+d^2)^(5/2)+1/15*(e*x+d)/d/e/(-e^2*x^2+d^2)^(3/2) +2/15*x/d^3/(-e^2*x^2+d^2)^(1/2)
Time = 0.35 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.62 \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (7 d^2-6 d e x+2 e^2 x^2\right )}{15 d^3 e (d-e x)^3} \] Input:
Integrate[(d + e*x)^3/(d^2 - e^2*x^2)^(7/2),x]
Output:
(Sqrt[d^2 - e^2*x^2]*(7*d^2 - 6*d*e*x + 2*e^2*x^2))/(15*d^3*e*(d - e*x)^3)
Time = 0.36 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.29, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {464, 461, 461, 460}
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.
\(\displaystyle \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx\) |
\(\Big \downarrow \) 464 |
\(\displaystyle \int \frac {1}{(d-e x)^3 \sqrt {d^2-e^2 x^2}}dx\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {2 \int \frac {1}{(d-e x)^2 \sqrt {d^2-e^2 x^2}}dx}{5 d}+\frac {\sqrt {d^2-e^2 x^2}}{5 d e (d-e x)^3}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {1}{(d-e x) \sqrt {d^2-e^2 x^2}}dx}{3 d}+\frac {\sqrt {d^2-e^2 x^2}}{3 d e (d-e x)^2}\right )}{5 d}+\frac {\sqrt {d^2-e^2 x^2}}{5 d e (d-e x)^3}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle \frac {2 \left (\frac {\sqrt {d^2-e^2 x^2}}{3 d^2 e (d-e x)}+\frac {\sqrt {d^2-e^2 x^2}}{3 d e (d-e x)^2}\right )}{5 d}+\frac {\sqrt {d^2-e^2 x^2}}{5 d e (d-e x)^3}\) |
Input:
Int[(d + e*x)^3/(d^2 - e^2*x^2)^(7/2),x]
Output:
Sqrt[d^2 - e^2*x^2]/(5*d*e*(d - e*x)^3) + (2*(Sqrt[d^2 - e^2*x^2]/(3*d*e*( d - e*x)^2) + Sqrt[d^2 - e^2*x^2]/(3*d^2*e*(d - e*x))))/(5*d)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(b*c*n)), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + 2*p + 2, 0]
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]
Time = 0.25 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.58
method | result | size |
trager | \(\frac {\left (2 e^{2} x^{2}-6 d e x +7 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{15 d^{3} \left (-e x +d \right )^{3} e}\) | \(50\) |
gosper | \(\frac {\left (e x +d \right )^{4} \left (-e x +d \right ) \left (2 e^{2} x^{2}-6 d e x +7 d^{2}\right )}{15 d^{3} e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\) | \(55\) |
orering | \(\frac {\left (e x +d \right )^{4} \left (-e x +d \right ) \left (2 e^{2} x^{2}-6 d e x +7 d^{2}\right )}{15 d^{3} e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}\) | \(55\) |
default | \(d^{3} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )+e^{3} \left (\frac {x^{2}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {2 d^{2}}{15 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\right )+3 d \,e^{2} \left (\frac {x}{4 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}-\frac {d^{2} \left (\frac {x}{5 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}+\frac {\frac {4 x}{15 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {8 x}{15 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}}{d^{2}}\right )}{4 e^{2}}\right )+\frac {3 d^{2}}{5 e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}\) | \(246\) |
Input:
int((e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/15*(2*e^2*x^2-6*d*e*x+7*d^2)/d^3/(-e*x+d)^3/e*(-e^2*x^2+d^2)^(1/2)
Time = 0.11 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.23 \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {7 \, e^{3} x^{3} - 21 \, d e^{2} x^{2} + 21 \, d^{2} e x - 7 \, d^{3} - {\left (2 \, e^{2} x^{2} - 6 \, d e x + 7 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{15 \, {\left (d^{3} e^{4} x^{3} - 3 \, d^{4} e^{3} x^{2} + 3 \, d^{5} e^{2} x - d^{6} e\right )}} \] Input:
integrate((e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="fricas")
Output:
1/15*(7*e^3*x^3 - 21*d*e^2*x^2 + 21*d^2*e*x - 7*d^3 - (2*e^2*x^2 - 6*d*e*x + 7*d^2)*sqrt(-e^2*x^2 + d^2))/(d^3*e^4*x^3 - 3*d^4*e^3*x^2 + 3*d^5*e^2*x - d^6*e)
\[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\int \frac {\left (d + e x\right )^{3}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {7}{2}}}\, dx \] Input:
integrate((e*x+d)**3/(-e**2*x**2+d**2)**(7/2),x)
Output:
Integral((d + e*x)**3/(-(-d + e*x)*(d + e*x))**(7/2), x)
Time = 0.03 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.17 \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {e x^{2}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {4 \, d x}{5 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}}} + \frac {7 \, d^{2}}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} e} + \frac {x}{15 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d} + \frac {2 \, x}{15 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{3}} \] Input:
integrate((e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="maxima")
Output:
1/3*e*x^2/(-e^2*x^2 + d^2)^(5/2) + 4/5*d*x/(-e^2*x^2 + d^2)^(5/2) + 7/15*d ^2/((-e^2*x^2 + d^2)^(5/2)*e) + 1/15*x/((-e^2*x^2 + d^2)^(3/2)*d) + 2/15*x /(sqrt(-e^2*x^2 + d^2)*d^3)
Leaf count of result is larger than twice the leaf count of optimal. 165 vs. \(2 (74) = 148\).
Time = 0.15 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.92 \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=-\frac {2 \, {\left (\frac {20 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} - \frac {40 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + \frac {30 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{6} x^{3}} - \frac {15 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{8} x^{4}} - 7\right )}}{15 \, d^{3} {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{5} {\left | e \right |}} \] Input:
integrate((e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x, algorithm="giac")
Output:
-2/15*(20*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) - 40*(d*e + sqrt(-e^ 2*x^2 + d^2)*abs(e))^2/(e^4*x^2) + 30*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^ 3/(e^6*x^3) - 15*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^4/(e^8*x^4) - 7)/(d^3 *((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^2*x) - 1)^5*abs(e))
Time = 6.41 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.57 \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {\sqrt {d^2-e^2\,x^2}\,\left (7\,d^2-6\,d\,e\,x+2\,e^2\,x^2\right )}{15\,d^3\,e\,{\left (d-e\,x\right )}^3} \] Input:
int((d + e*x)^3/(d^2 - e^2*x^2)^(7/2),x)
Output:
((d^2 - e^2*x^2)^(1/2)*(7*d^2 + 2*e^2*x^2 - 6*d*e*x))/(15*d^3*e*(d - e*x)^ 3)
Time = 0.22 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.40 \[ \int \frac {(d+e x)^3}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx=\frac {-\frac {2 \tan \left (\frac {\mathit {asin} \left (\frac {e x}{d}\right )}{2}\right )^{5}}{5}-\frac {4 \tan \left (\frac {\mathit {asin} \left (\frac {e x}{d}\right )}{2}\right )^{2}}{3}+\frac {2 \tan \left (\frac {\mathit {asin} \left (\frac {e x}{d}\right )}{2}\right )}{3}-\frac {8}{15}}{d^{3} e \left (\tan \left (\frac {\mathit {asin} \left (\frac {e x}{d}\right )}{2}\right )^{5}-5 \tan \left (\frac {\mathit {asin} \left (\frac {e x}{d}\right )}{2}\right )^{4}+10 \tan \left (\frac {\mathit {asin} \left (\frac {e x}{d}\right )}{2}\right )^{3}-10 \tan \left (\frac {\mathit {asin} \left (\frac {e x}{d}\right )}{2}\right )^{2}+5 \tan \left (\frac {\mathit {asin} \left (\frac {e x}{d}\right )}{2}\right )-1\right )} \] Input:
int((e*x+d)^3/(-e^2*x^2+d^2)^(7/2),x)
Output:
(2*( - 3*tan(asin((e*x)/d)/2)**5 - 10*tan(asin((e*x)/d)/2)**2 + 5*tan(asin ((e*x)/d)/2) - 4))/(15*d**3*e*(tan(asin((e*x)/d)/2)**5 - 5*tan(asin((e*x)/ d)/2)**4 + 10*tan(asin((e*x)/d)/2)**3 - 10*tan(asin((e*x)/d)/2)**2 + 5*tan (asin((e*x)/d)/2) - 1))