Integrand size = 29, antiderivative size = 191 \[ \int \frac {1}{(d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=-\frac {1}{4 c d e (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}-\frac {5}{16 c d^2 e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}+\frac {15 \sqrt {d+e x}}{32 c d^3 e \sqrt {c d^2-c e^2 x^2}}-\frac {15 \text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{32 \sqrt {2} c^{3/2} d^{7/2} e} \]
-15/64*arctanh(1/2*(-c*e^2*x^2+c*d^2)^(1/2)*2^(1/2)/c^(1/2)/d^(1/2)/(e*x+d )^(1/2))/c^(3/2)/d^(7/2)/e*2^(1/2)-1/4/c/d/e/(e*x+d)^(3/2)/(-c*e^2*x^2+c*d ^2)^(1/2)-5/16/c/d^2/e/(e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(1/2)+15/32*(e*x+d )^(1/2)/c/d^3/e/(-c*e^2*x^2+c*d^2)^(1/2)
Time = 0.89 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.72 \[ \int \frac {1}{(d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {d} \left (-3 d^2+20 d e x+15 e^2 x^2\right )-15 \sqrt {2} (d+e x)^{3/2} \sqrt {d^2-e^2 x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{64 c d^{7/2} e (d+e x)^{3/2} \sqrt {c \left (d^2-e^2 x^2\right )}} \]
(2*Sqrt[d]*(-3*d^2 + 20*d*e*x + 15*e^2*x^2) - 15*Sqrt[2]*(d + e*x)^(3/2)*S qrt[d^2 - e^2*x^2]*ArcTanh[(Sqrt[2]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[d^2 - e^2* x^2]])/(64*c*d^(7/2)*e*(d + e*x)^(3/2)*Sqrt[c*(d^2 - e^2*x^2)])
Time = 0.29 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.06, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {470, 470, 467, 471, 221}
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.
| \(\displaystyle \int \frac {1}{(d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 470 |
| \(\displaystyle \frac {5 \int \frac {1}{\sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}}dx}{8 d}-\frac {1}{4 c d e (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}\) |
| \(\Big \downarrow \) 470 |
| \(\displaystyle \frac {5 \left (\frac {3 \int \frac {\sqrt {d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}}dx}{4 d}-\frac {1}{2 c d e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}\right )}{8 d}-\frac {1}{4 c d e (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}\) |
| \(\Big \downarrow \) 467 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}dx}{2 c d}+\frac {\sqrt {d+e x}}{c d e \sqrt {c d^2-c e^2 x^2}}\right )}{4 d}-\frac {1}{2 c d e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}\right )}{8 d}-\frac {1}{4 c d e (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}\) |
| \(\Big \downarrow \) 471 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {e \int \frac {1}{\frac {e^2 \left (c d^2-c e^2 x^2\right )}{d+e x}-2 c d e^2}d\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {d+e x}}}{c d}+\frac {\sqrt {d+e x}}{c d e \sqrt {c d^2-c e^2 x^2}}\right )}{4 d}-\frac {1}{2 c d e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}\right )}{8 d}-\frac {1}{4 c d e (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {5 \left (\frac {3 \left (\frac {\sqrt {d+e x}}{c d e \sqrt {c d^2-c e^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{\sqrt {2} c^{3/2} d^{3/2} e}\right )}{4 d}-\frac {1}{2 c d e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}\right )}{8 d}-\frac {1}{4 c d e (d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}\) |
-1/4*1/(c*d*e*(d + e*x)^(3/2)*Sqrt[c*d^2 - c*e^2*x^2]) + (5*(-1/2*1/(c*d*e *Sqrt[d + e*x]*Sqrt[c*d^2 - c*e^2*x^2]) + (3*(Sqrt[d + e*x]/(c*d*e*Sqrt[c* d^2 - c*e^2*x^2]) - ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[ d]*Sqrt[d + e*x])]/(Sqrt[2]*c^(3/2)*d^(3/2)*e)))/(4*d)))/(8*d)
3.9.92.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-c)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*a*d*(p + 1))), x] + Simp[c*((n + 2 *p + 2)/(2*a*(p + 1))) Int[(c + d*x)^(n - 1)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[p, -1] && LtQ[0, n, 1] && IntegerQ[2*p]
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]
Int[1/(Sqrt[(c_) + (d_.)*(x_)]*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> Sim p[2*d Subst[Int[1/(2*b*c + d^2*x^2), x], x, Sqrt[a + b*x^2]/Sqrt[c + d*x] ], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0]
Time = 2.23 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.06
| method | result | size |
| default | \(-\frac {\sqrt {c \left (-x^{2} e^{2}+d^{2}\right )}\, \left (15 \sqrt {c \left (-e x +d \right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) e^{2} x^{2}+30 \sqrt {c \left (-e x +d \right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) d e x +15 \sqrt {c \left (-e x +d \right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) d^{2}-30 \sqrt {c d}\, e^{2} x^{2}-40 \sqrt {c d}\, d e x +6 \sqrt {c d}\, d^{2}\right )}{64 \left (e x +d \right )^{\frac {5}{2}} c^{2} \left (-e x +d \right ) e \,d^{3} \sqrt {c d}}\) | \(202\) |
-1/64/(e*x+d)^(5/2)*(c*(-e^2*x^2+d^2))^(1/2)/c^2*(15*(c*(-e*x+d))^(1/2)*2^ (1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^(1/2)/(c*d)^(1/2))*e^2*x^2+30*(c*(- e*x+d))^(1/2)*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^(1/2)/(c*d)^(1/2))* d*e*x+15*(c*(-e*x+d))^(1/2)*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^(1/2) /(c*d)^(1/2))*d^2-30*(c*d)^(1/2)*e^2*x^2-40*(c*d)^(1/2)*d*e*x+6*(c*d)^(1/2 )*d^2)/(-e*x+d)/e/d^3/(c*d)^(1/2)
Time = 0.30 (sec) , antiderivative size = 412, normalized size of antiderivative = 2.16 \[ \int \frac {1}{(d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\left [\frac {15 \, \sqrt {2} {\left (e^{4} x^{4} + 2 \, d e^{3} x^{3} - 2 \, d^{3} e x - d^{4}\right )} \sqrt {c d} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 2 \, \sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {c d} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 4 \, \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (15 \, d e^{2} x^{2} + 20 \, d^{2} e x - 3 \, d^{3}\right )} \sqrt {e x + d}}{128 \, {\left (c^{2} d^{4} e^{5} x^{4} + 2 \, c^{2} d^{5} e^{4} x^{3} - 2 \, c^{2} d^{7} e^{2} x - c^{2} d^{8} e\right )}}, -\frac {15 \, \sqrt {2} {\left (e^{4} x^{4} + 2 \, d e^{3} x^{3} - 2 \, d^{3} e x - d^{4}\right )} \sqrt {-c d} \arctan \left (\frac {\sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {-c d} \sqrt {e x + d}}{c e^{2} x^{2} - c d^{2}}\right ) + 2 \, \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (15 \, d e^{2} x^{2} + 20 \, d^{2} e x - 3 \, d^{3}\right )} \sqrt {e x + d}}{64 \, {\left (c^{2} d^{4} e^{5} x^{4} + 2 \, c^{2} d^{5} e^{4} x^{3} - 2 \, c^{2} d^{7} e^{2} x - c^{2} d^{8} e\right )}}\right ] \]
[1/128*(15*sqrt(2)*(e^4*x^4 + 2*d*e^3*x^3 - 2*d^3*e*x - d^4)*sqrt(c*d)*log (-(c*e^2*x^2 - 2*c*d*e*x - 3*c*d^2 + 2*sqrt(2)*sqrt(-c*e^2*x^2 + c*d^2)*sq rt(c*d)*sqrt(e*x + d))/(e^2*x^2 + 2*d*e*x + d^2)) - 4*sqrt(-c*e^2*x^2 + c* d^2)*(15*d*e^2*x^2 + 20*d^2*e*x - 3*d^3)*sqrt(e*x + d))/(c^2*d^4*e^5*x^4 + 2*c^2*d^5*e^4*x^3 - 2*c^2*d^7*e^2*x - c^2*d^8*e), -1/64*(15*sqrt(2)*(e^4* x^4 + 2*d*e^3*x^3 - 2*d^3*e*x - d^4)*sqrt(-c*d)*arctan(sqrt(2)*sqrt(-c*e^2 *x^2 + c*d^2)*sqrt(-c*d)*sqrt(e*x + d)/(c*e^2*x^2 - c*d^2)) + 2*sqrt(-c*e^ 2*x^2 + c*d^2)*(15*d*e^2*x^2 + 20*d^2*e*x - 3*d^3)*sqrt(e*x + d))/(c^2*d^4 *e^5*x^4 + 2*c^2*d^5*e^4*x^3 - 2*c^2*d^7*e^2*x - c^2*d^8*e)]
\[ \int \frac {1}{(d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\int \frac {1}{\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \left (d + e x\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {1}{(d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac {3}{2}} {\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \]
Time = 0.31 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.68 \[ \int \frac {1}{(d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\frac {15 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{64 \, \sqrt {-c d} c d^{3} e} + \frac {1}{4 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c d^{3} e} - \frac {18 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c d - 7 \, {\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}}}{32 \, {\left (e x + d\right )}^{2} c^{3} d^{3} e} \]
15/64*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-(e*x + d)*c + 2*c*d)/sqrt(-c*d))/(s qrt(-c*d)*c*d^3*e) + 1/4/(sqrt(-(e*x + d)*c + 2*c*d)*c*d^3*e) - 1/32*(18*s qrt(-(e*x + d)*c + 2*c*d)*c*d - 7*(-(e*x + d)*c + 2*c*d)^(3/2))/((e*x + d) ^2*c^3*d^3*e)
Timed out. \[ \int \frac {1}{(d+e x)^{3/2} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\int \frac {1}{{\left (c\,d^2-c\,e^2\,x^2\right )}^{3/2}\,{\left (d+e\,x\right )}^{3/2}} \,d x \]