Integrand size = 29, antiderivative size = 104 \[ \int \frac {\sqrt {d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {d+e x}}{c d e \sqrt {c d^2-c e^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-c e^2 x^2}}\right )}{\sqrt {2} c^{3/2} d^{3/2} e} \] Output:
(e*x+d)^(1/2)/c/d/e/(-c*e^2*x^2+c*d^2)^(1/2)-1/2*arctanh(2^(1/2)*c^(1/2)*d ^(1/2)*(e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(1/2))*2^(1/2)/c^(3/2)/d^(3/2)/e
Time = 0.42 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {d} (d+e x)-\sqrt {2} \sqrt {d+e x} \sqrt {d^2-e^2 x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{2 c d^{3/2} e \sqrt {d+e x} \sqrt {c \left (d^2-e^2 x^2\right )}} \] Input:
Integrate[Sqrt[d + e*x]/(c*d^2 - c*e^2*x^2)^(3/2),x]
Output:
(2*Sqrt[d]*(d + e*x) - Sqrt[2]*Sqrt[d + e*x]*Sqrt[d^2 - e^2*x^2]*ArcTanh[( Sqrt[2]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[d^2 - e^2*x^2]])/(2*c*d^(3/2)*e*Sqrt[d + e*x]*Sqrt[c*(d^2 - e^2*x^2)])
Time = 0.33 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {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.
\(\displaystyle \int \frac {\sqrt {d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 467 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}dx}{2 c d}+\frac {\sqrt {d+e x}}{c d e \sqrt {c d^2-c e^2 x^2}}\) |
\(\Big \downarrow \) 471 |
\(\displaystyle \frac {e \int \frac {1}{\frac {e^2 \left (c d^2-c e^2 x^2\right )}{d+e x}-2 c d e^2}d\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {d+e x}}}{c d}+\frac {\sqrt {d+e x}}{c d e \sqrt {c d^2-c e^2 x^2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\sqrt {d+e x}}{c d e \sqrt {c d^2-c e^2 x^2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{\sqrt {2} c^{3/2} d^{3/2} e}\) |
Input:
Int[Sqrt[d + e*x]/(c*d^2 - c*e^2*x^2)^(3/2),x]
Output:
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)
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[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 = 0.20 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.88
method | result | size |
default | \(-\frac {\sqrt {c \left (-e^{2} x^{2}+d^{2}\right )}\, \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) \sqrt {c \left (-e x +d \right )}-2 \sqrt {c d}\right )}{2 \sqrt {e x +d}\, c^{2} \left (-e x +d \right ) e d \sqrt {c d}}\) | \(91\) |
Input:
int((e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/2/(e*x+d)^(1/2)*(c*(-e^2*x^2+d^2))^(1/2)*(2^(1/2)*arctanh(1/2*(c*(-e*x+ d))^(1/2)*2^(1/2)/(c*d)^(1/2))*(c*(-e*x+d))^(1/2)-2*(c*d)^(1/2))/c^2/(-e*x +d)/e/d/(c*d)^(1/2)
Time = 0.10 (sec) , antiderivative size = 281, normalized size of antiderivative = 2.70 \[ \int \frac {\sqrt {d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\left [\frac {\sqrt {2} {\left (e^{2} x^{2} - d^{2}\right )} \sqrt {c d} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 2 \, \sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {c d} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 4 \, \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d}{4 \, {\left (c^{2} d^{2} e^{3} x^{2} - c^{2} d^{4} e\right )}}, \frac {\sqrt {2} {\left (e^{2} x^{2} - d^{2}\right )} \sqrt {-c d} \arctan \left (\frac {\sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {-c d} \sqrt {e x + d}}{2 \, {\left (c d e x + c d^{2}\right )}}\right ) - 2 \, \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d}{2 \, {\left (c^{2} d^{2} e^{3} x^{2} - c^{2} d^{4} e\right )}}\right ] \] Input:
integrate((e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(3/2),x, algorithm="fricas")
Output:
[1/4*(sqrt(2)*(e^2*x^2 - d^2)*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)*sqrt(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)*sqrt(e*x + d)*d)/(c^2*d^2* e^3*x^2 - c^2*d^4*e), 1/2*(sqrt(2)*(e^2*x^2 - d^2)*sqrt(-c*d)*arctan(1/2*s qrt(2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(-c*d)*sqrt(e*x + d)/(c*d*e*x + c*d^2) ) - 2*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)*d)/(c^2*d^2*e^3*x^2 - c^2*d^4 *e)]
\[ \int \frac {\sqrt {d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\int \frac {\sqrt {d + e x}}{\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((e*x+d)**(1/2)/(-c*e**2*x**2+c*d**2)**(3/2),x)
Output:
Integral(sqrt(d + e*x)/(-c*(-d + e*x)*(d + e*x))**(3/2), x)
\[ \int \frac {\sqrt {d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\int { \frac {\sqrt {e x + d}}{{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(3/2),x, algorithm="maxima")
Output:
integrate(sqrt(e*x + d)/(-c*e^2*x^2 + c*d^2)^(3/2), x)
Time = 0.12 (sec) , antiderivative size = 72, normalized size of antiderivative = 0.69 \[ \int \frac {\sqrt {d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d} c d} + \frac {2}{\sqrt {-{\left (e x + d\right )} c + 2 \, c d} c d}}{2 \, e} \] Input:
integrate((e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(3/2),x, algorithm="giac")
Output:
1/2*(sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-(e*x + d)*c + 2*c*d)/sqrt(-c*d))/(sq rt(-c*d)*c*d) + 2/(sqrt(-(e*x + d)*c + 2*c*d)*c*d))/e
Timed out. \[ \int \frac {\sqrt {d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\int \frac {\sqrt {d+e\,x}}{{\left (c\,d^2-c\,e^2\,x^2\right )}^{3/2}} \,d x \] Input:
int((d + e*x)^(1/2)/(c*d^2 - c*e^2*x^2)^(3/2),x)
Output:
int((d + e*x)^(1/2)/(c*d^2 - c*e^2*x^2)^(3/2), x)
Time = 0.22 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.77 \[ \int \frac {\sqrt {d+e x}}{\left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {c}\, \left (\sqrt {d}\, \sqrt {-e x +d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}-\sqrt {d}\, \sqrt {2}\right )-\sqrt {d}\, \sqrt {-e x +d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}+\sqrt {d}\, \sqrt {2}\right )+4 d \right )}{4 \sqrt {-e x +d}\, c^{2} d^{2} e} \] Input:
int((e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(3/2),x)
Output:
(sqrt(c)*(sqrt(d)*sqrt(d - e*x)*sqrt(2)*log(sqrt(d - e*x) - sqrt(d)*sqrt(2 )) - sqrt(d)*sqrt(d - e*x)*sqrt(2)*log(sqrt(d - e*x) + sqrt(d)*sqrt(2)) + 4*d))/(4*sqrt(d - e*x)*c**2*d**2*e)