Integrand size = 29, antiderivative size = 99 \[ \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{3/2}} \, dx=\frac {2 \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}-\frac {2 \sqrt {2} \sqrt {c} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-c e^2 x^2}}\right )}{e} \] Output:
2*(-c*e^2*x^2+c*d^2)^(1/2)/e/(e*x+d)^(1/2)-2*2^(1/2)*c^(1/2)*d^(1/2)*arcta nh(2^(1/2)*c^(1/2)*d^(1/2)*(e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(1/2))/e
Time = 0.18 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.99 \[ \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{3/2}} \, dx=\frac {2 \sqrt {c \left (d^2-e^2 x^2\right )} \left (\frac {1}{\sqrt {d+e x}}-\frac {\sqrt {2} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{\sqrt {d^2-e^2 x^2}}\right )}{e} \] Input:
Integrate[Sqrt[c*d^2 - c*e^2*x^2]/(d + e*x)^(3/2),x]
Output:
(2*Sqrt[c*(d^2 - e^2*x^2)]*(1/Sqrt[d + e*x] - (Sqrt[2]*Sqrt[d]*ArcTanh[(Sq rt[2]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[d^2 - e^2*x^2]])/Sqrt[d^2 - e^2*x^2]))/e
Time = 0.35 (sec) , antiderivative size = 99, 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 = {466, 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.
\(\displaystyle \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 466 |
\(\displaystyle 2 c d \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}dx+\frac {2 \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}\) |
\(\Big \downarrow \) 471 |
\(\displaystyle 4 c d e \int \frac {1}{\frac {e^2 \left (c d^2-c e^2 x^2\right )}{d+e x}-2 c d e^2}d\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {d+e x}}+\frac {2 \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}-\frac {2 \sqrt {2} \sqrt {c} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{e}\) |
Input:
Int[Sqrt[c*d^2 - c*e^2*x^2]/(d + e*x)^(3/2),x]
Output:
(2*Sqrt[c*d^2 - c*e^2*x^2])/(e*Sqrt[d + e*x]) - (2*Sqrt[2]*Sqrt[c]*Sqrt[d] *ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])]) /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 + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 2*(n + 2*p + 1))) Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 ] || EqQ[n + p + 1, 0]) && NeQ[n + 2*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 = 0.21 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.90
method | result | size |
default | \(-\frac {2 \sqrt {c \left (-e^{2} x^{2}+d^{2}\right )}\, \left (c d \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right )-\sqrt {c \left (-e x +d \right )}\, \sqrt {c d}\right )}{\sqrt {e x +d}\, \sqrt {c \left (-e x +d \right )}\, e \sqrt {c d}}\) | \(89\) |
risch | \(\frac {2 \left (-e x +d \right ) \sqrt {-\frac {c \left (e^{2} x^{2}-d^{2}\right )}{e x +d}}\, \sqrt {e x +d}\, c}{e \sqrt {-c \left (e x -d \right )}\, \sqrt {-c \left (e^{2} x^{2}-d^{2}\right )}}-\frac {2 d \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-c e x +c d}\, \sqrt {2}}{2 \sqrt {c d}}\right ) \sqrt {-\frac {c \left (e^{2} x^{2}-d^{2}\right )}{e x +d}}\, \sqrt {e x +d}\, c}{e \sqrt {c d}\, \sqrt {-c \left (e^{2} x^{2}-d^{2}\right )}}\) | \(163\) |
Input:
int((-c*e^2*x^2+c*d^2)^(1/2)/(e*x+d)^(3/2),x,method=_RETURNVERBOSE)
Output:
-2*(c*(-e^2*x^2+d^2))^(1/2)/(e*x+d)^(1/2)*(c*d*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)*(c*d)^(1/2))/(c*(-e*x+ d))^(1/2)/e/(c*d)^(1/2)
Time = 0.11 (sec) , antiderivative size = 233, normalized size of antiderivative = 2.35 \[ \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{3/2}} \, dx=\left [\frac {\sqrt {2} \sqrt {c d} {\left (e x + d\right )} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 2 \, \sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {c d} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}}{e^{2} x + d e}, \frac {2 \, {\left (\sqrt {2} \sqrt {-c d} {\left (e x + d\right )} \arctan \left (\frac {\sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {-c d} \sqrt {e x + d}}{2 \, {\left (c d e x + c d^{2}\right )}}\right ) + \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}\right )}}{e^{2} x + d e}\right ] \] Input:
integrate((-c*e^2*x^2+c*d^2)^(1/2)/(e*x+d)^(3/2),x, algorithm="fricas")
Output:
[(sqrt(2)*sqrt(c*d)*(e*x + d)*log(-(c*e^2*x^2 - 2*c*d*e*x - 3*c*d^2 + 2*sq rt(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)) + 2*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d))/(e^2*x + d*e), 2*(sqr t(2)*sqrt(-c*d)*(e*x + d)*arctan(1/2*sqrt(2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt (-c*d)*sqrt(e*x + d)/(c*d*e*x + c*d^2)) + sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e* x + d))/(e^2*x + d*e)]
\[ \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{3/2}} \, dx=\int \frac {\sqrt {- c \left (- d + e x\right ) \left (d + e x\right )}}{\left (d + e x\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((-c*e**2*x**2+c*d**2)**(1/2)/(e*x+d)**(3/2),x)
Output:
Integral(sqrt(-c*(-d + e*x)*(d + e*x))/(d + e*x)**(3/2), x)
\[ \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{3/2}} \, dx=\int { \frac {\sqrt {-c e^{2} x^{2} + c d^{2}}}{{\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((-c*e^2*x^2+c*d^2)^(1/2)/(e*x+d)^(3/2),x, algorithm="maxima")
Output:
integrate(sqrt(-c*e^2*x^2 + c*d^2)/(e*x + d)^(3/2), x)
Time = 0.12 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{3/2}} \, dx=\frac {2 \, {\left (\frac {\sqrt {2} d \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d}} + \frac {\sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{c}\right )} c}{e} \] Input:
integrate((-c*e^2*x^2+c*d^2)^(1/2)/(e*x+d)^(3/2),x, algorithm="giac")
Output:
2*(sqrt(2)*d*arctan(1/2*sqrt(2)*sqrt(-(e*x + d)*c + 2*c*d)/sqrt(-c*d))/sqr t(-c*d) + sqrt(-(e*x + d)*c + 2*c*d)/c)*c/e
Timed out. \[ \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{3/2}} \, dx=\int \frac {\sqrt {c\,d^2-c\,e^2\,x^2}}{{\left (d+e\,x\right )}^{3/2}} \,d x \] Input:
int((c*d^2 - c*e^2*x^2)^(1/2)/(d + e*x)^(3/2),x)
Output:
int((c*d^2 - c*e^2*x^2)^(1/2)/(d + e*x)^(3/2), x)
Time = 0.22 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.46 \[ \int \frac {\sqrt {c d^2-c e^2 x^2}}{(d+e x)^{3/2}} \, dx=\frac {2 \sqrt {c}\, \left (\sqrt {-e x +d}+\sqrt {d}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +d}}{\sqrt {d}\, \sqrt {2}}\right )}{2}\right )\right )-\sqrt {d}\, \sqrt {2}\right )}{e} \] Input:
int((-c*e^2*x^2+c*d^2)^(1/2)/(e*x+d)^(3/2),x)
Output:
(2*sqrt(c)*(sqrt(d - e*x) + sqrt(d)*sqrt(2)*log(tan(asin(sqrt(d + e*x)/(sq rt(d)*sqrt(2)))/2)) - sqrt(d)*sqrt(2)))/e