Integrand size = 29, antiderivative size = 109 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx=-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-c e^2 x^2}}\right )}{2 \sqrt {2} \sqrt {c} d^{3/2} e} \] Output:
-1/2*(-c*e^2*x^2+c*d^2)^(1/2)/c/d/e/(e*x+d)^(3/2)-1/4*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^(1/2)/d^(3/2 )/e
Time = 0.22 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\frac {-2 \sqrt {d} (d-e x)-\sqrt {2} \sqrt {d+e x} \sqrt {d^2-e^2 x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{4 d^{3/2} e \sqrt {d+e x} \sqrt {c \left (d^2-e^2 x^2\right )}} \] Input:
Integrate[1/((d + e*x)^(3/2)*Sqrt[c*d^2 - c*e^2*x^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]])/(4*d^(3/2)*e*Sqrt[d + e*x]*Sqrt[c*(d^2 - e^2*x^2)])
Time = 0.33 (sec) , antiderivative size = 109, 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 = {470, 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.
\(\displaystyle \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {\int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}dx}{4 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}\) |
\(\Big \downarrow \) 471 |
\(\displaystyle \frac {e \int \frac {1}{\frac {e^2 \left (c d^2-c e^2 x^2\right )}{d+e x}-2 c d e^2}d\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {d+e x}}}{2 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {\text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{2 \sqrt {2} \sqrt {c} d^{3/2} e}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}\) |
Input:
Int[1/((d + e*x)^(3/2)*Sqrt[c*d^2 - c*e^2*x^2]),x]
Output:
-1/2*Sqrt[c*d^2 - c*e^2*x^2]/(c*d*e*(d + e*x)^(3/2)) - ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])]/(2*Sqrt[2]*Sqrt[c]*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[ (-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 = 0.21 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.13
method | result | size |
default | \(-\frac {\sqrt {c \left (-e^{2} x^{2}+d^{2}\right )}\, \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c e x +c d \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right )+2 \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}\right )}{4 \left (e x +d \right )^{\frac {3}{2}} c \sqrt {c \left (-e x +d \right )}\, e d \sqrt {c d}}\) | \(123\) |
Input:
int(1/(e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/4/(e*x+d)^(3/2)*(c*(-e^2*x^2+d^2))^(1/2)/c*(2^(1/2)*arctanh(1/2*(c*(-e* x+d))^(1/2)*2^(1/2)/(c*d)^(1/2))*c*e*x+c*d*2^(1/2)*arctanh(1/2*(c*(-e*x+d) )^(1/2)*2^(1/2)/(c*d)^(1/2))+2*(c*(-e*x+d))^(1/2)*(c*d)^(1/2))/(c*(-e*x+d) )^(1/2)/e/d/(c*d)^(1/2)
Time = 0.08 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.72 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\left [\frac {\sqrt {2} {\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} \sqrt {c d} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 2 \, \sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {c d} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 4 \, \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d}{8 \, {\left (c d^{2} e^{3} x^{2} + 2 \, c d^{3} e^{2} x + c d^{4} e\right )}}, \frac {\sqrt {2} {\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} \sqrt {-c d} \arctan \left (\frac {\sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {-c d} \sqrt {e x + d}}{2 \, {\left (c d e x + c d^{2}\right )}}\right ) - 2 \, \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d}{4 \, {\left (c d^{2} e^{3} x^{2} + 2 \, c d^{3} e^{2} x + c d^{4} e\right )}}\right ] \] Input:
integrate(1/(e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="fricas")
Output:
[1/8*(sqrt(2)*(e^2*x^2 + 2*d*e*x + 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*d^2*e^3*x^2 + 2*c*d^3*e^2*x + c*d^4*e), 1/4*(sqrt(2)*(e^2*x^2 + 2*d*e* x + d^2)*sqrt(-c*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)) - 2*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)*d)/(c*d^2*e^3*x^2 + 2*c*d^3*e^2*x + c*d^4*e)]
\[ \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\int \frac {1}{\sqrt {- c \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(1/(e*x+d)**(3/2)/(-c*e**2*x**2+c*d**2)**(1/2),x)
Output:
Integral(1/(sqrt(-c*(-d + e*x)*(d + e*x))*(d + e*x)**(3/2)), x)
\[ \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\int { \frac {1}{\sqrt {-c e^{2} x^{2} + c d^{2}} {\left (e x + d\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(1/(e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="maxima")
Output:
integrate(1/(sqrt(-c*e^2*x^2 + c*d^2)*(e*x + d)^(3/2)), x)
Time = 0.13 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\frac {\frac {\sqrt {2} c \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d} d} - \frac {2 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{{\left (e x + d\right )} d}}{4 \, c e} \] Input:
integrate(1/(e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="giac")
Output:
1/4*(sqrt(2)*c*arctan(1/2*sqrt(2)*sqrt(-(e*x + d)*c + 2*c*d)/sqrt(-c*d))/( sqrt(-c*d)*d) - 2*sqrt(-(e*x + d)*c + 2*c*d)/((e*x + d)*d))/(c*e)
Timed out. \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\int \frac {1}{\sqrt {c\,d^2-c\,e^2\,x^2}\,{\left (d+e\,x\right )}^{3/2}} \,d x \] Input:
int(1/((c*d^2 - c*e^2*x^2)^(1/2)*(d + e*x)^(3/2)),x)
Output:
int(1/((c*d^2 - c*e^2*x^2)^(1/2)*(d + e*x)^(3/2)), x)
Time = 0.23 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.07 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\frac {\sqrt {c}\, \left (-4 \sqrt {-e x +d}\, d +\sqrt {d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}-\sqrt {d}\, \sqrt {2}\right ) d +\sqrt {d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}-\sqrt {d}\, \sqrt {2}\right ) e x -\sqrt {d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}+\sqrt {d}\, \sqrt {2}\right ) d -\sqrt {d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}+\sqrt {d}\, \sqrt {2}\right ) e x \right )}{8 c \,d^{2} e \left (e x +d \right )} \] Input:
int(1/(e*x+d)^(3/2)/(-c*e^2*x^2+c*d^2)^(1/2),x)
Output:
(sqrt(c)*( - 4*sqrt(d - e*x)*d + sqrt(d)*sqrt(2)*log(sqrt(d - e*x) - sqrt( d)*sqrt(2))*d + sqrt(d)*sqrt(2)*log(sqrt(d - e*x) - sqrt(d)*sqrt(2))*e*x - sqrt(d)*sqrt(2)*log(sqrt(d - e*x) + sqrt(d)*sqrt(2))*d - sqrt(d)*sqrt(2)* log(sqrt(d - e*x) + sqrt(d)*sqrt(2))*e*x))/(8*c*d**2*e*(d + e*x))