\(\int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx\) [883]

Optimal result

Integrand size = 29, antiderivative size = 65 \[ \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{\sqrt {c} \sqrt {d} e} \]

[Out]

-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))*2^(1/2)/e/c^(1/2)/d^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {675, 214} \[ \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx=-\frac {\sqrt {2} \text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{\sqrt {c} \sqrt {d} e} \]

[In]

Int[1/(Sqrt[d + e*x]*Sqrt[c*d^2 - c*e^2*x^2]),x]

[Out]

-((Sqrt[2]*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(Sqrt[c]*Sqrt[d]*e))

Rule 214

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]

Rule 675

Int[1/(Sqrt[(d_) + (e_.)*(x_)]*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e, Subst[Int[1/(2*c*d + e^2*x^2
), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0]

Rubi steps \begin{align*} \text {integral}& = (2 e) \text {Subst}\left (\int \frac {1}{-2 c d e^2+e^2 x^2} \, dx,x,\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {d+e x}}\right ) \\ & = -\frac {\sqrt {2} \tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{\sqrt {c} \sqrt {d} e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.18 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.32 \[ \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx=-\frac {\sqrt {2} \sqrt {d^2-e^2 x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{\sqrt {d} e \sqrt {c \left (d^2-e^2 x^2\right )}} \]

[In]

Integrate[1/(Sqrt[d + e*x]*Sqrt[c*d^2 - c*e^2*x^2]),x]

[Out]

-((Sqrt[2]*Sqrt[d^2 - e^2*x^2]*ArcTanh[(Sqrt[2]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[d^2 - e^2*x^2]])/(Sqrt[d]*e*Sqrt[c
*(d^2 - e^2*x^2)]))

Maple [A] (verified)

Time = 2.34 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.05

method	result	size
default	\(-\frac {\sqrt {c \left (-x^{2} e^{2}+d^{2}\right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right )}{\sqrt {e x +d}\, \sqrt {c \left (-e x +d \right )}\, e \sqrt {c d}}\)	\(68\)

[In]

int(1/(e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/(e*x+d)^(1/2)*(c*(-e^2*x^2+d^2))^(1/2)/(c*(-e*x+d))^(1/2)/e*2^(1/2)/(c*d)^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1
/2)*2^(1/2)/(c*d)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 169, normalized size of antiderivative = 2.60 \[ \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx=\left [\frac {\sqrt {2} \sqrt {\frac {1}{c d}} \log \left (-\frac {e^{2} x^{2} - 2 \, d e x + 2 \, \sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d \sqrt {\frac {1}{c d}} - 3 \, d^{2}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right )}{2 \, e}, -\frac {\sqrt {2} \sqrt {-\frac {1}{c d}} \arctan \left (\frac {\sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d \sqrt {-\frac {1}{c d}}}{e^{2} x^{2} - d^{2}}\right )}{e}\right ] \]

[In]

integrate(1/(e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

[1/2*sqrt(2)*sqrt(1/(c*d))*log(-(e^2*x^2 - 2*d*e*x + 2*sqrt(2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)*d*sqrt(1
/(c*d)) - 3*d^2)/(e^2*x^2 + 2*d*e*x + d^2))/e, -sqrt(2)*sqrt(-1/(c*d))*arctan(sqrt(2)*sqrt(-c*e^2*x^2 + c*d^2)
*sqrt(e*x + d)*d*sqrt(-1/(c*d))/(e^2*x^2 - d^2))/e]

Sympy [F]

\[ \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx=\int \frac {1}{\sqrt {- c \left (- d + e x\right ) \left (d + e x\right )} \sqrt {d + e x}}\, dx \]

[In]

integrate(1/(e*x+d)**(1/2)/(-c*e**2*x**2+c*d**2)**(1/2),x)

[Out]

Integral(1/(sqrt(-c*(-d + e*x)*(d + e*x))*sqrt(d + e*x)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx=\int { \frac {1}{\sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d}} \,d x } \]

[In]

integrate(1/(e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/(sqrt(-c*e^2*x^2 + c*d^2)*sqrt(e*x + d)), x)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx=\frac {\frac {\sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d}} - \frac {\sqrt {2} \arctan \left (\frac {\sqrt {c d}}{\sqrt {-c d}}\right )}{\sqrt {-c d}}}{e} \]

[In]

integrate(1/(e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="giac")

[Out]

(sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-(e*x + d)*c + 2*c*d)/sqrt(-c*d))/sqrt(-c*d) - sqrt(2)*arctan(sqrt(c*d)/sqrt(
-c*d))/sqrt(-c*d))/e

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx=\int \frac {1}{\sqrt {c\,d^2-c\,e^2\,x^2}\,\sqrt {d+e\,x}} \,d x \]

[In]

int(1/((c*d^2 - c*e^2*x^2)^(1/2)*(d + e*x)^(1/2)),x)

[Out]

int(1/((c*d^2 - c*e^2*x^2)^(1/2)*(d + e*x)^(1/2)), x)