\(\int \frac {1}{\sqrt {d+e x} (c d^2-c e^2 x^2)^{3/2}} \, dx\) [891]

Optimal result

Integrand size = 29, antiderivative size = 150 \[ \int \frac {1}{\sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=-\frac {1}{2 c d e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}+\frac {3 \sqrt {d+e x}}{4 c d^2 e \sqrt {c d^2-c e^2 x^2}}-\frac {3 \text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{4 \sqrt {2} c^{3/2} d^{5/2} e} \]

[Out]

-3/8*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^(5/2)/e*2^(1/2)-1/2
/c/d/e/(e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(1/2)+3/4*(e*x+d)^(1/2)/c/d^2/e/(-c*e^2*x^2+c*d^2)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {687, 681, 675, 214} \[ \int \frac {1}{\sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=-\frac {3 \text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{4 \sqrt {2} c^{3/2} d^{5/2} e}+\frac {3 \sqrt {d+e x}}{4 c d^2 e \sqrt {c d^2-c e^2 x^2}}-\frac {1}{2 c d e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \]

[In]

Int[1/(Sqrt[d + e*x]*(c*d^2 - c*e^2*x^2)^(3/2)),x]

[Out]

-1/2*1/(c*d*e*Sqrt[d + e*x]*Sqrt[c*d^2 - c*e^2*x^2]) + (3*Sqrt[d + e*x])/(4*c*d^2*e*Sqrt[c*d^2 - c*e^2*x^2]) -
(3*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(4*Sqrt[2]*c^(3/2)*d^(5/2)*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]

Rule 681

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-d)*(d + e*x)^m*((a + c*x^2)^(p +
1)/(2*a*e*(p + 1))), x] + Dist[d*((m + 2*p + 2)/(2*a*(p + 1))), Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x],
x] /; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && LtQ[0, m, 1] && IntegerQ[2*p]

Rule 687

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-e)*(d + e*x)^m*((a + c*x^2)^(p +
1)/(2*c*d*(m + p + 1))), x] + Dist[(m + 2*p + 2)/(2*d*(m + p + 1)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x
] /; FreeQ[{a, c, d, e, p}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[m, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*p]

Rubi steps \begin{align*} \text {integral}& = -\frac {1}{2 c d e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}+\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}}+\frac {3 \sqrt {d+e x}}{4 c d^2 e \sqrt {c d^2-c e^2 x^2}}+\frac {3 \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx}{8 c d^2} \\ & = -\frac {1}{2 c d e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}+\frac {3 \sqrt {d+e x}}{4 c d^2 e \sqrt {c d^2-c e^2 x^2}}+\frac {(3 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 )}{4 c d^2} \\ & = -\frac {1}{2 c d e \sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}+\frac {3 \sqrt {d+e x}}{4 c d^2 e \sqrt {c d^2-c e^2 x^2}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{4 \sqrt {2} c^{3/2} d^{5/2} e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\frac {2 \sqrt {d} (d+3 e x)-3 \sqrt {2} \sqrt {d+e x} \sqrt {d^2-e^2 x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{8 c d^{5/2} e \sqrt {d+e x} \sqrt {c \left (d^2-e^2 x^2\right )}} \]

[In]

Integrate[1/(Sqrt[d + e*x]*(c*d^2 - c*e^2*x^2)^(3/2)),x]

[Out]

(2*Sqrt[d]*(d + 3*e*x) - 3*Sqrt[2]*Sqrt[d + e*x]*Sqrt[d^2 - e^2*x^2]*ArcTanh[(Sqrt[2]*Sqrt[d]*Sqrt[d + e*x])/S
qrt[d^2 - e^2*x^2]])/(8*c*d^(5/2)*e*Sqrt[d + e*x]*Sqrt[c*(d^2 - e^2*x^2)])

Maple [A] (verified)

Time = 2.22 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.94

method	result	size
default	\(-\frac {\sqrt {c \left (-x^{2} e^{2}+d^{2}\right )}\, \left (3 \sqrt {c \left (-e x +d \right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) e x +3 \sqrt {c \left (-e x +d \right )}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) d -6 \sqrt {c d}\, e x -2 \sqrt {c d}\, d \right )}{8 \left (e x +d \right )^{\frac {3}{2}} c^{2} \left (-e x +d \right ) e \,d^{2} \sqrt {c d}}\)	\(141\)

[In]

int(1/(e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/8/(e*x+d)^(3/2)*(c*(-e^2*x^2+d^2))^(1/2)/c^2*(3*(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*x+3*(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-6*(
c*d)^(1/2)*e*x-2*(c*d)^(1/2)*d)/(-e*x+d)/e/d^2/(c*d)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.52 (sec) , antiderivative size = 382, normalized size of antiderivative = 2.55 \[ \int \frac {1}{\sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (e^{3} x^{3} + d e^{2} x^{2} - d^{2} e x - d^{3}\right )} \sqrt {c d} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 2 \, \sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {c d} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 4 \, \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (3 \, d e x + d^{2}\right )} \sqrt {e x + d}}{16 \, {\left (c^{2} d^{3} e^{4} x^{3} + c^{2} d^{4} e^{3} x^{2} - c^{2} d^{5} e^{2} x - c^{2} d^{6} e\right )}}, -\frac {3 \, \sqrt {2} {\left (e^{3} x^{3} + d e^{2} x^{2} - d^{2} e x - d^{3}\right )} \sqrt {-c d} \arctan \left (\frac {\sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {-c d} \sqrt {e x + d}}{c e^{2} x^{2} - c d^{2}}\right ) + 2 \, \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (3 \, d e x + d^{2}\right )} \sqrt {e x + d}}{8 \, {\left (c^{2} d^{3} e^{4} x^{3} + c^{2} d^{4} e^{3} x^{2} - c^{2} d^{5} e^{2} x - c^{2} d^{6} e\right )}}\right ] \]

[In]

integrate(1/(e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(3/2),x, algorithm="fricas")

[Out]

[1/16*(3*sqrt(2)*(e^3*x^3 + d*e^2*x^2 - d^2*e*x - d^3)*sqrt(c*d)*log(-(c*e^2*x^2 - 2*c*d*e*x - 3*c*d^2 + 2*sqr
t(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)
*(3*d*e*x + d^2)*sqrt(e*x + d))/(c^2*d^3*e^4*x^3 + c^2*d^4*e^3*x^2 - c^2*d^5*e^2*x - c^2*d^6*e), -1/8*(3*sqrt(
2)*(e^3*x^3 + d*e^2*x^2 - d^2*e*x - d^3)*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)*(3*d*e*x + d^2)*sqrt(e*x + d))/(c^2*d^3*e^4*x^3 + c^2
*d^4*e^3*x^2 - c^2*d^5*e^2*x - c^2*d^6*e)]

Sympy [F]

\[ \int \frac {1}{\sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\int \frac {1}{\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}} \sqrt {d + e x}}\, dx \]

[In]

integrate(1/(e*x+d)**(1/2)/(-c*e**2*x**2+c*d**2)**(3/2),x)

[Out]

Integral(1/((-c*(-d + e*x)*(d + e*x))**(3/2)*sqrt(d + e*x)), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\int { \frac {1}{{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac {3}{2}} \sqrt {e x + d}} \,d x } \]

[In]

integrate(1/(e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((-c*e^2*x^2 + c*d^2)^(3/2)*sqrt(e*x + d)), x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\frac {3 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{8 \, \sqrt {-c d} c d^{2} e} + \frac {3 \, {\left (e x + d\right )} c - 2 \, c d}{4 \, {\left (2 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c d - {\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}}\right )} c d^{2} e} \]

[In]

integrate(1/(e*x+d)^(1/2)/(-c*e^2*x^2+c*d^2)^(3/2),x, algorithm="giac")

[Out]

3/8*sqrt(2)*arctan(1/2*sqrt(2)*sqrt(-(e*x + d)*c + 2*c*d)/sqrt(-c*d))/(sqrt(-c*d)*c*d^2*e) + 1/4*(3*(e*x + d)*
c - 2*c*d)/((2*sqrt(-(e*x + d)*c + 2*c*d)*c*d - (-(e*x + d)*c + 2*c*d)^(3/2))*c*d^2*e)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {d+e x} \left (c d^2-c e^2 x^2\right )^{3/2}} \, dx=\int \frac {1}{{\left (c\,d^2-c\,e^2\,x^2\right )}^{3/2}\,\sqrt {d+e\,x}} \,d x \]

[In]

int(1/((c*d^2 - c*e^2*x^2)^(3/2)*(d + e*x)^(1/2)),x)

[Out]

int(1/((c*d^2 - c*e^2*x^2)^(3/2)*(d + e*x)^(1/2)), x)