\(\int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx\) [884]

Optimal result

Integrand size = 29, antiderivative size = 109 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx=-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}-\frac {\text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{2 \sqrt {2} \sqrt {c} d^{3/2} e} \]

[Out]

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

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {687, 675, 214} \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx=-\frac {\text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{2 \sqrt {2} \sqrt {c} d^{3/2} e}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}} \]

[In]

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

[Out]

-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)

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 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 {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}+\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}}+\frac {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 )}{2 d} \\ & = -\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}-\frac {\tanh ^{-1}\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} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\frac {-2 \sqrt {d} (d-e x)-\sqrt {2} \sqrt {d+e x} \sqrt {d^2-e^2 x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{4 d^{3/2} e \sqrt {d+e x} \sqrt {c \left (d^2-e^2 x^2\right )}} \]

[In]

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

[Out]

(-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)])

Maple [A] (verified)

Time = 2.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.13

method	result	size
default	\(-\frac {\sqrt {c \left (-x^{2} e^{2}+d^{2}\right )}\, \left (\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c e x +c d \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right )+2 \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}\right )}{4 \left (e x +d \right )^{\frac {3}{2}} c \sqrt {c \left (-e x +d \right )}\, e d \sqrt {c d}}\)	\(123\)

[In]

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

[Out]

-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)

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.75 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\left [\frac {\sqrt {2} {\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} \sqrt {c d} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 2 \, \sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {c d} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 4 \, \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d}{8 \, {\left (c d^{2} e^{3} x^{2} + 2 \, c d^{3} e^{2} x + c d^{4} e\right )}}, -\frac {\sqrt {2} {\left (e^{2} x^{2} + 2 \, d e x + d^{2}\right )} \sqrt {-c d} \arctan \left (\frac {\sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {-c d} \sqrt {e x + d}}{c e^{2} x^{2} - c d^{2}}\right ) + 2 \, \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {e x + d} d}{4 \, {\left (c d^{2} e^{3} x^{2} + 2 \, c d^{3} e^{2} x + c d^{4} e\right )}}\right ] \]

[In]

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

[Out]

[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(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)*sqrt(e*x
+ d)*d)/(c*d^2*e^3*x^2 + 2*c*d^3*e^2*x + c*d^4*e)]

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\frac {\frac {\sqrt {2} c \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d} d} - \frac {2 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{{\left (e x + d\right )} d}}{4 \, c e} \]

[In]

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

[Out]

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)

Mupad [F(-1)]

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

[In]

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

[Out]

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