\(\int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx\) [885]

Optimal result

Integrand size = 29, antiderivative size = 150 \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx=-\frac {\sqrt {c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}-\frac {3 \sqrt {c d^2-c e^2 x^2}}{16 c d^2 e (d+e x)^{3/2}}-\frac {3 \text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{16 \sqrt {2} \sqrt {c} d^{5/2} e} \]

[Out]

-3/32*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^(5/2)/e*2^(1/2)/c^(1/2)-1/
4*(-c*e^2*x^2+c*d^2)^(1/2)/c/d/e/(e*x+d)^(5/2)-3/16*(-c*e^2*x^2+c*d^2)^(1/2)/c/d^2/e/(e*x+d)^(3/2)

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 = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {687, 675, 214} \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx=-\frac {3 \text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{16 \sqrt {2} \sqrt {c} d^{5/2} e}-\frac {3 \sqrt {c d^2-c e^2 x^2}}{16 c d^2 e (d+e x)^{3/2}}-\frac {\sqrt {c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}} \]

[In]

Int[1/((d + e*x)^(5/2)*Sqrt[c*d^2 - c*e^2*x^2]),x]

[Out]

-1/4*Sqrt[c*d^2 - c*e^2*x^2]/(c*d*e*(d + e*x)^(5/2)) - (3*Sqrt[c*d^2 - c*e^2*x^2])/(16*c*d^2*e*(d + e*x)^(3/2)
) - (3*ArcTanh[Sqrt[c*d^2 - c*e^2*x^2]/(Sqrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])])/(16*Sqrt[2]*Sqrt[c]*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 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}}{4 c d e (d+e x)^{5/2}}+\frac {3 \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}} \, dx}{8 d} \\ & = -\frac {\sqrt {c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}-\frac {3 \sqrt {c d^2-c e^2 x^2}}{16 c d^2 e (d+e x)^{3/2}}+\frac {3 \int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}} \, dx}{32 d^2} \\ & = -\frac {\sqrt {c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}-\frac {3 \sqrt {c d^2-c e^2 x^2}}{16 c d^2 e (d+e x)^{3/2}}+\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 )}{16 d^2} \\ & = -\frac {\sqrt {c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}-\frac {3 \sqrt {c d^2-c e^2 x^2}}{16 c d^2 e (d+e x)^{3/2}}-\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 )}{16 \sqrt {2} \sqrt {c} d^{5/2} e} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.36 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\frac {2 \sqrt {d} \left (-7 d^2+4 d e x+3 e^2 x^2\right )-3 \sqrt {2} (d+e x)^{3/2} \sqrt {d^2-e^2 x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{32 d^{5/2} e (d+e x)^{3/2} \sqrt {c \left (d^2-e^2 x^2\right )}} \]

[In]

Integrate[1/((d + e*x)^(5/2)*Sqrt[c*d^2 - c*e^2*x^2]),x]

[Out]

(2*Sqrt[d]*(-7*d^2 + 4*d*e*x + 3*e^2*x^2) - 3*Sqrt[2]*(d + e*x)^(3/2)*Sqrt[d^2 - e^2*x^2]*ArcTanh[(Sqrt[2]*Sqr
t[d]*Sqrt[d + e*x])/Sqrt[d^2 - e^2*x^2]])/(32*d^(5/2)*e*(d + e*x)^(3/2)*Sqrt[c*(d^2 - e^2*x^2)])

Maple [A] (verified)

Time = 2.36 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.21

method	result	size
default	\(-\frac {\sqrt {c \left (-x^{2} e^{2}+d^{2}\right )}\, \left (3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,e^{2} x^{2}+6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c d e x +3 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,d^{2}+6 e x \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}+14 \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}\, d \right )}{32 \left (e x +d \right )^{\frac {5}{2}} c \sqrt {c \left (-e x +d \right )}\, e \,d^{2} \sqrt {c d}}\)	\(181\)

[In]

int(1/(e*x+d)^(5/2)/(-c*e^2*x^2+c*d^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/32/(e*x+d)^(5/2)*(c*(-e^2*x^2+d^2))^(1/2)/c*(3*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^(1/2)/(c*d)^(1/2))*
c*e^2*x^2+6*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)*2^(1/2)/(c*d)^(1/2))*c*d*e*x+3*2^(1/2)*arctanh(1/2*(c*(-e*x
+d))^(1/2)*2^(1/2)/(c*d)^(1/2))*c*d^2+6*e*x*(c*(-e*x+d))^(1/2)*(c*d)^(1/2)+14*(c*(-e*x+d))^(1/2)*(c*d)^(1/2)*d
)/(c*(-e*x+d))^(1/2)/e/d^2/(c*d)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 368, normalized size of antiderivative = 2.45 \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \sqrt {c d} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 2 \, \sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {c d} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 4 \, \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (3 \, d e x + 7 \, d^{2}\right )} \sqrt {e x + d}}{64 \, {\left (c d^{3} e^{4} x^{3} + 3 \, c d^{4} e^{3} x^{2} + 3 \, c d^{5} e^{2} x + c d^{6} e\right )}}, -\frac {3 \, \sqrt {2} {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \sqrt {-c d} \arctan \left (\frac {\sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {-c d} \sqrt {e x + d}}{c e^{2} x^{2} - c d^{2}}\right ) + 2 \, \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (3 \, d e x + 7 \, d^{2}\right )} \sqrt {e x + d}}{32 \, {\left (c d^{3} e^{4} x^{3} + 3 \, c d^{4} e^{3} x^{2} + 3 \, c d^{5} e^{2} x + c d^{6} e\right )}}\right ] \]

[In]

integrate(1/(e*x+d)^(5/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="fricas")

[Out]

[1/64*(3*sqrt(2)*(e^3*x^3 + 3*d*e^2*x^2 + 3*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
*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)*(3*d*e*x + 7*d^2)*sqrt(e*x + d))/(c*d^3*e^4*x^3 + 3*c*d^4*e^3*x^2 + 3*c*d^5*e^2*x + c*d^6*e), -1/32*(3*sq
rt(2)*(e^3*x^3 + 3*d*e^2*x^2 + 3*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 + 7*d^2)*sqrt(e*x + d))/(c*d^3*e^4*x^
3 + 3*c*d^4*e^3*x^2 + 3*c*d^5*e^2*x + c*d^6*e)]

Sympy [F]

\[ \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\int \frac {1}{\sqrt {- c \left (- d + e x\right ) \left (d + e x\right )} \left (d + e x\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(1/(e*x+d)**(5/2)/(-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)**(5/2)), x)

Maxima [F]

\[ \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\int { \frac {1}{\sqrt {-c e^{2} x^{2} + c d^{2}} {\left (e x + d\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(1/(e*x+d)^(5/2)/(-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)^(5/2)), x)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\frac {\frac {3 \, \sqrt {2} c \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d} d^{2}} - \frac {2 \, {\left (10 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c^{2} d - 3 \, {\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}} c\right )}}{{\left (e x + d\right )}^{2} c^{2} d^{2}}}{32 \, c e} \]

[In]

integrate(1/(e*x+d)^(5/2)/(-c*e^2*x^2+c*d^2)^(1/2),x, algorithm="giac")

[Out]

1/32*(3*sqrt(2)*c*arctan(1/2*sqrt(2)*sqrt(-(e*x + d)*c + 2*c*d)/sqrt(-c*d))/(sqrt(-c*d)*d^2) - 2*(10*sqrt(-(e*
x + d)*c + 2*c*d)*c^2*d - 3*(-(e*x + d)*c + 2*c*d)^(3/2)*c)/((e*x + d)^2*c^2*d^2))/(c*e)

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\int \frac {1}{\sqrt {c\,d^2-c\,e^2\,x^2}\,{\left (d+e\,x\right )}^{5/2}} \,d x \]

[In]

int(1/((c*d^2 - c*e^2*x^2)^(1/2)*(d + e*x)^(5/2)),x)

[Out]

int(1/((c*d^2 - c*e^2*x^2)^(1/2)*(d + e*x)^(5/2)), x)