\(\int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx\) [188]

Optimal result

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

-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)-3/32*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^(5/2)/e
 

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.90 \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\frac {2 \sqrt {d} \left (-7 d^2+4 d e x+3 e^2 x^2\right )-3 \sqrt {2} (d+e x)^{3/2} \sqrt {d^2-e^2 x^2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{32 d^{5/2} e (d+e x)^{3/2} \sqrt {c \left (d^2-e^2 x^2\right )}} \] Input:

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

Output:

(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]*Sqrt[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)])
 

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.05, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {470, 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.

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

\(\Big \downarrow \) 470

\(\displaystyle \frac {3 \int \frac {1}{(d+e x)^{3/2} \sqrt {c d^2-c e^2 x^2}}dx}{8 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}\)

\(\Big \downarrow \) 470

\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {d+e x} \sqrt {c d^2-c e^2 x^2}}dx}{4 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}\right )}{8 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}\)

\(\Big \downarrow \) 471

\(\displaystyle \frac {3 \left (\frac {e \int \frac {1}{\frac {e^2 \left (c d^2-c e^2 x^2\right )}{d+e x}-2 c d e^2}d\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {d+e x}}}{2 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}\right )}{8 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {3 \left (-\frac {\text {arctanh}\left (\frac {\sqrt {c d^2-c e^2 x^2}}{\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}\right )}{2 \sqrt {2} \sqrt {c} d^{3/2} e}-\frac {\sqrt {c d^2-c e^2 x^2}}{2 c d e (d+e x)^{3/2}}\right )}{8 d}-\frac {\sqrt {c d^2-c e^2 x^2}}{4 c d e (d+e x)^{5/2}}\)

Input:

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

Output:

-1/4*Sqrt[c*d^2 - c*e^2*x^2]/(c*d*e*(d + e*x)^(5/2)) + (3*(-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]/(S 
qrt[2]*Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])]/(2*Sqrt[2]*Sqrt[c]*d^(3/2)*e)))/(8* 
d)
 

Defintions of rubi rules used

rule 221

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 470

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]
 

rule 471

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]
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.21

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

Input:

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

Output:

-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)+1 
4*(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)

Time = 0.09 (sec) , antiderivative size = 365, normalized size of antiderivative = 2.43 \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\left [\frac {3 \, \sqrt {2} {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \sqrt {c d} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 2 \, \sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {c d} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - 4 \, \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (3 \, d e x + 7 \, d^{2}\right )} \sqrt {e x + d}}{64 \, {\left (c d^{3} e^{4} x^{3} + 3 \, c d^{4} e^{3} x^{2} + 3 \, c d^{5} e^{2} x + c d^{6} e\right )}}, \frac {3 \, \sqrt {2} {\left (e^{3} x^{3} + 3 \, d e^{2} x^{2} + 3 \, d^{2} e x + d^{3}\right )} \sqrt {-c d} \arctan \left (\frac {\sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {-c d} \sqrt {e x + d}}{2 \, {\left (c d e x + c d^{2}\right )}}\right ) - 2 \, \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (3 \, d e x + 7 \, d^{2}\right )} \sqrt {e x + d}}{32 \, {\left (c d^{3} e^{4} x^{3} + 3 \, c d^{4} e^{3} x^{2} + 3 \, c d^{5} e^{2} x + c d^{6} e\right )}}\right ] \] Input:

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

Output:

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

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

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

Maxima [F]

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

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

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.71 \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\frac {\frac {3 \, \sqrt {2} c \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d} d^{2}} - \frac {2 \, {\left (10 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c^{2} d - 3 \, {\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}} c\right )}}{{\left (e x + d\right )}^{2} c^{2} d^{2}}}{32 \, c e} \] Input:

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

Output:

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 203, normalized size of antiderivative = 1.35 \[ \int \frac {1}{(d+e x)^{5/2} \sqrt {c d^2-c e^2 x^2}} \, dx=\frac {\sqrt {c}\, \left (-28 \sqrt {-e x +d}\, d^{2}-12 \sqrt {-e x +d}\, d e x +3 \sqrt {d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}-\sqrt {d}\, \sqrt {2}\right ) d^{2}+6 \sqrt {d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}-\sqrt {d}\, \sqrt {2}\right ) d e x +3 \sqrt {d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}-\sqrt {d}\, \sqrt {2}\right ) e^{2} x^{2}-3 \sqrt {d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}+\sqrt {d}\, \sqrt {2}\right ) d^{2}-6 \sqrt {d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}+\sqrt {d}\, \sqrt {2}\right ) d e x -3 \sqrt {d}\, \sqrt {2}\, \mathrm {log}\left (\sqrt {-e x +d}+\sqrt {d}\, \sqrt {2}\right ) e^{2} x^{2}\right )}{64 c \,d^{3} e \left (e^{2} x^{2}+2 d e x +d^{2}\right )} \] Input:

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

Output:

(sqrt(c)*( - 28*sqrt(d - e*x)*d**2 - 12*sqrt(d - e*x)*d*e*x + 3*sqrt(d)*sq 
rt(2)*log(sqrt(d - e*x) - sqrt(d)*sqrt(2))*d**2 + 6*sqrt(d)*sqrt(2)*log(sq 
rt(d - e*x) - sqrt(d)*sqrt(2))*d*e*x + 3*sqrt(d)*sqrt(2)*log(sqrt(d - e*x) 
 - sqrt(d)*sqrt(2))*e**2*x**2 - 3*sqrt(d)*sqrt(2)*log(sqrt(d - e*x) + sqrt 
(d)*sqrt(2))*d**2 - 6*sqrt(d)*sqrt(2)*log(sqrt(d - e*x) + sqrt(d)*sqrt(2)) 
*d*e*x - 3*sqrt(d)*sqrt(2)*log(sqrt(d - e*x) + sqrt(d)*sqrt(2))*e**2*x**2) 
)/(64*c*d**3*e*(d**2 + 2*d*e*x + e**2*x**2))