\(\int \frac {(c d^2-c e^2 x^2)^{3/2}}{(d+e x)^{5/2}} \, dx\) [177]

Optimal result

Integrand size = 29, antiderivative size = 136 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {4 c d \sqrt {c d^2-c e^2 x^2}}{e \sqrt {d+e x}}+\frac {2 \left (c d^2-c e^2 x^2\right )^{3/2}}{3 e (d+e x)^{3/2}}-\frac {4 \sqrt {2} c^{3/2} d^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-c e^2 x^2}}\right )}{e} \] Output:

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

Mathematica [A] (verified)

Time = 0.56 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.81 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 c \sqrt {c \left (d^2-e^2 x^2\right )} \left (\frac {7 d-e x}{\sqrt {d+e x}}-\frac {6 \sqrt {2} d^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {d} \sqrt {d+e x}}{\sqrt {d^2-e^2 x^2}}\right )}{\sqrt {d^2-e^2 x^2}}\right )}{3 e} \] Input:

Integrate[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(5/2),x]
 

Output:

(2*c*Sqrt[c*(d^2 - e^2*x^2)]*((7*d - e*x)/Sqrt[d + e*x] - (6*Sqrt[2]*d^(3/ 
2)*ArcTanh[(Sqrt[2]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[d^2 - e^2*x^2]])/Sqrt[d^2 
- e^2*x^2]))/(3*e)
 

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {466, 466, 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.

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 471

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

\(\Big \downarrow \) 221

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

Input:

Int[(c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(5/2),x]
 

Output:

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

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 466

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 
2*(n + 2*p + 1)))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr 
eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 
] || EqQ[n + p + 1, 0]) && NeQ[n + 2*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 = 0.23 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.82

method	result	size
default	\(-\frac {2 \sqrt {c \left (-e^{2} x^{2}+d^{2}\right )}\, c \left (6 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {c \left (-e x +d \right )}\, \sqrt {2}}{2 \sqrt {c d}}\right ) c \,d^{2}+e x \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}-7 \sqrt {c \left (-e x +d \right )}\, \sqrt {c d}\, d \right )}{3 \sqrt {e x +d}\, \sqrt {c \left (-e x +d \right )}\, e \sqrt {c d}}\)	\(112\)
risch	\(\frac {2 \left (-e x +7 d \right ) \left (-e x +d \right ) \sqrt {-\frac {c \left (e^{2} x^{2}-d^{2}\right )}{e x +d}}\, \sqrt {e x +d}\, c^{2}}{3 e \sqrt {-c \left (e x -d \right )}\, \sqrt {-c \left (e^{2} x^{2}-d^{2}\right )}}-\frac {4 d^{2} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {-c e x +c d}\, \sqrt {2}}{2 \sqrt {c d}}\right ) \sqrt {-\frac {c \left (e^{2} x^{2}-d^{2}\right )}{e x +d}}\, \sqrt {e x +d}\, c^{2}}{e \sqrt {c d}\, \sqrt {-c \left (e^{2} x^{2}-d^{2}\right )}}\)	\(177\)

Input:

int((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-2/3*(c*(-e^2*x^2+d^2))^(1/2)*c*(6*2^(1/2)*arctanh(1/2*(c*(-e*x+d))^(1/2)* 
2^(1/2)/(c*d)^(1/2))*c*d^2+e*x*(c*(-e*x+d))^(1/2)*(c*d)^(1/2)-7*(c*(-e*x+d 
))^(1/2)*(c*d)^(1/2)*d)/(e*x+d)^(1/2)/(c*(-e*x+d))^(1/2)/e/(c*d)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.96 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\left [\frac {2 \, {\left (3 \, \sqrt {2} {\left (c d e x + c d^{2}\right )} \sqrt {c d} \log \left (-\frac {c e^{2} x^{2} - 2 \, c d e x - 3 \, c d^{2} + 2 \, \sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {c d} \sqrt {e x + d}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) - \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (c e x - 7 \, c d\right )} \sqrt {e x + d}\right )}}{3 \, {\left (e^{2} x + d e\right )}}, \frac {2 \, {\left (6 \, \sqrt {2} {\left (c d e x + c d^{2}\right )} \sqrt {-c d} \arctan \left (\frac {\sqrt {2} \sqrt {-c e^{2} x^{2} + c d^{2}} \sqrt {-c d} \sqrt {e x + d}}{2 \, {\left (c d e x + c d^{2}\right )}}\right ) - \sqrt {-c e^{2} x^{2} + c d^{2}} {\left (c e x - 7 \, c d\right )} \sqrt {e x + d}\right )}}{3 \, {\left (e^{2} x + d e\right )}}\right ] \] Input:

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(5/2),x, algorithm="fricas")
 

Output:

[2/3*(3*sqrt(2)*(c*d*e*x + c*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)) - sqrt(-c*e^2*x^2 + c*d^2)*(c*e*x - 7*c*d)*sqrt(e*x 
 + d))/(e^2*x + d*e), 2/3*(6*sqrt(2)*(c*d*e*x + c*d^2)*sqrt(-c*d)*arctan(1 
/2*sqrt(2)*sqrt(-c*e^2*x^2 + c*d^2)*sqrt(-c*d)*sqrt(e*x + d)/(c*d*e*x + c* 
d^2)) - sqrt(-c*e^2*x^2 + c*d^2)*(c*e*x - 7*c*d)*sqrt(e*x + d))/(e^2*x + d 
*e)]
 

Sympy [F]

\[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\int \frac {\left (- c \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {3}{2}}}{\left (d + e x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((-c*e**2*x**2+c*d**2)**(3/2)/(e*x+d)**(5/2),x)
 

Output:

Integral((-c*(-d + e*x)*(d + e*x))**(3/2)/(d + e*x)**(5/2), x)
 

Maxima [F]

\[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\int { \frac {{\left (-c e^{2} x^{2} + c d^{2}\right )}^{\frac {3}{2}}}{{\left (e x + d\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(5/2),x, algorithm="maxima")
 

Output:

integrate((-c*e^2*x^2 + c*d^2)^(3/2)/(e*x + d)^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.69 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \, {\left (\frac {6 \, \sqrt {2} c d^{2} \arctan \left (\frac {\sqrt {2} \sqrt {-{\left (e x + d\right )} c + 2 \, c d}}{2 \, \sqrt {-c d}}\right )}{\sqrt {-c d}} + \frac {6 \, \sqrt {-{\left (e x + d\right )} c + 2 \, c d} c^{3} d + {\left (-{\left (e x + d\right )} c + 2 \, c d\right )}^{\frac {3}{2}} c^{2}}{c^{3}}\right )} c}{3 \, e} \] Input:

integrate((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(5/2),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\int \frac {{\left (c\,d^2-c\,e^2\,x^2\right )}^{3/2}}{{\left (d+e\,x\right )}^{5/2}} \,d x \] Input:

int((c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(5/2),x)
 

Output:

int((c*d^2 - c*e^2*x^2)^(3/2)/(d + e*x)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.47 \[ \int \frac {\left (c d^2-c e^2 x^2\right )^{3/2}}{(d+e x)^{5/2}} \, dx=\frac {2 \sqrt {c}\, c \left (7 \sqrt {-e x +d}\, d -\sqrt {-e x +d}\, e x +6 \sqrt {d}\, \sqrt {2}\, \mathrm {log}\left (\tan \left (\frac {\mathit {asin} \left (\frac {\sqrt {e x +d}}{\sqrt {d}\, \sqrt {2}}\right )}{2}\right )\right ) d -8 \sqrt {d}\, \sqrt {2}\, d \right )}{3 e} \] Input:

int((-c*e^2*x^2+c*d^2)^(3/2)/(e*x+d)^(5/2),x)
 

Output:

(2*sqrt(c)*c*(7*sqrt(d - e*x)*d - sqrt(d - e*x)*e*x + 6*sqrt(d)*sqrt(2)*lo 
g(tan(asin(sqrt(d + e*x)/(sqrt(d)*sqrt(2)))/2))*d - 8*sqrt(d)*sqrt(2)*d))/ 
(3*e)