\(\int \frac {(d+e x)^5}{(d^2-e^2 x^2)^{5/2}} \, dx\) [128]

Optimal result

Integrand size = 24, antiderivative size = 109 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {2 (d+e x)^4}{3 e \left (d^2-e^2 x^2\right )^{3/2}}-\frac {20 d (d+e x)}{3 e \sqrt {d^2-e^2 x^2}}-\frac {5 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {5 d \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \] Output:

2/3*(e*x+d)^4/e/(-e^2*x^2+d^2)^(3/2)-20/3*d*(e*x+d)/e/(-e^2*x^2+d^2)^(1/2) 
-5/3*(-e^2*x^2+d^2)^(1/2)/e+5*d*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e
 

Mathematica [A] (verified)

Time = 0.39 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {\left (-23 d^2+34 d e x-3 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{3 e (-d+e x)^2}-\frac {5 d \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{\sqrt {-e^2}} \] Input:

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

Output:

((-23*d^2 + 34*d*e*x - 3*e^2*x^2)*Sqrt[d^2 - e^2*x^2])/(3*e*(-d + e*x)^2) 
- (5*d*Log[-(Sqrt[-e^2]*x) + Sqrt[d^2 - e^2*x^2]])/Sqrt[-e^2]
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {468, 462, 455, 224, 216}

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.

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

\(\Big \downarrow \) 468

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

\(\Big \downarrow \) 462

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, p}, x] &&  !LeQ[p, -1]
 

rule 462

Int[((c_) + (d_.)*(x_))^(n_)/((a_) + (b_.)*(x_)^2)^(3/2), x_Symbol] :> Simp 
[(-2^(n - 1))*d*c^(n - 2)*((c + d*x)/(b*Sqrt[a + b*x^2])), x] + Simp[d^2/b 
  Int[(1/Sqrt[a + b*x^2])*ExpandToSum[(2^(n - 1)*c^(n - 1) - (c + d*x)^(n - 
 1))/(c - d*x), x], x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 
0] && IGtQ[n, 2]
 

rule 468

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] - Simp[d^2*((n + 
p)/(b*(p + 1)))   Int[(c + d*x)^(n - 2)*(a + b*x^2)^(p + 1), x], x] /; Free 
Q[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[p, -1] && GtQ[n, 1] && I 
ntegerQ[2*p]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.33

method	result	size
risch	\(-\frac {\sqrt {-e^{2} x^{2}+d^{2}}}{e}+\frac {5 d \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}+\frac {28 d \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{3 e^{2} \left (x -\frac {d}{e}\right )}+\frac {8 d^{2} \sqrt {-\left (x -\frac {d}{e}\right )^{2} e^{2}-2 d e \left (x -\frac {d}{e}\right )}}{3 e^{3} \left (x -\frac {d}{e}\right )^{2}}\)	\(145\)
default	\(d^{5} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )+e^{5} \left (-\frac {x^{4}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {4 d^{2} \left (\frac {x^{2}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {2 d^{2}}{3 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}\right )}{e^{2}}\right )+5 d \,e^{4} \left (\frac {x^{3}}{3 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{e^{2} \sqrt {-e^{2} x^{2}+d^{2}}}-\frac {\arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{e^{2} \sqrt {e^{2}}}}{e^{2}}\right )+10 d^{2} e^{3} \left (\frac {x^{2}}{e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {2 d^{2}}{3 e^{4} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}\right )+10 d^{3} e^{2} \left (\frac {x}{2 e^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}-\frac {d^{2} \left (\frac {x}{3 d^{2} \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 d^{4} \sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2}}\right )+\frac {5 d^{4}}{3 e \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}\)	\(364\)

Input:

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

Output:

-(-e^2*x^2+d^2)^(1/2)/e+5*d/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2 
)^(1/2))+28/3*d/e^2/(x-d/e)*(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)+8/3*d^2/e 
^3/(x-d/e)^2*(-(x-d/e)^2*e^2-2*d*e*(x-d/e))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.17 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=-\frac {23 \, d e^{2} x^{2} - 46 \, d^{2} e x + 23 \, d^{3} + 30 \, {\left (d e^{2} x^{2} - 2 \, d^{2} e x + d^{3}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (3 \, e^{2} x^{2} - 34 \, d e x + 23 \, d^{2}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{3 \, {\left (e^{3} x^{2} - 2 \, d e^{2} x + d^{2} e\right )}} \] Input:

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

Output:

-1/3*(23*d*e^2*x^2 - 46*d^2*e*x + 23*d^3 + 30*(d*e^2*x^2 - 2*d^2*e*x + d^3 
)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (3*e^2*x^2 - 34*d*e*x + 23*d 
^2)*sqrt(-e^2*x^2 + d^2))/(e^3*x^2 - 2*d*e^2*x + d^2*e)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.65 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {5}{3} \, d e^{4} x {\left (\frac {3 \, x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{2}} - \frac {2 \, d^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e^{4}}\right )} - \frac {e^{3} x^{4}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {14 \, d^{2} e x^{2}}{{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} + \frac {11 \, d^{3} x}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}}} - \frac {23 \, d^{4}}{3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} e} - \frac {13 \, d x}{3 \, \sqrt {-e^{2} x^{2} + d^{2}}} + \frac {5 \, d \arcsin \left (\frac {e^{2} x}{d \sqrt {e^{2}}}\right )}{\sqrt {e^{2}}} \] Input:

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

Output:

5/3*d*e^4*x*(3*x^2/((-e^2*x^2 + d^2)^(3/2)*e^2) - 2*d^2/((-e^2*x^2 + d^2)^ 
(3/2)*e^4)) - e^3*x^4/(-e^2*x^2 + d^2)^(3/2) + 14*d^2*e*x^2/(-e^2*x^2 + d^ 
2)^(3/2) + 11/3*d^3*x/(-e^2*x^2 + d^2)^(3/2) - 23/3*d^4/((-e^2*x^2 + d^2)^ 
(3/2)*e) - 13/3*d*x/sqrt(-e^2*x^2 + d^2) + 5*d*arcsin(e^2*x/(d*sqrt(e^2))) 
/sqrt(e^2)
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.30 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {5 \, d \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{{\left | e \right |}} - \frac {\sqrt {-e^{2} x^{2} + d^{2}}}{e} - \frac {8 \, {\left (5 \, d - \frac {12 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )} d}{e^{2} x} + \frac {3 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2} d}{e^{4} x^{2}}\right )}}{3 \, {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{3} {\left | e \right |}} \] Input:

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

Output:

5*d*arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) - sqrt(-e^2*x^2 + d^2)/e - 8/3*(5*d 
 - 12*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))*d/(e^2*x) + 3*(d*e + sqrt(-e^2*x 
^2 + d^2)*abs(e))^2*d/(e^4*x^2))/(((d*e + sqrt(-e^2*x^2 + d^2)*abs(e))/(e^ 
2*x) - 1)^3*abs(e))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.17 \[ \int \frac {(d+e x)^5}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx=\frac {15 \sqrt {-e^{2} x^{2}+d^{2}}\, \mathit {asin} \left (\frac {e x}{d}\right ) d^{2}-15 \sqrt {-e^{2} x^{2}+d^{2}}\, \mathit {asin} \left (\frac {e x}{d}\right ) d e x -15 \mathit {asin} \left (\frac {e x}{d}\right ) d^{3}+30 \mathit {asin} \left (\frac {e x}{d}\right ) d^{2} e x -15 \mathit {asin} \left (\frac {e x}{d}\right ) d \,e^{2} x^{2}+8 \sqrt {-e^{2} x^{2}+d^{2}}\, d^{2}-19 \sqrt {-e^{2} x^{2}+d^{2}}\, d e x +3 \sqrt {-e^{2} x^{2}+d^{2}}\, e^{2} x^{2}-8 d^{3}-19 d^{2} e x +46 d \,e^{2} x^{2}-3 e^{3} x^{3}}{3 e \left (\sqrt {-e^{2} x^{2}+d^{2}}\, d -\sqrt {-e^{2} x^{2}+d^{2}}\, e x -d^{2}+2 d e x -e^{2} x^{2}\right )} \] Input:

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

Output:

(15*sqrt(d**2 - e**2*x**2)*asin((e*x)/d)*d**2 - 15*sqrt(d**2 - e**2*x**2)* 
asin((e*x)/d)*d*e*x - 15*asin((e*x)/d)*d**3 + 30*asin((e*x)/d)*d**2*e*x - 
15*asin((e*x)/d)*d*e**2*x**2 + 8*sqrt(d**2 - e**2*x**2)*d**2 - 19*sqrt(d** 
2 - e**2*x**2)*d*e*x + 3*sqrt(d**2 - e**2*x**2)*e**2*x**2 - 8*d**3 - 19*d* 
*2*e*x + 46*d*e**2*x**2 - 3*e**3*x**3)/(3*e*(sqrt(d**2 - e**2*x**2)*d - sq 
rt(d**2 - e**2*x**2)*e*x - d**2 + 2*d*e*x - e**2*x**2))