\(\int \frac {(d+e x)^{10}}{(d^2-e^2 x^2)^{11/2}} \, dx\) [149]

Optimal result
Mathematica [A] (verified)
Rubi [A] (verified)
Maple [B] (verified)
Fricas [A] (verification not implemented)
Sympy [F]
Maxima [B] (verification not implemented)
Giac [A] (verification not implemented)
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 24, antiderivative size = 172 \[ \int \frac {(d+e x)^{10}}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\frac {2 (d+e x)^9}{9 e \left (d^2-e^2 x^2\right )^{9/2}}-\frac {2 (d+e x)^7}{7 e \left (d^2-e^2 x^2\right )^{7/2}}+\frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}}-\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e} \] Output:

2/9*(e*x+d)^9/e/(-e^2*x^2+d^2)^(9/2)-2/7*(e*x+d)^7/e/(-e^2*x^2+d^2)^(7/2)+ 
2/5*(e*x+d)^5/e/(-e^2*x^2+d^2)^(5/2)-2/3*(e*x+d)^3/e/(-e^2*x^2+d^2)^(3/2)+ 
2*(e*x+d)/e/(-e^2*x^2+d^2)^(1/2)-arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e
 

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.65 \[ \int \frac {(d+e x)^{10}}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=-\frac {2 \sqrt {d^2-e^2 x^2} \left (263 d^4-1000 d^3 e x+1974 d^2 e^2 x^2-1240 d e^3 x^3+563 e^4 x^4\right )}{315 e (-d+e x)^5}+\frac {2 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{e} \] Input:

Integrate[(d + e*x)^10/(d^2 - e^2*x^2)^(11/2),x]
 

Output:

(-2*Sqrt[d^2 - e^2*x^2]*(263*d^4 - 1000*d^3*e*x + 1974*d^2*e^2*x^2 - 1240* 
d*e^3*x^3 + 563*e^4*x^4))/(315*e*(-d + e*x)^5) + (2*ArcTan[(e*x)/(Sqrt[d^2 
] - Sqrt[d^2 - e^2*x^2])])/e
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.292, Rules used = {468, 468, 468, 468, 457, 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.

\(\displaystyle \int \frac {(d+e x)^{10}}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx\)

\(\Big \downarrow \) 468

\(\displaystyle \frac {2 (d+e x)^9}{9 e \left (d^2-e^2 x^2\right )^{9/2}}-\int \frac {(d+e x)^8}{\left (d^2-e^2 x^2\right )^{9/2}}dx\)

\(\Big \downarrow \) 468

\(\displaystyle \int \frac {(d+e x)^6}{\left (d^2-e^2 x^2\right )^{7/2}}dx+\frac {2 (d+e x)^9}{9 e \left (d^2-e^2 x^2\right )^{9/2}}-\frac {2 (d+e x)^7}{7 e \left (d^2-e^2 x^2\right )^{7/2}}\)

\(\Big \downarrow \) 468

\(\displaystyle -\int \frac {(d+e x)^4}{\left (d^2-e^2 x^2\right )^{5/2}}dx+\frac {2 (d+e x)^9}{9 e \left (d^2-e^2 x^2\right )^{9/2}}-\frac {2 (d+e x)^7}{7 e \left (d^2-e^2 x^2\right )^{7/2}}+\frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 468

\(\displaystyle \int \frac {(d+e x)^2}{\left (d^2-e^2 x^2\right )^{3/2}}dx+\frac {2 (d+e x)^9}{9 e \left (d^2-e^2 x^2\right )^{9/2}}-\frac {2 (d+e x)^7}{7 e \left (d^2-e^2 x^2\right )^{7/2}}+\frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}\)

\(\Big \downarrow \) 457

\(\displaystyle -\int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {2 (d+e x)^9}{9 e \left (d^2-e^2 x^2\right )^{9/2}}-\frac {2 (d+e x)^7}{7 e \left (d^2-e^2 x^2\right )^{7/2}}+\frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}}\)

\(\Big \downarrow \) 224

\(\displaystyle -\int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {2 (d+e x)^9}{9 e \left (d^2-e^2 x^2\right )^{9/2}}-\frac {2 (d+e x)^7}{7 e \left (d^2-e^2 x^2\right )^{7/2}}+\frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}}\)

\(\Big \downarrow \) 216

\(\displaystyle -\frac {\arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{e}+\frac {2 (d+e x)^9}{9 e \left (d^2-e^2 x^2\right )^{9/2}}-\frac {2 (d+e x)^7}{7 e \left (d^2-e^2 x^2\right )^{7/2}}+\frac {2 (d+e x)^5}{5 e \left (d^2-e^2 x^2\right )^{5/2}}-\frac {2 (d+e x)^3}{3 e \left (d^2-e^2 x^2\right )^{3/2}}+\frac {2 (d+e x)}{e \sqrt {d^2-e^2 x^2}}\)

Input:

Int[(d + e*x)^10/(d^2 - e^2*x^2)^(11/2),x]
 

Output:

(2*(d + e*x)^9)/(9*e*(d^2 - e^2*x^2)^(9/2)) - (2*(d + e*x)^7)/(7*e*(d^2 - 
e^2*x^2)^(7/2)) + (2*(d + e*x)^5)/(5*e*(d^2 - e^2*x^2)^(5/2)) - (2*(d + e* 
x)^3)/(3*e*(d^2 - e^2*x^2)^(3/2)) + (2*(d + e*x))/(e*Sqrt[d^2 - e^2*x^2]) 
- ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]]/e
 

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 457
Int[((c_) + (d_.)*(x_))^2*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*( 
c + d*x)*((a + b*x^2)^(p + 1)/(b*(p + 1))), x] - Simp[d^2*((p + 2)/(b*(p + 
1)))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && EqQ[ 
b*c^2 + a*d^2, 0] && LtQ[p, -1]
 

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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1519\) vs. \(2(152)=304\).

Time = 1.16 (sec) , antiderivative size = 1520, normalized size of antiderivative = 8.84

method result size
default \(\text {Expression too large to display}\) \(1520\)

Input:

int((e*x+d)^10/(-e^2*x^2+d^2)^(11/2),x,method=_RETURNVERBOSE)
 

Output:

d^10*(1/9*x/d^2/(-e^2*x^2+d^2)^(9/2)+8/9/d^2*(1/7*x/d^2/(-e^2*x^2+d^2)^(7/ 
2)+6/7/d^2*(1/5*x/d^2/(-e^2*x^2+d^2)^(5/2)+4/5/d^2*(1/3*x/d^2/(-e^2*x^2+d^ 
2)^(3/2)+2/3*x/d^4/(-e^2*x^2+d^2)^(1/2)))))+e^10*(1/9*x^9/e^2/(-e^2*x^2+d^ 
2)^(9/2)-1/e^2*(1/7*x^7/e^2/(-e^2*x^2+d^2)^(7/2)-1/e^2*(1/5*x^5/e^2/(-e^2* 
x^2+d^2)^(5/2)-1/e^2*(1/3*x^3/e^2/(-e^2*x^2+d^2)^(3/2)-1/e^2*(x/e^2/(-e^2* 
x^2+d^2)^(1/2)-1/e^2/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2) 
))))))+10*d*e^9*(x^8/e^2/(-e^2*x^2+d^2)^(9/2)-8*d^2/e^2*(1/3*x^6/e^2/(-e^2 
*x^2+d^2)^(9/2)-2*d^2/e^2*(1/5*x^4/e^2/(-e^2*x^2+d^2)^(9/2)-4/5*d^2/e^2*(1 
/7*x^2/e^2/(-e^2*x^2+d^2)^(9/2)-2/63*d^2/e^4/(-e^2*x^2+d^2)^(9/2)))))+45*d 
^2*e^8*(1/2*x^7/e^2/(-e^2*x^2+d^2)^(9/2)-7/2*d^2/e^2*(1/4*x^5/e^2/(-e^2*x^ 
2+d^2)^(9/2)-5/4*d^2/e^2*(1/6*x^3/e^2/(-e^2*x^2+d^2)^(9/2)-1/2*d^2/e^2*(1/ 
8*x/e^2/(-e^2*x^2+d^2)^(9/2)-1/8*d^2/e^2*(1/9*x/d^2/(-e^2*x^2+d^2)^(9/2)+8 
/9/d^2*(1/7*x/d^2/(-e^2*x^2+d^2)^(7/2)+6/7/d^2*(1/5*x/d^2/(-e^2*x^2+d^2)^( 
5/2)+4/5/d^2*(1/3*x/d^2/(-e^2*x^2+d^2)^(3/2)+2/3*x/d^4/(-e^2*x^2+d^2)^(1/2 
)))))))))+120*d^3*e^7*(1/3*x^6/e^2/(-e^2*x^2+d^2)^(9/2)-2*d^2/e^2*(1/5*x^4 
/e^2/(-e^2*x^2+d^2)^(9/2)-4/5*d^2/e^2*(1/7*x^2/e^2/(-e^2*x^2+d^2)^(9/2)-2/ 
63*d^2/e^4/(-e^2*x^2+d^2)^(9/2))))+210*d^4*e^6*(1/4*x^5/e^2/(-e^2*x^2+d^2) 
^(9/2)-5/4*d^2/e^2*(1/6*x^3/e^2/(-e^2*x^2+d^2)^(9/2)-1/2*d^2/e^2*(1/8*x/e^ 
2/(-e^2*x^2+d^2)^(9/2)-1/8*d^2/e^2*(1/9*x/d^2/(-e^2*x^2+d^2)^(9/2)+8/9/d^2 
*(1/7*x/d^2/(-e^2*x^2+d^2)^(7/2)+6/7/d^2*(1/5*x/d^2/(-e^2*x^2+d^2)^(5/2...
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.44 \[ \int \frac {(d+e x)^{10}}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\frac {2 \, {\left (263 \, e^{5} x^{5} - 1315 \, d e^{4} x^{4} + 2630 \, d^{2} e^{3} x^{3} - 2630 \, d^{3} e^{2} x^{2} + 1315 \, d^{4} e x - 263 \, d^{5} + 315 \, {\left (e^{5} x^{5} - 5 \, d e^{4} x^{4} + 10 \, d^{2} e^{3} x^{3} - 10 \, d^{3} e^{2} x^{2} + 5 \, d^{4} e x - d^{5}\right )} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) - {\left (563 \, e^{4} x^{4} - 1240 \, d e^{3} x^{3} + 1974 \, d^{2} e^{2} x^{2} - 1000 \, d^{3} e x + 263 \, d^{4}\right )} \sqrt {-e^{2} x^{2} + d^{2}}\right )}}{315 \, {\left (e^{6} x^{5} - 5 \, d e^{5} x^{4} + 10 \, d^{2} e^{4} x^{3} - 10 \, d^{3} e^{3} x^{2} + 5 \, d^{4} e^{2} x - d^{5} e\right )}} \] Input:

integrate((e*x+d)^10/(-e^2*x^2+d^2)^(11/2),x, algorithm="fricas")
 

Output:

2/315*(263*e^5*x^5 - 1315*d*e^4*x^4 + 2630*d^2*e^3*x^3 - 2630*d^3*e^2*x^2 
+ 1315*d^4*e*x - 263*d^5 + 315*(e^5*x^5 - 5*d*e^4*x^4 + 10*d^2*e^3*x^3 - 1 
0*d^3*e^2*x^2 + 5*d^4*e*x - d^5)*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) 
 - (563*e^4*x^4 - 1240*d*e^3*x^3 + 1974*d^2*e^2*x^2 - 1000*d^3*e*x + 263*d 
^4)*sqrt(-e^2*x^2 + d^2))/(e^6*x^5 - 5*d*e^5*x^4 + 10*d^2*e^4*x^3 - 10*d^3 
*e^3*x^2 + 5*d^4*e^2*x - d^5*e)
 

Sympy [F]

\[ \int \frac {(d+e x)^{10}}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\int \frac {\left (d + e x\right )^{10}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac {11}{2}}}\, dx \] Input:

integrate((e*x+d)**10/(-e**2*x**2+d**2)**(11/2),x)
 

Output:

Integral((d + e*x)**10/(-(-d + e*x)*(d + e*x))**(11/2), x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 741 vs. \(2 (152) = 304\).

Time = 0.17 (sec) , antiderivative size = 741, normalized size of antiderivative = 4.31 \[ \int \frac {(d+e x)^{10}}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx =\text {Too large to display} \] Input:

integrate((e*x+d)^10/(-e^2*x^2+d^2)^(11/2),x, algorithm="maxima")
 

Output:

1/315*(315*x^8/((-e^2*x^2 + d^2)^(9/2)*e^2) - 840*d^2*x^6/((-e^2*x^2 + d^2 
)^(9/2)*e^4) + 1008*d^4*x^4/((-e^2*x^2 + d^2)^(9/2)*e^6) - 576*d^6*x^2/((- 
e^2*x^2 + d^2)^(9/2)*e^8) + 128*d^8/((-e^2*x^2 + d^2)^(9/2)*e^10))*e^10*x 
+ 10*d*e^7*x^8/(-e^2*x^2 + d^2)^(9/2) + 45/2*d^2*e^6*x^7/(-e^2*x^2 + d^2)^ 
(9/2) - 1/35*e^8*x*(35*x^6/((-e^2*x^2 + d^2)^(7/2)*e^2) - 70*d^2*x^4/((-e^ 
2*x^2 + d^2)^(7/2)*e^4) + 56*d^4*x^2/((-e^2*x^2 + d^2)^(7/2)*e^6) - 16*d^6 
/((-e^2*x^2 + d^2)^(7/2)*e^8)) + 40/3*d^3*e^5*x^6/(-e^2*x^2 + d^2)^(9/2) + 
 105/8*d^4*e^4*x^5/(-e^2*x^2 + d^2)^(9/2) + 1/15*e^6*x*(15*x^4/((-e^2*x^2 
+ d^2)^(5/2)*e^2) - 20*d^2*x^2/((-e^2*x^2 + d^2)^(5/2)*e^4) + 8*d^4/((-e^2 
*x^2 + d^2)^(5/2)*e^6)) + 172/5*d^5*e^3*x^4/(-e^2*x^2 + d^2)^(9/2) + 4/3*d 
^2*e^4*x^5/(-e^2*x^2 + d^2)^(7/2) + 385/16*d^6*e^2*x^3/(-e^2*x^2 + d^2)^(9 
/2) - 1/3*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)) - 88/35*d^7*e*x^2/(-e^2*x^2 + d^2)^(9/2) - 37/15*d^4*e^2*x 
^3/(-e^2*x^2 + d^2)^(7/2) - 419/144*d^8*x/(-e^2*x^2 + d^2)^(9/2) - d^2*e^2 
*x^3/(-e^2*x^2 + d^2)^(5/2) + 526/315*d^9/((-e^2*x^2 + d^2)^(9/2)*e) + 916 
7/5040*d^6*x/(-e^2*x^2 + d^2)^(7/2) + 979/840*d^4*x/(-e^2*x^2 + d^2)^(5/2) 
 + 181/630*d^2*x/(-e^2*x^2 + d^2)^(3/2) - 134/315*x/sqrt(-e^2*x^2 + d^2) - 
 arcsin(e^2*x/(d*sqrt(e^2)))/sqrt(e^2)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.77 \[ \int \frac {(d+e x)^{10}}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=-\frac {\arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{{\left | e \right |}} - \frac {4 \, {\left (\frac {2052 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}}{e^{2} x} - \frac {8208 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{2}}{e^{4} x^{2}} + \frac {14532 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{3}}{e^{6} x^{3}} - \frac {21798 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{4}}{e^{8} x^{4}} + \frac {11340 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{5}}{e^{10} x^{5}} - \frac {7560 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{6}}{e^{12} x^{6}} + \frac {1260 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{7}}{e^{14} x^{7}} - \frac {315 \, {\left (d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}\right )}^{8}}{e^{16} x^{8}} - 263\right )}}{315 \, {\left (\frac {d e + \sqrt {-e^{2} x^{2} + d^{2}} {\left | e \right |}}{e^{2} x} - 1\right )}^{9} {\left | e \right |}} \] Input:

integrate((e*x+d)^10/(-e^2*x^2+d^2)^(11/2),x, algorithm="giac")
 

Output:

-arcsin(e*x/d)*sgn(d)*sgn(e)/abs(e) - 4/315*(2052*(d*e + sqrt(-e^2*x^2 + d 
^2)*abs(e))/(e^2*x) - 8208*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^2/(e^4*x^2) 
 + 14532*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^3/(e^6*x^3) - 21798*(d*e + sq 
rt(-e^2*x^2 + d^2)*abs(e))^4/(e^8*x^4) + 11340*(d*e + sqrt(-e^2*x^2 + d^2) 
*abs(e))^5/(e^10*x^5) - 7560*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^6/(e^12*x 
^6) + 1260*(d*e + sqrt(-e^2*x^2 + d^2)*abs(e))^7/(e^14*x^7) - 315*(d*e + s 
qrt(-e^2*x^2 + d^2)*abs(e))^8/(e^16*x^8) - 263)/(((d*e + sqrt(-e^2*x^2 + d 
^2)*abs(e))/(e^2*x) - 1)^9*abs(e))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(d+e x)^{10}}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\int \frac {{\left (d+e\,x\right )}^{10}}{{\left (d^2-e^2\,x^2\right )}^{11/2}} \,d x \] Input:

int((d + e*x)^10/(d^2 - e^2*x^2)^(11/2),x)
 

Output:

int((d + e*x)^10/(d^2 - e^2*x^2)^(11/2), x)
 

Reduce [F]

\[ \int \frac {(d+e x)^{10}}{\left (d^2-e^2 x^2\right )^{11/2}} \, dx=\int \frac {\left (e x +d \right )^{10}}{\left (-e^{2} x^{2}+d^{2}\right )^{\frac {11}{2}}}d x \] Input:

int((e*x+d)^10/(-e^2*x^2+d^2)^(11/2),x)
 

Output:

int((e*x+d)^10/(-e^2*x^2+d^2)^(11/2),x)