3.2.4 \(\int \frac {x^3 (d^2-e^2 x^2)^{5/2}}{d+e x} \, dx\) [104]

3.2.4.1 Optimal result
3.2.4.2 Mathematica [A] (verified)
3.2.4.3 Rubi [A] (verified)
3.2.4.4 Maple [A] (verified)
3.2.4.5 Fricas [A] (verification not implemented)
3.2.4.6 Sympy [A] (verification not implemented)
3.2.4.7 Maxima [C] (verification not implemented)
3.2.4.8 Giac [A] (verification not implemented)
3.2.4.9 Mupad [F(-1)]

3.2.4.1 Optimal result

Integrand size = 27, antiderivative size = 172 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=-\frac {3 d^6 x \sqrt {d^2-e^2 x^2}}{128 e^3}-\frac {d^4 x \left (d^2-e^2 x^2\right )^{3/2}}{64 e^3}-\frac {d x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e^2}+\frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d^2 (32 d-35 e x) \left (d^2-e^2 x^2\right )^{5/2}}{560 e^4}-\frac {3 d^8 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{128 e^4} \]

output
-1/64*d^4*x*(-e^2*x^2+d^2)^(3/2)/e^3-1/7*d*x^2*(-e^2*x^2+d^2)^(5/2)/e^2+1/ 
8*x^3*(-e^2*x^2+d^2)^(5/2)/e-1/560*d^2*(-35*e*x+32*d)*(-e^2*x^2+d^2)^(5/2) 
/e^4-3/128*d^8*arctan(e*x/(-e^2*x^2+d^2)^(1/2))/e^4-3/128*d^6*x*(-e^2*x^2+ 
d^2)^(1/2)/e^3
 
3.2.4.2 Mathematica [A] (verified)

Time = 0.37 (sec) , antiderivative size = 136, normalized size of antiderivative = 0.79 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {\sqrt {d^2-e^2 x^2} \left (-256 d^7+105 d^6 e x-128 d^5 e^2 x^2+70 d^4 e^3 x^3+1024 d^3 e^4 x^4-840 d^2 e^5 x^5-640 d e^6 x^6+560 e^7 x^7\right )+210 d^8 \arctan \left (\frac {e x}{\sqrt {d^2}-\sqrt {d^2-e^2 x^2}}\right )}{4480 e^4} \]

input
Integrate[(x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x),x]
 
output
(Sqrt[d^2 - e^2*x^2]*(-256*d^7 + 105*d^6*e*x - 128*d^5*e^2*x^2 + 70*d^4*e^ 
3*x^3 + 1024*d^3*e^4*x^4 - 840*d^2*e^5*x^5 - 640*d*e^6*x^6 + 560*e^7*x^7) 
+ 210*d^8*ArcTan[(e*x)/(Sqrt[d^2] - Sqrt[d^2 - e^2*x^2])])/(4480*e^4)
 
3.2.4.3 Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.23, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.519, Rules used = {562, 533, 25, 27, 533, 25, 27, 533, 27, 455, 211, 211, 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.

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

\(\Big \downarrow \) 562

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

\(\Big \downarrow \) 533

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 533

\(\displaystyle \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d \left (\frac {\int -d e x (16 d-21 e x) \left (d^2-e^2 x^2\right )^{3/2}dx}{7 e^2}+\frac {8 x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e}\right )}{8 e}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d \left (\frac {8 x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e}-\frac {\int d e x (16 d-21 e x) \left (d^2-e^2 x^2\right )^{3/2}dx}{7 e^2}\right )}{8 e}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d \left (\frac {8 x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e}-\frac {d \int x (16 d-21 e x) \left (d^2-e^2 x^2\right )^{3/2}dx}{7 e}\right )}{8 e}\)

\(\Big \downarrow \) 533

\(\displaystyle \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d \left (\frac {8 x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e}-\frac {d \left (\frac {\int -3 d e (7 d-32 e x) \left (d^2-e^2 x^2\right )^{3/2}dx}{6 e^2}+\frac {7 x \left (d^2-e^2 x^2\right )^{5/2}}{2 e}\right )}{7 e}\right )}{8 e}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d \left (\frac {8 x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e}-\frac {d \left (\frac {7 x \left (d^2-e^2 x^2\right )^{5/2}}{2 e}-\frac {d \int (7 d-32 e x) \left (d^2-e^2 x^2\right )^{3/2}dx}{2 e}\right )}{7 e}\right )}{8 e}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d \left (\frac {8 x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e}-\frac {d \left (\frac {7 x \left (d^2-e^2 x^2\right )^{5/2}}{2 e}-\frac {d \left (7 d \int \left (d^2-e^2 x^2\right )^{3/2}dx+\frac {32 \left (d^2-e^2 x^2\right )^{5/2}}{5 e}\right )}{2 e}\right )}{7 e}\right )}{8 e}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d \left (\frac {8 x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e}-\frac {d \left (\frac {7 x \left (d^2-e^2 x^2\right )^{5/2}}{2 e}-\frac {d \left (7 d \left (\frac {3}{4} d^2 \int \sqrt {d^2-e^2 x^2}dx+\frac {1}{4} x \left (d^2-e^2 x^2\right )^{3/2}\right )+\frac {32 \left (d^2-e^2 x^2\right )^{5/2}}{5 e}\right )}{2 e}\right )}{7 e}\right )}{8 e}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d \left (\frac {8 x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e}-\frac {d \left (\frac {7 x \left (d^2-e^2 x^2\right )^{5/2}}{2 e}-\frac {d \left (7 d \left (\frac {3}{4} d^2 \left (\frac {1}{2} d^2 \int \frac {1}{\sqrt {d^2-e^2 x^2}}dx+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )+\frac {1}{4} x \left (d^2-e^2 x^2\right )^{3/2}\right )+\frac {32 \left (d^2-e^2 x^2\right )^{5/2}}{5 e}\right )}{2 e}\right )}{7 e}\right )}{8 e}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d \left (\frac {8 x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e}-\frac {d \left (\frac {7 x \left (d^2-e^2 x^2\right )^{5/2}}{2 e}-\frac {d \left (7 d \left (\frac {3}{4} d^2 \left (\frac {1}{2} d^2 \int \frac {1}{\frac {e^2 x^2}{d^2-e^2 x^2}+1}d\frac {x}{\sqrt {d^2-e^2 x^2}}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )+\frac {1}{4} x \left (d^2-e^2 x^2\right )^{3/2}\right )+\frac {32 \left (d^2-e^2 x^2\right )^{5/2}}{5 e}\right )}{2 e}\right )}{7 e}\right )}{8 e}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{8 e}-\frac {d \left (\frac {8 x^2 \left (d^2-e^2 x^2\right )^{5/2}}{7 e}-\frac {d \left (\frac {7 x \left (d^2-e^2 x^2\right )^{5/2}}{2 e}-\frac {d \left (7 d \left (\frac {3}{4} d^2 \left (\frac {d^2 \arctan \left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}+\frac {1}{2} x \sqrt {d^2-e^2 x^2}\right )+\frac {1}{4} x \left (d^2-e^2 x^2\right )^{3/2}\right )+\frac {32 \left (d^2-e^2 x^2\right )^{5/2}}{5 e}\right )}{2 e}\right )}{7 e}\right )}{8 e}\)

input
Int[(x^3*(d^2 - e^2*x^2)^(5/2))/(d + e*x),x]
 
output
(x^3*(d^2 - e^2*x^2)^(5/2))/(8*e) - (d*((8*x^2*(d^2 - e^2*x^2)^(5/2))/(7*e 
) - (d*((7*x*(d^2 - e^2*x^2)^(5/2))/(2*e) - (d*((32*(d^2 - e^2*x^2)^(5/2)) 
/(5*e) + 7*d*((x*(d^2 - e^2*x^2)^(3/2))/4 + (3*d^2*((x*Sqrt[d^2 - e^2*x^2] 
)/2 + (d^2*ArcTan[(e*x)/Sqrt[d^2 - e^2*x^2]])/(2*e)))/4)))/(2*e)))/(7*e))) 
/(8*e)
 

3.2.4.3.1 Defintions of rubi rules used

rule 25
Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27
Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 211
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 533
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* 
p + 2))   Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], 
 x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer 
Q[2*p]
 

rule 562
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> Simp[c^(2*n)/a^n   Int[x^m*((a + b*x^2)^(n + p)/(c - d*x)^n), x], x] 
/; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && IGtQ[m, 0] && ILtQ[n, 
 0] && IGtQ[n + p + 1/2, 0]
 
3.2.4.4 Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.76

method result size
risch \(-\frac {\left (-560 e^{7} x^{7}+640 d \,e^{6} x^{6}+840 d^{2} e^{5} x^{5}-1024 d^{3} e^{4} x^{4}-70 d^{4} e^{3} x^{3}+128 d^{5} e^{2} x^{2}-105 d^{6} e x +256 d^{7}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{4480 e^{4}}-\frac {3 d^{8} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{128 e^{3} \sqrt {e^{2}}}\) \(130\)
default \(\frac {-\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{8 e^{2}}+\frac {d^{2} \left (\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{6}+\frac {5 d^{2} \left (\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{6}\right )}{8 e^{2}}}{e}+\frac {d^{2} \left (\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {5}{2}}}{6}+\frac {5 d^{2} \left (\frac {x \left (-e^{2} x^{2}+d^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 d^{2} \left (\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )}{6}\right )}{e^{3}}+\frac {d \left (-e^{2} x^{2}+d^{2}\right )^{\frac {7}{2}}}{7 e^{4}}-\frac {d^{3} \left (\frac {\left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+d e \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \left (-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 e^{2}}+\frac {3 d^{2} \left (-\frac {\left (-2 \left (x +\frac {d}{e}\right ) e^{2}+2 d e \right ) \sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}{4 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-\left (x +\frac {d}{e}\right )^{2} e^{2}+2 d e \left (x +\frac {d}{e}\right )}}\right )}{2 \sqrt {e^{2}}}\right )}{4}\right )\right )}{e^{4}}\) \(449\)

input
int(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d),x,method=_RETURNVERBOSE)
 
output
-1/4480*(-560*e^7*x^7+640*d*e^6*x^6+840*d^2*e^5*x^5-1024*d^3*e^4*x^4-70*d^ 
4*e^3*x^3+128*d^5*e^2*x^2-105*d^6*e*x+256*d^7)/e^4*(-e^2*x^2+d^2)^(1/2)-3/ 
128*d^8/e^3/(e^2)^(1/2)*arctan((e^2)^(1/2)*x/(-e^2*x^2+d^2)^(1/2))
 
3.2.4.5 Fricas [A] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.74 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {210 \, d^{8} \arctan \left (-\frac {d - \sqrt {-e^{2} x^{2} + d^{2}}}{e x}\right ) + {\left (560 \, e^{7} x^{7} - 640 \, d e^{6} x^{6} - 840 \, d^{2} e^{5} x^{5} + 1024 \, d^{3} e^{4} x^{4} + 70 \, d^{4} e^{3} x^{3} - 128 \, d^{5} e^{2} x^{2} + 105 \, d^{6} e x - 256 \, d^{7}\right )} \sqrt {-e^{2} x^{2} + d^{2}}}{4480 \, e^{4}} \]

input
integrate(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d),x, algorithm="fricas")
 
output
1/4480*(210*d^8*arctan(-(d - sqrt(-e^2*x^2 + d^2))/(e*x)) + (560*e^7*x^7 - 
 640*d*e^6*x^6 - 840*d^2*e^5*x^5 + 1024*d^3*e^4*x^4 + 70*d^4*e^3*x^3 - 128 
*d^5*e^2*x^2 + 105*d^6*e*x - 256*d^7)*sqrt(-e^2*x^2 + d^2))/e^4
 
3.2.4.6 Sympy [A] (verification not implemented)

Time = 1.37 (sec) , antiderivative size = 400, normalized size of antiderivative = 2.33 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=d^{3} \left (\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {2 d^{4}}{15 e^{4}} - \frac {d^{2} x^{2}}{15 e^{2}} + \frac {x^{4}}{5}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{4} \sqrt {d^{2}}}{4} & \text {otherwise} \end {cases}\right ) - d^{2} e \left (\begin {cases} \frac {d^{6} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{16 e^{4}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {d^{4} x}{16 e^{4}} - \frac {d^{2} x^{3}}{24 e^{2}} + \frac {x^{5}}{6}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{5} \sqrt {d^{2}}}{5} & \text {otherwise} \end {cases}\right ) - d e^{2} \left (\begin {cases} \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {8 d^{6}}{105 e^{6}} - \frac {4 d^{4} x^{2}}{105 e^{4}} - \frac {d^{2} x^{4}}{35 e^{2}} + \frac {x^{6}}{7}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{6} \sqrt {d^{2}}}{6} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} \frac {5 d^{8} \left (\begin {cases} \frac {\log {\left (- 2 e^{2} x + 2 \sqrt {- e^{2}} \sqrt {d^{2} - e^{2} x^{2}} \right )}}{\sqrt {- e^{2}}} & \text {for}\: d^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- e^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{128 e^{6}} + \sqrt {d^{2} - e^{2} x^{2}} \left (- \frac {5 d^{6} x}{128 e^{6}} - \frac {5 d^{4} x^{3}}{192 e^{4}} - \frac {d^{2} x^{5}}{48 e^{2}} + \frac {x^{7}}{8}\right ) & \text {for}\: e^{2} \neq 0 \\\frac {x^{7} \sqrt {d^{2}}}{7} & \text {otherwise} \end {cases}\right ) \]

input
integrate(x**3*(-e**2*x**2+d**2)**(5/2)/(e*x+d),x)
 
output
d**3*Piecewise((sqrt(d**2 - e**2*x**2)*(-2*d**4/(15*e**4) - d**2*x**2/(15* 
e**2) + x**4/5), Ne(e**2, 0)), (x**4*sqrt(d**2)/4, True)) - d**2*e*Piecewi 
se((d**6*Piecewise((log(-2*e**2*x + 2*sqrt(-e**2)*sqrt(d**2 - e**2*x**2))/ 
sqrt(-e**2), Ne(d**2, 0)), (x*log(x)/sqrt(-e**2*x**2), True))/(16*e**4) + 
sqrt(d**2 - e**2*x**2)*(-d**4*x/(16*e**4) - d**2*x**3/(24*e**2) + x**5/6), 
 Ne(e**2, 0)), (x**5*sqrt(d**2)/5, True)) - d*e**2*Piecewise((sqrt(d**2 - 
e**2*x**2)*(-8*d**6/(105*e**6) - 4*d**4*x**2/(105*e**4) - d**2*x**4/(35*e* 
*2) + x**6/7), Ne(e**2, 0)), (x**6*sqrt(d**2)/6, True)) + e**3*Piecewise(( 
5*d**8*Piecewise((log(-2*e**2*x + 2*sqrt(-e**2)*sqrt(d**2 - e**2*x**2))/sq 
rt(-e**2), Ne(d**2, 0)), (x*log(x)/sqrt(-e**2*x**2), True))/(128*e**6) + s 
qrt(d**2 - e**2*x**2)*(-5*d**6*x/(128*e**6) - 5*d**4*x**3/(192*e**4) - d** 
2*x**5/(48*e**2) + x**7/8), Ne(e**2, 0)), (x**7*sqrt(d**2)/7, True))
 
3.2.4.7 Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.30 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.28 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=\frac {3 i \, d^{8} \arcsin \left (\frac {e x}{d} + 2\right )}{8 \, e^{4}} + \frac {45 \, d^{8} \arcsin \left (\frac {e x}{d}\right )}{128 \, e^{4}} - \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{6} x}{8 \, e^{3}} + \frac {45 \, \sqrt {-e^{2} x^{2} + d^{2}} d^{6} x}{128 \, e^{3}} - \frac {3 \, \sqrt {e^{2} x^{2} + 4 \, d e x + 3 \, d^{2}} d^{7}}{4 \, e^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {3}{2}} d^{4} x}{64 \, e^{3}} + \frac {3 \, {\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{2} x}{16 \, e^{3}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {5}{2}} d^{3}}{5 \, e^{4}} - \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} x}{8 \, e^{3}} + \frac {{\left (-e^{2} x^{2} + d^{2}\right )}^{\frac {7}{2}} d}{7 \, e^{4}} \]

input
integrate(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d),x, algorithm="maxima")
 
output
3/8*I*d^8*arcsin(e*x/d + 2)/e^4 + 45/128*d^8*arcsin(e*x/d)/e^4 - 3/8*sqrt( 
e^2*x^2 + 4*d*e*x + 3*d^2)*d^6*x/e^3 + 45/128*sqrt(-e^2*x^2 + d^2)*d^6*x/e 
^3 - 3/4*sqrt(e^2*x^2 + 4*d*e*x + 3*d^2)*d^7/e^4 - 1/64*(-e^2*x^2 + d^2)^( 
3/2)*d^4*x/e^3 + 3/16*(-e^2*x^2 + d^2)^(5/2)*d^2*x/e^3 - 1/5*(-e^2*x^2 + d 
^2)^(5/2)*d^3/e^4 - 1/8*(-e^2*x^2 + d^2)^(7/2)*x/e^3 + 1/7*(-e^2*x^2 + d^2 
)^(7/2)*d/e^4
 
3.2.4.8 Giac [A] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.70 \[ \int \frac {x^3 \left (d^2-e^2 x^2\right )^{5/2}}{d+e x} \, dx=-\frac {3 \, d^{8} \arcsin \left (\frac {e x}{d}\right ) \mathrm {sgn}\left (d\right ) \mathrm {sgn}\left (e\right )}{128 \, e^{3} {\left | e \right |}} - \frac {1}{4480} \, \sqrt {-e^{2} x^{2} + d^{2}} {\left (\frac {256 \, d^{7}}{e^{4}} - {\left (\frac {105 \, d^{6}}{e^{3}} - 2 \, {\left (\frac {64 \, d^{5}}{e^{2}} - {\left (\frac {35 \, d^{4}}{e} + 4 \, {\left (128 \, d^{3} - 5 \, {\left (21 \, d^{2} e - 2 \, {\left (7 \, e^{3} x - 8 \, d e^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \]

input
integrate(x^3*(-e^2*x^2+d^2)^(5/2)/(e*x+d),x, algorithm="giac")
 
output
-3/128*d^8*arcsin(e*x/d)*sgn(d)*sgn(e)/(e^3*abs(e)) - 1/4480*sqrt(-e^2*x^2 
 + d^2)*(256*d^7/e^4 - (105*d^6/e^3 - 2*(64*d^5/e^2 - (35*d^4/e + 4*(128*d 
^3 - 5*(21*d^2*e - 2*(7*e^3*x - 8*d*e^2)*x)*x)*x)*x)*x)*x)
 
3.2.4.9 Mupad [F(-1)]

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

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