\(\int \frac {x^5}{(c+d x)^2 (c^2-d^2 x^2)^{3/2}} \, dx\) [411]

Optimal result

Integrand size = 27, antiderivative size = 145 \[ \int \frac {x^5}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {2 c (30 c-23 d x)}{15 d^6 \sqrt {c^2-d^2 x^2}}+\frac {c^4}{5 d^6 (c+d x)^2 \sqrt {c^2-d^2 x^2}}-\frac {22 c^3}{15 d^6 (c+d x) \sqrt {c^2-d^2 x^2}}+\frac {\sqrt {c^2-d^2 x^2}}{d^6}+\frac {2 c \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^6} \] Output:

2/15*c*(-23*d*x+30*c)/d^6/(-d^2*x^2+c^2)^(1/2)+1/5*c^4/d^6/(d*x+c)^2/(-d^2 
*x^2+c^2)^(1/2)-22/15*c^3/d^6/(d*x+c)/(-d^2*x^2+c^2)^(1/2)+(-d^2*x^2+c^2)^ 
(1/2)/d^6+2*c*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^6
 

Mathematica [A] (verified)

Time = 0.50 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.80 \[ \int \frac {x^5}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (56 c^4+82 c^3 d x-32 c^2 d^2 x^2-76 c d^3 x^3-15 d^4 x^4\right )}{(c-d x) (c+d x)^3}-60 c \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{15 d^6} \] Input:

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

Output:

((Sqrt[c^2 - d^2*x^2]*(56*c^4 + 82*c^3*d*x - 32*c^2*d^2*x^2 - 76*c*d^3*x^3 
 - 15*d^4*x^4))/((c - d*x)*(c + d*x)^3) - 60*c*ArcTan[(d*x)/(Sqrt[c^2] - S 
qrt[c^2 - d^2*x^2])])/(15*d^6)
 

Rubi [A] (verified)

Time = 0.88 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.17, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {570, 529, 25, 2166, 2345, 27, 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.

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

\(\Big \downarrow \) 570

\(\displaystyle \int \frac {x^5 (c-d x)^2}{\left (c^2-d^2 x^2\right )^{7/2}}dx\)

\(\Big \downarrow \) 529

\(\displaystyle \frac {c^4 (c-d x)^2}{5 d^6 \left (c^2-d^2 x^2\right )^{5/2}}-\frac {\int -\frac {(c-d x) \left (\frac {2 c^5}{d^5}-\frac {5 x c^4}{d^4}+\frac {5 x^2 c^3}{d^3}-\frac {5 x^3 c^2}{d^2}+\frac {5 x^4 c}{d}\right )}{\left (c^2-d^2 x^2\right )^{5/2}}dx}{5 c}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {(c-d x) \left (\frac {2 c^5}{d^5}-\frac {5 x c^4}{d^4}+\frac {5 x^2 c^3}{d^3}-\frac {5 x^3 c^2}{d^2}+\frac {5 x^4 c}{d}\right )}{\left (c^2-d^2 x^2\right )^{5/2}}dx}{5 c}+\frac {c^4 (c-d x)^2}{5 d^6 \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 2166

\(\displaystyle \frac {-\frac {\int \frac {\frac {16 c^5}{d^5}-\frac {45 x c^4}{d^4}+\frac {30 x^2 c^3}{d^3}-\frac {15 x^3 c^2}{d^2}}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{3 c}-\frac {22 c^4 (c-d x)}{3 d^6 \left (c^2-d^2 x^2\right )^{3/2}}}{5 c}+\frac {c^4 (c-d x)^2}{5 d^6 \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {-\frac {-\frac {\int \frac {15 c^4 (2 c-d x)}{d^5 \sqrt {c^2-d^2 x^2}}dx}{c^2}-\frac {2 c^3 (30 c-23 d x)}{d^6 \sqrt {c^2-d^2 x^2}}}{3 c}-\frac {22 c^4 (c-d x)}{3 d^6 \left (c^2-d^2 x^2\right )^{3/2}}}{5 c}+\frac {c^4 (c-d x)^2}{5 d^6 \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {-\frac {15 c^2 \int \frac {2 c-d x}{\sqrt {c^2-d^2 x^2}}dx}{d^5}-\frac {2 c^3 (30 c-23 d x)}{d^6 \sqrt {c^2-d^2 x^2}}}{3 c}-\frac {22 c^4 (c-d x)}{3 d^6 \left (c^2-d^2 x^2\right )^{3/2}}}{5 c}+\frac {c^4 (c-d x)^2}{5 d^6 \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {-\frac {-\frac {15 c^2 \left (2 c \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {\sqrt {c^2-d^2 x^2}}{d}\right )}{d^5}-\frac {2 c^3 (30 c-23 d x)}{d^6 \sqrt {c^2-d^2 x^2}}}{3 c}-\frac {22 c^4 (c-d x)}{3 d^6 \left (c^2-d^2 x^2\right )^{3/2}}}{5 c}+\frac {c^4 (c-d x)^2}{5 d^6 \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {-\frac {-\frac {15 c^2 \left (2 c \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {\sqrt {c^2-d^2 x^2}}{d}\right )}{d^5}-\frac {2 c^3 (30 c-23 d x)}{d^6 \sqrt {c^2-d^2 x^2}}}{3 c}-\frac {22 c^4 (c-d x)}{3 d^6 \left (c^2-d^2 x^2\right )^{3/2}}}{5 c}+\frac {c^4 (c-d x)^2}{5 d^6 \left (c^2-d^2 x^2\right )^{5/2}}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {-\frac {-\frac {15 c^2 \left (\frac {2 c \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d}+\frac {\sqrt {c^2-d^2 x^2}}{d}\right )}{d^5}-\frac {2 c^3 (30 c-23 d x)}{d^6 \sqrt {c^2-d^2 x^2}}}{3 c}-\frac {22 c^4 (c-d x)}{3 d^6 \left (c^2-d^2 x^2\right )^{3/2}}}{5 c}+\frac {c^4 (c-d x)^2}{5 d^6 \left (c^2-d^2 x^2\right )^{5/2}}\)

Input:

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

Output:

(c^4*(c - d*x)^2)/(5*d^6*(c^2 - d^2*x^2)^(5/2)) + ((-22*c^4*(c - d*x))/(3* 
d^6*(c^2 - d^2*x^2)^(3/2)) - ((-2*c^3*(30*c - 23*d*x))/(d^6*Sqrt[c^2 - d^2 
*x^2]) - (15*c^2*(Sqrt[c^2 - d^2*x^2]/d + (2*c*ArcTan[(d*x)/Sqrt[c^2 - d^2 
*x^2]])/d))/d^5)/(3*c))/(5*c)
 

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 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 529

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m, a*d + b*c*x, x], R = PolynomialRem 
ainder[x^m, a*d + b*c*x, x]}, Simp[(-c)*R*(c + d*x)^n*((a + b*x^2)^(p + 1)/ 
(2*a*d*(p + 1))), x] + Simp[c/(2*a*(p + 1))   Int[(c + d*x)^(n - 1)*(a + b* 
x^2)^(p + 1)*ExpandToSum[2*a*d*(p + 1)*Qx + R*(n + 2*p + 2), x], x], x]] /; 
 FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 1] && LtQ[p, -1] && EqQ[b* 
c^2 + a*d^2, 0]
 

rule 570

Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), 
x_Symbol] :> Simp[c^(2*n)/a^n   Int[(e*x)^m*((a + b*x^2)^(n + p)/(c - d*x)^ 
n), x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[b*c^2 + a*d^2, 0] && I 
LtQ[n, -1] &&  !(IGtQ[m, 0] && ILtQ[m + n, 0] &&  !GtQ[p, 1])
 

rule 2166

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[Pq, a*e + b*d*x, x], R = PolynomialRemainde 
r[Pq, a*e + b*d*x, x]}, Simp[(-d)*R*(d + e*x)^m*((a + b*x^2)^(p + 1)/(2*a*e 
*(p + 1))), x] + Simp[d/(2*a*(p + 1))   Int[(d + e*x)^(m - 1)*(a + b*x^2)^( 
p + 1)*ExpandToSum[2*a*e*(p + 1)*Qx + R*(m + 2*p + 2), x], x], x]] /; FreeQ 
[{a, b, d, e}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && ILtQ[p + 1/2, 
 0] && GtQ[m, 0]
 

rule 2345

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot 
ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 
 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b 
*f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   In 
t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] 
/; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.60

method	result	size
risch	\(\frac {\sqrt {-d^{2} x^{2}+c^{2}}}{d^{6}}+\frac {2 c \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{5} \sqrt {d^{2}}}-\frac {c \sqrt {-d^{2} \left (x -\frac {c}{d}\right )^{2}-2 c d \left (x -\frac {c}{d}\right )}}{8 d^{7} \left (x -\frac {c}{d}\right )}+\frac {383 c \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{120 d^{7} \left (x +\frac {c}{d}\right )}-\frac {41 c^{2} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{60 d^{8} \left (x +\frac {c}{d}\right )^{2}}+\frac {c^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{10 d^{9} \left (x +\frac {c}{d}\right )^{3}}\)	\(232\)
default	\(\frac {-\frac {x^{2}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {2 c^{2}}{d^{4} \sqrt {-d^{2} x^{2}+c^{2}}}}{d^{2}}-\frac {4 c x}{d^{5} \sqrt {-d^{2} x^{2}+c^{2}}}-\frac {2 c \left (\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}\right )}{d^{3}}+\frac {3 c^{2}}{d^{6} \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {5 c^{4} \left (-\frac {1}{3 c d \left (x +\frac {c}{d}\right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d \,c^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{d^{6}}-\frac {c^{5} \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right )^{2} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}+\frac {3 d \left (-\frac {1}{3 c d \left (x +\frac {c}{d}\right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d \,c^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{d^{7}}\)	\(418\)

Input:

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

Output:

(-d^2*x^2+c^2)^(1/2)/d^6+2*c/d^5/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^ 
2+c^2)^(1/2))-1/8*c/d^7/(x-c/d)*(-d^2*(x-c/d)^2-2*c*d*(x-c/d))^(1/2)+383/1 
20*c/d^7/(x+c/d)*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)-41/60*c^2/d^8/(x+c/d 
)^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)+1/10*c^3/d^9/(x+c/d)^3*(-d^2*(x+c 
/d)^2+2*c*d*(x+c/d))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.30 \[ \int \frac {x^5}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {56 \, c d^{4} x^{4} + 112 \, c^{2} d^{3} x^{3} - 112 \, c^{4} d x - 56 \, c^{5} - 60 \, {\left (c d^{4} x^{4} + 2 \, c^{2} d^{3} x^{3} - 2 \, c^{4} d x - c^{5}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (15 \, d^{4} x^{4} + 76 \, c d^{3} x^{3} + 32 \, c^{2} d^{2} x^{2} - 82 \, c^{3} d x - 56 \, c^{4}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{15 \, {\left (d^{10} x^{4} + 2 \, c d^{9} x^{3} - 2 \, c^{3} d^{7} x - c^{4} d^{6}\right )}} \] Input:

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

Output:

1/15*(56*c*d^4*x^4 + 112*c^2*d^3*x^3 - 112*c^4*d*x - 56*c^5 - 60*(c*d^4*x^ 
4 + 2*c^2*d^3*x^3 - 2*c^4*d*x - c^5)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d 
*x)) + (15*d^4*x^4 + 76*c*d^3*x^3 + 32*c^2*d^2*x^2 - 82*c^3*d*x - 56*c^4)* 
sqrt(-d^2*x^2 + c^2))/(d^10*x^4 + 2*c*d^9*x^3 - 2*c^3*d^7*x - c^4*d^6)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.35 \[ \int \frac {x^5}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {c^{4}}{5 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} d^{8} x^{2} + 2 \, \sqrt {-d^{2} x^{2} + c^{2}} c d^{7} x + \sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{6}\right )}} - \frac {22 \, c^{3}}{15 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} d^{7} x + \sqrt {-d^{2} x^{2} + c^{2}} c d^{6}\right )}} - \frac {x^{2}}{\sqrt {-d^{2} x^{2} + c^{2}} d^{4}} - \frac {46 \, c x}{15 \, \sqrt {-d^{2} x^{2} + c^{2}} d^{5}} + \frac {2 \, c \arcsin \left (\frac {d x}{c}\right )}{d^{6}} + \frac {5 \, c^{2}}{\sqrt {-d^{2} x^{2} + c^{2}} d^{6}} \] Input:

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

Output:

1/5*c^4/(sqrt(-d^2*x^2 + c^2)*d^8*x^2 + 2*sqrt(-d^2*x^2 + c^2)*c*d^7*x + s 
qrt(-d^2*x^2 + c^2)*c^2*d^6) - 22/15*c^3/(sqrt(-d^2*x^2 + c^2)*d^7*x + sqr 
t(-d^2*x^2 + c^2)*c*d^6) - x^2/(sqrt(-d^2*x^2 + c^2)*d^4) - 46/15*c*x/(sqr 
t(-d^2*x^2 + c^2)*d^5) + 2*c*arcsin(d*x/c)/d^6 + 5*c^2/(sqrt(-d^2*x^2 + c^ 
2)*d^6)
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.59 \[ \int \frac {x^5}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {\frac {480 \, c \arctan \left (\sqrt {\frac {2 \, c}{d x + c} - 1}\right )}{d^{5} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )} - \frac {15 \, {\left (17 \, c {\left (\frac {2 \, c}{d x + c} - 1\right )} + c\right )}}{{\left ({\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} + \sqrt {\frac {2 \, c}{d x + c} - 1}\right )} d^{5} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )} - \frac {3 \, c d^{20} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{4} \mathrm {sgn}\left (d\right )^{4} - 35 \, c d^{20} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{4} \mathrm {sgn}\left (d\right )^{4} + 345 \, c d^{20} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{4} \mathrm {sgn}\left (d\right )^{4}}{d^{25} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{5} \mathrm {sgn}\left (d\right )^{5}}}{120 \, {\left | d \right |}} \] Input:

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

Output:

-1/120*(480*c*arctan(sqrt(2*c/(d*x + c) - 1))/(d^5*sgn(1/(d*x + c))*sgn(d) 
) - 15*(17*c*(2*c/(d*x + c) - 1) + c)/(((2*c/(d*x + c) - 1)^(3/2) + sqrt(2 
*c/(d*x + c) - 1))*d^5*sgn(1/(d*x + c))*sgn(d)) - (3*c*d^20*(2*c/(d*x + c) 
 - 1)^(5/2)*sgn(1/(d*x + c))^4*sgn(d)^4 - 35*c*d^20*(2*c/(d*x + c) - 1)^(3 
/2)*sgn(1/(d*x + c))^4*sgn(d)^4 + 345*c*d^20*sqrt(2*c/(d*x + c) - 1)*sgn(1 
/(d*x + c))^4*sgn(d)^4)/(d^25*sgn(1/(d*x + c))^5*sgn(d)^5))/abs(d)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.52 \[ \int \frac {x^5}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {30 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{3}+60 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{2} d x +30 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c \,d^{2} x^{2}+41 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3}+82 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d x +41 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{2} x^{2}+56 c^{4}+82 c^{3} d x -32 c^{2} d^{2} x^{2}-76 c \,d^{3} x^{3}-15 d^{4} x^{4}}{15 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{6} \left (d^{2} x^{2}+2 c d x +c^{2}\right )} \] Input:

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

Output:

(30*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**3 + 60*sqrt(c**2 - d**2*x**2)* 
asin((d*x)/c)*c**2*d*x + 30*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c*d**2*x* 
*2 + 41*sqrt(c**2 - d**2*x**2)*c**3 + 82*sqrt(c**2 - d**2*x**2)*c**2*d*x + 
 41*sqrt(c**2 - d**2*x**2)*c*d**2*x**2 + 56*c**4 + 82*c**3*d*x - 32*c**2*d 
**2*x**2 - 76*c*d**3*x**3 - 15*d**4*x**4)/(15*sqrt(c**2 - d**2*x**2)*d**6* 
(c**2 + 2*c*d*x + d**2*x**2))