\(\int \frac {x}{(c+d x)^2 (c^2-d^2 x^2)^{5/2}} \, dx\) [451]

Optimal result

Integrand size = 25, antiderivative size = 118 \[ \int \frac {x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {8 x}{105 c^3 d \left (c^2-d^2 x^2\right )^{3/2}}+\frac {1}{7 d^2 (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {2}{35 c d^2 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}+\frac {16 x}{105 c^5 d \sqrt {c^2-d^2 x^2}} \] Output:

8/105*x/c^3/d/(-d^2*x^2+c^2)^(3/2)+1/7/d^2/(d*x+c)^2/(-d^2*x^2+c^2)^(3/2)- 
2/35/c/d^2/(d*x+c)/(-d^2*x^2+c^2)^(3/2)+16/105*x/c^5/d/(-d^2*x^2+c^2)^(1/2 
)
 

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.79 \[ \int \frac {x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (9 c^5+18 c^4 d x+48 c^3 d^2 x^2+8 c^2 d^3 x^3-32 c d^4 x^4-16 d^5 x^5\right )}{105 c^5 d^2 (c-d x)^2 (c+d x)^4} \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(9*c^5 + 18*c^4*d*x + 48*c^3*d^2*x^2 + 8*c^2*d^3*x^3 
- 32*c*d^4*x^4 - 16*d^5*x^5))/(105*c^5*d^2*(c - d*x)^2*(c + d*x)^4)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {571, 470, 209, 208}

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.

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

\(\Big \downarrow \) 571

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

\(\Big \downarrow \) 470

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

\(\Big \downarrow \) 209

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

\(\Big \downarrow \) 208

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

Input:

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

Output:

1/(7*d^2*(c + d*x)^2*(c^2 - d^2*x^2)^(3/2)) + (2*(-1/5*1/(c*d*(c + d*x)*(c 
^2 - d^2*x^2)^(3/2)) + (4*(x/(3*c^2*(c^2 - d^2*x^2)^(3/2)) + (2*x)/(3*c^4* 
Sqrt[c^2 - d^2*x^2])))/(5*c)))/(7*d)
 

Defintions of rubi rules used

rule 208

Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), 
x] /; FreeQ[{a, b}, x]
 

rule 209

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 0]
 

rule 470

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

rule 571

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

Maple [A] (verified)

Time = 0.21 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.75

method	result	size
gosper	\(\frac {\left (-d x +c \right ) \left (-16 x^{5} d^{5}-32 x^{4} c \,d^{4}+8 c^{2} d^{3} x^{3}+48 c^{3} d^{2} x^{2}+18 c^{4} d x +9 c^{5}\right )}{105 \left (d x +c \right ) c^{5} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(88\)
orering	\(\frac {\left (-d x +c \right ) \left (-16 x^{5} d^{5}-32 x^{4} c \,d^{4}+8 c^{2} d^{3} x^{3}+48 c^{3} d^{2} x^{2}+18 c^{4} d x +9 c^{5}\right )}{105 \left (d x +c \right ) c^{5} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\)	\(88\)
trager	\(\frac {\left (-16 x^{5} d^{5}-32 x^{4} c \,d^{4}+8 c^{2} d^{3} x^{3}+48 c^{3} d^{2} x^{2}+18 c^{4} d x +9 c^{5}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{105 c^{5} \left (d x +c \right )^{4} \left (-d x +c \right )^{2} d^{2}}\)	\(90\)
default	\(\frac {-\frac {1}{5 c d \left (x +\frac {c}{d}\right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {4 d \left (-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{6 c^{2} d^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d^{2} c^{4} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}}{d^{2}}-\frac {c \left (-\frac {1}{7 c d \left (x +\frac {c}{d}\right )^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {5 d \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {4 d \left (-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{6 c^{2} d^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d^{2} c^{4} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{7 c}\right )}{d^{3}}\)	\(382\)

Input:

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

Output:

1/105*(-d*x+c)*(-16*d^5*x^5-32*c*d^4*x^4+8*c^2*d^3*x^3+48*c^3*d^2*x^2+18*c 
^4*d*x+9*c^5)/(d*x+c)/c^5/d^2/(-d^2*x^2+c^2)^(5/2)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 206 vs. \(2 (102) = 204\).

Time = 0.11 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.75 \[ \int \frac {x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {9 \, d^{6} x^{6} + 18 \, c d^{5} x^{5} - 9 \, c^{2} d^{4} x^{4} - 36 \, c^{3} d^{3} x^{3} - 9 \, c^{4} d^{2} x^{2} + 18 \, c^{5} d x + 9 \, c^{6} - {\left (16 \, d^{5} x^{5} + 32 \, c d^{4} x^{4} - 8 \, c^{2} d^{3} x^{3} - 48 \, c^{3} d^{2} x^{2} - 18 \, c^{4} d x - 9 \, c^{5}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{105 \, {\left (c^{5} d^{8} x^{6} + 2 \, c^{6} d^{7} x^{5} - c^{7} d^{6} x^{4} - 4 \, c^{8} d^{5} x^{3} - c^{9} d^{4} x^{2} + 2 \, c^{10} d^{3} x + c^{11} d^{2}\right )}} \] Input:

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

Output:

1/105*(9*d^6*x^6 + 18*c*d^5*x^5 - 9*c^2*d^4*x^4 - 36*c^3*d^3*x^3 - 9*c^4*d 
^2*x^2 + 18*c^5*d*x + 9*c^6 - (16*d^5*x^5 + 32*c*d^4*x^4 - 8*c^2*d^3*x^3 - 
 48*c^3*d^2*x^2 - 18*c^4*d*x - 9*c^5)*sqrt(-d^2*x^2 + c^2))/(c^5*d^8*x^6 + 
 2*c^6*d^7*x^5 - c^7*d^6*x^4 - 4*c^8*d^5*x^3 - c^9*d^4*x^2 + 2*c^10*d^3*x 
+ c^11*d^2)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.36 \[ \int \frac {x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {1}{7 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{4} x^{2} + 2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{3} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d^{2}\right )}} - \frac {2}{35 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{3} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2} d^{2}\right )}} + \frac {8 \, x}{105 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{3} d} + \frac {16 \, x}{105 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{5} d} \] Input:

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

Output:

1/7/((-d^2*x^2 + c^2)^(3/2)*d^4*x^2 + 2*(-d^2*x^2 + c^2)^(3/2)*c*d^3*x + ( 
-d^2*x^2 + c^2)^(3/2)*c^2*d^2) - 2/35/((-d^2*x^2 + c^2)^(3/2)*c*d^3*x + (- 
d^2*x^2 + c^2)^(3/2)*c^2*d^2) + 8/105*x/((-d^2*x^2 + c^2)^(3/2)*c^3*d) + 1 
6/105*x/(sqrt(-d^2*x^2 + c^2)*c^5*d)
 

Giac [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.88 \[ \int \frac {x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {-\frac {512 i \, \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )}{c^{5}} - \frac {70 \, {\left (\frac {9 \, c}{d x + c} - 4\right )}}{c^{5} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )} - \frac {15 \, c^{30} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{6} \mathrm {sgn}\left (d\right )^{6} + 63 \, c^{30} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{6} \mathrm {sgn}\left (d\right )^{6} + 70 \, c^{30} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{6} \mathrm {sgn}\left (d\right )^{6} - 210 \, c^{30} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{6} \mathrm {sgn}\left (d\right )^{6}}{c^{35} \mathrm {sgn}\left (\frac {1}{d x + c}\right )^{7} \mathrm {sgn}\left (d\right )^{7}}}{3360 \, d {\left | d \right |}} \] Input:

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

Output:

-1/3360*(-512*I*sgn(1/(d*x + c))*sgn(d)/c^5 - 70*(9*c/(d*x + c) - 4)/(c^5* 
(2*c/(d*x + c) - 1)^(3/2)*sgn(1/(d*x + c))*sgn(d)) - (15*c^30*(2*c/(d*x + 
c) - 1)^(7/2)*sgn(1/(d*x + c))^6*sgn(d)^6 + 63*c^30*(2*c/(d*x + c) - 1)^(5 
/2)*sgn(1/(d*x + c))^6*sgn(d)^6 + 70*c^30*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/ 
(d*x + c))^6*sgn(d)^6 - 210*c^30*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))^ 
6*sgn(d)^6)/(c^35*sgn(1/(d*x + c))^7*sgn(d)^7))/(d*abs(d))
 

Mupad [B] (verification not implemented)

Time = 7.14 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.23 \[ \int \frac {x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {1}{35\,c^2\,d^2}+\frac {23\,x}{420\,c^3\,d}\right )}{{\left (c+d\,x\right )}^2\,{\left (c-d\,x\right )}^2}+\frac {\sqrt {c^2-d^2\,x^2}}{28\,c^2\,d^2\,{\left (c+d\,x\right )}^4}+\frac {3\,\sqrt {c^2-d^2\,x^2}}{140\,c^3\,d^2\,{\left (c+d\,x\right )}^3}+\frac {16\,x\,\sqrt {c^2-d^2\,x^2}}{105\,c^5\,d\,\left (c+d\,x\right )\,\left (c-d\,x\right )} \] Input:

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

Output:

((c^2 - d^2*x^2)^(1/2)*(1/(35*c^2*d^2) + (23*x)/(420*c^3*d)))/((c + d*x)^2 
*(c - d*x)^2) + (c^2 - d^2*x^2)^(1/2)/(28*c^2*d^2*(c + d*x)^4) + (3*(c^2 - 
 d^2*x^2)^(1/2))/(140*c^3*d^2*(c + d*x)^3) + (16*x*(c^2 - d^2*x^2)^(1/2))/ 
(105*c^5*d*(c + d*x)*(c - d*x))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.58 \[ \int \frac {x}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {9 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4}+18 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d x -18 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{3} x^{3}-9 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{4} x^{4}+9 c^{5}+18 c^{4} d x +48 c^{3} d^{2} x^{2}+8 c^{2} d^{3} x^{3}-32 c \,d^{4} x^{4}-16 d^{5} x^{5}}{105 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{5} d^{2} \left (-d^{4} x^{4}-2 c \,d^{3} x^{3}+2 c^{3} d x +c^{4}\right )} \] Input:

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

Output:

(9*sqrt(c**2 - d**2*x**2)*c**4 + 18*sqrt(c**2 - d**2*x**2)*c**3*d*x - 18*s 
qrt(c**2 - d**2*x**2)*c*d**3*x**3 - 9*sqrt(c**2 - d**2*x**2)*d**4*x**4 + 9 
*c**5 + 18*c**4*d*x + 48*c**3*d**2*x**2 + 8*c**2*d**3*x**3 - 32*c*d**4*x** 
4 - 16*d**5*x**5)/(105*sqrt(c**2 - d**2*x**2)*c**5*d**2*(c**4 + 2*c**3*d*x 
 - 2*c*d**3*x**3 - d**4*x**4))