\(\int \frac {x^5}{(c+d x) (c^2-d^2 x^2)^{3/2}} \, dx\) [401]

Optimal result

Integrand size = 27, antiderivative size = 144 \[ \int \frac {x^5}{(c+d x) \left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {c^2 (6 c-7 d x)}{3 d^6 \sqrt {c^2-d^2 x^2}}+\frac {c^4}{3 d^6 (c+d x) \sqrt {c^2-d^2 x^2}}-\frac {c \sqrt {c^2-d^2 x^2}}{d^6}+\frac {x \sqrt {c^2-d^2 x^2}}{2 d^5}-\frac {5 c^2 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^6} \] Output:

-1/3*c^2*(-7*d*x+6*c)/d^6/(-d^2*x^2+c^2)^(1/2)+1/3*c^4/d^6/(d*x+c)/(-d^2*x 
^2+c^2)^(1/2)-c*(-d^2*x^2+c^2)^(1/2)/d^6+1/2*x*(-d^2*x^2+c^2)^(1/2)/d^5-5/ 
2*c^2*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^6
 

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.83 \[ \int \frac {x^5}{(c+d x) \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (16 c^4+c^3 d x-23 c^2 d^2 x^2-3 c d^3 x^3+3 d^4 x^4\right )}{6 d^6 (-c+d x) (c+d x)^2}+\frac {5 c^2 \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{d^6} \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(16*c^4 + c^3*d*x - 23*c^2*d^2*x^2 - 3*c*d^3*x^3 + 3* 
d^4*x^4))/(6*d^6*(-c + d*x)*(c + d*x)^2) + (5*c^2*ArcTan[(d*x)/(Sqrt[c^2] 
- Sqrt[c^2 - d^2*x^2])])/d^6
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {568, 530, 25, 2346, 25, 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.

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

\(\Big \downarrow \) 568

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

\(\Big \downarrow \) 530

\(\displaystyle \frac {x^4}{3 d^2 (c+d x) \sqrt {c^2-d^2 x^2}}-\frac {\frac {c^2 (4 c-5 d x)}{d^4 \sqrt {c^2-d^2 x^2}}-\frac {\int -\frac {\frac {5 c^4}{d^3}-\frac {4 x c^3}{d^2}+\frac {5 x^2 c^2}{d}}{\sqrt {c^2-d^2 x^2}}dx}{c^2}}{3 d^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {x^4}{3 d^2 (c+d x) \sqrt {c^2-d^2 x^2}}-\frac {\frac {\int \frac {\frac {5 c^4}{d^3}-\frac {4 x c^3}{d^2}+\frac {5 x^2 c^2}{d}}{\sqrt {c^2-d^2 x^2}}dx}{c^2}+\frac {c^2 (4 c-5 d x)}{d^4 \sqrt {c^2-d^2 x^2}}}{3 d^2}\)

\(\Big \downarrow \) 2346

\(\displaystyle \frac {x^4}{3 d^2 (c+d x) \sqrt {c^2-d^2 x^2}}-\frac {\frac {-\frac {\int -\frac {c^3 (15 c-8 d x)}{d \sqrt {c^2-d^2 x^2}}dx}{2 d^2}-\frac {5 c^2 x \sqrt {c^2-d^2 x^2}}{2 d^3}}{c^2}+\frac {c^2 (4 c-5 d x)}{d^4 \sqrt {c^2-d^2 x^2}}}{3 d^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {x^4}{3 d^2 (c+d x) \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {\int \frac {c^3 (15 c-8 d x)}{d \sqrt {c^2-d^2 x^2}}dx}{2 d^2}-\frac {5 c^2 x \sqrt {c^2-d^2 x^2}}{2 d^3}}{c^2}+\frac {c^2 (4 c-5 d x)}{d^4 \sqrt {c^2-d^2 x^2}}}{3 d^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^4}{3 d^2 (c+d x) \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {c^3 \int \frac {15 c-8 d x}{\sqrt {c^2-d^2 x^2}}dx}{2 d^3}-\frac {5 c^2 x \sqrt {c^2-d^2 x^2}}{2 d^3}}{c^2}+\frac {c^2 (4 c-5 d x)}{d^4 \sqrt {c^2-d^2 x^2}}}{3 d^2}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {x^4}{3 d^2 (c+d x) \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {c^3 \left (15 c \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {8 \sqrt {c^2-d^2 x^2}}{d}\right )}{2 d^3}-\frac {5 c^2 x \sqrt {c^2-d^2 x^2}}{2 d^3}}{c^2}+\frac {c^2 (4 c-5 d x)}{d^4 \sqrt {c^2-d^2 x^2}}}{3 d^2}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {x^4}{3 d^2 (c+d x) \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {c^3 \left (15 c \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {8 \sqrt {c^2-d^2 x^2}}{d}\right )}{2 d^3}-\frac {5 c^2 x \sqrt {c^2-d^2 x^2}}{2 d^3}}{c^2}+\frac {c^2 (4 c-5 d x)}{d^4 \sqrt {c^2-d^2 x^2}}}{3 d^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {x^4}{3 d^2 (c+d x) \sqrt {c^2-d^2 x^2}}-\frac {\frac {\frac {c^3 \left (\frac {15 c \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d}+\frac {8 \sqrt {c^2-d^2 x^2}}{d}\right )}{2 d^3}-\frac {5 c^2 x \sqrt {c^2-d^2 x^2}}{2 d^3}}{c^2}+\frac {c^2 (4 c-5 d x)}{d^4 \sqrt {c^2-d^2 x^2}}}{3 d^2}\)

Input:

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

Output:

x^4/(3*d^2*(c + d*x)*Sqrt[c^2 - d^2*x^2]) - ((c^2*(4*c - 5*d*x))/(d^4*Sqrt 
[c^2 - d^2*x^2]) + ((-5*c^2*x*Sqrt[c^2 - d^2*x^2])/(2*d^3) + (c^3*((8*Sqrt 
[c^2 - d^2*x^2])/d + (15*c*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/d))/(2*d^3)) 
/c^2)/(3*d^2)
 

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 530

Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symb 
ol] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Co 
eff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Po 
lynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x 
)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(a 
 + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Qx + e*(2*p + 3), x], x], x]] /; 
FreeQ[{a, b, c, d}, x] && IGtQ[n, 0] && IGtQ[m, 0] && LtQ[p, -1] && EqQ[n, 
1] && IntegerQ[2*p]
 

rule 568

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

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.29 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.40

method	result	size
risch	\(-\frac {\left (-d x +2 c \right ) \sqrt {-d^{2} x^{2}+c^{2}}}{2 d^{6}}+\frac {c^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{6 d^{8} \left (x +\frac {c}{d}\right )^{2}}-\frac {25 c^{2} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{12 d^{7} \left (x +\frac {c}{d}\right )}-\frac {5 c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 d^{5} \sqrt {d^{2}}}-\frac {c^{2} \sqrt {-d^{2} \left (x -\frac {c}{d}\right )^{2}-2 c d \left (x -\frac {c}{d}\right )}}{4 d^{7} \left (x -\frac {c}{d}\right )}\)	\(202\)
default	\(\frac {c^{2} x}{d^{5} \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {-\frac {x^{3}}{2 d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {3 c^{2} \left (\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}\right )}{2 d^{2}}}{d}+\frac {c^{2} \left (\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}\right )}{d^{3}}-\frac {c \left (-\frac {x^{2}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {2 c^{2}}{d^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2}}-\frac {c^{5} \left (-\frac {1}{3 c d \left (x +\frac {c}{d}\right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d \,c^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{d^{6}}-\frac {c^{3}}{d^{6} \sqrt {-d^{2} x^{2}+c^{2}}}\)	\(350\)

Input:

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

Output:

-1/2*(-d*x+2*c)/d^6*(-d^2*x^2+c^2)^(1/2)+1/6*c^3/d^8/(x+c/d)^2*(-d^2*(x+c/ 
d)^2+2*c*d*(x+c/d))^(1/2)-25/12/d^7*c^2/(x+c/d)*(-d^2*(x+c/d)^2+2*c*d*(x+c 
/d))^(1/2)-5/2*c^2/d^5/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/ 
2))-1/4*c^2/d^7/(x-c/d)*(-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 = 190, normalized size of antiderivative = 1.32 \[ \int \frac {x^5}{(c+d x) \left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {16 \, c^{2} d^{3} x^{3} + 16 \, c^{3} d^{2} x^{2} - 16 \, c^{4} d x - 16 \, c^{5} - 30 \, {\left (c^{2} d^{3} x^{3} + c^{3} d^{2} x^{2} - c^{4} d x - c^{5}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (3 \, d^{4} x^{4} - 3 \, c d^{3} x^{3} - 23 \, c^{2} d^{2} x^{2} + c^{3} d x + 16 \, c^{4}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{6 \, {\left (d^{9} x^{3} + c d^{8} x^{2} - c^{2} d^{7} x - c^{3} d^{6}\right )}} \] Input:

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

Output:

-1/6*(16*c^2*d^3*x^3 + 16*c^3*d^2*x^2 - 16*c^4*d*x - 16*c^5 - 30*(c^2*d^3* 
x^3 + c^3*d^2*x^2 - c^4*d*x - c^5)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x 
)) - (3*d^4*x^4 - 3*c*d^3*x^3 - 23*c^2*d^2*x^2 + c^3*d*x + 16*c^4)*sqrt(-d 
^2*x^2 + c^2))/(d^9*x^3 + c*d^8*x^2 - c^2*d^7*x - c^3*d^6)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.05 \[ \int \frac {x^5}{(c+d x) \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {c^{4}}{3 \, {\left (\sqrt {-d^{2} x^{2} + c^{2}} d^{7} x + \sqrt {-d^{2} x^{2} + c^{2}} c d^{6}\right )}} - \frac {x^{3}}{2 \, \sqrt {-d^{2} x^{2} + c^{2}} d^{3}} + \frac {c x^{2}}{\sqrt {-d^{2} x^{2} + c^{2}} d^{4}} + \frac {17 \, c^{2} x}{6 \, \sqrt {-d^{2} x^{2} + c^{2}} d^{5}} - \frac {5 \, c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{6}} - \frac {3 \, c^{3}}{\sqrt {-d^{2} x^{2} + c^{2}} d^{6}} \] Input:

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

Output:

1/3*c^4/(sqrt(-d^2*x^2 + c^2)*d^7*x + sqrt(-d^2*x^2 + c^2)*c*d^6) - 1/2*x^ 
3/(sqrt(-d^2*x^2 + c^2)*d^3) + c*x^2/(sqrt(-d^2*x^2 + c^2)*d^4) + 17/6*c^2 
*x/(sqrt(-d^2*x^2 + c^2)*d^5) - 5/2*c^2*arcsin(d*x/c)/d^6 - 3*c^3/(sqrt(-d 
^2*x^2 + c^2)*d^6)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.10 \[ \int \frac {x^5}{(c+d x) \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {-15 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{3}-15 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{2} d x -10 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3}-10 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d x -16 c^{4}-c^{3} d x +23 c^{2} d^{2} x^{2}+3 c \,d^{3} x^{3}-3 d^{4} x^{4}}{6 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{6} \left (d x +c \right )} \] Input:

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

Output:

( - 15*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**3 - 15*sqrt(c**2 - d**2*x** 
2)*asin((d*x)/c)*c**2*d*x - 10*sqrt(c**2 - d**2*x**2)*c**3 - 10*sqrt(c**2 
- d**2*x**2)*c**2*d*x - 16*c**4 - c**3*d*x + 23*c**2*d**2*x**2 + 3*c*d**3* 
x**3 - 3*d**4*x**4)/(6*sqrt(c**2 - d**2*x**2)*d**6*(c + d*x))