\(\int \frac {x^6}{(c+d x) (c^2-d^2 x^2)^{5/2}} \, dx\) [434]

Optimal result

Integrand size = 27, antiderivative size = 147 \[ \int \frac {x^6}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {c^3 (15 c-11 d x)}{15 d^7 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {c^5}{5 d^7 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}-\frac {c (45 c-23 d x)}{15 d^7 \sqrt {c^2-d^2 x^2}}-\frac {\sqrt {c^2-d^2 x^2}}{d^7}-\frac {c \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^7} \] Output:

1/15*c^3*(-11*d*x+15*c)/d^7/(-d^2*x^2+c^2)^(3/2)-1/5*c^5/d^7/(d*x+c)/(-d^2 
*x^2+c^2)^(3/2)-1/15*c*(-23*d*x+45*c)/d^7/(-d^2*x^2+c^2)^(1/2)-(-d^2*x^2+c 
^2)^(1/2)/d^7-c*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^7
 

Mathematica [A] (verified)

Time = 0.53 (sec) , antiderivative size = 130, normalized size of antiderivative = 0.88 \[ \int \frac {x^6}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (-48 c^5-33 c^4 d x+87 c^3 d^2 x^2+52 c^2 d^3 x^3-38 c d^4 x^4-15 d^5 x^5\right )}{15 d^7 (-c+d x)^2 (c+d x)^3}+\frac {2 c \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{d^7} \] Input:

Integrate[x^6/((c + d*x)*(c^2 - d^2*x^2)^(5/2)),x]
 

Output:

(Sqrt[c^2 - d^2*x^2]*(-48*c^5 - 33*c^4*d*x + 87*c^3*d^2*x^2 + 52*c^2*d^3*x 
^3 - 38*c*d^4*x^4 - 15*d^5*x^5))/(15*d^7*(-c + d*x)^2*(c + d*x)^3) + (2*c* 
ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/d^7
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 172, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {568, 530, 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.

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

\(\Big \downarrow \) 568

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

\(\Big \downarrow \) 530

\(\displaystyle \frac {x^5}{5 d^2 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}-\frac {-\frac {\int \frac {\frac {5 c^5}{d^4}-\frac {18 x c^4}{d^3}+\frac {15 x^2 c^3}{d^2}-\frac {18 x^3 c^2}{d}}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{3 c^2}-\frac {c^3 (6 c-5 d x)}{3 d^5 \left (c^2-d^2 x^2\right )^{3/2}}}{5 d^2}\)

\(\Big \downarrow \) 2345

\(\displaystyle \frac {x^5}{5 d^2 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}-\frac {-\frac {-\frac {\int \frac {3 c^4 (5 c-6 d x)}{d^4 \sqrt {c^2-d^2 x^2}}dx}{c^2}-\frac {4 c^3 (9 c-5 d x)}{d^5 \sqrt {c^2-d^2 x^2}}}{3 c^2}-\frac {c^3 (6 c-5 d x)}{3 d^5 \left (c^2-d^2 x^2\right )^{3/2}}}{5 d^2}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {x^5}{5 d^2 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}-\frac {-\frac {-\frac {3 c^2 \int \frac {5 c-6 d x}{\sqrt {c^2-d^2 x^2}}dx}{d^4}-\frac {4 c^3 (9 c-5 d x)}{d^5 \sqrt {c^2-d^2 x^2}}}{3 c^2}-\frac {c^3 (6 c-5 d x)}{3 d^5 \left (c^2-d^2 x^2\right )^{3/2}}}{5 d^2}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {x^5}{5 d^2 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}-\frac {-\frac {-\frac {3 c^2 \left (5 c \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {6 \sqrt {c^2-d^2 x^2}}{d}\right )}{d^4}-\frac {4 c^3 (9 c-5 d x)}{d^5 \sqrt {c^2-d^2 x^2}}}{3 c^2}-\frac {c^3 (6 c-5 d x)}{3 d^5 \left (c^2-d^2 x^2\right )^{3/2}}}{5 d^2}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {x^5}{5 d^2 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}-\frac {-\frac {-\frac {3 c^2 \left (5 c \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {6 \sqrt {c^2-d^2 x^2}}{d}\right )}{d^4}-\frac {4 c^3 (9 c-5 d x)}{d^5 \sqrt {c^2-d^2 x^2}}}{3 c^2}-\frac {c^3 (6 c-5 d x)}{3 d^5 \left (c^2-d^2 x^2\right )^{3/2}}}{5 d^2}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {x^5}{5 d^2 (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}-\frac {-\frac {-\frac {3 c^2 \left (\frac {5 c \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d}+\frac {6 \sqrt {c^2-d^2 x^2}}{d}\right )}{d^4}-\frac {4 c^3 (9 c-5 d x)}{d^5 \sqrt {c^2-d^2 x^2}}}{3 c^2}-\frac {c^3 (6 c-5 d x)}{3 d^5 \left (c^2-d^2 x^2\right )^{3/2}}}{5 d^2}\)

Input:

Int[x^6/((c + d*x)*(c^2 - d^2*x^2)^(5/2)),x]
 

Output:

x^5/(5*d^2*(c + d*x)*(c^2 - d^2*x^2)^(3/2)) - (-1/3*(c^3*(6*c - 5*d*x))/(d 
^5*(c^2 - d^2*x^2)^(3/2)) - ((-4*c^3*(9*c - 5*d*x))/(d^5*Sqrt[c^2 - d^2*x^ 
2]) - (3*c^2*((6*Sqrt[c^2 - d^2*x^2])/d + (5*c*ArcTan[(d*x)/Sqrt[c^2 - d^2 
*x^2]])/d))/d^4)/(3*c^2))/(5*d^2)
 

Defintions of rubi rules used

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 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 [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(280\) vs. \(2(131)=262\).

Time = 0.31 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.91

method	result	size
risch	\(-\frac {\sqrt {-d^{2} x^{2}+c^{2}}}{d^{7}}-\frac {c \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{6} \sqrt {d^{2}}}+\frac {25 c \sqrt {-d^{2} \left (x -\frac {c}{d}\right )^{2}-2 c d \left (x -\frac {c}{d}\right )}}{48 d^{8} \left (x -\frac {c}{d}\right )}-\frac {493 c \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{240 d^{8} \left (x +\frac {c}{d}\right )}+\frac {c^{2} \sqrt {-d^{2} \left (x -\frac {c}{d}\right )^{2}-2 c d \left (x -\frac {c}{d}\right )}}{24 d^{9} \left (x -\frac {c}{d}\right )^{2}}+\frac {23 c^{2} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{60 d^{9} \left (x +\frac {c}{d}\right )^{2}}-\frac {c^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{20 d^{10} \left (x +\frac {c}{d}\right )^{3}}\)	\(281\)
default	\(\frac {-\frac {x^{4}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {4 c^{2} \left (\frac {x^{2}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {2 c^{2}}{3 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\right )}{d^{2}}}{d}+\frac {c^{2} \left (\frac {x^{2}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {2 c^{2}}{3 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\right )}{d^{3}}+\frac {c^{4}}{3 d^{7} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {c^{6} \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {4 d \left (-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{6 c^{2} d^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d^{2} c^{4} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{d^{7}}-\frac {c^{5} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{6}}-\frac {c \left (\frac {x^{3}}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}\right )}{d^{2}}-\frac {c^{3} \left (\frac {x}{2 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {c^{2} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 d^{2}}\right )}{d^{4}}\)	\(533\)

Input:

int(x^6/(d*x+c)/(-d^2*x^2+c^2)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-(-d^2*x^2+c^2)^(1/2)/d^7-c/d^6/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2 
+c^2)^(1/2))+25/48*c/d^8/(x-c/d)*(-d^2*(x-c/d)^2-2*c*d*(x-c/d))^(1/2)-493/ 
240*c/d^8/(x+c/d)*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(1/2)+1/24*c^2/d^9/(x-c/d 
)^2*(-d^2*(x-c/d)^2-2*c*d*(x-c/d))^(1/2)+23/60*c^2/d^9/(x+c/d)^2*(-d^2*(x+ 
c/d)^2+2*c*d*(x+c/d))^(1/2)-1/20*c^3/d^10/(x+c/d)^3*(-d^2*(x+c/d)^2+2*c*d* 
(x+c/d))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.76 \[ \int \frac {x^6}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {48 \, c d^{5} x^{5} + 48 \, c^{2} d^{4} x^{4} - 96 \, c^{3} d^{3} x^{3} - 96 \, c^{4} d^{2} x^{2} + 48 \, c^{5} d x + 48 \, c^{6} - 30 \, {\left (c d^{5} x^{5} + c^{2} d^{4} x^{4} - 2 \, c^{3} d^{3} x^{3} - 2 \, c^{4} d^{2} x^{2} + c^{5} d x + c^{6}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (15 \, d^{5} x^{5} + 38 \, c d^{4} x^{4} - 52 \, c^{2} d^{3} x^{3} - 87 \, c^{3} d^{2} x^{2} + 33 \, c^{4} d x + 48 \, c^{5}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{15 \, {\left (d^{12} x^{5} + c d^{11} x^{4} - 2 \, c^{2} d^{10} x^{3} - 2 \, c^{3} d^{9} x^{2} + c^{4} d^{8} x + c^{5} d^{7}\right )}} \] Input:

integrate(x^6/(d*x+c)/(-d^2*x^2+c^2)^(5/2),x, algorithm="fricas")
 

Output:

-1/15*(48*c*d^5*x^5 + 48*c^2*d^4*x^4 - 96*c^3*d^3*x^3 - 96*c^4*d^2*x^2 + 4 
8*c^5*d*x + 48*c^6 - 30*(c*d^5*x^5 + c^2*d^4*x^4 - 2*c^3*d^3*x^3 - 2*c^4*d 
^2*x^2 + c^5*d*x + c^6)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + (15*d^ 
5*x^5 + 38*c*d^4*x^4 - 52*c^2*d^3*x^3 - 87*c^3*d^2*x^2 + 33*c^4*d*x + 48*c 
^5)*sqrt(-d^2*x^2 + c^2))/(d^12*x^5 + c*d^11*x^4 - 2*c^2*d^10*x^3 - 2*c^3* 
d^9*x^2 + c^4*d^8*x + c^5*d^7)
 

Sympy [F]

\[ \int \frac {x^6}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {x^{6}}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}} \left (c + d x\right )}\, dx \] Input:

integrate(x**6/(d*x+c)/(-d**2*x**2+c**2)**(5/2),x)
 

Output:

Integral(x**6/((-(-c + d*x)*(c + d*x))**(5/2)*(c + d*x)), x)
 

Maxima [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.76 \[ \int \frac {x^6}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {c^{5}}{5 \, {\left ({\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{8} x + {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c d^{7}\right )}} - \frac {x^{4}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{3}} - \frac {5 \, c x^{3}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{4}} + \frac {20 \, c^{2} x^{2}}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{5}} + \frac {64 \, c^{3} x}{15 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{6}} + \frac {x^{2}}{3 \, \sqrt {-d^{2} x^{2} + c^{2}} d^{5}} - \frac {14 \, c^{4}}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{7}} - \frac {52 \, c x}{15 \, \sqrt {-d^{2} x^{2} + c^{2}} d^{6}} - \frac {c \arcsin \left (\frac {d x}{c}\right )}{d^{7}} + \frac {4 \, c^{2}}{3 \, \sqrt {-d^{2} x^{2} + c^{2}} d^{7}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}}}{3 \, d^{7}} \] Input:

integrate(x^6/(d*x+c)/(-d^2*x^2+c^2)^(5/2),x, algorithm="maxima")
 

Output:

-1/5*c^5/((-d^2*x^2 + c^2)^(3/2)*d^8*x + (-d^2*x^2 + c^2)^(3/2)*c*d^7) - x 
^4/((-d^2*x^2 + c^2)^(3/2)*d^3) - 5*c*x^3/((-d^2*x^2 + c^2)^(3/2)*d^4) + 2 
0/3*c^2*x^2/((-d^2*x^2 + c^2)^(3/2)*d^5) + 64/15*c^3*x/((-d^2*x^2 + c^2)^( 
3/2)*d^6) + 1/3*x^2/(sqrt(-d^2*x^2 + c^2)*d^5) - 14/3*c^4/((-d^2*x^2 + c^2 
)^(3/2)*d^7) - 52/15*c*x/(sqrt(-d^2*x^2 + c^2)*d^6) - c*arcsin(d*x/c)/d^7 
+ 4/3*c^2/(sqrt(-d^2*x^2 + c^2)*d^7) + 1/3*sqrt(-d^2*x^2 + c^2)/d^7
 

Giac [F]

\[ \int \frac {x^6}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int { \frac {x^{6}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} {\left (d x + c\right )}} \,d x } \] Input:

integrate(x^6/(d*x+c)/(-d^2*x^2+c^2)^(5/2),x, algorithm="giac")
 

Output:

integrate(x^6/((-d^2*x^2 + c^2)^(5/2)*(d*x + c)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^6}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {x^6}{{\left (c^2-d^2\,x^2\right )}^{5/2}\,\left (c+d\,x\right )} \,d x \] Input:

int(x^6/((c^2 - d^2*x^2)^(5/2)*(c + d*x)),x)
 

Output:

int(x^6/((c^2 - d^2*x^2)^(5/2)*(c + d*x)), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 297, normalized size of antiderivative = 2.02 \[ \int \frac {x^6}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {-15 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{4}-15 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{3} d x +15 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c^{2} d^{2} x^{2}+15 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) c \,d^{3} x^{3}-33 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4}-33 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d x +33 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{2} x^{2}+33 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{3} x^{3}-48 c^{5}-33 c^{4} d x +87 c^{3} d^{2} x^{2}+52 c^{2} d^{3} x^{3}-38 c \,d^{4} x^{4}-15 d^{5} x^{5}}{15 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{7} \left (-d^{3} x^{3}-c \,d^{2} x^{2}+c^{2} d x +c^{3}\right )} \] Input:

int(x^6/(d*x+c)/(-d^2*x^2+c^2)^(5/2),x)
 

Output:

( - 15*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**4 - 15*sqrt(c**2 - d**2*x** 
2)*asin((d*x)/c)*c**3*d*x + 15*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c**2*d 
**2*x**2 + 15*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*c*d**3*x**3 - 33*sqrt(c 
**2 - d**2*x**2)*c**4 - 33*sqrt(c**2 - d**2*x**2)*c**3*d*x + 33*sqrt(c**2 
- d**2*x**2)*c**2*d**2*x**2 + 33*sqrt(c**2 - d**2*x**2)*c*d**3*x**3 - 48*c 
**5 - 33*c**4*d*x + 87*c**3*d**2*x**2 + 52*c**2*d**3*x**3 - 38*c*d**4*x**4 
 - 15*d**5*x**5)/(15*sqrt(c**2 - d**2*x**2)*d**7*(c**3 + c**2*d*x - c*d**2 
*x**2 - d**3*x**3))