\(\int \frac {(A+B x) (c^2-d^2 x^2)^{5/2}}{c+d x} \, dx\) [29]

Optimal result

Integrand size = 29, antiderivative size = 169 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=-\frac {c^3 (B c-6 A d) x \sqrt {c^2-d^2 x^2}}{16 d}-\frac {c (B c-6 A d) x \left (c^2-d^2 x^2\right )^{3/2}}{24 d}-\frac {(B c-6 A d) \left (c^2-d^2 x^2\right )^{5/2}}{30 d^2}-\frac {B \left (c^2-d^2 x^2\right )^{7/2}}{6 d^2 (c+d x)}-\frac {c^5 (B c-6 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{16 d^2} \] Output:

-1/16*c^3*(-6*A*d+B*c)*x*(-d^2*x^2+c^2)^(1/2)/d-1/24*c*(-6*A*d+B*c)*x*(-d^ 
2*x^2+c^2)^(3/2)/d-1/30*(-6*A*d+B*c)*(-d^2*x^2+c^2)^(5/2)/d^2-1/6*B*(-d^2* 
x^2+c^2)^(7/2)/d^2/(d*x+c)-1/16*c^5*(-6*A*d+B*c)*arctan(d*x/(-d^2*x^2+c^2) 
^(1/2))/d^2
 

Mathematica [A] (verified)

Time = 0.92 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.01 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (6 A d \left (8 c^4+25 c^3 d x-16 c^2 d^2 x^2-10 c d^3 x^3+8 d^4 x^4\right )+B \left (-48 c^5+15 c^4 d x+96 c^3 d^2 x^2-70 c^2 d^3 x^3-48 c d^4 x^4+40 d^5 x^5\right )\right )+30 c^5 (B c-6 A d) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{240 d^2} \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(6*A*d*(8*c^4 + 25*c^3*d*x - 16*c^2*d^2*x^2 - 10*c*d^ 
3*x^3 + 8*d^4*x^4) + B*(-48*c^5 + 15*c^4*d*x + 96*c^3*d^2*x^2 - 70*c^2*d^3 
*x^3 - 48*c*d^4*x^4 + 40*d^5*x^5)) + 30*c^5*(B*c - 6*A*d)*ArcTan[(d*x)/(Sq 
rt[c^2] - Sqrt[c^2 - d^2*x^2])])/(240*d^2)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.91, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {667, 676, 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.

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

\(\Big \downarrow \) 667

\(\displaystyle \int (A+B x) (c-d x) \left (c^2-d^2 x^2\right )^{3/2}dx\)

\(\Big \downarrow \) 676

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

Defintions of rubi rules used

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 667

Int[(((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*( 
x_)), x_Symbol] :> Int[(a/d + c*(x/e))*(f + g*x)^n*(a + c*x^2)^(p - 1), x] 
/; FreeQ[{a, c, d, e, f, g, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 0]
 

rule 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.38 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.01

method	result	size
risch	\(\frac {\left (40 B \,d^{5} x^{5}+48 A \,d^{5} x^{4}-48 B c \,d^{4} x^{4}-60 A c \,d^{4} x^{3}-70 B \,c^{2} d^{3} x^{3}-96 A \,c^{2} d^{3} x^{2}+96 B \,c^{3} d^{2} x^{2}+150 A \,c^{3} d^{2} x +15 B \,c^{4} d x +48 A \,c^{4} d -48 B \,c^{5}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{240 d^{2}}+\frac {c^{5} \left (6 A d -B c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{16 d \sqrt {d^{2}}}\)	\(171\)
default	\(\frac {B \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{6}+\frac {5 c^{2} \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )}{6}\right )}{d}+\frac {\left (A d -B c \right ) \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{5}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{8 d^{2}}+\frac {3 c^{2} \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )\right )}{d^{2}}\)	\(302\)

Input:

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

Output:

1/240/d^2*(40*B*d^5*x^5+48*A*d^5*x^4-48*B*c*d^4*x^4-60*A*c*d^4*x^3-70*B*c^ 
2*d^3*x^3-96*A*c^2*d^3*x^2+96*B*c^3*d^2*x^2+150*A*c^3*d^2*x+15*B*c^4*d*x+4 
8*A*c^4*d-48*B*c^5)*(-d^2*x^2+c^2)^(1/2)+1/16*c^5/d*(6*A*d-B*c)/(d^2)^(1/2 
)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=\frac {30 \, {\left (B c^{6} - 6 \, A c^{5} d\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (40 \, B d^{5} x^{5} - 48 \, B c^{5} + 48 \, A c^{4} d - 48 \, {\left (B c d^{4} - A d^{5}\right )} x^{4} - 10 \, {\left (7 \, B c^{2} d^{3} + 6 \, A c d^{4}\right )} x^{3} + 96 \, {\left (B c^{3} d^{2} - A c^{2} d^{3}\right )} x^{2} + 15 \, {\left (B c^{4} d + 10 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{240 \, d^{2}} \] Input:

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

Output:

1/240*(30*(B*c^6 - 6*A*c^5*d)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + 
(40*B*d^5*x^5 - 48*B*c^5 + 48*A*c^4*d - 48*(B*c*d^4 - A*d^5)*x^4 - 10*(7*B 
*c^2*d^3 + 6*A*c*d^4)*x^3 + 96*(B*c^3*d^2 - A*c^2*d^3)*x^2 + 15*(B*c^4*d + 
 10*A*c^3*d^2)*x)*sqrt(-d^2*x^2 + c^2))/d^2
 

Sympy [A] (verification not implemented)

Time = 2.39 (sec) , antiderivative size = 672, normalized size of antiderivative = 3.98 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx =\text {Too large to display} \] Input:

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

Output:

A*c**3*Piecewise((c**2*Piecewise((log(-2*d**2*x + 2*sqrt(-d**2)*sqrt(c**2 
- d**2*x**2))/sqrt(-d**2), Ne(c**2, 0)), (x*log(x)/sqrt(-d**2*x**2), True) 
)/2 + x*sqrt(c**2 - d**2*x**2)/2, Ne(d**2, 0)), (x*sqrt(c**2), True)) - A* 
c**2*d*Piecewise((sqrt(c**2 - d**2*x**2)*(-c**2/(3*d**2) + x**2/3), Ne(d** 
2, 0)), (x**2*sqrt(c**2)/2, True)) - A*c*d**2*Piecewise((c**4*Piecewise((l 
og(-2*d**2*x + 2*sqrt(-d**2)*sqrt(c**2 - d**2*x**2))/sqrt(-d**2), Ne(c**2, 
 0)), (x*log(x)/sqrt(-d**2*x**2), True))/(8*d**2) + sqrt(c**2 - d**2*x**2) 
*(-c**2*x/(8*d**2) + x**3/4), Ne(d**2, 0)), (x**3*sqrt(c**2)/3, True)) + A 
*d**3*Piecewise((sqrt(c**2 - d**2*x**2)*(-2*c**4/(15*d**4) - c**2*x**2/(15 
*d**2) + x**4/5), Ne(d**2, 0)), (x**4*sqrt(c**2)/4, True)) + B*c**3*Piecew 
ise((sqrt(c**2 - d**2*x**2)*(-c**2/(3*d**2) + x**2/3), Ne(d**2, 0)), (x**2 
*sqrt(c**2)/2, True)) - B*c**2*d*Piecewise((c**4*Piecewise((log(-2*d**2*x 
+ 2*sqrt(-d**2)*sqrt(c**2 - d**2*x**2))/sqrt(-d**2), Ne(c**2, 0)), (x*log( 
x)/sqrt(-d**2*x**2), True))/(8*d**2) + sqrt(c**2 - d**2*x**2)*(-c**2*x/(8* 
d**2) + x**3/4), Ne(d**2, 0)), (x**3*sqrt(c**2)/3, True)) - B*c*d**2*Piece 
wise((sqrt(c**2 - d**2*x**2)*(-2*c**4/(15*d**4) - c**2*x**2/(15*d**2) + x* 
*4/5), Ne(d**2, 0)), (x**4*sqrt(c**2)/4, True)) + B*d**3*Piecewise((c**6*P 
iecewise((log(-2*d**2*x + 2*sqrt(-d**2)*sqrt(c**2 - d**2*x**2))/sqrt(-d**2 
), Ne(c**2, 0)), (x*log(x)/sqrt(-d**2*x**2), True))/(16*d**4) + sqrt(c**2 
- d**2*x**2)*(-c**4*x/(16*d**4) - c**2*x**3/(24*d**2) + x**5/6), Ne(d**...
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.14 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.76 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=\frac {3 i \, B c^{6} \arcsin \left (\frac {d x}{c} + 2\right )}{8 \, d^{2}} - \frac {3 i \, A c^{5} \arcsin \left (\frac {d x}{c} + 2\right )}{8 \, d} + \frac {5 \, B c^{6} \arcsin \left (\frac {d x}{c}\right )}{16 \, d^{2}} + \frac {3}{8} \, \sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} A c^{3} x - \frac {3 \, \sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} B c^{4} x}{8 \, d} + \frac {5 \, \sqrt {-d^{2} x^{2} + c^{2}} B c^{4} x}{16 \, d} - \frac {3 \, \sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} B c^{5}}{4 \, d^{2}} + \frac {3 \, \sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} A c^{4}}{4 \, d} + \frac {1}{4} \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A c x - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c^{2} x}{24 \, d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B x}{6 \, d} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B c}{5 \, d^{2}} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} A}{5 \, d} \] Input:

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

Output:

3/8*I*B*c^6*arcsin(d*x/c + 2)/d^2 - 3/8*I*A*c^5*arcsin(d*x/c + 2)/d + 5/16 
*B*c^6*arcsin(d*x/c)/d^2 + 3/8*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*A*c^3*x - 3 
/8*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*B*c^4*x/d + 5/16*sqrt(-d^2*x^2 + c^2)*B 
*c^4*x/d - 3/4*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*B*c^5/d^2 + 3/4*sqrt(d^2*x^ 
2 + 4*c*d*x + 3*c^2)*A*c^4/d + 1/4*(-d^2*x^2 + c^2)^(3/2)*A*c*x - 1/24*(-d 
^2*x^2 + c^2)^(3/2)*B*c^2*x/d + 1/6*(-d^2*x^2 + c^2)^(5/2)*B*x/d - 1/5*(-d 
^2*x^2 + c^2)^(5/2)*B*c/d^2 + 1/5*(-d^2*x^2 + c^2)^(5/2)*A/d
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=-\frac {{\left (B c^{6} - 6 \, A c^{5} d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{16 \, d {\left | d \right |}} + \frac {1}{240} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, B d^{3} x - \frac {6 \, {\left (B c d^{10} - A d^{11}\right )}}{d^{8}}\right )} x - \frac {5 \, {\left (7 \, B c^{2} d^{9} + 6 \, A c d^{10}\right )}}{d^{8}}\right )} x + \frac {48 \, {\left (B c^{3} d^{8} - A c^{2} d^{9}\right )}}{d^{8}}\right )} x + \frac {15 \, {\left (B c^{4} d^{7} + 10 \, A c^{3} d^{8}\right )}}{d^{8}}\right )} x - \frac {48 \, {\left (B c^{5} d^{6} - A c^{4} d^{7}\right )}}{d^{8}}\right )} \] Input:

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

Output:

-1/16*(B*c^6 - 6*A*c^5*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) + 1/240*s 
qrt(-d^2*x^2 + c^2)*((2*((4*(5*B*d^3*x - 6*(B*c*d^10 - A*d^11)/d^8)*x - 5* 
(7*B*c^2*d^9 + 6*A*c*d^10)/d^8)*x + 48*(B*c^3*d^8 - A*c^2*d^9)/d^8)*x + 15 
*(B*c^4*d^7 + 10*A*c^3*d^8)/d^8)*x - 48*(B*c^5*d^6 - A*c^4*d^7)/d^8)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.74 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{c+d x} \, dx=\frac {90 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{5} d -15 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{6}+48 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{4} d +150 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d^{2} x -96 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{3} x^{2}-60 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{4} x^{3}+48 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{5} x^{4}-48 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{5}+15 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4} d x +96 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d^{2} x^{2}-70 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{3} x^{3}-48 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{4} x^{4}+40 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{5} x^{5}-48 a \,c^{5} d +48 b \,c^{6}}{240 d^{2}} \] Input:

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

Output:

(90*asin((d*x)/c)*a*c**5*d - 15*asin((d*x)/c)*b*c**6 + 48*sqrt(c**2 - d**2 
*x**2)*a*c**4*d + 150*sqrt(c**2 - d**2*x**2)*a*c**3*d**2*x - 96*sqrt(c**2 
- d**2*x**2)*a*c**2*d**3*x**2 - 60*sqrt(c**2 - d**2*x**2)*a*c*d**4*x**3 + 
48*sqrt(c**2 - d**2*x**2)*a*d**5*x**4 - 48*sqrt(c**2 - d**2*x**2)*b*c**5 + 
 15*sqrt(c**2 - d**2*x**2)*b*c**4*d*x + 96*sqrt(c**2 - d**2*x**2)*b*c**3*d 
**2*x**2 - 70*sqrt(c**2 - d**2*x**2)*b*c**2*d**3*x**3 - 48*sqrt(c**2 - d** 
2*x**2)*b*c*d**4*x**4 + 40*sqrt(c**2 - d**2*x**2)*b*d**5*x**5 - 48*a*c**5* 
d + 48*b*c**6)/(240*d**2)