\(\int \frac {(A+B x) (c^2-d^2 x^2)^{5/2}}{(c+d x)^3} \, dx\) [31]

Optimal result

Integrand size = 29, antiderivative size = 156 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^3} \, dx=-\frac {5 c (3 B c-4 A d) x \sqrt {c^2-d^2 x^2}}{8 d}-\frac {(3 B c-4 A d) (8 c-3 d x) \left (c^2-d^2 x^2\right )^{3/2}}{12 c d^2}-\frac {(B c-A d) \left (c^2-d^2 x^2\right )^{7/2}}{c d^2 (c+d x)^3}-\frac {5 c^3 (3 B c-4 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d^2} \] Output:

-5/8*c*(-4*A*d+3*B*c)*x*(-d^2*x^2+c^2)^(1/2)/d-1/12*(-4*A*d+3*B*c)*(-3*d*x 
+8*c)*(-d^2*x^2+c^2)^(3/2)/c/d^2-(-A*d+B*c)*(-d^2*x^2+c^2)^(7/2)/c/d^2/(d* 
x+c)^3-5/8*c^3*(-4*A*d+3*B*c)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^2
 

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.81 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^3} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (4 A d \left (22 c^2-9 c d x+2 d^2 x^2\right )+B \left (-72 c^3+45 c^2 d x-24 c d^2 x^2+6 d^3 x^3\right )\right )+30 c^3 (3 B c-4 A d) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{24 d^2} \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(4*A*d*(22*c^2 - 9*c*d*x + 2*d^2*x^2) + B*(-72*c^3 + 
45*c^2*d*x - 24*c*d^2*x^2 + 6*d^3*x^3)) + 30*c^3*(3*B*c - 4*A*d)*ArcTan[(d 
*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/(24*d^2)
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {671, 464, 469, 455, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 464

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

\(\Big \downarrow \) 469

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

-(((B*c - A*d)*(c^2 - d^2*x^2)^(7/2))/(c*d^2*(c + d*x)^3)) - ((3*B*c - 4*A 
*d)*(((c - d*x)*(c^2 - d^2*x^2)^(3/2))/(4*d) + (5*c*((c^2 - d^2*x^2)^(3/2) 
/(3*d) + c*((x*Sqrt[c^2 - d^2*x^2])/2 + (c^2*ArcTan[(d*x)/Sqrt[c^2 - d^2*x 
^2]])/(2*d))))/4))/(c*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 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 464

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

rule 469

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

rule 671

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[(d*g - e*f)*(d + e*x)^m*((a + c*x^2)^(p + 1)/(2*c*d*(m 
 + p + 1))), x] + Simp[(m*(g*c*d + c*e*f) + 2*e*c*f*(p + 1))/(e*(2*c*d)*(m 
+ p + 1))   Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, 
e, f, g, m, p}, x] && EqQ[c*d^2 + a*e^2, 0] && ((LtQ[m, -1] &&  !IGtQ[m + p 
 + 1, 0]) || (LtQ[m, 0] && LtQ[p, -1]) || EqQ[m + 2*p + 2, 0]) && NeQ[m + p 
 + 1, 0]
 

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.79

method	result	size
risch	\(\frac {\left (6 B \,d^{3} x^{3}+8 A \,d^{3} x^{2}-24 B c \,d^{2} x^{2}-36 A c \,d^{2} x +45 B \,c^{2} d x +88 A \,c^{2} d -72 B \,c^{3}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{24 d^{2}}+\frac {5 c^{3} \left (4 A d -3 B c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{8 d \sqrt {d^{2}}}\)	\(123\)
default	\(\frac {B \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{3 c d \left (x +\frac {c}{d}\right )^{2}}+\frac {5 d \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 )}{3 c}\right )}{d^{3}}+\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 {7}{2}}}{c d \left (x +\frac {c}{d}\right )^{3}}+\frac {4 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{3 c d \left (x +\frac {c}{d}\right )^{2}}+\frac {5 d \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 )}{3 c}\right )}{c}\right )}{d^{4}}\)	\(548\)

Input:

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

Output:

1/24/d^2*(6*B*d^3*x^3+8*A*d^3*x^2-24*B*c*d^2*x^2-36*A*c*d^2*x+45*B*c^2*d*x 
+88*A*c^2*d-72*B*c^3)*(-d^2*x^2+c^2)^(1/2)+5/8*c^3/d*(4*A*d-3*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 = 0.08 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.79 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^3} \, dx=\frac {30 \, {\left (3 \, B c^{4} - 4 \, A c^{3} d\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (6 \, B d^{3} x^{3} - 72 \, B c^{3} + 88 \, A c^{2} d - 8 \, {\left (3 \, B c d^{2} - A d^{3}\right )} x^{2} + 9 \, {\left (5 \, B c^{2} d - 4 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{24 \, d^{2}} \] Input:

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

Output:

1/24*(30*(3*B*c^4 - 4*A*c^3*d)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + 
 (6*B*d^3*x^3 - 72*B*c^3 + 88*A*c^2*d - 8*(3*B*c*d^2 - A*d^3)*x^2 + 9*(5*B 
*c^2*d - 4*A*c*d^2)*x)*sqrt(-d^2*x^2 + c^2))/d^2
 

Sympy [F]

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

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

Output:

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

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.12 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.24 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^3} \, dx=-\frac {5 i \, B c^{4} \arcsin \left (\frac {d x}{c} + 2\right )}{8 \, d^{2}} - \frac {5 \, B c^{4} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{2}} + \frac {5 \, A c^{3} \arcsin \left (\frac {d x}{c}\right )}{2 \, d} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B c}{3 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} - \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c^{2}}{6 \, {\left (d^{3} x + c d^{2}\right )}} + \frac {5 \, \sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} B c^{2} x}{8 \, d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} A}{3 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B}{4 \, {\left (d^{3} x + c d^{2}\right )}} + \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A c}{6 \, {\left (d^{2} x + c d\right )}} + \frac {5 \, \sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} B c^{3}}{4 \, d^{2}} - \frac {5 \, \sqrt {-d^{2} x^{2} + c^{2}} B c^{3}}{2 \, d^{2}} + \frac {5 \, \sqrt {-d^{2} x^{2} + c^{2}} A c^{2}}{2 \, d} + \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c}{12 \, d^{2}} \] Input:

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

Output:

-5/8*I*B*c^4*arcsin(d*x/c + 2)/d^2 - 5/2*B*c^4*arcsin(d*x/c)/d^2 + 5/2*A*c 
^3*arcsin(d*x/c)/d - 1/3*(-d^2*x^2 + c^2)^(5/2)*B*c/(d^4*x^2 + 2*c*d^3*x + 
 c^2*d^2) - 5/6*(-d^2*x^2 + c^2)^(3/2)*B*c^2/(d^3*x + c*d^2) + 5/8*sqrt(d^ 
2*x^2 + 4*c*d*x + 3*c^2)*B*c^2*x/d + 1/3*(-d^2*x^2 + c^2)^(5/2)*A/(d^3*x^2 
 + 2*c*d^2*x + c^2*d) + 1/4*(-d^2*x^2 + c^2)^(5/2)*B/(d^3*x + c*d^2) + 5/6 
*(-d^2*x^2 + c^2)^(3/2)*A*c/(d^2*x + c*d) + 5/4*sqrt(d^2*x^2 + 4*c*d*x + 3 
*c^2)*B*c^3/d^2 - 5/2*sqrt(-d^2*x^2 + c^2)*B*c^3/d^2 + 5/2*sqrt(-d^2*x^2 + 
 c^2)*A*c^2/d + 5/12*(-d^2*x^2 + c^2)^(3/2)*B*c/d^2
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^3} \, dx=-\frac {5 \, {\left (3 \, B c^{4} - 4 \, A c^{3} d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{8 \, d {\left | d \right |}} + \frac {1}{24} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left (3 \, B d x - \frac {4 \, {\left (3 \, B c d^{4} - A d^{5}\right )}}{d^{4}}\right )} x + \frac {9 \, {\left (5 \, B c^{2} d^{3} - 4 \, A c d^{4}\right )}}{d^{4}}\right )} x - \frac {8 \, {\left (9 \, B c^{3} d^{2} - 11 \, A c^{2} d^{3}\right )}}{d^{4}}\right )} \] Input:

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

Output:

-5/8*(3*B*c^4 - 4*A*c^3*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) + 1/24*s 
qrt(-d^2*x^2 + c^2)*((2*(3*B*d*x - 4*(3*B*c*d^4 - A*d^5)/d^4)*x + 9*(5*B*c 
^2*d^3 - 4*A*c*d^4)/d^4)*x - 8*(9*B*c^3*d^2 - 11*A*c^2*d^3)/d^4)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.24 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^3} \, dx=\frac {60 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{3} d -45 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{4}+88 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d -36 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{2} x +8 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{3} x^{2}-72 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3}+45 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d x -24 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{2} x^{2}+6 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{3} x^{3}-88 a \,c^{3} d +72 b \,c^{4}}{24 d^{2}} \] Input:

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

Output:

(60*asin((d*x)/c)*a*c**3*d - 45*asin((d*x)/c)*b*c**4 + 88*sqrt(c**2 - d**2 
*x**2)*a*c**2*d - 36*sqrt(c**2 - d**2*x**2)*a*c*d**2*x + 8*sqrt(c**2 - d** 
2*x**2)*a*d**3*x**2 - 72*sqrt(c**2 - d**2*x**2)*b*c**3 + 45*sqrt(c**2 - d* 
*2*x**2)*b*c**2*d*x - 24*sqrt(c**2 - d**2*x**2)*b*c*d**2*x**2 + 6*sqrt(c** 
2 - d**2*x**2)*b*d**3*x**3 - 88*a*c**3*d + 72*b*c**4)/(24*d**2)