\(\int \frac {(A+B x) (c^2-d^2 x^2)^{3/2}}{(c+d x)^4} \, dx\) [19]

Optimal result

Integrand size = 29, antiderivative size = 132 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^4} \, dx=-\frac {B \sqrt {c^2-d^2 x^2}}{d^2}-\frac {2 (3 B c-A d) \sqrt {c^2-d^2 x^2}}{d^2 (c+d x)}+\frac {2 (B c-A d) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^2 (c+d x)^3}-\frac {(4 B c-A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^4} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (-19 B c^2+4 A c d-26 B c d x+8 A d^2 x-3 B d^2 x^2\right )}{3 d^2 (c+d x)^2}+\frac {\sqrt {-d^2} (-4 B c+A d) \log \left (-\sqrt {-d^2} x+\sqrt {c^2-d^2 x^2}\right )}{d^3} \] Input:

Integrate[((A + B*x)*(c^2 - d^2*x^2)^(3/2))/(c + d*x)^4,x]
 

Output:

(Sqrt[c^2 - d^2*x^2]*(-19*B*c^2 + 4*A*c*d - 26*B*c*d*x + 8*A*d^2*x - 3*B*d 
^2*x^2))/(3*d^2*(c + d*x)^2) + (Sqrt[-d^2]*(-4*B*c + A*d)*Log[-(Sqrt[-d^2] 
*x) + Sqrt[c^2 - d^2*x^2]])/d^3
 

Rubi [A] (verified)

Time = 0.42 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {671, 463, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 463

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

Int[((A + B*x)*(c^2 - d^2*x^2)^(3/2))/(c + d*x)^4,x]
 

Output:

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

Defintions of rubi rules used

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 463

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-(-c)^(-n - 2))*d^(2*n + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b^(n + 2)*(c + d*x 
))), x] - Simp[d^(2*n + 2)/b^(n + 1)   Int[(1/Sqrt[a + b*x^2])*ExpandToSum[ 
(2^(-n - 1)*(-c)^(-n - 1) - (-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; F 
reeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[n, 0] && EqQ[n + p, 
-3/2]
 

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 = 0.46 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.63

method	result	size
risch	\(-\frac {B \sqrt {-d^{2} x^{2}+c^{2}}}{d^{2}}+\frac {\left (A d -4 B c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d \sqrt {d^{2}}}+\frac {4 \left (A d -2 B c \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{d^{3} \left (x +\frac {c}{d}\right )}+\frac {4 c^{2} \left (A d -B c \right ) \left (-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c d \left (x +\frac {c}{d}\right )^{2}}-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c^{2} \left (x +\frac {c}{d}\right )}\right )}{d^{3}}\)	\(215\)
default	\(\frac {B \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{c d \left (x +\frac {c}{d}\right )^{3}}-\frac {2 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}+\frac {3 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )\right )}{c}\right )}{c}\right )}{d^{4}}+\frac {\left (A d -B c \right ) \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{3 c d \left (x +\frac {c}{d}\right )^{4}}-\frac {d \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{c d \left (x +\frac {c}{d}\right )^{3}}-\frac {2 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}+\frac {3 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )\right )}{c}\right )}{c}\right )}{3 c}\right )}{d^{5}}\)	\(539\)

Input:

int((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^4,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.56 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^4} \, dx=-\frac {19 \, B c^{3} - 4 \, A c^{2} d + {\left (19 \, B c d^{2} - 4 \, A d^{3}\right )} x^{2} + 2 \, {\left (19 \, B c^{2} d - 4 \, A c d^{2}\right )} x - 6 \, {\left (4 \, B c^{3} - A c^{2} d + {\left (4 \, B c d^{2} - A d^{3}\right )} x^{2} + 2 \, {\left (4 \, B c^{2} d - A c d^{2}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (3 \, B d^{2} x^{2} + 19 \, B c^{2} - 4 \, A c d + 2 \, {\left (13 \, B c d - 4 \, A d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{3 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^4,x, algorithm="fricas")
 

Output:

-1/3*(19*B*c^3 - 4*A*c^2*d + (19*B*c*d^2 - 4*A*d^3)*x^2 + 2*(19*B*c^2*d - 
4*A*c*d^2)*x - 6*(4*B*c^3 - A*c^2*d + (4*B*c*d^2 - A*d^3)*x^2 + 2*(4*B*c^2 
*d - A*c*d^2)*x)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + (3*B*d^2*x^2 
+ 19*B*c^2 - 4*A*c*d + 2*(13*B*c*d - 4*A*d^2)*x)*sqrt(-d^2*x^2 + c^2))/(d^ 
4*x^2 + 2*c*d^3*x + c^2*d^2)
 

Sympy [F]

\[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^4} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {3}{2}} \left (A + B x\right )}{\left (c + d x\right )^{4}}\, dx \] Input:

integrate((B*x+A)*(-d**2*x**2+c**2)**(3/2)/(d*x+c)**4,x)
 

Output:

Integral((-(-c + d*x)*(c + d*x))**(3/2)*(A + B*x)/(c + d*x)**4, x)
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 313 vs. \(2 (122) = 244\).

Time = 0.13 (sec) , antiderivative size = 313, normalized size of antiderivative = 2.37 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^4} \, dx=\frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c}{3 \, {\left (d^{5} x^{3} + 3 \, c d^{4} x^{2} + 3 \, c^{2} d^{3} x + c^{3} d^{2}\right )}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B c^{2}}{3 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A}{3 \, {\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )}} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B}{d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} A c}{3 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}} - \frac {25 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{3 \, {\left (d^{3} x + c d^{2}\right )}} - \frac {4 \, B c \arcsin \left (\frac {d x}{c}\right )}{d^{2}} + \frac {A \arcsin \left (\frac {d x}{c}\right )}{d} + \frac {7 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{3 \, {\left (d^{2} x + c d\right )}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^4,x, algorithm="maxima")
 

Output:

1/3*(-d^2*x^2 + c^2)^(3/2)*B*c/(d^5*x^3 + 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3* 
d^2) + 2/3*sqrt(-d^2*x^2 + c^2)*B*c^2/(d^4*x^2 + 2*c*d^3*x + c^2*d^2) - 1/ 
3*(-d^2*x^2 + c^2)^(3/2)*A/(d^4*x^3 + 3*c*d^3*x^2 + 3*c^2*d^2*x + c^3*d) + 
 (-d^2*x^2 + c^2)^(3/2)*B/(d^4*x^2 + 2*c*d^3*x + c^2*d^2) - 2/3*sqrt(-d^2* 
x^2 + c^2)*A*c/(d^3*x^2 + 2*c*d^2*x + c^2*d) - 25/3*sqrt(-d^2*x^2 + c^2)*B 
*c/(d^3*x + c*d^2) - 4*B*c*arcsin(d*x/c)/d^2 + A*arcsin(d*x/c)/d + 7/3*sqr 
t(-d^2*x^2 + c^2)*A/(d^2*x + c*d)
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.47 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^4} \, dx=-\frac {{\left (4 \, B c - A d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{d {\left | d \right |}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} B}{d^{2}} + \frac {8 \, {\left (4 \, B c - A d + \frac {9 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} B c}{d^{2} x} - \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} A}{d x} + \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} B c}{d^{4} x^{2}}\right )}}{3 \, d {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )}^{3} {\left | d \right |}} \] Input:

integrate((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^4,x, algorithm="giac")
 

Output:

-(4*B*c - A*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) - sqrt(-d^2*x^2 + c^ 
2)*B/d^2 + 8/3*(4*B*c - A*d + 9*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*B*c/(d 
^2*x) - 3*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*A/(d*x) + 3*(c*d + sqrt(-d^2 
*x^2 + c^2)*abs(d))^2*B*c/(d^4*x^2))/(d*((c*d + sqrt(-d^2*x^2 + c^2)*abs(d 
))/(d^2*x) + 1)^3*abs(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^4} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{3/2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^4} \,d x \] Input:

int(((c^2 - d^2*x^2)^(3/2)*(A + B*x))/(c + d*x)^4,x)
 

Output:

int(((c^2 - d^2*x^2)^(3/2)*(A + B*x))/(c + d*x)^4, x)
 

Reduce [B] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 382, normalized size of antiderivative = 2.89 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^4} \, dx=\frac {3 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a c d +3 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a \,d^{2} x -12 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2}-12 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b c d x -3 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d -6 \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{2} x -3 \mathit {asin} \left (\frac {d x}{c}\right ) a \,d^{3} x^{2}+12 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3}+24 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d x +12 \mathit {asin} \left (\frac {d x}{c}\right ) b c \,d^{2} x^{2}-4 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{2} x +8 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2}+15 \sqrt {-d^{2} x^{2}+c^{2}}\, b c d x +3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{2} x^{2}-4 a c \,d^{2} x -12 a \,d^{3} x^{2}-8 b \,c^{3}+15 b \,c^{2} d x +34 b c \,d^{2} x^{2}+3 b \,d^{3} x^{3}}{3 d^{2} \left (\sqrt {-d^{2} x^{2}+c^{2}}\, c +\sqrt {-d^{2} x^{2}+c^{2}}\, d x -c^{2}-2 c d x -d^{2} x^{2}\right )} \] Input:

int((B*x+A)*(-d^2*x^2+c^2)^(3/2)/(d*x+c)^4,x)
 

Output:

(3*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*a*c*d + 3*sqrt(c**2 - d**2*x**2)*a 
sin((d*x)/c)*a*d**2*x - 12*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*b*c**2 - 1 
2*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*b*c*d*x - 3*asin((d*x)/c)*a*c**2*d 
- 6*asin((d*x)/c)*a*c*d**2*x - 3*asin((d*x)/c)*a*d**3*x**2 + 12*asin((d*x) 
/c)*b*c**3 + 24*asin((d*x)/c)*b*c**2*d*x + 12*asin((d*x)/c)*b*c*d**2*x**2 
- 4*sqrt(c**2 - d**2*x**2)*a*d**2*x + 8*sqrt(c**2 - d**2*x**2)*b*c**2 + 15 
*sqrt(c**2 - d**2*x**2)*b*c*d*x + 3*sqrt(c**2 - d**2*x**2)*b*d**2*x**2 - 4 
*a*c*d**2*x - 12*a*d**3*x**2 - 8*b*c**3 + 15*b*c**2*d*x + 34*b*c*d**2*x**2 
 + 3*b*d**3*x**3)/(3*d**2*(sqrt(c**2 - d**2*x**2)*c + sqrt(c**2 - d**2*x** 
2)*d*x - c**2 - 2*c*d*x - d**2*x**2))