\(\int \frac {(A+B x) (c^2-d^2 x^2)^{5/2}}{(c+d x)^2} \, dx\) [30]

Optimal result

Integrand size = 29, antiderivative size = 171 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=-\frac {c^2 (2 B c-5 A d) x \sqrt {c^2-d^2 x^2}}{8 d}-\frac {(2 B c-5 A d) x \left (c^2-d^2 x^2\right )^{3/2}}{12 d}+\frac {B \left (c^2-d^2 x^2\right )^{5/2}}{5 d^2}-\frac {2 (B c-A d) \left (c^2-d^2 x^2\right )^{5/2}}{3 d^2 (c+d x)}-\frac {c^4 (2 B c-5 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.61 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\frac {d \sqrt {c^2-d^2 x^2} \left (5 A d \left (16 c^3+9 c^2 d x-16 c d^2 x^2+6 d^3 x^3\right )+B \left (-56 c^4+30 c^3 d x+32 c^2 d^2 x^2-60 c d^3 x^3+24 d^4 x^4\right )\right )+15 c^4 \sqrt {-d^2} (-2 B c+5 A d) \log \left (-\sqrt {-d^2} x+\sqrt {c^2-d^2 x^2}\right )}{120 d^3} \] Input:

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

Output:

(d*Sqrt[c^2 - d^2*x^2]*(5*A*d*(16*c^3 + 9*c^2*d*x - 16*c*d^2*x^2 + 6*d^3*x 
^3) + B*(-56*c^4 + 30*c^3*d*x + 32*c^2*d^2*x^2 - 60*c*d^3*x^3 + 24*d^4*x^4 
)) + 15*c^4*Sqrt[-d^2]*(-2*B*c + 5*A*d)*Log[-(Sqrt[-d^2]*x) + Sqrt[c^2 - d 
^2*x^2]])/(120*d^3)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {671, 466, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

-1/3*((B*c - A*d)*(c^2 - d^2*x^2)^(7/2))/(c*d^2*(c + d*x)^2) - ((2*B*c - 5 
*A*d)*((c^2 - d^2*x^2)^(5/2)/(5*d) + c*((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)))/(3*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 466

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(c + d*x)^(n + 1)*((a + b*x^2)^p/(d*(n + 2*p + 1))), x] - Simp[2*b*c*(p/(d^ 
2*(n + 2*p + 1)))   Int[(c + d*x)^(n + 1)*(a + b*x^2)^(p - 1), x], x] /; Fr 
eeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[p, 0] && (LeQ[-2, n, 0 
] || EqQ[n + p + 1, 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 = 0.40 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.86

method	result	size
risch	\(\frac {\left (24 B \,d^{4} x^{4}+30 A \,d^{4} x^{3}-60 B c \,d^{3} x^{3}-80 A c \,d^{3} x^{2}+32 x^{2} c^{2} B \,d^{2}+45 A \,c^{2} d^{2} x +30 B \,c^{3} d x +80 A \,c^{3} d -56 B \,c^{4}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{120 d^{2}}+\frac {c^{4} \left (5 A d -2 B c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{8 d \sqrt {d^{2}}}\)	\(147\)
default	\(\frac {B \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{5}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{8 d^{2}}+\frac {3 c^{2} \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )\right )}{d^{2}}+\frac {\left (A d -B c \right ) \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{3 c d \left (x +\frac {c}{d}\right )^{2}}+\frac {5 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{5}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{8 d^{2}}+\frac {3 c^{2} \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )\right )}{3 c}\right )}{d^{3}}\)	\(445\)

Input:

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

Output:

1/120/d^2*(24*B*d^4*x^4+30*A*d^4*x^3-60*B*c*d^3*x^3-80*A*c*d^3*x^2+32*B*c^ 
2*d^2*x^2+45*A*c^2*d^2*x+30*B*c^3*d*x+80*A*c^3*d-56*B*c^4)*(-d^2*x^2+c^2)^ 
(1/2)+1/8*c^4/d*(5*A*d-2*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 = 148, normalized size of antiderivative = 0.87 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\frac {30 \, {\left (2 \, B c^{5} - 5 \, A c^{4} d\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (24 \, B d^{4} x^{4} - 56 \, B c^{4} + 80 \, A c^{3} d - 30 \, {\left (2 \, B c d^{3} - A d^{4}\right )} x^{3} + 16 \, {\left (2 \, B c^{2} d^{2} - 5 \, A c d^{3}\right )} x^{2} + 15 \, {\left (2 \, B c^{3} d + 3 \, A c^{2} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{120 \, d^{2}} \] Input:

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

Output:

1/120*(30*(2*B*c^5 - 5*A*c^4*d)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) 
+ (24*B*d^4*x^4 - 56*B*c^4 + 80*A*c^3*d - 30*(2*B*c*d^3 - A*d^4)*x^3 + 16* 
(2*B*c^2*d^2 - 5*A*c*d^3)*x^2 + 15*(2*B*c^3*d + 3*A*c^2*d^2)*x)*sqrt(-d^2* 
x^2 + c^2))/d^2
 

Sympy [A] (verification not implemented)

Time = 3.02 (sec) , antiderivative size = 481, normalized size of antiderivative = 2.81 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx =\text {Too large to display} \] Input:

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

Output:

A*c**2*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)) - 2* 
A*c*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*d**2*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* 
*2*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)) - 2*B*c*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*d**2 
*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))
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 297, normalized size of antiderivative = 1.74 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\frac {i \, B c^{5} \arcsin \left (\frac {d x}{c} + 2\right )}{4 \, d^{2}} - \frac {5 i \, A c^{4} \arcsin \left (\frac {d x}{c} + 2\right )}{8 \, d} + \frac {5}{8} \, \sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} A c^{2} x - \frac {\sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} B c^{3} x}{4 \, d} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B c}{4 \, {\left (d^{3} x + c d^{2}\right )}} - \frac {\sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} B c^{4}}{2 \, d^{2}} + \frac {5 \, \sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} A c^{3}}{4 \, d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c x}{4 \, d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} A}{4 \, {\left (d^{2} x + c d\right )}} - \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c^{2}}{12 \, d^{2}} + \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A c}{12 \, d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B}{5 \, d^{2}} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 409 vs. \(2 (151) = 302\).

Time = 0.19 (sec) , antiderivative size = 409, normalized size of antiderivative = 2.39 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\frac {{\left (480 \, {\left (2 \, B c^{6} d^{6} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 5 \, A c^{5} d^{7} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )\right )} \arctan \left (\sqrt {\frac {2 \, c}{d x + c} - 1}\right ) + \frac {{\left (30 \, B c^{6} d^{6} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {9}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 75 \, A c^{5} d^{7} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {9}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 500 \, B c^{6} d^{6} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + 290 \, A c^{5} d^{7} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 256 \, B c^{6} d^{6} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + 640 \, A c^{5} d^{7} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 140 \, B c^{6} d^{6} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + 350 \, A c^{5} d^{7} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 30 \, B c^{6} d^{6} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + 75 \, A c^{5} d^{7} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )\right )} {\left (d x + c\right )}^{5}}{c^{5}}\right )} {\left | d \right |}}{1920 \, c d^{9}} \] Input:

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

Output:

1/1920*(480*(2*B*c^6*d^6*sgn(1/(d*x + c))*sgn(d) - 5*A*c^5*d^7*sgn(1/(d*x 
+ c))*sgn(d))*arctan(sqrt(2*c/(d*x + c) - 1)) + (30*B*c^6*d^6*(2*c/(d*x + 
c) - 1)^(9/2)*sgn(1/(d*x + c))*sgn(d) - 75*A*c^5*d^7*(2*c/(d*x + c) - 1)^( 
9/2)*sgn(1/(d*x + c))*sgn(d) - 500*B*c^6*d^6*(2*c/(d*x + c) - 1)^(7/2)*sgn 
(1/(d*x + c))*sgn(d) + 290*A*c^5*d^7*(2*c/(d*x + c) - 1)^(7/2)*sgn(1/(d*x 
+ c))*sgn(d) - 256*B*c^6*d^6*(2*c/(d*x + c) - 1)^(5/2)*sgn(1/(d*x + c))*sg 
n(d) + 640*A*c^5*d^7*(2*c/(d*x + c) - 1)^(5/2)*sgn(1/(d*x + c))*sgn(d) - 1 
40*B*c^6*d^6*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/(d*x + c))*sgn(d) + 350*A*c^5 
*d^7*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/(d*x + c))*sgn(d) - 30*B*c^6*d^6*sqrt 
(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) + 75*A*c^5*d^7*sqrt(2*c/(d*x + 
 c) - 1)*sgn(1/(d*x + c))*sgn(d))*(d*x + c)^5/c^5)*abs(d)/(c*d^9)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.43 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\frac {75 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{4} d -30 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{5}+80 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d +45 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{2} x -80 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{3} x^{2}+30 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{4} x^{3}-56 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4}+30 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d x +32 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{2} x^{2}-60 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{3} x^{3}+24 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{4} x^{4}-80 a \,c^{4} d +56 b \,c^{5}}{120 d^{2}} \] Input:

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

Output:

(75*asin((d*x)/c)*a*c**4*d - 30*asin((d*x)/c)*b*c**5 + 80*sqrt(c**2 - d**2 
*x**2)*a*c**3*d + 45*sqrt(c**2 - d**2*x**2)*a*c**2*d**2*x - 80*sqrt(c**2 - 
 d**2*x**2)*a*c*d**3*x**2 + 30*sqrt(c**2 - d**2*x**2)*a*d**4*x**3 - 56*sqr 
t(c**2 - d**2*x**2)*b*c**4 + 30*sqrt(c**2 - d**2*x**2)*b*c**3*d*x + 32*sqr 
t(c**2 - d**2*x**2)*b*c**2*d**2*x**2 - 60*sqrt(c**2 - d**2*x**2)*b*c*d**3* 
x**3 + 24*sqrt(c**2 - d**2*x**2)*b*d**4*x**4 - 80*a*c**4*d + 56*b*c**5)/(1 
20*d**2)