\(\int \frac {(A+B x) (c^2-d^2 x^2)^{5/2}}{(c+d x)^5} \, dx\) [33]

Optimal result

Integrand size = 29, antiderivative size = 161 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^5} \, dx=-\frac {5 (5 B c-2 A d) (4 c-d x) \sqrt {c^2-d^2 x^2}}{6 c d^2}+\frac {2 (B c-A d) \left (c^2-d^2 x^2\right )^{5/2}}{3 d^2 (c+d x)^4}-\frac {(11 B c-5 A d) \left (c^2-d^2 x^2\right )^{5/2}}{3 c d^2 (c+d x)^3}-\frac {5 c (5 B c-2 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^2} \] Output:

-5/6*(-2*A*d+5*B*c)*(-d*x+4*c)*(-d^2*x^2+c^2)^(1/2)/c/d^2+2/3*(-A*d+B*c)*( 
-d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)^4-1/3*(-5*A*d+11*B*c)*(-d^2*x^2+c^2)^(5/2) 
/c/d^2/(d*x+c)^3-5/2*c*(-2*A*d+5*B*c)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^2
 

Mathematica [A] (verified)

Time = 0.80 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.83 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^5} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (2 A d \left (23 c^2+34 c d x+3 d^2 x^2\right )-B \left (118 c^3+161 c^2 d x+24 c d^2 x^2-3 d^3 x^3\right )\right )}{(c+d x)^2}+30 c (5 B c-2 A d) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{6 d^2} \] Input:

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

Output:

((Sqrt[c^2 - d^2*x^2]*(2*A*d*(23*c^2 + 34*c*d*x + 3*d^2*x^2) - B*(118*c^3 
+ 161*c^2*d*x + 24*c*d^2*x^2 - 3*d^3*x^3)))/(c + d*x)^2 + 30*c*(5*B*c - 2* 
A*d)*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/(6*d^2)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 167, normalized size of antiderivative = 1.04, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {671, 463, 25, 2346, 25, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 463

\(\displaystyle \frac {(5 B c-2 A d) \left (\int -\frac {7 c^2-4 d x c+d^2 x^2}{\sqrt {c^2-d^2 x^2}}dx-\frac {8 c^2 \sqrt {c^2-d^2 x^2}}{d (c+d x)}\right )}{3 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{3 c d^2 (c+d x)^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(5 B c-2 A d) \left (-\int \frac {7 c^2-4 d x c+d^2 x^2}{\sqrt {c^2-d^2 x^2}}dx-\frac {8 c^2 \sqrt {c^2-d^2 x^2}}{d (c+d x)}\right )}{3 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{3 c d^2 (c+d x)^5}\)

\(\Big \downarrow \) 2346

\(\displaystyle \frac {(5 B c-2 A d) \left (\frac {\int -\frac {c d^2 (15 c-8 d x)}{\sqrt {c^2-d^2 x^2}}dx}{2 d^2}-\frac {8 c^2 \sqrt {c^2-d^2 x^2}}{d (c+d x)}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )}{3 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{3 c d^2 (c+d x)^5}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {(5 B c-2 A d) \left (-\frac {\int \frac {c d^2 (15 c-8 d x)}{\sqrt {c^2-d^2 x^2}}dx}{2 d^2}-\frac {8 c^2 \sqrt {c^2-d^2 x^2}}{d (c+d x)}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )}{3 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{3 c d^2 (c+d x)^5}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(5 B c-2 A d) \left (-\frac {1}{2} c \int \frac {15 c-8 d x}{\sqrt {c^2-d^2 x^2}}dx-\frac {8 c^2 \sqrt {c^2-d^2 x^2}}{d (c+d x)}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )}{3 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{3 c d^2 (c+d x)^5}\)

\(\Big \downarrow \) 455

\(\displaystyle \frac {(5 B c-2 A d) \left (-\frac {1}{2} c \left (15 c \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {8 \sqrt {c^2-d^2 x^2}}{d}\right )-\frac {8 c^2 \sqrt {c^2-d^2 x^2}}{d (c+d x)}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )}{3 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{3 c d^2 (c+d x)^5}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {(5 B c-2 A d) \left (-\frac {1}{2} c \left (15 c \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {8 \sqrt {c^2-d^2 x^2}}{d}\right )-\frac {8 c^2 \sqrt {c^2-d^2 x^2}}{d (c+d x)}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )}{3 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{3 c d^2 (c+d x)^5}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {(5 B c-2 A d) \left (-\frac {1}{2} c \left (\frac {15 c \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d}+\frac {8 \sqrt {c^2-d^2 x^2}}{d}\right )-\frac {8 c^2 \sqrt {c^2-d^2 x^2}}{d (c+d x)}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )}{3 c d}+\frac {\left (c^2-d^2 x^2\right )^{7/2} (B c-A d)}{3 c d^2 (c+d x)^5}\)

Input:

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

Output:

((B*c - A*d)*(c^2 - d^2*x^2)^(7/2))/(3*c*d^2*(c + d*x)^5) + ((5*B*c - 2*A* 
d)*((x*Sqrt[c^2 - d^2*x^2])/2 - (8*c^2*Sqrt[c^2 - d^2*x^2])/(d*(c + d*x)) 
- (c*((8*Sqrt[c^2 - d^2*x^2])/d + (15*c*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]]) 
/d))/2))/(3*c*d)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 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]
 

rule 2346

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], 
e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( 
q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1))   Int[(a + b*x^2)^p*ExpandToS 
um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], 
x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.56 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.60

method	result	size
risch	\(\frac {\left (B d x +2 A d -10 B c \right ) \sqrt {-d^{2} x^{2}+c^{2}}}{2 d^{2}}+\frac {c \left (\frac {10 A d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}-\frac {25 B c \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}+\frac {8 \left (3 A d -5 B c \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{d^{2} \left (x +\frac {c}{d}\right )}+\frac {16 c^{2} \left (A d -B c \right ) \left (-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c d \left (x +\frac {c}{d}\right )^{2}}-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c^{2} \left (x +\frac {c}{d}\right )}\right )}{d^{2}}\right )}{2 d}\)	\(258\)
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}}}{c d \left (x +\frac {c}{d}\right )^{4}}-\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 {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 )}{c}\right )}{d^{5}}+\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 )^{5}}-\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 {7}{2}}}{c d \left (x +\frac {c}{d}\right )^{4}}-\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 {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 )}{c}\right )}{3 c}\right )}{d^{6}}\)	\(755\)

Input:

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

Output:

1/2*(B*d*x+2*A*d-10*B*c)/d^2*(-d^2*x^2+c^2)^(1/2)+1/2*c/d*(10*A*d/(d^2)^(1 
/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))-25*B*c/(d^2)^(1/2)*arctan(( 
d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))+8*(3*A*d-5*B*c)/d^2/(x+c/d)*(-d^2*(x+c/ 
d)^2+2*c*d*(x+c/d))^(1/2)+16*c^2*(A*d-B*c)/d^2*(-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 = 242, normalized size of antiderivative = 1.50 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^5} \, dx=-\frac {118 \, B c^{4} - 46 \, A c^{3} d + 2 \, {\left (59 \, B c^{2} d^{2} - 23 \, A c d^{3}\right )} x^{2} + 4 \, {\left (59 \, B c^{3} d - 23 \, A c^{2} d^{2}\right )} x - 30 \, {\left (5 \, B c^{4} - 2 \, A c^{3} d + {\left (5 \, B c^{2} d^{2} - 2 \, A c d^{3}\right )} x^{2} + 2 \, {\left (5 \, B c^{3} d - 2 \, A c^{2} d^{2}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (3 \, B d^{3} x^{3} - 118 \, B c^{3} + 46 \, A c^{2} d - 6 \, {\left (4 \, B c d^{2} - A d^{3}\right )} x^{2} - {\left (161 \, B c^{2} d - 68 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{6 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} \] Input:

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

Output:

-1/6*(118*B*c^4 - 46*A*c^3*d + 2*(59*B*c^2*d^2 - 23*A*c*d^3)*x^2 + 4*(59*B 
*c^3*d - 23*A*c^2*d^2)*x - 30*(5*B*c^4 - 2*A*c^3*d + (5*B*c^2*d^2 - 2*A*c* 
d^3)*x^2 + 2*(5*B*c^3*d - 2*A*c^2*d^2)*x)*arctan(-(c - sqrt(-d^2*x^2 + c^2 
))/(d*x)) - (3*B*d^3*x^3 - 118*B*c^3 + 46*A*c^2*d - 6*(4*B*c*d^2 - A*d^3)* 
x^2 - (161*B*c^2*d - 68*A*c*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]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 503 vs. \(2 (144) = 288\).

Time = 0.13 (sec) , antiderivative size = 503, normalized size of antiderivative = 3.12 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^5} \, dx=-\frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B c}{d^{6} x^{4} + 4 \, c d^{5} x^{3} + 6 \, c^{2} d^{4} x^{2} + 4 \, c^{3} d^{3} x + c^{4} d^{2}} + \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c^{2}}{3 \, {\left (d^{5} x^{3} + 3 \, c d^{4} x^{2} + 3 \, c^{2} d^{3} x + c^{3} d^{2}\right )}} + \frac {10 \, \sqrt {-d^{2} x^{2} + c^{2}} B c^{3}}{3 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} A}{d^{5} x^{4} + 4 \, c d^{4} x^{3} + 6 \, c^{2} d^{3} x^{2} + 4 \, c^{3} d^{2} x + c^{4} d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B}{2 \, {\left (d^{5} x^{3} + 3 \, c d^{4} x^{2} + 3 \, c^{2} d^{3} x + c^{3} d^{2}\right )}} - \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A c}{3 \, {\left (d^{4} x^{3} + 3 \, c d^{3} x^{2} + 3 \, c^{2} d^{2} x + c^{3} d\right )}} + \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c}{2 \, {\left (d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}\right )}} - \frac {10 \, \sqrt {-d^{2} x^{2} + c^{2}} A c^{2}}{3 \, {\left (d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d\right )}} - \frac {80 \, \sqrt {-d^{2} x^{2} + c^{2}} B c^{2}}{3 \, {\left (d^{3} x + c d^{2}\right )}} - \frac {25 \, B c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{2}} + \frac {5 \, A c \arcsin \left (\frac {d x}{c}\right )}{d} + \frac {35 \, \sqrt {-d^{2} x^{2} + c^{2}} A c}{3 \, {\left (d^{2} x + c d\right )}} \] Input:

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

Output:

-(-d^2*x^2 + c^2)^(5/2)*B*c/(d^6*x^4 + 4*c*d^5*x^3 + 6*c^2*d^4*x^2 + 4*c^3 
*d^3*x + c^4*d^2) + 5/3*(-d^2*x^2 + c^2)^(3/2)*B*c^2/(d^5*x^3 + 3*c*d^4*x^ 
2 + 3*c^2*d^3*x + c^3*d^2) + 10/3*sqrt(-d^2*x^2 + c^2)*B*c^3/(d^4*x^2 + 2* 
c*d^3*x + c^2*d^2) + (-d^2*x^2 + c^2)^(5/2)*A/(d^5*x^4 + 4*c*d^4*x^3 + 6*c 
^2*d^3*x^2 + 4*c^3*d^2*x + c^4*d) + 1/2*(-d^2*x^2 + c^2)^(5/2)*B/(d^5*x^3 
+ 3*c*d^4*x^2 + 3*c^2*d^3*x + c^3*d^2) - 5/3*(-d^2*x^2 + c^2)^(3/2)*A*c/(d 
^4*x^3 + 3*c*d^3*x^2 + 3*c^2*d^2*x + c^3*d) + 5/2*(-d^2*x^2 + c^2)^(3/2)*B 
*c/(d^4*x^2 + 2*c*d^3*x + c^2*d^2) - 10/3*sqrt(-d^2*x^2 + c^2)*A*c^2/(d^3* 
x^2 + 2*c*d^2*x + c^2*d) - 80/3*sqrt(-d^2*x^2 + c^2)*B*c^2/(d^3*x + c*d^2) 
 - 25/2*B*c^2*arcsin(d*x/c)/d^2 + 5*A*c*arcsin(d*x/c)/d + 35/3*sqrt(-d^2*x 
^2 + c^2)*A*c/(d^2*x + c*d)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 344 vs. \(2 (144) = 288\).

Time = 0.19 (sec) , antiderivative size = 344, normalized size of antiderivative = 2.14 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^5} \, dx=\frac {{\left (16 \, B c^{3} d^{3} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 16 \, A c^{2} d^{4} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 192 \, B c^{3} d^{3} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + 96 \, A c^{2} d^{4} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + 60 \, {\left (5 \, B c^{3} d^{3} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 2 \, A c^{2} d^{4} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )\right )} \arctan \left (\sqrt {\frac {2 \, c}{d x + c} - 1}\right ) - \frac {3 \, {\left (11 \, B c^{3} d^{3} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 2 \, A c^{2} d^{4} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + 9 \, B c^{3} d^{3} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 2 \, A c^{2} d^{4} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )\right )} {\left (d x + c\right )}^{2}}{c^{2}}\right )} {\left | d \right |}}{12 \, c d^{6}} \] Input:

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

Output:

1/12*(16*B*c^3*d^3*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/(d*x + c))*sgn(d) - 16* 
A*c^2*d^4*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/(d*x + c))*sgn(d) - 192*B*c^3*d^ 
3*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) + 96*A*c^2*d^4*sqrt(2*c/ 
(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) + 60*(5*B*c^3*d^3*sgn(1/(d*x + c))* 
sgn(d) - 2*A*c^2*d^4*sgn(1/(d*x + c))*sgn(d))*arctan(sqrt(2*c/(d*x + c) - 
1)) - 3*(11*B*c^3*d^3*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/(d*x + c))*sgn(d) - 
2*A*c^2*d^4*(2*c/(d*x + c) - 1)^(3/2)*sgn(1/(d*x + c))*sgn(d) + 9*B*c^3*d^ 
3*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d) - 2*A*c^2*d^4*sqrt(2*c/( 
d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d))*(d*x + c)^2/c^2)*abs(d)/(c*d^6)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 491, normalized size of antiderivative = 3.05 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^5} \, dx=\frac {-30 \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{3} x^{2}+75 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d^{2} x^{2}-30 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{3} d -16 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d -6 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{3} x^{2}-3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{3} x^{3}-75 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3}-38 a \,c^{2} d^{2} x -92 a c \,d^{3} x^{2}+93 b \,c^{3} d x +205 b \,c^{2} d^{2} x^{2}+27 b c \,d^{3} x^{3}+30 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d -60 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d^{2} x +150 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3} d x -6 a \,d^{4} x^{3}-3 b \,d^{4} x^{4}-50 b \,c^{4}+75 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{4}+50 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3}+16 a \,c^{3} d -38 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{2} x +93 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d x +24 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{2} x^{2}+30 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{2} x -75 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d x}{6 d^{2} \left (\sqrt {-d^{2} x^{2}+c^{2}}\, c +\sqrt {-d^{2} x^{2}+c^{2}}\, d x -c^{2}-2 c d x -d^{2} x^{2}\right )} \] Input:

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

Output:

(30*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*a*c**2*d + 30*sqrt(c**2 - d**2*x* 
*2)*asin((d*x)/c)*a*c*d**2*x - 75*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*b*c 
**3 - 75*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*b*c**2*d*x - 30*asin((d*x)/c 
)*a*c**3*d - 60*asin((d*x)/c)*a*c**2*d**2*x - 30*asin((d*x)/c)*a*c*d**3*x* 
*2 + 75*asin((d*x)/c)*b*c**4 + 150*asin((d*x)/c)*b*c**3*d*x + 75*asin((d*x 
)/c)*b*c**2*d**2*x**2 - 16*sqrt(c**2 - d**2*x**2)*a*c**2*d - 38*sqrt(c**2 
- d**2*x**2)*a*c*d**2*x - 6*sqrt(c**2 - d**2*x**2)*a*d**3*x**2 + 50*sqrt(c 
**2 - d**2*x**2)*b*c**3 + 93*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 - 3*sqrt(c**2 - d**2*x**2)*b*d**3*x**3 + 16 
*a*c**3*d - 38*a*c**2*d**2*x - 92*a*c*d**3*x**2 - 6*a*d**4*x**3 - 50*b*c** 
4 + 93*b*c**3*d*x + 205*b*c**2*d**2*x**2 + 27*b*c*d**3*x**3 - 3*b*d**4*x** 
4)/(6*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))