\(\int \frac {(A+B x) (c+d x)^5}{(c^2-d^2 x^2)^{5/2}} \, dx\) [55]

Optimal result

Integrand size = 29, antiderivative size = 153 \[ \int \frac {(A+B x) (c+d x)^5}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {2 (B c+A d) (c+d x)^4}{3 d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {2 (11 B c+5 A d) (c+d x)^2}{3 d^2 \sqrt {c^2-d^2 x^2}}-\frac {(2 (13 B c+5 A d)+B d x) \sqrt {c^2-d^2 x^2}}{2 d^2}+\frac {5 c (5 B c+2 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^2} \] Output:

2/3*(A*d+B*c)*(d*x+c)^4/d^2/(-d^2*x^2+c^2)^(3/2)-2/3*(5*A*d+11*B*c)*(d*x+c 
)^2/d^2/(-d^2*x^2+c^2)^(1/2)-1/2*(B*d*x+10*A*d+26*B*c)*(-d^2*x^2+c^2)^(1/2 
)/d^2+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 = 0.82 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.87 \[ \int \frac {(A+B x) (c+d x)^5}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {\frac {\sqrt {c^2-d^2 x^2} \left (2 A d \left (23 c^2-34 c d x+3 d^2 x^2\right )+B \left (118 c^3-161 c^2 d x+24 c d^2 x^2+3 d^3 x^3\right )\right )}{(c-d x)^2}+30 c (5 B c+2 A d) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{6 d^2} \] Input:

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

Output:

-1/6*((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])])/d^2
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.07, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {669, 462, 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.

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

\(\Big \downarrow \) 669

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

\(\Big \downarrow \) 462

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

\(\Big \downarrow \) 2346

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

((B*c + A*d)*(c + d*x)^5)/(3*c*d^2*(c^2 - d^2*x^2)^(3/2)) - ((5*B*c + 2*A* 
d)*((8*c^2*(c + d*x))/(d*Sqrt[c^2 - d^2*x^2]) + (x*Sqrt[c^2 - d^2*x^2])/2 
- (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 462

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

rule 669

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* 
(p + 1))), x] - Simp[e*((m*(d*g + e*f) + 2*e*f*(p + 1))/(2*c*d*(p + 1))) 
Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, 
g}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 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 = 0.74 (sec) , antiderivative size = 266, normalized size of antiderivative = 1.74

method	result	size
risch	\(-\frac {\left (B d x +2 A d +10 B c \right ) \sqrt {-d^{2} x^{2}+c^{2}}}{2 d^{2}}+\frac {c \left (\frac {10 A d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}+\frac {25 B c \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}+\frac {8 \left (3 A d +5 B c \right ) \sqrt {-d^{2} \left (x -\frac {c}{d}\right )^{2}-2 c d \left (x -\frac {c}{d}\right )}}{d^{2} \left (x -\frac {c}{d}\right )}+\frac {16 c^{2} \left (A d +B c \right ) \left (\frac {\sqrt {-d^{2} \left (x -\frac {c}{d}\right )^{2}-2 c d \left (x -\frac {c}{d}\right )}}{3 c d \left (x -\frac {c}{d}\right )^{2}}-\frac {\sqrt {-d^{2} \left (x -\frac {c}{d}\right )^{2}-2 c d \left (x -\frac {c}{d}\right )}}{3 c^{2} \left (x -\frac {c}{d}\right )}\right )}{d^{2}}\right )}{2 d}\)	\(266\)
default	\(A \,c^{5} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )+d^{4} \left (A d +5 B c \right ) \left (-\frac {x^{4}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {4 c^{2} \left (\frac {x^{2}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {2 c^{2}}{3 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\right )}{d^{2}}\right )+5 c \,d^{3} \left (A d +2 B c \right ) \left (\frac {x^{3}}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}\right )+10 c^{2} d^{2} \left (A d +B c \right ) \left (\frac {x^{2}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {2 c^{2}}{3 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\right )+5 d \,c^{3} \left (2 A d +B c \right ) \left (\frac {x}{2 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {c^{2} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 d^{2}}\right )+\frac {c^{4} \left (5 A d +B c \right )}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+B \,d^{5} \left (-\frac {x^{5}}{2 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {5 c^{2} \left (\frac {x^{3}}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}}{d^{2}}\right )}{2 d^{2}}\right )\)	\(518\)

Input:

int((B*x+A)*(d*x+c)^5/(-d^2*x^2+c^2)^(5/2),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 = 240, normalized size of antiderivative = 1.57 \[ \int \frac {(A+B x) (c+d x)^5}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {118 \, B c^{4} + 46 \, A c^{3} d + 2 \, {\left (59 \, B c^{2} d^{2} + 23 \, A c d^{3}\right )} x^{2} - 4 \, {\left (59 \, B c^{3} d + 23 \, A c^{2} d^{2}\right )} x + 30 \, {\left (5 \, B c^{4} + 2 \, A c^{3} d + {\left (5 \, B c^{2} d^{2} + 2 \, A c d^{3}\right )} x^{2} - 2 \, {\left (5 \, B c^{3} d + 2 \, A c^{2} d^{2}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (3 \, B d^{3} x^{3} + 118 \, B c^{3} + 46 \, A c^{2} d + 6 \, {\left (4 \, B c d^{2} + A d^{3}\right )} x^{2} - {\left (161 \, B c^{2} d + 68 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{6 \, {\left (d^{4} x^{2} - 2 \, c d^{3} x + c^{2} d^{2}\right )}} \] Input:

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

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 594 vs. \(2 (134) = 268\).

Time = 0.12 (sec) , antiderivative size = 594, normalized size of antiderivative = 3.88 \[ \int \frac {(A+B x) (c+d x)^5}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {5}{6} \, B c^{2} d^{3} x {\left (\frac {3 \, x^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} - \frac {2 \, c^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{4}}\right )} - \frac {B d^{3} x^{5}}{2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}}} + \frac {A c^{3} x}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}}} + \frac {5}{3} \, {\left (2 \, B c^{2} d^{3} + A c d^{4}\right )} x {\left (\frac {3 \, x^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} - \frac {2 \, c^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{4}}\right )} + \frac {B c^{5}}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {5 \, A c^{4}}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d} + \frac {2 \, A c x}{3 \, \sqrt {-d^{2} x^{2} + c^{2}}} - \frac {5 \, B c^{2} x}{6 \, \sqrt {-d^{2} x^{2} + c^{2}} d} + \frac {5 \, B c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{2}} - \frac {{\left (5 \, B c d^{4} + A d^{5}\right )} x^{4}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {4 \, {\left (5 \, B c d^{4} + A d^{5}\right )} c^{2} x^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{4}} + \frac {10 \, {\left (B c^{3} d^{2} + A c^{2} d^{3}\right )} x^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {5 \, {\left (B c^{4} d + 2 \, A c^{3} d^{2}\right )} x}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} - \frac {8 \, {\left (5 \, B c d^{4} + A d^{5}\right )} c^{4}}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{6}} - \frac {20 \, {\left (B c^{3} d^{2} + A c^{2} d^{3}\right )} c^{2}}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{4}} - \frac {5 \, {\left (2 \, B c^{2} d^{3} + A c d^{4}\right )} x}{3 \, \sqrt {-d^{2} x^{2} + c^{2}} d^{4}} - \frac {5 \, {\left (B c^{4} d + 2 \, A c^{3} d^{2}\right )} x}{3 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{2}} + \frac {5 \, {\left (2 \, B c^{2} d^{3} + A c d^{4}\right )} \arcsin \left (\frac {d x}{c}\right )}{d^{5}} \] Input:

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

Output:

5/6*B*c^2*d^3*x*(3*x^2/((-d^2*x^2 + c^2)^(3/2)*d^2) - 2*c^2/((-d^2*x^2 + c 
^2)^(3/2)*d^4)) - 1/2*B*d^3*x^5/(-d^2*x^2 + c^2)^(3/2) + 1/3*A*c^3*x/(-d^2 
*x^2 + c^2)^(3/2) + 5/3*(2*B*c^2*d^3 + A*c*d^4)*x*(3*x^2/((-d^2*x^2 + c^2) 
^(3/2)*d^2) - 2*c^2/((-d^2*x^2 + c^2)^(3/2)*d^4)) + 1/3*B*c^5/((-d^2*x^2 + 
 c^2)^(3/2)*d^2) + 5/3*A*c^4/((-d^2*x^2 + c^2)^(3/2)*d) + 2/3*A*c*x/sqrt(- 
d^2*x^2 + c^2) - 5/6*B*c^2*x/(sqrt(-d^2*x^2 + c^2)*d) + 5/2*B*c^2*arcsin(d 
*x/c)/d^2 - (5*B*c*d^4 + A*d^5)*x^4/((-d^2*x^2 + c^2)^(3/2)*d^2) + 4*(5*B* 
c*d^4 + A*d^5)*c^2*x^2/((-d^2*x^2 + c^2)^(3/2)*d^4) + 10*(B*c^3*d^2 + A*c^ 
2*d^3)*x^2/((-d^2*x^2 + c^2)^(3/2)*d^2) + 5/3*(B*c^4*d + 2*A*c^3*d^2)*x/(( 
-d^2*x^2 + c^2)^(3/2)*d^2) - 8/3*(5*B*c*d^4 + A*d^5)*c^4/((-d^2*x^2 + c^2) 
^(3/2)*d^6) - 20/3*(B*c^3*d^2 + A*c^2*d^3)*c^2/((-d^2*x^2 + c^2)^(3/2)*d^4 
) - 5/3*(2*B*c^2*d^3 + A*c*d^4)*x/(sqrt(-d^2*x^2 + c^2)*d^4) - 5/3*(B*c^4* 
d + 2*A*c^3*d^2)*x/(sqrt(-d^2*x^2 + c^2)*c^2*d^2) + 5*(2*B*c^2*d^3 + A*c*d 
^4)*arcsin(d*x/c)/d^5
 

Giac [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.69 \[ \int \frac {(A+B x) (c+d x)^5}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {5 \, {\left (5 \, B c^{2} + 2 \, A c d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{2 \, d {\left | d \right |}} - \frac {1}{2} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (\frac {B x}{d} + \frac {2 \, {\left (5 \, B c d^{2} + A d^{3}\right )}}{d^{4}}\right )} - \frac {8 \, {\left (11 \, B c^{2} + 5 \, A c d - \frac {24 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} B c^{2}}{d^{2} x} - \frac {12 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} A c}{d x} + \frac {9 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} B c^{2}}{d^{4} x^{2}} + \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} A c}{d^{3} x^{2}}\right )}}{3 \, d {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} - 1\right )}^{3} {\left | d \right |}} \] Input:

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

Output:

5/2*(5*B*c^2 + 2*A*c*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) - 1/2*sqrt( 
-d^2*x^2 + c^2)*(B*x/d + 2*(5*B*c*d^2 + A*d^3)/d^4) - 8/3*(11*B*c^2 + 5*A* 
c*d - 24*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*B*c^2/(d^2*x) - 12*(c*d + sqr 
t(-d^2*x^2 + c^2)*abs(d))*A*c/(d*x) + 9*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d) 
)^2*B*c^2/(d^4*x^2) + 3*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*A*c/(d^3*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. \[ \int \frac {(A+B x) (c+d x)^5}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (c+d\,x\right )}^5}{{\left (c^2-d^2\,x^2\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 492, normalized size of antiderivative = 3.22 \[ \int \frac {(A+B x) (c+d x)^5}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {-30 \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{3} x^{2}-75 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d^{2} x^{2}-30 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{3} d +16 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d +6 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{3} x^{2}+3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{3} x^{3}+75 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3}-38 a \,c^{2} d^{2} x +92 a c \,d^{3} x^{2}-93 b \,c^{3} d x +205 b \,c^{2} d^{2} x^{2}-27 b c \,d^{3} x^{3}+30 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d +60 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d^{2} x +150 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3} d x -6 a \,d^{4} x^{3}-3 b \,d^{4} x^{4}-50 b \,c^{4}-75 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{4}+50 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3}-16 a \,c^{3} d -38 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{2} x -93 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d x +24 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{2} x^{2}-30 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{2} x -75 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d x}{6 d^{2} \left (\sqrt {-d^{2} x^{2}+c^{2}}\, c -\sqrt {-d^{2} x^{2}+c^{2}}\, d x -c^{2}+2 c d x -d^{2} x^{2}\right )} \] Input:

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