\(\int (A+B x) (c+d x) (c^2-d^2 x^2)^{5/2} \, dx\) [27]

Optimal result

Integrand size = 27, antiderivative size = 202 \[ \int (A+B x) (c+d x) \left (c^2-d^2 x^2\right )^{5/2} \, dx=\frac {5 c^5 (B c+8 A d) x \sqrt {c^2-d^2 x^2}}{128 d}+\frac {5 c^3 (B c+8 A d) x \left (c^2-d^2 x^2\right )^{3/2}}{192 d}+\frac {c (B c+8 A d) x \left (c^2-d^2 x^2\right )^{5/2}}{48 d}-\frac {(B c+8 A d) \left (c^2-d^2 x^2\right )^{7/2}}{56 d^2}-\frac {B (c+d x) \left (c^2-d^2 x^2\right )^{7/2}}{8 d^2}+\frac {5 c^7 (B c+8 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{128 d^2} \] Output:

5/128*c^5*(8*A*d+B*c)*x*(-d^2*x^2+c^2)^(1/2)/d+5/192*c^3*(8*A*d+B*c)*x*(-d 
^2*x^2+c^2)^(3/2)/d+1/48*c*(8*A*d+B*c)*x*(-d^2*x^2+c^2)^(5/2)/d-1/56*(8*A* 
d+B*c)*(-d^2*x^2+c^2)^(7/2)/d^2-1/8*B*(d*x+c)*(-d^2*x^2+c^2)^(7/2)/d^2+5/1 
28*c^7*(8*A*d+B*c)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^2
 

Mathematica [A] (verified)

Time = 1.19 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.06 \[ \int (A+B x) (c+d x) \left (c^2-d^2 x^2\right )^{5/2} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (8 A d \left (-48 c^6+231 c^5 d x+144 c^4 d^2 x^2-182 c^3 d^3 x^3-144 c^2 d^4 x^4+56 c d^5 x^5+48 d^6 x^6\right )+B \left (-384 c^7-105 c^6 d x+1152 c^5 d^2 x^2+826 c^4 d^3 x^3-1152 c^3 d^4 x^4-952 c^2 d^5 x^5+384 c d^6 x^6+336 d^7 x^7\right )\right )-210 c^7 (B c+8 A d) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{2688 d^2} \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(8*A*d*(-48*c^6 + 231*c^5*d*x + 144*c^4*d^2*x^2 - 182 
*c^3*d^3*x^3 - 144*c^2*d^4*x^4 + 56*c*d^5*x^5 + 48*d^6*x^6) + B*(-384*c^7 
- 105*c^6*d*x + 1152*c^5*d^2*x^2 + 826*c^4*d^3*x^3 - 1152*c^3*d^4*x^4 - 95 
2*c^2*d^5*x^5 + 384*c*d^6*x^6 + 336*d^7*x^7)) - 210*c^7*(B*c + 8*A*d)*ArcT 
an[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/(2688*d^2)
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.90, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {676, 211, 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.

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

\(\Big \downarrow \) 676

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

\(\Big \downarrow \) 211

\(\displaystyle \frac {c (8 A d+B c) \left (\frac {5}{6} c^2 \int \left (c^2-d^2 x^2\right )^{3/2}dx+\frac {1}{6} x \left (c^2-d^2 x^2\right )^{5/2}\right )}{8 d}-\frac {\left (c^2-d^2 x^2\right )^{7/2} (A d+B c)}{7 d^2}-\frac {B x \left (c^2-d^2 x^2\right )^{7/2}}{8 d}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {c (8 A d+B c) \left (\frac {5}{6} c^2 \left (\frac {3}{4} c^2 \int \sqrt {c^2-d^2 x^2}dx+\frac {1}{4} x \left (c^2-d^2 x^2\right )^{3/2}\right )+\frac {1}{6} x \left (c^2-d^2 x^2\right )^{5/2}\right )}{8 d}-\frac {\left (c^2-d^2 x^2\right )^{7/2} (A d+B c)}{7 d^2}-\frac {B x \left (c^2-d^2 x^2\right )^{7/2}}{8 d}\)

\(\Big \downarrow \) 211

\(\displaystyle \frac {c (8 A d+B c) \left (\frac {5}{6} c^2 \left (\frac {3}{4} c^2 \left (\frac {1}{2} c^2 \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {1}{4} x \left (c^2-d^2 x^2\right )^{3/2}\right )+\frac {1}{6} x \left (c^2-d^2 x^2\right )^{5/2}\right )}{8 d}-\frac {\left (c^2-d^2 x^2\right )^{7/2} (A d+B c)}{7 d^2}-\frac {B x \left (c^2-d^2 x^2\right )^{7/2}}{8 d}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {c (8 A d+B c) \left (\frac {5}{6} c^2 \left (\frac {3}{4} c^2 \left (\frac {1}{2} c^2 \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {1}{4} x \left (c^2-d^2 x^2\right )^{3/2}\right )+\frac {1}{6} x \left (c^2-d^2 x^2\right )^{5/2}\right )}{8 d}-\frac {\left (c^2-d^2 x^2\right )^{7/2} (A d+B c)}{7 d^2}-\frac {B x \left (c^2-d^2 x^2\right )^{7/2}}{8 d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {c (8 A d+B c) \left (\frac {5}{6} c^2 \left (\frac {3}{4} c^2 \left (\frac {c^2 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {1}{4} x \left (c^2-d^2 x^2\right )^{3/2}\right )+\frac {1}{6} x \left (c^2-d^2 x^2\right )^{5/2}\right )}{8 d}-\frac {\left (c^2-d^2 x^2\right )^{7/2} (A d+B c)}{7 d^2}-\frac {B x \left (c^2-d^2 x^2\right )^{7/2}}{8 d}\)

Input:

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

Output:

-1/7*((B*c + A*d)*(c^2 - d^2*x^2)^(7/2))/d^2 - (B*x*(c^2 - d^2*x^2)^(7/2)) 
/(8*d) + (c*(B*c + 8*A*d)*((x*(c^2 - d^2*x^2)^(5/2))/6 + (5*c^2*((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))/6))/(8*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 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.08

method	result	size
risch	\(-\frac {\left (-336 B \,d^{7} x^{7}-384 A \,d^{7} x^{6}-384 B c \,d^{6} x^{6}-448 A c \,d^{6} x^{5}+952 B \,c^{2} d^{5} x^{5}+1152 A \,c^{2} d^{5} x^{4}+1152 B \,c^{3} d^{4} x^{4}+1456 A \,c^{3} d^{4} x^{3}-826 B \,c^{4} d^{3} x^{3}-1152 c^{4} x^{2} A \,d^{3}-1152 c^{5} x^{2} B \,d^{2}-1848 A \,c^{5} d^{2} x +105 B \,c^{6} d x +384 A \,c^{6} d +384 B \,c^{7}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{2688 d^{2}}+\frac {5 c^{7} \left (8 A d +B c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{128 d \sqrt {d^{2}}}\)	\(218\)
default	\(A c \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{6}+\frac {5 c^{2} \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )}{6}\right )-\frac {\left (A d +B c \right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {7}{2}}}{7 d^{2}}+B d \left (-\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {7}{2}}}{8 d^{2}}+\frac {c^{2} \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{6}+\frac {5 c^{2} \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )}{6}\right )}{8 d^{2}}\right )\)	\(255\)

Input:

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

Output:

-1/2688/d^2*(-336*B*d^7*x^7-384*A*d^7*x^6-384*B*c*d^6*x^6-448*A*c*d^6*x^5+ 
952*B*c^2*d^5*x^5+1152*A*c^2*d^5*x^4+1152*B*c^3*d^4*x^4+1456*A*c^3*d^4*x^3 
-826*B*c^4*d^3*x^3-1152*A*c^4*d^3*x^2-1152*B*c^5*d^2*x^2-1848*A*c^5*d^2*x+ 
105*B*c^6*d*x+384*A*c^6*d+384*B*c^7)*(-d^2*x^2+c^2)^(1/2)+5/128*c^7/d*(8*A 
*d+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 = 214, normalized size of antiderivative = 1.06 \[ \int (A+B x) (c+d x) \left (c^2-d^2 x^2\right )^{5/2} \, dx=-\frac {210 \, {\left (B c^{8} + 8 \, A c^{7} d\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (336 \, B d^{7} x^{7} - 384 \, B c^{7} - 384 \, A c^{6} d + 384 \, {\left (B c d^{6} + A d^{7}\right )} x^{6} - 56 \, {\left (17 \, B c^{2} d^{5} - 8 \, A c d^{6}\right )} x^{5} - 1152 \, {\left (B c^{3} d^{4} + A c^{2} d^{5}\right )} x^{4} + 14 \, {\left (59 \, B c^{4} d^{3} - 104 \, A c^{3} d^{4}\right )} x^{3} + 1152 \, {\left (B c^{5} d^{2} + A c^{4} d^{3}\right )} x^{2} - 21 \, {\left (5 \, B c^{6} d - 88 \, A c^{5} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{2688 \, d^{2}} \] Input:

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

Output:

-1/2688*(210*(B*c^8 + 8*A*c^7*d)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) 
 - (336*B*d^7*x^7 - 384*B*c^7 - 384*A*c^6*d + 384*(B*c*d^6 + A*d^7)*x^6 - 
56*(17*B*c^2*d^5 - 8*A*c*d^6)*x^5 - 1152*(B*c^3*d^4 + A*c^2*d^5)*x^4 + 14* 
(59*B*c^4*d^3 - 104*A*c^3*d^4)*x^3 + 1152*(B*c^5*d^2 + A*c^4*d^3)*x^2 - 21 
*(5*B*c^6*d - 88*A*c^5*d^2)*x)*sqrt(-d^2*x^2 + c^2))/d^2
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 632 vs. \(2 (182) = 364\).

Time = 0.76 (sec) , antiderivative size = 632, normalized size of antiderivative = 3.13 \[ \int (A+B x) (c+d x) \left (c^2-d^2 x^2\right )^{5/2} \, dx=\begin {cases} \sqrt {c^{2} - d^{2} x^{2}} \left (\frac {B d^{5} x^{7}}{8} - \frac {x^{6} \left (- A d^{7} - B c d^{6}\right )}{7 d^{2}} - \frac {x^{5} \left (- A c d^{6} + \frac {17 B c^{2} d^{5}}{8}\right )}{6 d^{2}} - \frac {x^{4} \cdot \left (3 A c^{2} d^{5} + 3 B c^{3} d^{4} + \frac {6 c^{2} \left (- A d^{7} - B c d^{6}\right )}{7 d^{2}}\right )}{5 d^{2}} - \frac {x^{3} \cdot \left (3 A c^{3} d^{4} - 3 B c^{4} d^{3} + \frac {5 c^{2} \left (- A c d^{6} + \frac {17 B c^{2} d^{5}}{8}\right )}{6 d^{2}}\right )}{4 d^{2}} - \frac {x^{2} \left (- 3 A c^{4} d^{3} - 3 B c^{5} d^{2} + \frac {4 c^{2} \cdot \left (3 A c^{2} d^{5} + 3 B c^{3} d^{4} + \frac {6 c^{2} \left (- A d^{7} - B c d^{6}\right )}{7 d^{2}}\right )}{5 d^{2}}\right )}{3 d^{2}} - \frac {x \left (- 3 A c^{5} d^{2} + B c^{6} d + \frac {3 c^{2} \cdot \left (3 A c^{3} d^{4} - 3 B c^{4} d^{3} + \frac {5 c^{2} \left (- A c d^{6} + \frac {17 B c^{2} d^{5}}{8}\right )}{6 d^{2}}\right )}{4 d^{2}}\right )}{2 d^{2}} - \frac {A c^{6} d + B c^{7} + \frac {2 c^{2} \left (- 3 A c^{4} d^{3} - 3 B c^{5} d^{2} + \frac {4 c^{2} \cdot \left (3 A c^{2} d^{5} + 3 B c^{3} d^{4} + \frac {6 c^{2} \left (- A d^{7} - B c d^{6}\right )}{7 d^{2}}\right )}{5 d^{2}}\right )}{3 d^{2}}}{d^{2}}\right ) + \left (A c^{7} + \frac {c^{2} \left (- 3 A c^{5} d^{2} + B c^{6} d + \frac {3 c^{2} \cdot \left (3 A c^{3} d^{4} - 3 B c^{4} d^{3} + \frac {5 c^{2} \left (- A c d^{6} + \frac {17 B c^{2} d^{5}}{8}\right )}{6 d^{2}}\right )}{4 d^{2}}\right )}{2 d^{2}}\right ) \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: d^{2} \neq 0 \\\left (A c x + \frac {B d x^{3}}{3} + \frac {x^{2} \left (A d + B c\right )}{2}\right ) \left (c^{2}\right )^{\frac {5}{2}} & \text {otherwise} \end {cases} \] Input:

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

Output:

Piecewise((sqrt(c**2 - d**2*x**2)*(B*d**5*x**7/8 - x**6*(-A*d**7 - B*c*d** 
6)/(7*d**2) - x**5*(-A*c*d**6 + 17*B*c**2*d**5/8)/(6*d**2) - x**4*(3*A*c** 
2*d**5 + 3*B*c**3*d**4 + 6*c**2*(-A*d**7 - B*c*d**6)/(7*d**2))/(5*d**2) - 
x**3*(3*A*c**3*d**4 - 3*B*c**4*d**3 + 5*c**2*(-A*c*d**6 + 17*B*c**2*d**5/8 
)/(6*d**2))/(4*d**2) - x**2*(-3*A*c**4*d**3 - 3*B*c**5*d**2 + 4*c**2*(3*A* 
c**2*d**5 + 3*B*c**3*d**4 + 6*c**2*(-A*d**7 - B*c*d**6)/(7*d**2))/(5*d**2) 
)/(3*d**2) - x*(-3*A*c**5*d**2 + B*c**6*d + 3*c**2*(3*A*c**3*d**4 - 3*B*c* 
*4*d**3 + 5*c**2*(-A*c*d**6 + 17*B*c**2*d**5/8)/(6*d**2))/(4*d**2))/(2*d** 
2) - (A*c**6*d + B*c**7 + 2*c**2*(-3*A*c**4*d**3 - 3*B*c**5*d**2 + 4*c**2* 
(3*A*c**2*d**5 + 3*B*c**3*d**4 + 6*c**2*(-A*d**7 - B*c*d**6)/(7*d**2))/(5* 
d**2))/(3*d**2))/d**2) + (A*c**7 + c**2*(-3*A*c**5*d**2 + B*c**6*d + 3*c** 
2*(3*A*c**3*d**4 - 3*B*c**4*d**3 + 5*c**2*(-A*c*d**6 + 17*B*c**2*d**5/8)/( 
6*d**2))/(4*d**2))/(2*d**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)), Ne(d**2, 0)), ((A*c*x + B*d*x**3/3 + x**2*(A*d + B*c)/2)*(c**2)** 
(5/2), True))
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.13 \[ \int (A+B x) (c+d x) \left (c^2-d^2 x^2\right )^{5/2} \, dx=\frac {5 \, B c^{8} \arcsin \left (\frac {d x}{c}\right )}{128 \, d^{2}} + \frac {5 \, A c^{7} \arcsin \left (\frac {d x}{c}\right )}{16 \, d} + \frac {5}{16} \, \sqrt {-d^{2} x^{2} + c^{2}} A c^{5} x + \frac {5 \, \sqrt {-d^{2} x^{2} + c^{2}} B c^{6} x}{128 \, d} + \frac {5}{24} \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A c^{3} x + \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c^{4} x}{192 \, d} + \frac {1}{6} \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} A c x + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B c^{2} x}{48 \, d} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {7}{2}} B x}{8 \, d} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {7}{2}} B c}{7 \, d^{2}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {7}{2}} A}{7 \, d} \] Input:

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

Output:

5/128*B*c^8*arcsin(d*x/c)/d^2 + 5/16*A*c^7*arcsin(d*x/c)/d + 5/16*sqrt(-d^ 
2*x^2 + c^2)*A*c^5*x + 5/128*sqrt(-d^2*x^2 + c^2)*B*c^6*x/d + 5/24*(-d^2*x 
^2 + c^2)^(3/2)*A*c^3*x + 5/192*(-d^2*x^2 + c^2)^(3/2)*B*c^4*x/d + 1/6*(-d 
^2*x^2 + c^2)^(5/2)*A*c*x + 1/48*(-d^2*x^2 + c^2)^(5/2)*B*c^2*x/d - 1/8*(- 
d^2*x^2 + c^2)^(7/2)*B*x/d - 1/7*(-d^2*x^2 + c^2)^(7/2)*B*c/d^2 - 1/7*(-d^ 
2*x^2 + c^2)^(7/2)*A/d
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.15 \[ \int (A+B x) (c+d x) \left (c^2-d^2 x^2\right )^{5/2} \, dx=\frac {5 \, {\left (B c^{8} + 8 \, A c^{7} d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{128 \, d {\left | d \right |}} + \frac {1}{2688} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left ({\left (4 \, {\left ({\left (6 \, {\left (7 \, B d^{5} x + \frac {8 \, {\left (B c d^{16} + A d^{17}\right )}}{d^{12}}\right )} x - \frac {7 \, {\left (17 \, B c^{2} d^{15} - 8 \, A c d^{16}\right )}}{d^{12}}\right )} x - \frac {144 \, {\left (B c^{3} d^{14} + A c^{2} d^{15}\right )}}{d^{12}}\right )} x + \frac {7 \, {\left (59 \, B c^{4} d^{13} - 104 \, A c^{3} d^{14}\right )}}{d^{12}}\right )} x + \frac {576 \, {\left (B c^{5} d^{12} + A c^{4} d^{13}\right )}}{d^{12}}\right )} x - \frac {21 \, {\left (5 \, B c^{6} d^{11} - 88 \, A c^{5} d^{12}\right )}}{d^{12}}\right )} x - \frac {384 \, {\left (B c^{7} d^{10} + A c^{6} d^{11}\right )}}{d^{12}}\right )} \] Input:

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

Output:

5/128*(B*c^8 + 8*A*c^7*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) + 1/2688* 
sqrt(-d^2*x^2 + c^2)*((2*((4*((6*(7*B*d^5*x + 8*(B*c*d^16 + A*d^17)/d^12)* 
x - 7*(17*B*c^2*d^15 - 8*A*c*d^16)/d^12)*x - 144*(B*c^3*d^14 + A*c^2*d^15) 
/d^12)*x + 7*(59*B*c^4*d^13 - 104*A*c^3*d^14)/d^12)*x + 576*(B*c^5*d^12 + 
A*c^4*d^13)/d^12)*x - 21*(5*B*c^6*d^11 - 88*A*c^5*d^12)/d^12)*x - 384*(B*c 
^7*d^10 + A*c^6*d^11)/d^12)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 394, normalized size of antiderivative = 1.95 \[ \int (A+B x) (c+d x) \left (c^2-d^2 x^2\right )^{5/2} \, dx=\frac {840 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{7} d +105 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{8}-384 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{6} d +1848 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{5} d^{2} x +1152 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{4} d^{3} x^{2}-1456 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d^{4} x^{3}-1152 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{5} x^{4}+448 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{6} x^{5}+384 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{7} x^{6}-384 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{7}-105 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{6} d x +1152 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{5} d^{2} x^{2}+826 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4} d^{3} x^{3}-1152 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d^{4} x^{4}-952 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{5} x^{5}+384 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{6} x^{6}+336 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{7} x^{7}+384 a \,c^{7} d +384 b \,c^{8}}{2688 d^{2}} \] Input:

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

Output:

(840*asin((d*x)/c)*a*c**7*d + 105*asin((d*x)/c)*b*c**8 - 384*sqrt(c**2 - d 
**2*x**2)*a*c**6*d + 1848*sqrt(c**2 - d**2*x**2)*a*c**5*d**2*x + 1152*sqrt 
(c**2 - d**2*x**2)*a*c**4*d**3*x**2 - 1456*sqrt(c**2 - d**2*x**2)*a*c**3*d 
**4*x**3 - 1152*sqrt(c**2 - d**2*x**2)*a*c**2*d**5*x**4 + 448*sqrt(c**2 - 
d**2*x**2)*a*c*d**6*x**5 + 384*sqrt(c**2 - d**2*x**2)*a*d**7*x**6 - 384*sq 
rt(c**2 - d**2*x**2)*b*c**7 - 105*sqrt(c**2 - d**2*x**2)*b*c**6*d*x + 1152 
*sqrt(c**2 - d**2*x**2)*b*c**5*d**2*x**2 + 826*sqrt(c**2 - d**2*x**2)*b*c* 
*4*d**3*x**3 - 1152*sqrt(c**2 - d**2*x**2)*b*c**3*d**4*x**4 - 952*sqrt(c** 
2 - d**2*x**2)*b*c**2*d**5*x**5 + 384*sqrt(c**2 - d**2*x**2)*b*c*d**6*x**6 
 + 336*sqrt(c**2 - d**2*x**2)*b*d**7*x**7 + 384*a*c**7*d + 384*b*c**8)/(26 
88*d**2)