\(\int (A+B x) (c+d x)^2 (c^2-d^2 x^2)^{3/2} \, dx\) [13]

Optimal result

Integrand size = 29, antiderivative size = 183 \[ \int (A+B x) (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {c^4 (2 B c+7 A d) x \sqrt {c^2-d^2 x^2}}{16 d}+\frac {c^2 (2 B c+7 A d) x \left (c^2-d^2 x^2\right )^{3/2}}{24 d}-\frac {B (c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}}{7 d^2}-\frac {(2 B c+7 A d) (12 c+5 d x) \left (c^2-d^2 x^2\right )^{5/2}}{210 d^2}+\frac {c^6 (2 B c+7 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{16 d^2} \] Output:

1/16*c^4*(7*A*d+2*B*c)*x*(-d^2*x^2+c^2)^(1/2)/d+1/24*c^2*(7*A*d+2*B*c)*x*( 
-d^2*x^2+c^2)^(3/2)/d-1/7*B*(d*x+c)^2*(-d^2*x^2+c^2)^(5/2)/d^2-1/210*(7*A* 
d+2*B*c)*(5*d*x+12*c)*(-d^2*x^2+c^2)^(5/2)/d^2+1/16*c^6*(7*A*d+2*B*c)*arct 
an(d*x/(-d^2*x^2+c^2)^(1/2))/d^2
 

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 201, normalized size of antiderivative = 1.10 \[ \int (A+B x) (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {d \sqrt {c^2-d^2 x^2} \left (-7 A d \left (96 c^5-135 c^4 d x-192 c^3 d^2 x^2-10 c^2 d^3 x^3+96 c d^4 x^4+40 d^5 x^5\right )+B \left (-432 c^6-210 c^5 d x+624 c^4 d^2 x^2+980 c^3 d^3 x^3+48 c^2 d^4 x^4-560 c d^5 x^5-240 d^6 x^6\right )\right )+105 c^6 \sqrt {-d^2} (2 B c+7 A d) \log \left (-\sqrt {-d^2} x+\sqrt {c^2-d^2 x^2}\right )}{1680 d^3} \] Input:

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

Output:

(d*Sqrt[c^2 - d^2*x^2]*(-7*A*d*(96*c^5 - 135*c^4*d*x - 192*c^3*d^2*x^2 - 1 
0*c^2*d^3*x^3 + 96*c*d^4*x^4 + 40*d^5*x^5) + B*(-432*c^6 - 210*c^5*d*x + 6 
24*c^4*d^2*x^2 + 980*c^3*d^3*x^3 + 48*c^2*d^4*x^4 - 560*c*d^5*x^5 - 240*d^ 
6*x^6)) + 105*c^6*Sqrt[-d^2]*(2*B*c + 7*A*d)*Log[-(Sqrt[-d^2]*x) + Sqrt[c^ 
2 - d^2*x^2]])/(1680*d^3)
 

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {672, 469, 455, 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.

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

\(\Big \downarrow \) 672

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

\(\Big \downarrow \) 469

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

-1/7*(B*(c + d*x)^2*(c^2 - d^2*x^2)^(5/2))/d^2 + ((2*B*c + 7*A*d)*(-1/6*(( 
c + d*x)*(c^2 - d^2*x^2)^(5/2))/d + (7*c*(-1/5*(c^2 - d^2*x^2)^(5/2)/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*A 
rcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/(2*d)))/4)))/6))/(7*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 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 469

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* 
((n + p)/(n + 2*p + 1))   Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; Fr 
eeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[n, 0] && NeQ[n + 2* 
p + 1, 0] && IntegerQ[2*p]
 

rule 672

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_ 
), x_Symbol] :> Simp[g*(d + e*x)^m*((a + c*x^2)^(p + 1)/(c*(m + 2*p + 2))), 
 x] + Simp[(m*(d*g + e*f) + 2*e*f*(p + 1))/(e*(m + 2*p + 2))   Int[(d + e*x 
)^m*(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] && NeQ[m + 2*p + 2, 0]
 

Maple [A] (verified)

Time = 0.40 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.07

method	result	size
risch	\(-\frac {\left (240 B \,d^{6} x^{6}+280 A \,d^{6} x^{5}+560 B c \,d^{5} x^{5}+672 A c \,d^{5} x^{4}-48 B \,c^{2} d^{4} x^{4}-70 A \,c^{2} d^{4} x^{3}-980 B \,c^{3} d^{3} x^{3}-1344 A \,c^{3} d^{3} x^{2}-624 B \,c^{4} d^{2} x^{2}-945 A \,c^{4} d^{2} x +210 B \,c^{5} d x +672 A \,c^{5} d +432 B \,c^{6}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{1680 d^{2}}+\frac {c^{6} \left (7 A d +2 B c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{16 d \sqrt {d^{2}}}\)	\(195\)
default	\(A \,c^{2} \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )+d \left (A d +2 B c \right ) \left (-\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{6 d^{2}}+\frac {c^{2} \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )}{6 d^{2}}\right )-\frac {c \left (2 A d +B c \right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{5 d^{2}}+B \,d^{2} \left (-\frac {x^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{7 d^{2}}-\frac {2 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{35 d^{4}}\right )\)	\(270\)

Input:

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

Output:

-1/1680/d^2*(240*B*d^6*x^6+280*A*d^6*x^5+560*B*c*d^5*x^5+672*A*c*d^5*x^4-4 
8*B*c^2*d^4*x^4-70*A*c^2*d^4*x^3-980*B*c^3*d^3*x^3-1344*A*c^3*d^3*x^2-624* 
B*c^4*d^2*x^2-945*A*c^4*d^2*x+210*B*c^5*d*x+672*A*c^5*d+432*B*c^6)*(-d^2*x 
^2+c^2)^(1/2)+1/16*c^6/d*(7*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 = 0.08 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.05 \[ \int (A+B x) (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {210 \, {\left (2 \, B c^{7} + 7 \, A c^{6} d\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (240 \, B d^{6} x^{6} + 432 \, B c^{6} + 672 \, A c^{5} d + 280 \, {\left (2 \, B c d^{5} + A d^{6}\right )} x^{5} - 48 \, {\left (B c^{2} d^{4} - 14 \, A c d^{5}\right )} x^{4} - 70 \, {\left (14 \, B c^{3} d^{3} + A c^{2} d^{4}\right )} x^{3} - 48 \, {\left (13 \, B c^{4} d^{2} + 28 \, A c^{3} d^{3}\right )} x^{2} + 105 \, {\left (2 \, B c^{5} d - 9 \, A c^{4} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{1680 \, d^{2}} \] Input:

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

Output:

-1/1680*(210*(2*B*c^7 + 7*A*c^6*d)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x 
)) + (240*B*d^6*x^6 + 432*B*c^6 + 672*A*c^5*d + 280*(2*B*c*d^5 + A*d^6)*x^ 
5 - 48*(B*c^2*d^4 - 14*A*c*d^5)*x^4 - 70*(14*B*c^3*d^3 + A*c^2*d^4)*x^3 - 
48*(13*B*c^4*d^2 + 28*A*c^3*d^3)*x^2 + 105*(2*B*c^5*d - 9*A*c^4*d^2)*x)*sq 
rt(-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. 534 vs. \(2 (163) = 326\).

Time = 0.90 (sec) , antiderivative size = 534, normalized size of antiderivative = 2.92 \[ \int (A+B x) (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2} \, dx=\begin {cases} \sqrt {c^{2} - d^{2} x^{2}} \left (- \frac {B d^{4} x^{6}}{7} - \frac {x^{5} \left (A d^{6} + 2 B c d^{5}\right )}{6 d^{2}} - \frac {x^{4} \cdot \left (2 A c d^{5} - \frac {B c^{2} d^{4}}{7}\right )}{5 d^{2}} - \frac {x^{3} \left (- A c^{2} d^{4} - 4 B c^{3} d^{3} + \frac {5 c^{2} \left (A d^{6} + 2 B c d^{5}\right )}{6 d^{2}}\right )}{4 d^{2}} - \frac {x^{2} \left (- 4 A c^{3} d^{3} - B c^{4} d^{2} + \frac {4 c^{2} \cdot \left (2 A c d^{5} - \frac {B c^{2} d^{4}}{7}\right )}{5 d^{2}}\right )}{3 d^{2}} - \frac {x \left (- A c^{4} d^{2} + 2 B c^{5} d + \frac {3 c^{2} \left (- A c^{2} d^{4} - 4 B c^{3} d^{3} + \frac {5 c^{2} \left (A d^{6} + 2 B c d^{5}\right )}{6 d^{2}}\right )}{4 d^{2}}\right )}{2 d^{2}} - \frac {2 A c^{5} d + B c^{6} + \frac {2 c^{2} \left (- 4 A c^{3} d^{3} - B c^{4} d^{2} + \frac {4 c^{2} \cdot \left (2 A c d^{5} - \frac {B c^{2} d^{4}}{7}\right )}{5 d^{2}}\right )}{3 d^{2}}}{d^{2}}\right ) + \left (A c^{6} + \frac {c^{2} \left (- A c^{4} d^{2} + 2 B c^{5} d + \frac {3 c^{2} \left (- A c^{2} d^{4} - 4 B c^{3} d^{3} + \frac {5 c^{2} \left (A d^{6} + 2 B c d^{5}\right )}{6 d^{2}}\right )}{4 d^{2}}\right )}{2 d^{2}}\right ) \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: d^{2} \neq 0 \\\left (A c^{2} x + \frac {B d^{2} x^{4}}{4} + \frac {x^{3} \left (A d^{2} + 2 B c d\right )}{3} + \frac {x^{2} \cdot \left (2 A c d + B c^{2}\right )}{2}\right ) \left (c^{2}\right )^{\frac {3}{2}} & \text {otherwise} \end {cases} \] Input:

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.36 \[ \int (A+B x) (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {3 \, A c^{6} \arcsin \left (\frac {d x}{c}\right )}{8 \, d} + \frac {3}{8} \, \sqrt {-d^{2} x^{2} + c^{2}} A c^{4} x + \frac {1}{4} \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A c^{2} x - \frac {1}{7} \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B x^{2} + \frac {{\left (2 \, B c d + A d^{2}\right )} c^{6} \arcsin \left (\frac {d x}{c}\right )}{16 \, d^{3}} + \frac {\sqrt {-d^{2} x^{2} + c^{2}} {\left (2 \, B c d + A d^{2}\right )} c^{4} x}{16 \, d^{2}} - \frac {9 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B c^{2}}{35 \, d^{2}} - \frac {2 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} A c}{5 \, d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} {\left (2 \, B c d + A d^{2}\right )} c^{2} x}{24 \, d^{2}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} {\left (2 \, B c d + A d^{2}\right )} x}{6 \, d^{2}} \] Input:

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

Output:

3/8*A*c^6*arcsin(d*x/c)/d + 3/8*sqrt(-d^2*x^2 + c^2)*A*c^4*x + 1/4*(-d^2*x 
^2 + c^2)^(3/2)*A*c^2*x - 1/7*(-d^2*x^2 + c^2)^(5/2)*B*x^2 + 1/16*(2*B*c*d 
 + A*d^2)*c^6*arcsin(d*x/c)/d^3 + 1/16*sqrt(-d^2*x^2 + c^2)*(2*B*c*d + A*d 
^2)*c^4*x/d^2 - 9/35*(-d^2*x^2 + c^2)^(5/2)*B*c^2/d^2 - 2/5*(-d^2*x^2 + c^ 
2)^(5/2)*A*c/d + 1/24*(-d^2*x^2 + c^2)^(3/2)*(2*B*c*d + A*d^2)*c^2*x/d^2 - 
 1/6*(-d^2*x^2 + c^2)^(5/2)*(2*B*c*d + A*d^2)*x/d^2
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.15 \[ \int (A+B x) (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {{\left (2 \, B c^{7} + 7 \, A c^{6} d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{16 \, d {\left | d \right |}} - \frac {1}{1680} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, {\left (6 \, B d^{4} x + \frac {7 \, {\left (2 \, B c d^{13} + A d^{14}\right )}}{d^{10}}\right )} x - \frac {6 \, {\left (B c^{2} d^{12} - 14 \, A c d^{13}\right )}}{d^{10}}\right )} x - \frac {35 \, {\left (14 \, B c^{3} d^{11} + A c^{2} d^{12}\right )}}{d^{10}}\right )} x - \frac {24 \, {\left (13 \, B c^{4} d^{10} + 28 \, A c^{3} d^{11}\right )}}{d^{10}}\right )} x + \frac {105 \, {\left (2 \, B c^{5} d^{9} - 9 \, A c^{4} d^{10}\right )}}{d^{10}}\right )} x + \frac {48 \, {\left (9 \, B c^{6} d^{8} + 14 \, A c^{5} d^{9}\right )}}{d^{10}}\right )} \] Input:

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

Output:

1/16*(2*B*c^7 + 7*A*c^6*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) - 1/1680 
*sqrt(-d^2*x^2 + c^2)*((2*((4*(5*(6*B*d^4*x + 7*(2*B*c*d^13 + A*d^14)/d^10 
)*x - 6*(B*c^2*d^12 - 14*A*c*d^13)/d^10)*x - 35*(14*B*c^3*d^11 + A*c^2*d^1 
2)/d^10)*x - 24*(13*B*c^4*d^10 + 28*A*c^3*d^11)/d^10)*x + 105*(2*B*c^5*d^9 
 - 9*A*c^4*d^10)/d^10)*x + 48*(9*B*c^6*d^8 + 14*A*c^5*d^9)/d^10)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.88 \[ \int (A+B x) (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {735 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{6} d +210 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{7}-672 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{5} d +945 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{4} d^{2} x +1344 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d^{3} x^{2}+70 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{4} x^{3}-672 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{5} x^{4}-280 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{6} x^{5}-432 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{6}-210 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{5} d x +624 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4} d^{2} x^{2}+980 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d^{3} x^{3}+48 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{4} x^{4}-560 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{5} x^{5}-240 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{6} x^{6}+672 a \,c^{6} d +432 b \,c^{7}}{1680 d^{2}} \] Input:

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

Output:

(735*asin((d*x)/c)*a*c**6*d + 210*asin((d*x)/c)*b*c**7 - 672*sqrt(c**2 - d 
**2*x**2)*a*c**5*d + 945*sqrt(c**2 - d**2*x**2)*a*c**4*d**2*x + 1344*sqrt( 
c**2 - d**2*x**2)*a*c**3*d**3*x**2 + 70*sqrt(c**2 - d**2*x**2)*a*c**2*d**4 
*x**3 - 672*sqrt(c**2 - d**2*x**2)*a*c*d**5*x**4 - 280*sqrt(c**2 - d**2*x* 
*2)*a*d**6*x**5 - 432*sqrt(c**2 - d**2*x**2)*b*c**6 - 210*sqrt(c**2 - d**2 
*x**2)*b*c**5*d*x + 624*sqrt(c**2 - d**2*x**2)*b*c**4*d**2*x**2 + 980*sqrt 
(c**2 - d**2*x**2)*b*c**3*d**3*x**3 + 48*sqrt(c**2 - d**2*x**2)*b*c**2*d** 
4*x**4 - 560*sqrt(c**2 - d**2*x**2)*b*c*d**5*x**5 - 240*sqrt(c**2 - d**2*x 
**2)*b*d**6*x**6 + 672*a*c**6*d + 432*b*c**7)/(1680*d**2)