\(\int (A+B x) (c+d x)^3 \sqrt {c^2-d^2 x^2} \, dx\) [1]

Optimal result

Integrand size = 29, antiderivative size = 183 \[ \int (A+B x) (c+d x)^3 \sqrt {c^2-d^2 x^2} \, dx=\frac {7 c^3 (B c+2 A d) x \sqrt {c^2-d^2 x^2}}{16 d}-\frac {(B c+2 A d) (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}}{10 d^2}-\frac {B (c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}}{6 d^2}-\frac {7 c (B c+2 A d) (8 c+3 d x) \left (c^2-d^2 x^2\right )^{3/2}}{120 d^2}+\frac {7 c^5 (B c+2 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{16 d^2} \] Output:

7/16*c^3*(2*A*d+B*c)*x*(-d^2*x^2+c^2)^(1/2)/d-1/10*(2*A*d+B*c)*(d*x+c)^2*( 
-d^2*x^2+c^2)^(3/2)/d^2-1/6*B*(d*x+c)^3*(-d^2*x^2+c^2)^(3/2)/d^2-7/120*c*( 
2*A*d+B*c)*(3*d*x+8*c)*(-d^2*x^2+c^2)^(3/2)/d^2+7/16*c^5*(2*A*d+B*c)*arcta 
n(d*x/(-d^2*x^2+c^2)^(1/2))/d^2
 

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.93 \[ \int (A+B x) (c+d x)^3 \sqrt {c^2-d^2 x^2} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (2 A d \left (-136 c^4+15 c^3 d x+112 c^2 d^2 x^2+90 c d^3 x^3+24 d^4 x^4\right )+B \left (-176 c^5-105 c^4 d x+32 c^3 d^2 x^2+170 c^2 d^3 x^3+144 c d^4 x^4+40 d^5 x^5\right )\right )-210 c^5 (B c+2 A d) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{240 d^2} \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(2*A*d*(-136*c^4 + 15*c^3*d*x + 112*c^2*d^2*x^2 + 90* 
c*d^3*x^3 + 24*d^4*x^4) + B*(-176*c^5 - 105*c^4*d*x + 32*c^3*d^2*x^2 + 170 
*c^2*d^3*x^3 + 144*c*d^4*x^4 + 40*d^5*x^5)) - 210*c^5*(B*c + 2*A*d)*ArcTan 
[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/(240*d^2)
 

Rubi [A] (verified)

Time = 0.49 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {672, 469, 469, 455, 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.

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

\(\Big \downarrow \) 672

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

\(\Big \downarrow \) 469

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

\(\Big \downarrow \) 469

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

-1/6*(B*(c + d*x)^3*(c^2 - d^2*x^2)^(3/2))/d^2 + ((B*c + 2*A*d)*(-1/5*((c 
+ d*x)^2*(c^2 - d^2*x^2)^(3/2))/d + (7*c*(-1/4*((c + d*x)*(c^2 - d^2*x^2)^ 
(3/2))/d + (5*c*(-1/3*(c^2 - d^2*x^2)^(3/2)/d + c*((x*Sqrt[c^2 - d^2*x^2]) 
/2 + (c^2*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/(2*d))))/4))/5))/(2*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 = 0.51 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.93

method	result	size
risch	\(-\frac {\left (-40 B \,d^{5} x^{5}-48 A \,d^{5} x^{4}-144 B c \,d^{4} x^{4}-180 A c \,d^{4} x^{3}-170 B \,c^{2} d^{3} x^{3}-224 A \,c^{2} d^{3} x^{2}-32 B \,c^{3} d^{2} x^{2}-30 A \,c^{3} d^{2} x +105 B \,c^{4} d x +272 A \,c^{4} d +176 B \,c^{5}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{240 d^{2}}+\frac {7 c^{5} \left (2 A d +B c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{16 d \sqrt {d^{2}}}\)	\(170\)
default	\(A \,c^{3} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )+d^{2} \left (A d +3 B c \right ) \left (-\frac {x^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{5 d^{2}}-\frac {2 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{15 d^{4}}\right )+3 c d \left (A d +B c \right ) \left (-\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4 d^{2}}+\frac {c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4 d^{2}}\right )-\frac {c^{2} \left (3 A d +B c \right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{3 d^{2}}+B \,d^{3} \left (-\frac {x^{3} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{6 d^{2}}+\frac {c^{2} \left (-\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4 d^{2}}+\frac {c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4 d^{2}}\right )}{2 d^{2}}\right )\)	\(349\)

Input:

int((B*x+A)*(d*x+c)^3*(-d^2*x^2+c^2)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/240/d^2*(-40*B*d^5*x^5-48*A*d^5*x^4-144*B*c*d^4*x^4-180*A*c*d^4*x^3-170 
*B*c^2*d^3*x^3-224*A*c^2*d^3*x^2-32*B*c^3*d^2*x^2-30*A*c^3*d^2*x+105*B*c^4 
*d*x+272*A*c^4*d+176*B*c^5)*(-d^2*x^2+c^2)^(1/2)+7/16*c^5/d*(2*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 = 170, normalized size of antiderivative = 0.93 \[ \int (A+B x) (c+d x)^3 \sqrt {c^2-d^2 x^2} \, dx=-\frac {210 \, {\left (B c^{6} + 2 \, A c^{5} d\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (40 \, B d^{5} x^{5} - 176 \, B c^{5} - 272 \, A c^{4} d + 48 \, {\left (3 \, B c d^{4} + A d^{5}\right )} x^{4} + 10 \, {\left (17 \, B c^{2} d^{3} + 18 \, A c d^{4}\right )} x^{3} + 32 \, {\left (B c^{3} d^{2} + 7 \, A c^{2} d^{3}\right )} x^{2} - 15 \, {\left (7 \, B c^{4} d - 2 \, A c^{3} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{240 \, d^{2}} \] Input:

integrate((B*x+A)*(d*x+c)^3*(-d^2*x^2+c^2)^(1/2),x, algorithm="fricas")
 

Output:

-1/240*(210*(B*c^6 + 2*A*c^5*d)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) 
- (40*B*d^5*x^5 - 176*B*c^5 - 272*A*c^4*d + 48*(3*B*c*d^4 + A*d^5)*x^4 + 1 
0*(17*B*c^2*d^3 + 18*A*c*d^4)*x^3 + 32*(B*c^3*d^2 + 7*A*c^2*d^3)*x^2 - 15* 
(7*B*c^4*d - 2*A*c^3*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. 471 vs. \(2 (168) = 336\).

Time = 0.61 (sec) , antiderivative size = 471, normalized size of antiderivative = 2.57 \[ \int (A+B x) (c+d x)^3 \sqrt {c^2-d^2 x^2} \, dx=\begin {cases} \sqrt {c^{2} - d^{2} x^{2}} \left (\frac {B d^{3} x^{5}}{6} - \frac {x^{4} \left (- A d^{5} - 3 B c d^{4}\right )}{5 d^{2}} - \frac {x^{3} \left (- 3 A c d^{4} - \frac {17 B c^{2} d^{3}}{6}\right )}{4 d^{2}} - \frac {x^{2} \left (- 2 A c^{2} d^{3} + 2 B c^{3} d^{2} + \frac {4 c^{2} \left (- A d^{5} - 3 B c d^{4}\right )}{5 d^{2}}\right )}{3 d^{2}} - \frac {x \left (2 A c^{3} d^{2} + 3 B c^{4} d + \frac {3 c^{2} \left (- 3 A c d^{4} - \frac {17 B c^{2} d^{3}}{6}\right )}{4 d^{2}}\right )}{2 d^{2}} - \frac {3 A c^{4} d + B c^{5} + \frac {2 c^{2} \left (- 2 A c^{2} d^{3} + 2 B c^{3} d^{2} + \frac {4 c^{2} \left (- A d^{5} - 3 B c d^{4}\right )}{5 d^{2}}\right )}{3 d^{2}}}{d^{2}}\right ) + \left (A c^{5} + \frac {c^{2} \cdot \left (2 A c^{3} d^{2} + 3 B c^{4} d + \frac {3 c^{2} \left (- 3 A c d^{4} - \frac {17 B c^{2} d^{3}}{6}\right )}{4 d^{2}}\right )}{2 d^{2}}\right ) \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: d^{2} \neq 0 \\\left (A c^{3} x + \frac {B d^{3} x^{5}}{5} + \frac {x^{4} \left (A d^{3} + 3 B c d^{2}\right )}{4} + \frac {x^{3} \cdot \left (3 A c d^{2} + 3 B c^{2} d\right )}{3} + \frac {x^{2} \cdot \left (3 A c^{2} d + B c^{3}\right )}{2}\right ) \sqrt {c^{2}} & \text {otherwise} \end {cases} \] Input:

integrate((B*x+A)*(d*x+c)**3*(-d**2*x**2+c**2)**(1/2),x)
 

Output:

Piecewise((sqrt(c**2 - d**2*x**2)*(B*d**3*x**5/6 - x**4*(-A*d**5 - 3*B*c*d 
**4)/(5*d**2) - x**3*(-3*A*c*d**4 - 17*B*c**2*d**3/6)/(4*d**2) - x**2*(-2* 
A*c**2*d**3 + 2*B*c**3*d**2 + 4*c**2*(-A*d**5 - 3*B*c*d**4)/(5*d**2))/(3*d 
**2) - x*(2*A*c**3*d**2 + 3*B*c**4*d + 3*c**2*(-3*A*c*d**4 - 17*B*c**2*d** 
3/6)/(4*d**2))/(2*d**2) - (3*A*c**4*d + B*c**5 + 2*c**2*(-2*A*c**2*d**3 + 
2*B*c**3*d**2 + 4*c**2*(-A*d**5 - 3*B*c*d**4)/(5*d**2))/(3*d**2))/d**2) + 
(A*c**5 + c**2*(2*A*c**3*d**2 + 3*B*c**4*d + 3*c**2*(-3*A*c*d**4 - 17*B*c* 
*2*d**3/6)/(4*d**2))/(2*d**2))*Piecewise((log(-2*d**2*x + 2*sqrt(-d**2)*sq 
rt(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**3*x + B*d**3*x**5/5 + x**4*(A*d**3 + 3*B*c 
*d**2)/4 + x**3*(3*A*c*d**2 + 3*B*c**2*d)/3 + x**2*(3*A*c**2*d + B*c**3)/2 
)*sqrt(c**2), True))
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 336 vs. \(2 (163) = 326\).

Time = 0.12 (sec) , antiderivative size = 336, normalized size of antiderivative = 1.84 \[ \int (A+B x) (c+d x)^3 \sqrt {c^2-d^2 x^2} \, dx=-\frac {1}{6} \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B d x^{3} + \frac {B c^{6} \arcsin \left (\frac {d x}{c}\right )}{16 \, d^{2}} + \frac {A c^{5} \arcsin \left (\frac {d x}{c}\right )}{2 \, d} + \frac {1}{2} \, \sqrt {-d^{2} x^{2} + c^{2}} A c^{3} x + \frac {\sqrt {-d^{2} x^{2} + c^{2}} B c^{4} x}{16 \, d} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c^{2} x}{8 \, d} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c^{3}}{3 \, d^{2}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A c^{2}}{d} + \frac {3 \, {\left (B c^{2} d + A c d^{2}\right )} c^{4} \arcsin \left (\frac {d x}{c}\right )}{8 \, d^{3}} + \frac {3 \, {\left (B c^{2} d + A c d^{2}\right )} \sqrt {-d^{2} x^{2} + c^{2}} c^{2} x}{8 \, d^{2}} - \frac {{\left (3 \, B c d^{2} + A d^{3}\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} x^{2}}{5 \, d^{2}} - \frac {3 \, {\left (B c^{2} d + A c d^{2}\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} x}{4 \, d^{2}} - \frac {2 \, {\left (3 \, B c d^{2} + A d^{3}\right )} {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{2}}{15 \, d^{4}} \] Input:

integrate((B*x+A)*(d*x+c)^3*(-d^2*x^2+c^2)^(1/2),x, algorithm="maxima")
 

Output:

-1/6*(-d^2*x^2 + c^2)^(3/2)*B*d*x^3 + 1/16*B*c^6*arcsin(d*x/c)/d^2 + 1/2*A 
*c^5*arcsin(d*x/c)/d + 1/2*sqrt(-d^2*x^2 + c^2)*A*c^3*x + 1/16*sqrt(-d^2*x 
^2 + c^2)*B*c^4*x/d - 1/8*(-d^2*x^2 + c^2)^(3/2)*B*c^2*x/d - 1/3*(-d^2*x^2 
 + c^2)^(3/2)*B*c^3/d^2 - (-d^2*x^2 + c^2)^(3/2)*A*c^2/d + 3/8*(B*c^2*d + 
A*c*d^2)*c^4*arcsin(d*x/c)/d^3 + 3/8*(B*c^2*d + A*c*d^2)*sqrt(-d^2*x^2 + c 
^2)*c^2*x/d^2 - 1/5*(3*B*c*d^2 + A*d^3)*(-d^2*x^2 + c^2)^(3/2)*x^2/d^2 - 3 
/4*(B*c^2*d + A*c*d^2)*(-d^2*x^2 + c^2)^(3/2)*x/d^2 - 2/15*(3*B*c*d^2 + A* 
d^3)*(-d^2*x^2 + c^2)^(3/2)*c^2/d^4
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00 \[ \int (A+B x) (c+d x)^3 \sqrt {c^2-d^2 x^2} \, dx=\frac {7 \, {\left (B c^{6} + 2 \, A c^{5} d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{16 \, d {\left | d \right |}} + \frac {1}{240} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, B d^{3} x + \frac {6 \, {\left (3 \, B c d^{10} + A d^{11}\right )}}{d^{8}}\right )} x + \frac {5 \, {\left (17 \, B c^{2} d^{9} + 18 \, A c d^{10}\right )}}{d^{8}}\right )} x + \frac {16 \, {\left (B c^{3} d^{8} + 7 \, A c^{2} d^{9}\right )}}{d^{8}}\right )} x - \frac {15 \, {\left (7 \, B c^{4} d^{7} - 2 \, A c^{3} d^{8}\right )}}{d^{8}}\right )} x - \frac {16 \, {\left (11 \, B c^{5} d^{6} + 17 \, A c^{4} d^{7}\right )}}{d^{8}}\right )} \] Input:

integrate((B*x+A)*(d*x+c)^3*(-d^2*x^2+c^2)^(1/2),x, algorithm="giac")
 

Output:

7/16*(B*c^6 + 2*A*c^5*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) + 1/240*sq 
rt(-d^2*x^2 + c^2)*((2*((4*(5*B*d^3*x + 6*(3*B*c*d^10 + A*d^11)/d^8)*x + 5 
*(17*B*c^2*d^9 + 18*A*c*d^10)/d^8)*x + 16*(B*c^3*d^8 + 7*A*c^2*d^9)/d^8)*x 
 - 15*(7*B*c^4*d^7 - 2*A*c^3*d^8)/d^8)*x - 16*(11*B*c^5*d^6 + 17*A*c^4*d^7 
)/d^8)
 

Mupad [F(-1)]

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

int((c^2 - d^2*x^2)^(1/2)*(A + B*x)*(c + d*x)^3,x)
 

Output:

int((c^2 - d^2*x^2)^(1/2)*(A + B*x)*(c + d*x)^3, x)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.61 \[ \int (A+B x) (c+d x)^3 \sqrt {c^2-d^2 x^2} \, dx=\frac {210 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{5} d +105 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{6}-272 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{4} d +30 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d^{2} x +224 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{3} x^{2}+180 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{4} x^{3}+48 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{5} x^{4}-176 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{5}-105 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4} d x +32 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d^{2} x^{2}+170 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{3} x^{3}+144 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{4} x^{4}+40 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{5} x^{5}+272 a \,c^{5} d +176 b \,c^{6}}{240 d^{2}} \] Input:

int((B*x+A)*(d*x+c)^3*(-d^2*x^2+c^2)^(1/2),x)
 

Output:

(210*asin((d*x)/c)*a*c**5*d + 105*asin((d*x)/c)*b*c**6 - 272*sqrt(c**2 - d 
**2*x**2)*a*c**4*d + 30*sqrt(c**2 - d**2*x**2)*a*c**3*d**2*x + 224*sqrt(c* 
*2 - d**2*x**2)*a*c**2*d**3*x**2 + 180*sqrt(c**2 - d**2*x**2)*a*c*d**4*x** 
3 + 48*sqrt(c**2 - d**2*x**2)*a*d**5*x**4 - 176*sqrt(c**2 - d**2*x**2)*b*c 
**5 - 105*sqrt(c**2 - d**2*x**2)*b*c**4*d*x + 32*sqrt(c**2 - d**2*x**2)*b* 
c**3*d**2*x**2 + 170*sqrt(c**2 - d**2*x**2)*b*c**2*d**3*x**3 + 144*sqrt(c* 
*2 - d**2*x**2)*b*c*d**4*x**4 + 40*sqrt(c**2 - d**2*x**2)*b*d**5*x**5 + 27 
2*a*c**5*d + 176*b*c**6)/(240*d**2)