\(\int \frac {(A+B x) (c^2-d^2 x^2)^{3/2}}{(c+d x)^3} \, dx\) [18]

Optimal result

Integrand size = 29, antiderivative size = 119 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^3} \, dx=\frac {(3 B c-2 A d) (4 c-d x) \sqrt {c^2-d^2 x^2}}{2 c d^2}+\frac {(B c-A d) \left (c^2-d^2 x^2\right )^{5/2}}{c d^2 (c+d x)^3}+\frac {3 c (3 B c-2 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.59 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.92 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^3} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (-2 A d (5 c+d x)+B \left (14 c^2+5 c d x-d^2 x^2\right )\right )}{c+d x}-6 c (3 B c-2 A d) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{2 d^2} \] Input:

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

Output:

((Sqrt[c^2 - d^2*x^2]*(-2*A*d*(5*c + d*x) + B*(14*c^2 + 5*c*d*x - d^2*x^2) 
))/(c + d*x) - 6*c*(3*B*c - 2*A*d)*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^ 
2*x^2])])/(2*d^2)
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.15, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {671, 466, 466, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 466

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

Defintions of rubi rules used

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 466

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

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]
 

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.29

method	result	size
risch	\(-\frac {\left (B d x +2 A d -6 B c \right ) \sqrt {-d^{2} x^{2}+c^{2}}}{2 d^{2}}-\frac {c \left (\frac {6 A d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}-\frac {9 B c \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}+\frac {8 \left (A d -B c \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{d^{2} \left (x +\frac {c}{d}\right )}\right )}{2 d}\)	\(153\)
default	\(\frac {B \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}+\frac {3 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )\right )}{c}\right )}{d^{3}}+\frac {\left (A d -B c \right ) \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{c d \left (x +\frac {c}{d}\right )^{3}}-\frac {2 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}+\frac {3 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )\right )}{c}\right )}{c}\right )}{d^{4}}\)	\(435\)

Input:

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

Output:

-1/2*(B*d*x+2*A*d-6*B*c)/d^2*(-d^2*x^2+c^2)^(1/2)-1/2*c/d*(6*A*d/(d^2)^(1/ 
2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))-9*B*c/(d^2)^(1/2)*arctan((d^ 
2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))+8*(A*d-B*c)/d^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 = 158, normalized size of antiderivative = 1.33 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^3} \, dx=\frac {14 \, B c^{3} - 10 \, A c^{2} d + 2 \, {\left (7 \, B c^{2} d - 5 \, A c d^{2}\right )} x - 6 \, {\left (3 \, B c^{3} - 2 \, A c^{2} d + {\left (3 \, B c^{2} d - 2 \, A c d^{2}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (B d^{2} x^{2} - 14 \, B c^{2} + 10 \, A c d - {\left (5 \, B c d - 2 \, A d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{2 \, {\left (d^{3} x + c d^{2}\right )}} \] Input:

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

Output:

1/2*(14*B*c^3 - 10*A*c^2*d + 2*(7*B*c^2*d - 5*A*c*d^2)*x - 6*(3*B*c^3 - 2* 
A*c^2*d + (3*B*c^2*d - 2*A*c*d^2)*x)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d 
*x)) - (B*d^2*x^2 - 14*B*c^2 + 10*A*c*d - (5*B*c*d - 2*A*d^2)*x)*sqrt(-d^2 
*x^2 + c^2))/(d^3*x + c*d^2)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 224 vs. \(2 (108) = 216\).

Time = 0.12 (sec) , antiderivative size = 224, normalized size of antiderivative = 1.88 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^3} \, dx=-\frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c}{d^{4} x^{2} + 2 \, c d^{3} x + c^{2} d^{2}} + \frac {6 \, \sqrt {-d^{2} x^{2} + c^{2}} B c^{2}}{d^{3} x + c d^{2}} + \frac {9 \, B c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{2}} - \frac {3 \, A c \arcsin \left (\frac {d x}{c}\right )}{d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A}{d^{3} x^{2} + 2 \, c d^{2} x + c^{2} d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B}{2 \, {\left (d^{3} x + c d^{2}\right )}} - \frac {6 \, \sqrt {-d^{2} x^{2} + c^{2}} A c}{d^{2} x + c d} + \frac {3 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{2 \, d^{2}} \] Input:

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

Output:

-(-d^2*x^2 + c^2)^(3/2)*B*c/(d^4*x^2 + 2*c*d^3*x + c^2*d^2) + 6*sqrt(-d^2* 
x^2 + c^2)*B*c^2/(d^3*x + c*d^2) + 9/2*B*c^2*arcsin(d*x/c)/d^2 - 3*A*c*arc 
sin(d*x/c)/d + (-d^2*x^2 + c^2)^(3/2)*A/(d^3*x^2 + 2*c*d^2*x + c^2*d) + 1/ 
2*(-d^2*x^2 + c^2)^(3/2)*B/(d^3*x + c*d^2) - 6*sqrt(-d^2*x^2 + c^2)*A*c/(d 
^2*x + c*d) + 3/2*sqrt(-d^2*x^2 + c^2)*B*c/d^2
 

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.07 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^3} \, dx=\frac {3 \, {\left (3 \, B c^{2} - 2 \, A c d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{2 \, d {\left | d \right |}} - \frac {1}{2} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (\frac {B x}{d} - \frac {2 \, {\left (3 \, B c d^{2} - A d^{3}\right )}}{d^{4}}\right )} - \frac {8 \, {\left (B c^{2} - A c d\right )}}{d {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )} {\left | d \right |}} \] Input:

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

Output:

3/2*(3*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*(3*B*c*d^2 - A*d^3)/d^4) - 8*(B*c^2 - A*c*d)/(d 
*((c*d + sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*x) + 1)*abs(d))
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.25 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.44 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{(c+d x)^3} \, dx=\frac {-6 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a c d +9 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2}+6 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d +6 \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{2} x -9 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3}-9 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d x +16 \sqrt {-d^{2} x^{2}+c^{2}}\, a c d +2 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{2} x -18 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2}-5 \sqrt {-d^{2} x^{2}+c^{2}}\, b c d x +\sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{2} x^{2}-16 a \,c^{2} d +2 a c \,d^{2} x +2 a \,d^{3} x^{2}+18 b \,c^{3}-5 b \,c^{2} d x -6 b c \,d^{2} x^{2}+b \,d^{3} x^{3}}{2 d^{2} \left (\sqrt {-d^{2} x^{2}+c^{2}}-c -d x \right )} \] Input:

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

Output:

( - 6*sqrt(c**2 - d**2*x**2)*asin((d*x)/c)*a*c*d + 9*sqrt(c**2 - d**2*x**2 
)*asin((d*x)/c)*b*c**2 + 6*asin((d*x)/c)*a*c**2*d + 6*asin((d*x)/c)*a*c*d* 
*2*x - 9*asin((d*x)/c)*b*c**3 - 9*asin((d*x)/c)*b*c**2*d*x + 16*sqrt(c**2 
- d**2*x**2)*a*c*d + 2*sqrt(c**2 - d**2*x**2)*a*d**2*x - 18*sqrt(c**2 - d* 
*2*x**2)*b*c**2 - 5*sqrt(c**2 - d**2*x**2)*b*c*d*x + sqrt(c**2 - d**2*x**2 
)*b*d**2*x**2 - 16*a*c**2*d + 2*a*c*d**2*x + 2*a*d**3*x**2 + 18*b*c**3 - 5 
*b*c**2*d*x - 6*b*c*d**2*x**2 + b*d**3*x**3)/(2*d**2*(sqrt(c**2 - d**2*x** 
2) - c - d*x))