\(\int \frac {(A+B x) (c+d x)^3}{(c^2-d^2 x^2)^{3/2}} \, dx\) [48]

Optimal result

Integrand size = 29, antiderivative size = 112 \[ \int \frac {(A+B x) (c+d x)^3}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {2 (B c+A d) (c+d x)^2}{d^2 \sqrt {c^2-d^2 x^2}}+\frac {(2 (5 B c+3 A d)+B d x) \sqrt {c^2-d^2 x^2}}{2 d^2}-\frac {3 c (3 B c+2 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d^2} \] Output:

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

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.98 \[ \int \frac {(A+B x) (c+d x)^3}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\frac {\sqrt {c^2-d^2 x^2} \left (2 A d (-5 c+d x)+B \left (-14 c^2+5 c d x+d^2 x^2\right )\right )}{-c+d x}+6 c (3 B c+2 A d) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{2 d^2} \] Input:

Integrate[((A + B*x)*(c + d*x)^3)/(c^2 - d^2*x^2)^(3/2),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 = 0.57 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.18, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.241, Rules used = {665, 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.

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

\(\Big \downarrow \) 665

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

\(\Big \downarrow \) 2346

\(\displaystyle \frac {4 c (c+d x) (A d+B c)}{d^2 \sqrt {c^2-d^2 x^2}}-\frac {-\frac {\int -\frac {d^2 (3 c (3 B c+2 A d)+2 d (3 B c+A d) x)}{\sqrt {c^2-d^2 x^2}}dx}{2 d^2}-\frac {1}{2} B x \sqrt {c^2-d^2 x^2}}{d}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {4 c (c+d x) (A d+B c)}{d^2 \sqrt {c^2-d^2 x^2}}-\frac {\frac {\int \frac {d^2 (3 c (3 B c+2 A d)+2 d (3 B c+A d) x)}{\sqrt {c^2-d^2 x^2}}dx}{2 d^2}-\frac {1}{2} B x \sqrt {c^2-d^2 x^2}}{d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {4 c (c+d x) (A d+B c)}{d^2 \sqrt {c^2-d^2 x^2}}-\frac {\frac {1}{2} \int \frac {3 c (3 B c+2 A d)+2 d (3 B c+A d) x}{\sqrt {c^2-d^2 x^2}}dx-\frac {1}{2} B x \sqrt {c^2-d^2 x^2}}{d}\)

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

(4*c*(B*c + A*d)*(c + d*x))/(d^2*Sqrt[c^2 - d^2*x^2]) - (-1/2*(B*x*Sqrt[c^ 
2 - d^2*x^2]) + ((-2*(3*B*c + A*d)*Sqrt[c^2 - d^2*x^2])/d + (3*c*(3*B*c + 
2*A*d)*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/d)/2)/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 665

Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( 
x_)^2)^(3/2), x_Symbol] :> Simp[(-2^(m - 1))*d^(m - 2)*(e*f + d*g)^n*((d + 
e*x)/(c*e^(n - 1)*Sqrt[a + c*x^2])), x] + Simp[1/(c*e^(n - 2))   Int[Expand 
ToSum[(2^(m - 1)*d^(m - 1)*(e*f + d*g)^n - e^n*(d + e*x)^(m - 1)*(f + g*x)^ 
n)/(d - e*x), x]/Sqrt[a + c*x^2], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && 
 EqQ[c*d^2 + a*e^2, 0] && IGtQ[m, 0] && IGtQ[n, 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 = 0.50 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.38

method	result	size
risch	\(\frac {\left (B d x +2 A d +6 B c \right ) \sqrt {-d^{2} x^{2}+c^{2}}}{2 d^{2}}-\frac {c \left (\frac {6 A d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}+\frac {9 B c \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{\sqrt {d^{2}}}+\frac {8 \left (A d +B c \right ) \sqrt {-d^{2} \left (x -\frac {c}{d}\right )^{2}-2 c d \left (x -\frac {c}{d}\right )}}{d^{2} \left (x -\frac {c}{d}\right )}\right )}{2 d}\)	\(155\)
default	\(\frac {A c x}{\sqrt {-d^{2} x^{2}+c^{2}}}+d^{2} \left (A d +3 B c \right ) \left (-\frac {x^{2}}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {2 c^{2}}{d^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )+3 c d \left (A d +B c \right ) \left (\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}\right )+\frac {c^{2} \left (3 A d +B c \right )}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}+B \,d^{3} \left (-\frac {x^{3}}{2 d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {3 c^{2} \left (\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}\right )}{2 d^{2}}\right )\)	\(257\)

Input:

int((B*x+A)*(d*x+c)^3/(-d^2*x^2+c^2)^(3/2),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 = 159, normalized size of antiderivative = 1.42 \[ \int \frac {(A+B x) (c+d x)^3}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {14 \, B c^{3} + 10 \, A c^{2} d - 2 \, {\left (7 \, B c^{2} d + 5 \, A c d^{2}\right )} x + 6 \, {\left (3 \, B c^{3} + 2 \, A c^{2} d - {\left (3 \, B c^{2} d + 2 \, A c d^{2}\right )} x\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (B d^{2} x^{2} - 14 \, B c^{2} - 10 \, A c d + {\left (5 \, B c d + 2 \, A d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{2 \, {\left (d^{3} x - c d^{2}\right )}} \] Input:

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

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (99) = 198\).

Time = 0.11 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.26 \[ \int \frac {(A+B x) (c+d x)^3}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {B d x^{3}}{2 \, \sqrt {-d^{2} x^{2} + c^{2}}} + \frac {A c x}{\sqrt {-d^{2} x^{2} + c^{2}}} + \frac {3 \, B c^{2} x}{2 \, \sqrt {-d^{2} x^{2} + c^{2}} d} - \frac {3 \, B c^{2} \arcsin \left (\frac {d x}{c}\right )}{2 \, d^{2}} + \frac {B c^{3}}{\sqrt {-d^{2} x^{2} + c^{2}} d^{2}} + \frac {3 \, A c^{2}}{\sqrt {-d^{2} x^{2} + c^{2}} d} - \frac {{\left (3 \, B c d^{2} + A d^{3}\right )} x^{2}}{\sqrt {-d^{2} x^{2} + c^{2}} d^{2}} + \frac {3 \, {\left (B c^{2} d + A c d^{2}\right )} x}{\sqrt {-d^{2} x^{2} + c^{2}} d^{2}} - \frac {3 \, {\left (B c^{2} d + A c d^{2}\right )} \arcsin \left (\frac {d x}{c}\right )}{d^{3}} + \frac {2 \, {\left (3 \, B c d^{2} + A d^{3}\right )} c^{2}}{\sqrt {-d^{2} x^{2} + c^{2}} d^{4}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.12 \[ \int \frac {(A+B x) (c+d x)^3}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {3 \, {\left (3 \, B c^{2} + 2 \, A c d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{2 \, d {\left | d \right |}} + \frac {1}{2} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (\frac {B x}{d} + \frac {2 \, {\left (3 \, B c d^{2} + A d^{3}\right )}}{d^{4}}\right )} + \frac {8 \, {\left (B c^{2} + A c d\right )}}{d {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} - 1\right )} {\left | d \right |}} \] Input:

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 290, normalized size of antiderivative = 2.59 \[ \int \frac {(A+B x) (c+d x)^3}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {-6 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) a c d -9 \sqrt {-d^{2} x^{2}+c^{2}}\, \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2}+6 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{2} d -6 \mathit {asin} \left (\frac {d x}{c}\right ) a c \,d^{2} x +9 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{3}-9 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{2} d x -16 \sqrt {-d^{2} x^{2}+c^{2}}\, a c d +2 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{2} x -18 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2}+5 \sqrt {-d^{2} x^{2}+c^{2}}\, b c d x +\sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{2} x^{2}+16 a \,c^{2} d +2 a c \,d^{2} x -2 a \,d^{3} x^{2}+18 b \,c^{3}+5 b \,c^{2} d x -6 b c \,d^{2} x^{2}-b \,d^{3} x^{3}}{2 d^{2} \left (\sqrt {-d^{2} x^{2}+c^{2}}-c +d x \right )} \] Input:

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