\(\int \frac {(A+B x) (c^2-d^2 x^2)^{3/2}}{c+d x} \, dx\) [16]

Optimal result

Integrand size = 29, antiderivative size = 134 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{c+d x} \, dx=-\frac {c (B c-4 A d) x \sqrt {c^2-d^2 x^2}}{8 d}-\frac {(B c-4 A d) \left (c^2-d^2 x^2\right )^{3/2}}{12 d^2}-\frac {B \left (c^2-d^2 x^2\right )^{5/2}}{4 d^2 (c+d x)}-\frac {c^3 (B c-4 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d^2} \] Output:

-1/8*c*(-4*A*d+B*c)*x*(-d^2*x^2+c^2)^(1/2)/d-1/12*(-4*A*d+B*c)*(-d^2*x^2+c 
^2)^(3/2)/d^2-1/4*B*(-d^2*x^2+c^2)^(5/2)/d^2/(d*x+c)-1/8*c^3*(-4*A*d+B*c)* 
arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^2
 

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{c+d x} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (4 A d \left (2 c^2+3 c d x-2 d^2 x^2\right )+B \left (-8 c^3+3 c^2 d x+8 c d^2 x^2-6 d^3 x^3\right )\right )+6 c^3 (B c-4 A d) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{24 d^2} \] Input:

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

Output:

(Sqrt[c^2 - d^2*x^2]*(4*A*d*(2*c^2 + 3*c*d*x - 2*d^2*x^2) + B*(-8*c^3 + 3* 
c^2*d*x + 8*c*d^2*x^2 - 6*d^3*x^3)) + 6*c^3*(B*c - 4*A*d)*ArcTan[(d*x)/(Sq 
rt[c^2] - Sqrt[c^2 - d^2*x^2])])/(24*d^2)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.93, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {667, 676, 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.

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

\(\Big \downarrow \) 667

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

\(\Big \downarrow \) 676

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

-1/3*((B*c - A*d)*(c^2 - d^2*x^2)^(3/2))/d^2 + (B*x*(c^2 - d^2*x^2)^(3/2)) 
/(4*d) - (c*(B*c - 4*A*d)*((x*Sqrt[c^2 - d^2*x^2])/2 + (c^2*ArcTan[(d*x)/S 
qrt[c^2 - d^2*x^2]])/(2*d)))/(4*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 667

Int[(((f_.) + (g_.)*(x_))^(n_.)*((a_) + (c_.)*(x_)^2)^(p_))/((d_) + (e_.)*( 
x_)), x_Symbol] :> Int[(a/d + c*(x/e))*(f + g*x)^n*(a + c*x^2)^(p - 1), x] 
/; FreeQ[{a, c, d, e, f, g, n, p}, x] && EqQ[c*d^2 + a*e^2, 0] && GtQ[p, 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 = 0.38 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.92

method	result	size
risch	\(\frac {\left (-6 B \,d^{3} x^{3}-8 A \,d^{3} x^{2}+8 B c \,d^{2} x^{2}+12 A c \,d^{2} x +3 B \,c^{2} d x +8 A \,c^{2} d -8 B \,c^{3}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{24 d^{2}}+\frac {c^{3} \left (4 A d -B c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{8 d \sqrt {d^{2}}}\)	\(123\)
default	\(\frac {B \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )}{d}+\frac {\left (A d -B c \right ) \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )\right )}{d^{2}}\)	\(223\)

Input:

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

Output:

1/24/d^2*(-6*B*d^3*x^3-8*A*d^3*x^2+8*B*c*d^2*x^2+12*A*c*d^2*x+3*B*c^2*d*x+ 
8*A*c^2*d-8*B*c^3)*(-d^2*x^2+c^2)^(1/2)+1/8*c^3/d*(4*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 = 0.10 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.91 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{c+d x} \, dx=\frac {6 \, {\left (B c^{4} - 4 \, A c^{3} d\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (6 \, B d^{3} x^{3} + 8 \, B c^{3} - 8 \, A c^{2} d - 8 \, {\left (B c d^{2} - A d^{3}\right )} x^{2} - 3 \, {\left (B c^{2} d + 4 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{24 \, d^{2}} \] Input:

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

Output:

1/24*(6*(B*c^4 - 4*A*c^3*d)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) - (6 
*B*d^3*x^3 + 8*B*c^3 - 8*A*c^2*d - 8*(B*c*d^2 - A*d^3)*x^2 - 3*(B*c^2*d + 
4*A*c*d^2)*x)*sqrt(-d^2*x^2 + c^2))/d^2
 

Sympy [A] (verification not implemented)

Time = 2.05 (sec) , antiderivative size = 296, normalized size of antiderivative = 2.21 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{c+d x} \, dx=A c \left (\begin {cases} \frac {c^{2} \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {c^{2} - d^{2} x^{2}}}{2} & \text {for}\: d^{2} \neq 0 \\x \sqrt {c^{2}} & \text {otherwise} \end {cases}\right ) - A d \left (\begin {cases} \sqrt {c^{2} - d^{2} x^{2}} \left (- \frac {c^{2}}{3 d^{2}} + \frac {x^{2}}{3}\right ) & \text {for}\: d^{2} \neq 0 \\\frac {x^{2} \sqrt {c^{2}}}{2} & \text {otherwise} \end {cases}\right ) + B c \left (\begin {cases} \sqrt {c^{2} - d^{2} x^{2}} \left (- \frac {c^{2}}{3 d^{2}} + \frac {x^{2}}{3}\right ) & \text {for}\: d^{2} \neq 0 \\\frac {x^{2} \sqrt {c^{2}}}{2} & \text {otherwise} \end {cases}\right ) - B d \left (\begin {cases} \frac {c^{4} \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8 d^{2}} + \sqrt {c^{2} - d^{2} x^{2}} \left (- \frac {c^{2} x}{8 d^{2}} + \frac {x^{3}}{4}\right ) & \text {for}\: d^{2} \neq 0 \\\frac {x^{3} \sqrt {c^{2}}}{3} & \text {otherwise} \end {cases}\right ) \] Input:

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

Output:

A*c*Piecewise((c**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))/2 
 + x*sqrt(c**2 - d**2*x**2)/2, Ne(d**2, 0)), (x*sqrt(c**2), True)) - A*d*P 
iecewise((sqrt(c**2 - d**2*x**2)*(-c**2/(3*d**2) + x**2/3), Ne(d**2, 0)), 
(x**2*sqrt(c**2)/2, True)) + B*c*Piecewise((sqrt(c**2 - d**2*x**2)*(-c**2/ 
(3*d**2) + x**2/3), Ne(d**2, 0)), (x**2*sqrt(c**2)/2, True)) - B*d*Piecewi 
se((c**4*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))/(8*d**2) + s 
qrt(c**2 - d**2*x**2)*(-c**2*x/(8*d**2) + x**3/4), Ne(d**2, 0)), (x**3*sqr 
t(c**2)/3, True))
 

Maxima [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.13 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.87 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{c+d x} \, dx=\frac {i \, B c^{4} \arcsin \left (\frac {d x}{c} + 2\right )}{2 \, d^{2}} - \frac {i \, A c^{3} \arcsin \left (\frac {d x}{c} + 2\right )}{2 \, d} + \frac {3 \, B c^{4} \arcsin \left (\frac {d x}{c}\right )}{8 \, d^{2}} + \frac {1}{2} \, \sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} A c x - \frac {\sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} B c^{2} x}{2 \, d} + \frac {3 \, \sqrt {-d^{2} x^{2} + c^{2}} B c^{2} x}{8 \, d} - \frac {\sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} B c^{3}}{d^{2}} + \frac {\sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} A c^{2}}{d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B x}{4 \, d} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c}{3 \, d^{2}} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A}{3 \, d} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.93 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{c+d x} \, dx=-\frac {{\left (B c^{4} - 4 \, A c^{3} d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{8 \, d {\left | d \right |}} - \frac {1}{24} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left (3 \, B d x - \frac {4 \, {\left (B c d^{4} - A d^{5}\right )}}{d^{4}}\right )} x - \frac {3 \, {\left (B c^{2} d^{3} + 4 \, A c d^{4}\right )}}{d^{4}}\right )} x + \frac {8 \, {\left (B c^{3} d^{2} - A c^{2} d^{3}\right )}}{d^{4}}\right )} \] Input:

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

Output:

-1/8*(B*c^4 - 4*A*c^3*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) - 1/24*sqr 
t(-d^2*x^2 + c^2)*((2*(3*B*d*x - 4*(B*c*d^4 - A*d^5)/d^4)*x - 3*(B*c^2*d^3 
 + 4*A*c*d^4)/d^4)*x + 8*(B*c^3*d^2 - A*c^2*d^3)/d^4)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.45 \[ \int \frac {(A+B x) \left (c^2-d^2 x^2\right )^{3/2}}{c+d x} \, dx=\frac {12 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{3} d -3 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{4}+8 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d +12 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{2} x -8 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{3} x^{2}-8 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3}+3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d x +8 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{2} x^{2}-6 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{3} x^{3}-8 a \,c^{3} d +8 b \,c^{4}}{24 d^{2}} \] Input:

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

Output:

(12*asin((d*x)/c)*a*c**3*d - 3*asin((d*x)/c)*b*c**4 + 8*sqrt(c**2 - d**2*x 
**2)*a*c**2*d + 12*sqrt(c**2 - d**2*x**2)*a*c*d**2*x - 8*sqrt(c**2 - d**2* 
x**2)*a*d**3*x**2 - 8*sqrt(c**2 - d**2*x**2)*b*c**3 + 3*sqrt(c**2 - d**2*x 
**2)*b*c**2*d*x + 8*sqrt(c**2 - d**2*x**2)*b*c*d**2*x**2 - 6*sqrt(c**2 - d 
**2*x**2)*b*d**3*x**3 - 8*a*c**3*d + 8*b*c**4)/(24*d**2)