\(\int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^3} \, dx\) [7]

Optimal result

Integrand size = 29, antiderivative size = 107 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^3} \, dx=\frac {2 (B c-A d) \sqrt {c^2-d^2 x^2}}{3 d^2 (c+d x)^2}-\frac {(7 B c-A d) \sqrt {c^2-d^2 x^2}}{3 c d^2 (c+d x)}-\frac {B \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^2} \] Output:

2/3*(-A*d+B*c)*(-d^2*x^2+c^2)^(1/2)/d^2/(d*x+c)^2-1/3*(-A*d+7*B*c)*(-d^2*x 
^2+c^2)^(1/2)/c/d^2/(d*x+c)-B*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^2
 

Mathematica [A] (verified)

Time = 0.68 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.86 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^3} \, dx=\frac {-\frac {\sqrt {c^2-d^2 x^2} (A d (c-d x)+B c (5 c+7 d x))}{c (c+d x)^2}+6 B \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{3 d^2} \] Input:

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

Output:

(-((Sqrt[c^2 - d^2*x^2]*(A*d*(c - d*x) + B*c*(5*c + 7*d*x)))/(c*(c + d*x)^ 
2)) + 6*B*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/(3*d^2)
 

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {671, 463, 25, 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.

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

\(\Big \downarrow \) 671

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

\(\Big \downarrow \) 463

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], 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 463

Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ 
(-(-c)^(-n - 2))*d^(2*n + 3)*(Sqrt[a + b*x^2]/(2^(n + 1)*b^(n + 2)*(c + d*x 
))), x] - Simp[d^(2*n + 2)/b^(n + 1)   Int[(1/Sqrt[a + b*x^2])*ExpandToSum[ 
(2^(-n - 1)*(-c)^(-n - 1) - (-c + d*x)^(-n - 1))/(c + d*x), x], x], x] /; F 
reeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[n, 0] && EqQ[n + p, 
-3/2]
 

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 = 0.42 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.73

method	result	size
default	\(\frac {B \left (-\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{c d \left (x +\frac {c}{d}\right )^{2}}-\frac {d \left (\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}+\frac {c d \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{\sqrt {d^{2}}}\right )}{c}\right )}{d^{3}}-\frac {\left (A d -B c \right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{3 d^{5} c \left (x +\frac {c}{d}\right )^{3}}\)	\(185\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.57 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^3} \, dx=-\frac {5 \, B c^{3} + A c^{2} d + {\left (5 \, B c d^{2} + A d^{3}\right )} x^{2} + 2 \, {\left (5 \, B c^{2} d + A c d^{2}\right )} x - 6 \, {\left (B c d^{2} x^{2} + 2 \, B c^{2} d x + B c^{3}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + \sqrt {-d^{2} x^{2} + c^{2}} {\left (5 \, B c^{2} + A c d + {\left (7 \, B c d - A d^{2}\right )} x\right )}}{3 \, {\left (c d^{4} x^{2} + 2 \, c^{2} d^{3} x + c^{3} d^{2}\right )}} \] Input:

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

Output:

-1/3*(5*B*c^3 + A*c^2*d + (5*B*c*d^2 + A*d^3)*x^2 + 2*(5*B*c^2*d + A*c*d^2 
)*x - 6*(B*c*d^2*x^2 + 2*B*c^2*d*x + B*c^3)*arctan(-(c - sqrt(-d^2*x^2 + c 
^2))/(d*x)) + sqrt(-d^2*x^2 + c^2)*(5*B*c^2 + A*c*d + (7*B*c*d - A*d^2)*x) 
)/(c*d^4*x^2 + 2*c^2*d^3*x + c^3*d^2)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.59 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^3} \, dx=-\frac {B \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{d {\left | d \right |}} + \frac {2 \, {\left (5 \, B c + A d + \frac {12 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} B c}{d^{2} x} + \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} B c}{d^{4} x^{2}} + \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} A}{d^{3} x^{2}}\right )}}{3 \, c d {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )}^{3} {\left | d \right |}} \] Input:

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

Output:

-B*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) + 2/3*(5*B*c + A*d + 12*(c*d + s 
qrt(-d^2*x^2 + c^2)*abs(d))*B*c/(d^2*x) + 3*(c*d + sqrt(-d^2*x^2 + c^2)*ab 
s(d))^2*B*c/(d^4*x^2) + 3*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*A/(d^3*x^2 
))/(c*d*((c*d + sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2*x) + 1)^3*abs(d))
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^3} \, dx=\int \frac {\sqrt {c^2-d^2\,x^2}\,\left (A+B\,x\right )}{{\left (c+d\,x\right )}^3} \,d x \] 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 = 0.23 (sec) , antiderivative size = 193, normalized size of antiderivative = 1.80 \[ \int \frac {(A+B x) \sqrt {c^2-d^2 x^2}}{(c+d x)^3} \, dx=\frac {-3 \mathit {asin} \left (\frac {d x}{c}\right ) \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3} b c -9 \mathit {asin} \left (\frac {d x}{c}\right ) \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2} b c -9 \mathit {asin} \left (\frac {d x}{c}\right ) \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) b c -3 \mathit {asin} \left (\frac {d x}{c}\right ) b c +2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3} a d +2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3} b c +6 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) a d -18 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) b c -8 b c}{3 c \,d^{2} \left (\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3}+3 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2}+3 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )+1\right )} \] Input:

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

Output:

( - 3*asin((d*x)/c)*tan(asin((d*x)/c)/2)**3*b*c - 9*asin((d*x)/c)*tan(asin 
((d*x)/c)/2)**2*b*c - 9*asin((d*x)/c)*tan(asin((d*x)/c)/2)*b*c - 3*asin((d 
*x)/c)*b*c + 2*tan(asin((d*x)/c)/2)**3*a*d + 2*tan(asin((d*x)/c)/2)**3*b*c 
 + 6*tan(asin((d*x)/c)/2)*a*d - 18*tan(asin((d*x)/c)/2)*b*c - 8*b*c)/(3*c* 
d**2*(tan(asin((d*x)/c)/2)**3 + 3*tan(asin((d*x)/c)/2)**2 + 3*tan(asin((d* 
x)/c)/2) + 1))