Integrand size = 29, antiderivative size = 120 \[ \int \frac {A+B x}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx=\frac {(3 B c+2 A d) x}{15 c^3 d \sqrt {c^2-d^2 x^2}}-\frac {2 (3 B c+2 A d)}{15 c d^2 (c+d x) \sqrt {c^2-d^2 x^2}}+\frac {(B c-A d) \sqrt {c^2-d^2 x^2}}{5 c d^2 (c+d x)^3} \] Output:
1/15*(2*A*d+3*B*c)*x/c^3/d/(-d^2*x^2+c^2)^(1/2)-2/15*(2*A*d+3*B*c)/c/d^2/( d*x+c)/(-d^2*x^2+c^2)^(1/2)+1/5*(-A*d+B*c)*(-d^2*x^2+c^2)^(1/2)/c/d^2/(d*x +c)^3
Time = 0.60 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.63 \[ \int \frac {A+B x}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx=-\frac {\sqrt {c^2-d^2 x^2} \left (3 B c \left (c^2+3 c d x+d^2 x^2\right )+A d \left (7 c^2+6 c d x+2 d^2 x^2\right )\right )}{15 c^3 d^2 (c+d x)^3} \] Input:
Integrate[(A + B*x)/((c + d*x)^3*Sqrt[c^2 - d^2*x^2]),x]
Output:
-1/15*(Sqrt[c^2 - d^2*x^2]*(3*B*c*(c^2 + 3*c*d*x + d^2*x^2) + A*d*(7*c^2 + 6*c*d*x + 2*d^2*x^2)))/(c^3*d^2*(c + d*x)^3)
Time = 0.39 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.07, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {671, 461, 460}
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.
\(\displaystyle \int \frac {A+B x}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx\) |
\(\Big \downarrow \) 671 |
\(\displaystyle \frac {(2 A d+3 B c) \int \frac {1}{(c+d x)^2 \sqrt {c^2-d^2 x^2}}dx}{5 c d}+\frac {\sqrt {c^2-d^2 x^2} (B c-A d)}{5 c d^2 (c+d x)^3}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {(2 A d+3 B c) \left (\frac {\int \frac {1}{(c+d x) \sqrt {c^2-d^2 x^2}}dx}{3 c}-\frac {\sqrt {c^2-d^2 x^2}}{3 c d (c+d x)^2}\right )}{5 c d}+\frac {\sqrt {c^2-d^2 x^2} (B c-A d)}{5 c d^2 (c+d x)^3}\) |
\(\Big \downarrow \) 460 |
\(\displaystyle \frac {\sqrt {c^2-d^2 x^2} (B c-A d)}{5 c d^2 (c+d x)^3}+\frac {\left (-\frac {\sqrt {c^2-d^2 x^2}}{3 c^2 d (c+d x)}-\frac {\sqrt {c^2-d^2 x^2}}{3 c d (c+d x)^2}\right ) (2 A d+3 B c)}{5 c d}\) |
Input:
Int[(A + B*x)/((c + d*x)^3*Sqrt[c^2 - d^2*x^2]),x]
Output:
((B*c - A*d)*Sqrt[c^2 - d^2*x^2])/(5*c*d^2*(c + d*x)^3) + ((3*B*c + 2*A*d) *(-1/3*Sqrt[c^2 - d^2*x^2]/(c*d*(c + d*x)^2) - Sqrt[c^2 - d^2*x^2]/(3*c^2* d*(c + d*x))))/(5*c*d)
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(b*c*n)), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && EqQ[n + 2*p + 2, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ (-d)*(c + d*x)^n*((a + b*x^2)^(p + 1)/(2*b*c*(n + p + 1))), x] + Simp[Simpl ify[n + 2*p + 2]/(2*c*(n + p + 1)) Int[(c + d*x)^(n + 1)*(a + b*x^2)^p, x ], x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && ILtQ[Simp lify[n + 2*p + 2], 0] && (LtQ[n, -1] || GtQ[n + p, 0])
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]
Time = 0.42 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.66
method | result | size |
trager | \(-\frac {\left (2 A \,d^{3} x^{2}+3 B c \,d^{2} x^{2}+6 A c \,d^{2} x +9 B \,c^{2} d x +7 A \,c^{2} d +3 B \,c^{3}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{15 c^{3} \left (d x +c \right )^{3} d^{2}}\) | \(79\) |
gosper | \(-\frac {\left (-d x +c \right ) \left (2 A \,d^{3} x^{2}+3 B c \,d^{2} x^{2}+6 A c \,d^{2} x +9 B \,c^{2} d x +7 A \,c^{2} d +3 B \,c^{3}\right )}{15 \left (d x +c \right )^{2} c^{3} d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}\) | \(85\) |
orering | \(-\frac {\left (-d x +c \right ) \left (2 A \,d^{3} x^{2}+3 B c \,d^{2} x^{2}+6 A c \,d^{2} x +9 B \,c^{2} d x +7 A \,c^{2} d +3 B \,c^{3}\right )}{15 \left (d x +c \right )^{2} c^{3} d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}\) | \(85\) |
default | \(\frac {B \left (-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c d \left (x +\frac {c}{d}\right )^{2}}-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c^{2} \left (x +\frac {c}{d}\right )}\right )}{d^{3}}+\frac {\left (A d -B c \right ) \left (-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{5 c d \left (x +\frac {c}{d}\right )^{3}}+\frac {2 d \left (-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c d \left (x +\frac {c}{d}\right )^{2}}-\frac {\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{3 c^{2} \left (x +\frac {c}{d}\right )}\right )}{5 c}\right )}{d^{4}}\) | \(247\) |
Input:
int((B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/15*(2*A*d^3*x^2+3*B*c*d^2*x^2+6*A*c*d^2*x+9*B*c^2*d*x+7*A*c^2*d+3*B*c^3 )/c^3/(d*x+c)^3/d^2*(-d^2*x^2+c^2)^(1/2)
Time = 0.09 (sec) , antiderivative size = 181, normalized size of antiderivative = 1.51 \[ \int \frac {A+B x}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx=-\frac {3 \, B c^{4} + 7 \, A c^{3} d + {\left (3 \, B c d^{3} + 7 \, A d^{4}\right )} x^{3} + 3 \, {\left (3 \, B c^{2} d^{2} + 7 \, A c d^{3}\right )} x^{2} + 3 \, {\left (3 \, B c^{3} d + 7 \, A c^{2} d^{2}\right )} x + {\left (3 \, B c^{3} + 7 \, A c^{2} d + {\left (3 \, B c d^{2} + 2 \, A d^{3}\right )} x^{2} + 3 \, {\left (3 \, B c^{2} d + 2 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{15 \, {\left (c^{3} d^{5} x^{3} + 3 \, c^{4} d^{4} x^{2} + 3 \, c^{5} d^{3} x + c^{6} d^{2}\right )}} \] Input:
integrate((B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(1/2),x, algorithm="fricas")
Output:
-1/15*(3*B*c^4 + 7*A*c^3*d + (3*B*c*d^3 + 7*A*d^4)*x^3 + 3*(3*B*c^2*d^2 + 7*A*c*d^3)*x^2 + 3*(3*B*c^3*d + 7*A*c^2*d^2)*x + (3*B*c^3 + 7*A*c^2*d + (3 *B*c*d^2 + 2*A*d^3)*x^2 + 3*(3*B*c^2*d + 2*A*c*d^2)*x)*sqrt(-d^2*x^2 + c^2 ))/(c^3*d^5*x^3 + 3*c^4*d^4*x^2 + 3*c^5*d^3*x + c^6*d^2)
\[ \int \frac {A+B x}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx=\int \frac {A + B x}{\sqrt {- \left (- c + d x\right ) \left (c + d x\right )} \left (c + d x\right )^{3}}\, dx \] Input:
integrate((B*x+A)/(d*x+c)**3/(-d**2*x**2+c**2)**(1/2),x)
Output:
Integral((A + B*x)/(sqrt(-(-c + d*x)*(c + d*x))*(c + d*x)**3), x)
Leaf count of result is larger than twice the leaf count of optimal. 349 vs. \(2 (108) = 216\).
Time = 0.12 (sec) , antiderivative size = 349, normalized size of antiderivative = 2.91 \[ \int \frac {A+B x}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx=\frac {\sqrt {-d^{2} x^{2} + c^{2}} B c}{5 \, {\left (c d^{5} x^{3} + 3 \, c^{2} d^{4} x^{2} + 3 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{15 \, {\left (c^{2} d^{4} x^{2} + 2 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} + \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} B c}{15 \, {\left (c^{3} d^{3} x + c^{4} d^{2}\right )}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} A}{5 \, {\left (c d^{4} x^{3} + 3 \, c^{2} d^{3} x^{2} + 3 \, c^{3} d^{2} x + c^{4} d\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{15 \, {\left (c^{2} d^{3} x^{2} + 2 \, c^{3} d^{2} x + c^{4} d\right )}} - \frac {2 \, \sqrt {-d^{2} x^{2} + c^{2}} A}{15 \, {\left (c^{3} d^{2} x + c^{4} d\right )}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} B}{3 \, {\left (c d^{4} x^{2} + 2 \, c^{2} d^{3} x + c^{3} d^{2}\right )}} - \frac {\sqrt {-d^{2} x^{2} + c^{2}} B}{3 \, {\left (c^{2} d^{3} x + c^{3} d^{2}\right )}} \] Input:
integrate((B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(1/2),x, algorithm="maxima")
Output:
1/5*sqrt(-d^2*x^2 + c^2)*B*c/(c*d^5*x^3 + 3*c^2*d^4*x^2 + 3*c^3*d^3*x + c^ 4*d^2) + 2/15*sqrt(-d^2*x^2 + c^2)*B*c/(c^2*d^4*x^2 + 2*c^3*d^3*x + c^4*d^ 2) + 2/15*sqrt(-d^2*x^2 + c^2)*B*c/(c^3*d^3*x + c^4*d^2) - 1/5*sqrt(-d^2*x ^2 + c^2)*A/(c*d^4*x^3 + 3*c^2*d^3*x^2 + 3*c^3*d^2*x + c^4*d) - 2/15*sqrt( -d^2*x^2 + c^2)*A/(c^2*d^3*x^2 + 2*c^3*d^2*x + c^4*d) - 2/15*sqrt(-d^2*x^2 + c^2)*A/(c^3*d^2*x + c^4*d) - 1/3*sqrt(-d^2*x^2 + c^2)*B/(c*d^4*x^2 + 2* c^2*d^3*x + c^3*d^2) - 1/3*sqrt(-d^2*x^2 + c^2)*B/(c^2*d^3*x + c^3*d^2)
Leaf count of result is larger than twice the leaf count of optimal. 276 vs. \(2 (108) = 216\).
Time = 0.15 (sec) , antiderivative size = 276, normalized size of antiderivative = 2.30 \[ \int \frac {A+B x}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx=\frac {2 \, {\left (3 \, B c + 7 \, A d + \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} B c}{d^{2} x} + \frac {20 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} A}{d x} + \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} B c}{d^{4} x^{2}} + \frac {40 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} A}{d^{3} x^{2}} + \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} B c}{d^{6} x^{3}} + \frac {30 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{3} A}{d^{5} x^{3}} + \frac {15 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{4} A}{d^{7} x^{4}}\right )}}{15 \, c^{3} d {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} + 1\right )}^{5} {\left | d \right |}} \] Input:
integrate((B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(1/2),x, algorithm="giac")
Output:
2/15*(3*B*c + 7*A*d + 15*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*B*c/(d^2*x) + 20*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*A/(d*x) + 15*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2*B*c/(d^4*x^2) + 40*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^2* A/(d^3*x^2) + 15*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*B*c/(d^6*x^3) + 30* (c*d + sqrt(-d^2*x^2 + c^2)*abs(d))^3*A/(d^5*x^3) + 15*(c*d + sqrt(-d^2*x^ 2 + c^2)*abs(d))^4*A/(d^7*x^4))/(c^3*d*((c*d + sqrt(-d^2*x^2 + c^2)*abs(d) )/(d^2*x) + 1)^5*abs(d))
Time = 9.15 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.65 \[ \int \frac {A+B x}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx=-\frac {\sqrt {c^2-d^2\,x^2}\,\left (3\,B\,c^3+9\,B\,c^2\,d\,x+7\,A\,c^2\,d+3\,B\,c\,d^2\,x^2+6\,A\,c\,d^2\,x+2\,A\,d^3\,x^2\right )}{15\,c^3\,d^2\,{\left (c+d\,x\right )}^3} \] Input:
int((A + B*x)/((c^2 - d^2*x^2)^(1/2)*(c + d*x)^3),x)
Output:
-((c^2 - d^2*x^2)^(1/2)*(3*B*c^3 + 2*A*d^3*x^2 + 7*A*c^2*d + 6*A*c*d^2*x + 9*B*c^2*d*x + 3*B*c*d^2*x^2))/(15*c^3*d^2*(c + d*x)^3)
Time = 0.20 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.49 \[ \int \frac {A+B x}{(c+d x)^3 \sqrt {c^2-d^2 x^2}} \, dx=\frac {\frac {2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{5} a d}{5}-2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3} b c -\frac {4 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2} a d}{3}-2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2} b c -\frac {2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) a d}{3}-2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) b c -\frac {8 a d}{15}-\frac {2 b c}{5}}{c^{3} d^{2} \left (\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{5}+5 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{4}+10 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3}+10 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2}+5 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )+1\right )} \] Input:
int((B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(1/2),x)
Output:
(2*(3*tan(asin((d*x)/c)/2)**5*a*d - 15*tan(asin((d*x)/c)/2)**3*b*c - 10*ta n(asin((d*x)/c)/2)**2*a*d - 15*tan(asin((d*x)/c)/2)**2*b*c - 5*tan(asin((d *x)/c)/2)*a*d - 15*tan(asin((d*x)/c)/2)*b*c - 4*a*d - 3*b*c))/(15*c**3*d** 2*(tan(asin((d*x)/c)/2)**5 + 5*tan(asin((d*x)/c)/2)**4 + 10*tan(asin((d*x) /c)/2)**3 + 10*tan(asin((d*x)/c)/2)**2 + 5*tan(asin((d*x)/c)/2) + 1))