Integrand size = 29, antiderivative size = 81 \[ \int \frac {(A+B x) (c+d x)^2}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {2 (B c+A d) (c+d x)}{3 d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {3 B c^2+d (2 B c-A d) x}{3 c^2 d^2 \sqrt {c^2-d^2 x^2}} \] Output:
2/3*(A*d+B*c)*(d*x+c)/d^2/(-d^2*x^2+c^2)^(3/2)-1/3*(3*B*c^2+d*(-A*d+2*B*c) *x)/c^2/d^2/(-d^2*x^2+c^2)^(1/2)
Time = 0.46 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.67 \[ \int \frac {(A+B x) (c+d x)^2}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c^2-d^2 x^2} (B c (c-2 d x)+A d (-2 c+d x))}{3 c^2 d^2 (c-d x)^2} \] Input:
Integrate[((A + B*x)*(c + d*x)^2)/(c^2 - d^2*x^2)^(5/2),x]
Output:
-1/3*(Sqrt[c^2 - d^2*x^2]*(B*c*(c - 2*d*x) + A*d*(-2*c + d*x)))/(c^2*d^2*( c - d*x)^2)
Time = 0.32 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {669, 453}
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.
\(\displaystyle \int \frac {(A+B x) (c+d x)^2}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 669 |
\(\displaystyle \frac {(c+d x)^2 (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {(2 B c-A d) \int \frac {c+d x}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{3 c d}\) |
\(\Big \downarrow \) 453 |
\(\displaystyle \frac {(c+d x)^2 (A d+B c)}{3 c d^2 \left (c^2-d^2 x^2\right )^{3/2}}-\frac {(c+d x) (2 B c-A d)}{3 c^2 d^2 \sqrt {c^2-d^2 x^2}}\) |
Input:
Int[((A + B*x)*(c + d*x)^2)/(c^2 - d^2*x^2)^(5/2),x]
Output:
((B*c + A*d)*(c + d*x)^2)/(3*c*d^2*(c^2 - d^2*x^2)^(3/2)) - ((2*B*c - A*d) *(c + d*x))/(3*c^2*d^2*Sqrt[c^2 - d^2*x^2])
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[-(a* d - b*c*x)/(a*b*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b, c, d}, x]
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* (p + 1))), x] - Simp[e*((m*(d*g + e*f) + 2*e*f*(p + 1))/(2*c*d*(p + 1))) Int[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0]
Time = 0.39 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.69
method | result | size |
trager | \(\frac {\left (-A \,d^{2} x +2 B c d x +2 A c d -B \,c^{2}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{3 c^{2} d^{2} \left (-d x +c \right )^{2}}\) | \(56\) |
gosper | \(\frac {\left (d x +c \right )^{3} \left (-d x +c \right ) \left (-A \,d^{2} x +2 B c d x +2 A c d -B \,c^{2}\right )}{3 c^{2} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\) | \(61\) |
orering | \(\frac {\left (d x +c \right )^{3} \left (-d x +c \right ) \left (-A \,d^{2} x +2 B c d x +2 A c d -B \,c^{2}\right )}{3 c^{2} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\) | \(61\) |
default | \(A \,c^{2} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )+d \left (A d +2 B c \right ) \left (\frac {x}{2 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {c^{2} \left (\frac {x}{3 c^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+\frac {2 x}{3 c^{4} \sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 d^{2}}\right )+\frac {c \left (2 A d +B c \right )}{3 d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}+B \,d^{2} \left (\frac {x^{2}}{d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}-\frac {2 c^{2}}{3 d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\right )\) | \(205\) |
Input:
int((B*x+A)*(d*x+c)^2/(-d^2*x^2+c^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
1/3*(-A*d^2*x+2*B*c*d*x+2*A*c*d-B*c^2)/c^2/d^2/(-d*x+c)^2*(-d^2*x^2+c^2)^( 1/2)
Time = 0.08 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.47 \[ \int \frac {(A+B x) (c+d x)^2}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {B c^{3} - 2 \, A c^{2} d + {\left (B c d^{2} - 2 \, A d^{3}\right )} x^{2} - 2 \, {\left (B c^{2} d - 2 \, A c d^{2}\right )} x + \sqrt {-d^{2} x^{2} + c^{2}} {\left (B c^{2} - 2 \, A c d - {\left (2 \, B c d - A d^{2}\right )} x\right )}}{3 \, {\left (c^{2} d^{4} x^{2} - 2 \, c^{3} d^{3} x + c^{4} d^{2}\right )}} \] Input:
integrate((B*x+A)*(d*x+c)^2/(-d^2*x^2+c^2)^(5/2),x, algorithm="fricas")
Output:
-1/3*(B*c^3 - 2*A*c^2*d + (B*c*d^2 - 2*A*d^3)*x^2 - 2*(B*c^2*d - 2*A*c*d^2 )*x + sqrt(-d^2*x^2 + c^2)*(B*c^2 - 2*A*c*d - (2*B*c*d - A*d^2)*x))/(c^2*d ^4*x^2 - 2*c^3*d^3*x + c^4*d^2)
\[ \int \frac {(A+B x) (c+d x)^2}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {\left (A + B x\right ) \left (c + d x\right )^{2}}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((B*x+A)*(d*x+c)**2/(-d**2*x**2+c**2)**(5/2),x)
Output:
Integral((A + B*x)*(c + d*x)**2/(-(-c + d*x)*(c + d*x))**(5/2), x)
Leaf count of result is larger than twice the leaf count of optimal. 168 vs. \(2 (73) = 146\).
Time = 0.03 (sec) , antiderivative size = 168, normalized size of antiderivative = 2.07 \[ \int \frac {(A+B x) (c+d x)^2}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {B x^{2}}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}}} + \frac {A x}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}}} - \frac {B c^{2}}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} + \frac {2 \, A c}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d} + \frac {2 \, A x}{3 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{2}} + \frac {{\left (2 \, B c d + A d^{2}\right )} x}{3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} d^{2}} - \frac {{\left (2 \, B c d + A d^{2}\right )} x}{3 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{2} d^{2}} \] Input:
integrate((B*x+A)*(d*x+c)^2/(-d^2*x^2+c^2)^(5/2),x, algorithm="maxima")
Output:
B*x^2/(-d^2*x^2 + c^2)^(3/2) + 1/3*A*x/(-d^2*x^2 + c^2)^(3/2) - 1/3*B*c^2/ ((-d^2*x^2 + c^2)^(3/2)*d^2) + 2/3*A*c/((-d^2*x^2 + c^2)^(3/2)*d) + 2/3*A* x/(sqrt(-d^2*x^2 + c^2)*c^2) + 1/3*(2*B*c*d + A*d^2)*x/((-d^2*x^2 + c^2)^( 3/2)*d^2) - 1/3*(2*B*c*d + A*d^2)*x/(sqrt(-d^2*x^2 + c^2)*c^2*d^2)
Time = 0.15 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.79 \[ \int \frac {(A+B x) (c+d x)^2}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {2 \, {\left (B c - 2 \, A d - \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} B c}{d^{2} x} + \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )} A}{d x} - \frac {3 \, {\left (c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}\right )}^{2} A}{d^{3} x^{2}}\right )}}{3 \, c^{2} d {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} - 1\right )}^{3} {\left | d \right |}} \] Input:
integrate((B*x+A)*(d*x+c)^2/(-d^2*x^2+c^2)^(5/2),x, algorithm="giac")
Output:
-2/3*(B*c - 2*A*d - 3*(c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*B*c/(d^2*x) + 3* (c*d + sqrt(-d^2*x^2 + c^2)*abs(d))*A/(d*x) - 3*(c*d + sqrt(-d^2*x^2 + c^2 )*abs(d))^2*A/(d^3*x^2))/(c^2*d*((c*d + sqrt(-d^2*x^2 + c^2)*abs(d))/(d^2* x) - 1)^3*abs(d))
Time = 10.00 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.65 \[ \int \frac {(A+B x) (c+d x)^2}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c^2-d^2\,x^2}\,\left (B\,c^2-2\,A\,c\,d+A\,d^2\,x-2\,B\,c\,d\,x\right )}{3\,c^2\,d^2\,{\left (c-d\,x\right )}^2} \] Input:
int(((A + B*x)*(c + d*x)^2)/(c^2 - d^2*x^2)^(5/2),x)
Output:
-((c^2 - d^2*x^2)^(1/2)*(B*c^2 - 2*A*c*d + A*d^2*x - 2*B*c*d*x))/(3*c^2*d^ 2*(c - d*x)^2)
Time = 0.25 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.09 \[ \int \frac {(A+B x) (c+d x)^2}{\left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {-\frac {2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3} a d}{3}-2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) b c -\frac {2 a d}{3}+\frac {2 b c}{3}}{c^{2} d^{2} \left (\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{3}-3 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )^{2}+3 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )-1\right )} \] Input:
int((B*x+A)*(d*x+c)^2/(-d^2*x^2+c^2)^(5/2),x)
Output:
(2*( - tan(asin((d*x)/c)/2)**3*a*d - 3*tan(asin((d*x)/c)/2)*b*c - a*d + b* c))/(3*c**2*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))