Integrand size = 29, antiderivative size = 188 \[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {(B c+2 A d) x}{21 c^3 d \left (c^2-d^2 x^2\right )^{5/2}}-\frac {2 (B c+2 A d)}{21 c d^2 (c+d x) \left (c^2-d^2 x^2\right )^{5/2}}+\frac {4 (B c+2 A d) x}{63 c^5 d \left (c^2-d^2 x^2\right )^{3/2}}+\frac {B c-A d}{9 c d^2 (c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}}+\frac {8 (B c+2 A d) x}{63 c^7 d \sqrt {c^2-d^2 x^2}} \] Output:
1/21*(2*A*d+B*c)*x/c^3/d/(-d^2*x^2+c^2)^(5/2)-2/21*(2*A*d+B*c)/c/d^2/(d*x+ c)/(-d^2*x^2+c^2)^(5/2)+4/63*(2*A*d+B*c)*x/c^5/d/(-d^2*x^2+c^2)^(3/2)+1/9* (-A*d+B*c)/c/d^2/(d*x+c)^3/(-d^2*x^2+c^2)^(3/2)+8/63*(2*A*d+B*c)*x/c^7/d/( -d^2*x^2+c^2)^(1/2)
Time = 1.04 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.91 \[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (A d \left (-19 c^6+6 c^5 d x+66 c^4 d^2 x^2+56 c^3 d^3 x^3-24 c^2 d^4 x^4-48 c d^5 x^5-16 d^6 x^6\right )+B c \left (c^6+3 c^5 d x+33 c^4 d^2 x^2+28 c^3 d^3 x^3-12 c^2 d^4 x^4-24 c d^5 x^5-8 d^6 x^6\right )\right )}{63 c^7 d^2 (c-d x)^2 (c+d x)^5} \] Input:
Integrate[(A + B*x)/((c + d*x)^3*(c^2 - d^2*x^2)^(5/2)),x]
Output:
(Sqrt[c^2 - d^2*x^2]*(A*d*(-19*c^6 + 6*c^5*d*x + 66*c^4*d^2*x^2 + 56*c^3*d ^3*x^3 - 24*c^2*d^4*x^4 - 48*c*d^5*x^5 - 16*d^6*x^6) + B*c*(c^6 + 3*c^5*d* x + 33*c^4*d^2*x^2 + 28*c^3*d^3*x^3 - 12*c^2*d^4*x^4 - 24*c*d^5*x^5 - 8*d^ 6*x^6)))/(63*c^7*d^2*(c - d*x)^2*(c + d*x)^5)
Time = 0.46 (sec) , antiderivative size = 191, normalized size of antiderivative = 1.02, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {671, 461, 470, 209, 208}
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.
\(\displaystyle \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 671 |
\(\displaystyle \frac {(2 A d+B c) \int \frac {1}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{5/2}}dx}{3 c d}+\frac {B c-A d}{9 c d^2 (c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {(2 A d+B c) \left (\frac {5 \int \frac {1}{(c+d x) \left (c^2-d^2 x^2\right )^{5/2}}dx}{7 c}-\frac {1}{7 c d (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}}\right )}{3 c d}+\frac {B c-A d}{9 c d^2 (c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {(2 A d+B c) \left (\frac {5 \left (\frac {4 \int \frac {1}{\left (c^2-d^2 x^2\right )^{5/2}}dx}{5 c}-\frac {1}{5 c d (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}\right )}{7 c}-\frac {1}{7 c d (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}}\right )}{3 c d}+\frac {B c-A d}{9 c d^2 (c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 209 |
\(\displaystyle \frac {(2 A d+B c) \left (\frac {5 \left (\frac {4 \left (\frac {2 \int \frac {1}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{3 c^2}+\frac {x}{3 c^2 \left (c^2-d^2 x^2\right )^{3/2}}\right )}{5 c}-\frac {1}{5 c d (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}\right )}{7 c}-\frac {1}{7 c d (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}}\right )}{3 c d}+\frac {B c-A d}{9 c d^2 (c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {B c-A d}{9 c d^2 (c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}}+\frac {\left (\frac {5 \left (\frac {4 \left (\frac {x}{3 c^2 \left (c^2-d^2 x^2\right )^{3/2}}+\frac {2 x}{3 c^4 \sqrt {c^2-d^2 x^2}}\right )}{5 c}-\frac {1}{5 c d (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}\right )}{7 c}-\frac {1}{7 c d (c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}}\right ) (2 A d+B c)}{3 c d}\) |
Input:
Int[(A + B*x)/((c + d*x)^3*(c^2 - d^2*x^2)^(5/2)),x]
Output:
(B*c - A*d)/(9*c*d^2*(c + d*x)^3*(c^2 - d^2*x^2)^(3/2)) + ((B*c + 2*A*d)*( -1/7*1/(c*d*(c + d*x)^2*(c^2 - d^2*x^2)^(3/2)) + (5*(-1/5*1/(c*d*(c + d*x) *(c^2 - d^2*x^2)^(3/2)) + (4*(x/(3*c^2*(c^2 - d^2*x^2)^(3/2)) + (2*x)/(3*c ^4*Sqrt[c^2 - d^2*x^2])))/(5*c)))/(7*c)))/(3*c*d)
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && ILtQ[p + 3/2, 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[((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[(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, p}, x] && EqQ[b*c^2 + a*d^2, 0] && LtQ[n, 0] && NeQ[n + p + 1, 0] && IntegerQ[2*p]
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 = 181, normalized size of antiderivative = 0.96
method | result | size |
gosper | \(-\frac {\left (-d x +c \right ) \left (16 A \,d^{7} x^{6}+8 B c \,d^{6} x^{6}+48 A c \,d^{6} x^{5}+24 B \,c^{2} d^{5} x^{5}+24 A \,c^{2} d^{5} x^{4}+12 B \,c^{3} d^{4} x^{4}-56 A \,c^{3} d^{4} x^{3}-28 B \,c^{4} d^{3} x^{3}-66 c^{4} x^{2} A \,d^{3}-33 c^{5} x^{2} B \,d^{2}-6 A \,c^{5} d^{2} x -3 B \,c^{6} d x +19 A \,c^{6} d -B \,c^{7}\right )}{63 \left (d x +c \right )^{2} c^{7} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\) | \(181\) |
orering | \(-\frac {\left (-d x +c \right ) \left (16 A \,d^{7} x^{6}+8 B c \,d^{6} x^{6}+48 A c \,d^{6} x^{5}+24 B \,c^{2} d^{5} x^{5}+24 A \,c^{2} d^{5} x^{4}+12 B \,c^{3} d^{4} x^{4}-56 A \,c^{3} d^{4} x^{3}-28 B \,c^{4} d^{3} x^{3}-66 c^{4} x^{2} A \,d^{3}-33 c^{5} x^{2} B \,d^{2}-6 A \,c^{5} d^{2} x -3 B \,c^{6} d x +19 A \,c^{6} d -B \,c^{7}\right )}{63 \left (d x +c \right )^{2} c^{7} d^{2} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}\) | \(181\) |
trager | \(-\frac {\left (16 A \,d^{7} x^{6}+8 B c \,d^{6} x^{6}+48 A c \,d^{6} x^{5}+24 B \,c^{2} d^{5} x^{5}+24 A \,c^{2} d^{5} x^{4}+12 B \,c^{3} d^{4} x^{4}-56 A \,c^{3} d^{4} x^{3}-28 B \,c^{4} d^{3} x^{3}-66 c^{4} x^{2} A \,d^{3}-33 c^{5} x^{2} B \,d^{2}-6 A \,c^{5} d^{2} x -3 B \,c^{6} d x +19 A \,c^{6} d -B \,c^{7}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{63 c^{7} \left (d x +c \right )^{5} \left (-d x +c \right )^{2} d^{2}}\) | \(183\) |
default | \(\frac {B \left (-\frac {1}{7 c d \left (x +\frac {c}{d}\right )^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {5 d \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {4 d \left (-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{6 c^{2} d^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d^{2} c^{4} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{7 c}\right )}{d^{3}}+\frac {\left (A d -B c \right ) \left (-\frac {1}{9 c d \left (x +\frac {c}{d}\right )^{3} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {2 d \left (-\frac {1}{7 c d \left (x +\frac {c}{d}\right )^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {5 d \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}+\frac {4 d \left (-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{6 c^{2} d^{2} \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d^{2} c^{4} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{7 c}\right )}{3 c}\right )}{d^{4}}\) | \(493\) |
Input:
int((B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(5/2),x,method=_RETURNVERBOSE)
Output:
-1/63*(-d*x+c)*(16*A*d^7*x^6+8*B*c*d^6*x^6+48*A*c*d^6*x^5+24*B*c^2*d^5*x^5 +24*A*c^2*d^5*x^4+12*B*c^3*d^4*x^4-56*A*c^3*d^4*x^3-28*B*c^4*d^3*x^3-66*A* c^4*d^3*x^2-33*B*c^5*d^2*x^2-6*A*c^5*d^2*x-3*B*c^6*d*x+19*A*c^6*d-B*c^7)/( d*x+c)^2/c^7/d^2/(-d^2*x^2+c^2)^(5/2)
Leaf count of result is larger than twice the leaf count of optimal. 399 vs. \(2 (168) = 336\).
Time = 0.28 (sec) , antiderivative size = 399, normalized size of antiderivative = 2.12 \[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {B c^{8} - 19 \, A c^{7} d + {\left (B c d^{7} - 19 \, A d^{8}\right )} x^{7} + 3 \, {\left (B c^{2} d^{6} - 19 \, A c d^{7}\right )} x^{6} + {\left (B c^{3} d^{5} - 19 \, A c^{2} d^{6}\right )} x^{5} - 5 \, {\left (B c^{4} d^{4} - 19 \, A c^{3} d^{5}\right )} x^{4} - 5 \, {\left (B c^{5} d^{3} - 19 \, A c^{4} d^{4}\right )} x^{3} + {\left (B c^{6} d^{2} - 19 \, A c^{5} d^{3}\right )} x^{2} + 3 \, {\left (B c^{7} d - 19 \, A c^{6} d^{2}\right )} x + {\left (B c^{7} - 19 \, A c^{6} d - 8 \, {\left (B c d^{6} + 2 \, A d^{7}\right )} x^{6} - 24 \, {\left (B c^{2} d^{5} + 2 \, A c d^{6}\right )} x^{5} - 12 \, {\left (B c^{3} d^{4} + 2 \, A c^{2} d^{5}\right )} x^{4} + 28 \, {\left (B c^{4} d^{3} + 2 \, A c^{3} d^{4}\right )} x^{3} + 33 \, {\left (B c^{5} d^{2} + 2 \, A c^{4} d^{3}\right )} x^{2} + 3 \, {\left (B c^{6} d + 2 \, A c^{5} d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{63 \, {\left (c^{7} d^{9} x^{7} + 3 \, c^{8} d^{8} x^{6} + c^{9} d^{7} x^{5} - 5 \, c^{10} d^{6} x^{4} - 5 \, c^{11} d^{5} x^{3} + c^{12} d^{4} x^{2} + 3 \, c^{13} d^{3} x + c^{14} d^{2}\right )}} \] Input:
integrate((B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(5/2),x, algorithm="fricas")
Output:
1/63*(B*c^8 - 19*A*c^7*d + (B*c*d^7 - 19*A*d^8)*x^7 + 3*(B*c^2*d^6 - 19*A* c*d^7)*x^6 + (B*c^3*d^5 - 19*A*c^2*d^6)*x^5 - 5*(B*c^4*d^4 - 19*A*c^3*d^5) *x^4 - 5*(B*c^5*d^3 - 19*A*c^4*d^4)*x^3 + (B*c^6*d^2 - 19*A*c^5*d^3)*x^2 + 3*(B*c^7*d - 19*A*c^6*d^2)*x + (B*c^7 - 19*A*c^6*d - 8*(B*c*d^6 + 2*A*d^7 )*x^6 - 24*(B*c^2*d^5 + 2*A*c*d^6)*x^5 - 12*(B*c^3*d^4 + 2*A*c^2*d^5)*x^4 + 28*(B*c^4*d^3 + 2*A*c^3*d^4)*x^3 + 33*(B*c^5*d^2 + 2*A*c^4*d^3)*x^2 + 3* (B*c^6*d + 2*A*c^5*d^2)*x)*sqrt(-d^2*x^2 + c^2))/(c^7*d^9*x^7 + 3*c^8*d^8* x^6 + c^9*d^7*x^5 - 5*c^10*d^6*x^4 - 5*c^11*d^5*x^3 + c^12*d^4*x^2 + 3*c^1 3*d^3*x + c^14*d^2)
\[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int \frac {A + B x}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}} \left (c + d x\right )^{3}}\, dx \] Input:
integrate((B*x+A)/(d*x+c)**3/(-d**2*x**2+c**2)**(5/2),x)
Output:
Integral((A + B*x)/((-(-c + d*x)*(c + d*x))**(5/2)*(c + d*x)**3), x)
Leaf count of result is larger than twice the leaf count of optimal. 649 vs. \(2 (168) = 336\).
Time = 0.05 (sec) , antiderivative size = 649, normalized size of antiderivative = 3.45 \[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate((B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(5/2),x, algorithm="maxima")
Output:
1/9*B*c/((-d^2*x^2 + c^2)^(3/2)*c*d^5*x^3 + 3*(-d^2*x^2 + c^2)^(3/2)*c^2*d ^4*x^2 + 3*(-d^2*x^2 + c^2)^(3/2)*c^3*d^3*x + (-d^2*x^2 + c^2)^(3/2)*c^4*d ^2) + 2/21*B*c/((-d^2*x^2 + c^2)^(3/2)*c^2*d^4*x^2 + 2*(-d^2*x^2 + c^2)^(3 /2)*c^3*d^3*x + (-d^2*x^2 + c^2)^(3/2)*c^4*d^2) + 2/21*B*c/((-d^2*x^2 + c^ 2)^(3/2)*c^3*d^3*x + (-d^2*x^2 + c^2)^(3/2)*c^4*d^2) - 1/9*A/((-d^2*x^2 + c^2)^(3/2)*c*d^4*x^3 + 3*(-d^2*x^2 + c^2)^(3/2)*c^2*d^3*x^2 + 3*(-d^2*x^2 + c^2)^(3/2)*c^3*d^2*x + (-d^2*x^2 + c^2)^(3/2)*c^4*d) - 2/21*A/((-d^2*x^2 + c^2)^(3/2)*c^2*d^3*x^2 + 2*(-d^2*x^2 + c^2)^(3/2)*c^3*d^2*x + (-d^2*x^2 + c^2)^(3/2)*c^4*d) - 2/21*A/((-d^2*x^2 + c^2)^(3/2)*c^3*d^2*x + (-d^2*x^ 2 + c^2)^(3/2)*c^4*d) - 1/7*B/((-d^2*x^2 + c^2)^(3/2)*c*d^4*x^2 + 2*(-d^2* x^2 + c^2)^(3/2)*c^2*d^3*x + (-d^2*x^2 + c^2)^(3/2)*c^3*d^2) - 1/7*B/((-d^ 2*x^2 + c^2)^(3/2)*c^2*d^3*x + (-d^2*x^2 + c^2)^(3/2)*c^3*d^2) + 8/63*A*x/ ((-d^2*x^2 + c^2)^(3/2)*c^5) + 4/63*B*x/((-d^2*x^2 + c^2)^(3/2)*c^4*d) + 1 6/63*A*x/(sqrt(-d^2*x^2 + c^2)*c^7) + 8/63*B*x/(sqrt(-d^2*x^2 + c^2)*c^6*d )
\[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\int { \frac {B x + A}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} {\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(5/2),x, algorithm="giac")
Output:
integrate((B*x + A)/((-d^2*x^2 + c^2)^(5/2)*(d*x + c)^3), x)
Time = 10.82 (sec) , antiderivative size = 366, normalized size of antiderivative = 1.95 \[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {\sqrt {c^2-d^2\,x^2}\,\left (x\,\left (\frac {65\,B\,c^3+151\,A\,d\,c^2}{1008\,c^7\,d}+\frac {46\,A\,d^3+2\,B\,c\,d^2}{1008\,c^5\,d^3}\right )-\frac {5\,\left (31\,A\,d+5\,B\,c\right )}{1008\,c^4\,d^2}\right )}{{\left (c+d\,x\right )}^2\,{\left (c-d\,x\right )}^2}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {17\,B\,c^3+55\,A\,d\,c^2}{1680\,c^7\,d^2}+\frac {c\,\left (\frac {47\,A\,d-8\,B\,c}{1680\,c^6\,d}+\frac {13\,A\,d^3-4\,B\,c\,d^2}{1680\,c^6\,d^3}\right )}{d}\right )}{{\left (c+d\,x\right )}^3}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {A}{36\,c^3\,d}-\frac {B}{36\,c^2\,d^2}\right )}{{\left (c+d\,x\right )}^5}-\frac {\sqrt {c^2-d^2\,x^2}\,\left (\frac {5\,B\,c^3+31\,A\,d\,c^2}{1008\,c^6\,d^2}+\frac {c\,\left (\frac {17\,\left (A\,d-B\,c\right )}{1008\,c^5\,d}+\frac {4\,A\,d^3-4\,B\,c\,d^2}{1008\,c^5\,d^3}\right )}{d}\right )}{{\left (c+d\,x\right )}^4}+\frac {x\,\sqrt {c^2-d^2\,x^2}\,\left (16\,A\,d+8\,B\,c\right )}{63\,c^7\,d\,\left (c+d\,x\right )\,\left (c-d\,x\right )} \] Input:
int((A + B*x)/((c^2 - d^2*x^2)^(5/2)*(c + d*x)^3),x)
Output:
((c^2 - d^2*x^2)^(1/2)*(x*((65*B*c^3 + 151*A*c^2*d)/(1008*c^7*d) + (46*A*d ^3 + 2*B*c*d^2)/(1008*c^5*d^3)) - (5*(31*A*d + 5*B*c))/(1008*c^4*d^2)))/(( c + d*x)^2*(c - d*x)^2) - ((c^2 - d^2*x^2)^(1/2)*((17*B*c^3 + 55*A*c^2*d)/ (1680*c^7*d^2) + (c*((47*A*d - 8*B*c)/(1680*c^6*d) + (13*A*d^3 - 4*B*c*d^2 )/(1680*c^6*d^3)))/d))/(c + d*x)^3 - ((c^2 - d^2*x^2)^(1/2)*(A/(36*c^3*d) - B/(36*c^2*d^2)))/(c + d*x)^5 - ((c^2 - d^2*x^2)^(1/2)*((5*B*c^3 + 31*A*c ^2*d)/(1008*c^6*d^2) + (c*((17*(A*d - B*c))/(1008*c^5*d) + (4*A*d^3 - 4*B* c*d^2)/(1008*c^5*d^3)))/d))/(c + d*x)^4 + (x*(c^2 - d^2*x^2)^(1/2)*(16*A*d + 8*B*c))/(63*c^7*d*(c + d*x)*(c - d*x))
Time = 0.22 (sec) , antiderivative size = 494, normalized size of antiderivative = 2.63 \[ \int \frac {A+B x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{5/2}} \, dx=\frac {b \,c^{7}+\sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{6}-19 a \,c^{6} d +6 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{4} d^{2} x +4 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d^{3} x^{2}-4 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{4} x^{3}-6 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{5} x^{4}+3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{5} d x +2 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4} d^{2} x^{2}-2 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d^{3} x^{3}-3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{4} x^{4}-\sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{5} x^{5}+6 a \,c^{5} d^{2} x +66 a \,c^{4} d^{3} x^{2}+56 a \,c^{3} d^{4} x^{3}-24 a \,c^{2} d^{5} x^{4}-48 a c \,d^{6} x^{5}+3 b \,c^{6} d x +33 b \,c^{5} d^{2} x^{2}+28 b \,c^{4} d^{3} x^{3}-12 b \,c^{3} d^{4} x^{4}-24 b \,c^{2} d^{5} x^{5}-8 b c \,d^{6} x^{6}+2 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{5} d -2 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{6} x^{5}-16 a \,d^{7} x^{6}}{63 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{7} d^{2} \left (-d^{5} x^{5}-3 c \,d^{4} x^{4}-2 c^{2} d^{3} x^{3}+2 c^{3} d^{2} x^{2}+3 c^{4} d x +c^{5}\right )} \] Input:
int((B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(5/2),x)
Output:
(2*sqrt(c**2 - d**2*x**2)*a*c**5*d + 6*sqrt(c**2 - d**2*x**2)*a*c**4*d**2* x + 4*sqrt(c**2 - d**2*x**2)*a*c**3*d**3*x**2 - 4*sqrt(c**2 - d**2*x**2)*a *c**2*d**4*x**3 - 6*sqrt(c**2 - d**2*x**2)*a*c*d**5*x**4 - 2*sqrt(c**2 - d **2*x**2)*a*d**6*x**5 + sqrt(c**2 - d**2*x**2)*b*c**6 + 3*sqrt(c**2 - d**2 *x**2)*b*c**5*d*x + 2*sqrt(c**2 - d**2*x**2)*b*c**4*d**2*x**2 - 2*sqrt(c** 2 - d**2*x**2)*b*c**3*d**3*x**3 - 3*sqrt(c**2 - d**2*x**2)*b*c**2*d**4*x** 4 - sqrt(c**2 - d**2*x**2)*b*c*d**5*x**5 - 19*a*c**6*d + 6*a*c**5*d**2*x + 66*a*c**4*d**3*x**2 + 56*a*c**3*d**4*x**3 - 24*a*c**2*d**5*x**4 - 48*a*c* d**6*x**5 - 16*a*d**7*x**6 + b*c**7 + 3*b*c**6*d*x + 33*b*c**5*d**2*x**2 + 28*b*c**4*d**3*x**3 - 12*b*c**3*d**4*x**4 - 24*b*c**2*d**5*x**5 - 8*b*c*d **6*x**6)/(63*sqrt(c**2 - d**2*x**2)*c**7*d**2*(c**5 + 3*c**4*d*x + 2*c**3 *d**2*x**2 - 2*c**2*d**3*x**3 - 3*c*d**4*x**4 - d**5*x**5))