Integrand size = 39, antiderivative size = 233 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\left (10 c^2 C d+18 B c d^2+24 A d^3-3 c^3 D\right ) x}{105 c^5 d^3 \sqrt {c^2-d^2 x^2}}-\frac {c^2 C d-B c d^2+A d^3-c^3 D}{7 c d^4 (c+d x)^3 \sqrt {c^2-d^2 x^2}}+\frac {10 c^2 C d-3 B c d^2-4 A d^3-17 c^3 D}{35 c^2 d^4 (c+d x)^2 \sqrt {c^2-d^2 x^2}}-\frac {5 c^2 C d+9 B c d^2+12 A d^3-54 c^3 D}{105 c^3 d^4 (c+d x) \sqrt {c^2-d^2 x^2}} \] Output:
1/105*(24*A*d^3+18*B*c*d^2+10*C*c^2*d-3*D*c^3)*x/c^5/d^3/(-d^2*x^2+c^2)^(1 /2)-1/7*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)/c/d^4/(d*x+c)^3/(-d^2*x^2+c^2)^(1/2) +1/35*(-4*A*d^3-3*B*c*d^2+10*C*c^2*d-17*D*c^3)/c^2/d^4/(d*x+c)^2/(-d^2*x^2 +c^2)^(1/2)-1/105*(12*A*d^3+9*B*c*d^2+5*C*c^2*d-54*D*c^3)/c^3/d^4/(d*x+c)/ (-d^2*x^2+c^2)^(1/2)
Time = 1.49 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.81 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (18 c^7 D+24 A d^7 x^4+18 c d^6 x^3 (4 A+B x)+2 c^6 d (5 C+27 D x)+2 c^2 d^5 x^2 (30 A+x (27 B+5 C x))-3 c^5 d^2 (B-5 x (2 C+3 D x))-3 c^3 d^4 x \left (4 A+x \left (-15 B-10 C x+D x^2\right )\right )-c^4 d^3 \left (39 A+x \left (9 B-25 C x+9 D x^2\right )\right )\right )}{105 c^5 d^4 (c-d x) (c+d x)^4} \] Input:
Integrate[(A + B*x + C*x^2 + D*x^3)/((c + d*x)^3*(c^2 - d^2*x^2)^(3/2)),x]
Output:
(Sqrt[c^2 - d^2*x^2]*(18*c^7*D + 24*A*d^7*x^4 + 18*c*d^6*x^3*(4*A + B*x) + 2*c^6*d*(5*C + 27*D*x) + 2*c^2*d^5*x^2*(30*A + x*(27*B + 5*C*x)) - 3*c^5* d^2*(B - 5*x*(2*C + 3*D*x)) - 3*c^3*d^4*x*(4*A + x*(-15*B - 10*C*x + D*x^2 )) - c^4*d^3*(39*A + x*(9*B - 25*C*x + 9*D*x^2))))/(105*c^5*d^4*(c - d*x)* (c + d*x)^4)
Time = 1.08 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.18, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2170, 2170, 27, 671, 461, 470, 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.
\(\displaystyle \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 2170 |
\(\displaystyle \frac {\int \frac {2 B x d^5+(2 C d-3 c D) x^2 d^4+\left (D c^3+2 A d^3\right ) d^2}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}}dx}{2 d^5}+\frac {D}{2 d^4 (c+d x) \sqrt {c^2-d^2 x^2}}\) |
\(\Big \downarrow \) 2170 |
\(\displaystyle \frac {\frac {\int \frac {d^6 \left (-3 D c^3+4 C d c^2+6 A d^3-d \left (-3 D c^2+2 C d c-6 B d^2\right ) x\right )}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}}dx}{3 d^4}+\frac {d (2 C d-3 c D)}{3 (c+d x)^2 \sqrt {c^2-d^2 x^2}}}{2 d^5}+\frac {D}{2 d^4 (c+d x) \sqrt {c^2-d^2 x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {1}{3} d^2 \int \frac {-3 D c^3+4 C d c^2+6 A d^3-d \left (-3 D c^2+2 C d c-6 B d^2\right ) x}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}}dx+\frac {d (2 C d-3 c D)}{3 (c+d x)^2 \sqrt {c^2-d^2 x^2}}}{2 d^5}+\frac {D}{2 d^4 (c+d x) \sqrt {c^2-d^2 x^2}}\) |
\(\Big \downarrow \) 671 |
\(\displaystyle \frac {\frac {1}{3} d^2 \left (\frac {\left (24 A d^3+18 B c d^2-3 c^3 D+10 c^2 C d\right ) \int \frac {1}{(c+d x)^2 \left (c^2-d^2 x^2\right )^{3/2}}dx}{7 c}-\frac {6 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{7 c d (c+d x)^3 \sqrt {c^2-d^2 x^2}}\right )+\frac {d (2 C d-3 c D)}{3 (c+d x)^2 \sqrt {c^2-d^2 x^2}}}{2 d^5}+\frac {D}{2 d^4 (c+d x) \sqrt {c^2-d^2 x^2}}\) |
\(\Big \downarrow \) 461 |
\(\displaystyle \frac {\frac {1}{3} d^2 \left (\frac {\left (24 A d^3+18 B c d^2-3 c^3 D+10 c^2 C d\right ) \left (\frac {3 \int \frac {1}{(c+d x) \left (c^2-d^2 x^2\right )^{3/2}}dx}{5 c}-\frac {1}{5 c d (c+d x)^2 \sqrt {c^2-d^2 x^2}}\right )}{7 c}-\frac {6 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{7 c d (c+d x)^3 \sqrt {c^2-d^2 x^2}}\right )+\frac {d (2 C d-3 c D)}{3 (c+d x)^2 \sqrt {c^2-d^2 x^2}}}{2 d^5}+\frac {D}{2 d^4 (c+d x) \sqrt {c^2-d^2 x^2}}\) |
\(\Big \downarrow \) 470 |
\(\displaystyle \frac {\frac {1}{3} d^2 \left (\frac {\left (24 A d^3+18 B c d^2-3 c^3 D+10 c^2 C d\right ) \left (\frac {3 \left (\frac {2 \int \frac {1}{\left (c^2-d^2 x^2\right )^{3/2}}dx}{3 c}-\frac {1}{3 c d (c+d x) \sqrt {c^2-d^2 x^2}}\right )}{5 c}-\frac {1}{5 c d (c+d x)^2 \sqrt {c^2-d^2 x^2}}\right )}{7 c}-\frac {6 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{7 c d (c+d x)^3 \sqrt {c^2-d^2 x^2}}\right )+\frac {d (2 C d-3 c D)}{3 (c+d x)^2 \sqrt {c^2-d^2 x^2}}}{2 d^5}+\frac {D}{2 d^4 (c+d x) \sqrt {c^2-d^2 x^2}}\) |
\(\Big \downarrow \) 208 |
\(\displaystyle \frac {\frac {1}{3} d^2 \left (\frac {\left (\frac {3 \left (\frac {2 x}{3 c^3 \sqrt {c^2-d^2 x^2}}-\frac {1}{3 c d (c+d x) \sqrt {c^2-d^2 x^2}}\right )}{5 c}-\frac {1}{5 c d (c+d x)^2 \sqrt {c^2-d^2 x^2}}\right ) \left (24 A d^3+18 B c d^2-3 c^3 D+10 c^2 C d\right )}{7 c}-\frac {6 \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{7 c d (c+d x)^3 \sqrt {c^2-d^2 x^2}}\right )+\frac {d (2 C d-3 c D)}{3 (c+d x)^2 \sqrt {c^2-d^2 x^2}}}{2 d^5}+\frac {D}{2 d^4 (c+d x) \sqrt {c^2-d^2 x^2}}\) |
Input:
Int[(A + B*x + C*x^2 + D*x^3)/((c + d*x)^3*(c^2 - d^2*x^2)^(3/2)),x]
Output:
D/(2*d^4*(c + d*x)*Sqrt[c^2 - d^2*x^2]) + ((d*(2*C*d - 3*c*D))/(3*(c + d*x )^2*Sqrt[c^2 - d^2*x^2]) + (d^2*((-6*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D))/ (7*c*d*(c + d*x)^3*Sqrt[c^2 - d^2*x^2]) + ((10*c^2*C*d + 18*B*c*d^2 + 24*A *d^3 - 3*c^3*D)*(-1/5*1/(c*d*(c + d*x)^2*Sqrt[c^2 - d^2*x^2]) + (3*((2*x)/ (3*c^3*Sqrt[c^2 - d^2*x^2]) - 1/(3*c*d*(c + d*x)*Sqrt[c^2 - d^2*x^2])))/(5 *c)))/(7*c)))/3)/(2*d^5)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-3/2), x_Symbol] :> Simp[x/(a*Sqrt[a + b*x^2]), x] /; FreeQ[{a, b}, x]
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]
Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : > With[{q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(d + e*x) ^(m + q - 1)*((a + b*x^2)^(p + 1)/(b*e^(q - 1)*(m + q + 2*p + 1))), x] + Si mp[1/(b*e^q*(m + q + 2*p + 1)) Int[(d + e*x)^m*(a + b*x^2)^p*ExpandToSum[ b*e^q*(m + q + 2*p + 1)*Pq - b*f*(m + q + 2*p + 1)*(d + e*x)^q - 2*e*f*(m + p + q)*(d + e*x)^(q - 2)*(a*e - b*d*x), x], x], x] /; NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, d, e, m, p}, x] && PolyQ[Pq, x] && EqQ[b*d^2 + a*e^2, 0] && !IGtQ[m, 0]
Time = 0.47 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.04
method | result | size |
gosper | \(-\frac {\left (-d x +c \right ) \left (-24 A \,d^{7} x^{4}-18 B c \,d^{6} x^{4}-10 C \,c^{2} d^{5} x^{4}+3 D c^{3} d^{4} x^{4}-72 A c \,d^{6} x^{3}-54 B \,c^{2} d^{5} x^{3}-30 C \,c^{3} d^{4} x^{3}+9 D c^{4} d^{3} x^{3}-60 A \,c^{2} d^{5} x^{2}-45 B \,c^{3} d^{4} x^{2}-25 C \,c^{4} d^{3} x^{2}-45 D c^{5} d^{2} x^{2}+12 A \,c^{3} d^{4} x +9 B \,c^{4} d^{3} x -30 C \,c^{5} d^{2} x -54 D c^{6} d x +39 A \,c^{4} d^{3}+3 B \,c^{5} d^{2}-10 C \,c^{6} d -18 D c^{7}\right )}{105 \left (d x +c \right )^{2} c^{5} d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\) | \(243\) |
orering | \(-\frac {\left (-d x +c \right ) \left (-24 A \,d^{7} x^{4}-18 B c \,d^{6} x^{4}-10 C \,c^{2} d^{5} x^{4}+3 D c^{3} d^{4} x^{4}-72 A c \,d^{6} x^{3}-54 B \,c^{2} d^{5} x^{3}-30 C \,c^{3} d^{4} x^{3}+9 D c^{4} d^{3} x^{3}-60 A \,c^{2} d^{5} x^{2}-45 B \,c^{3} d^{4} x^{2}-25 C \,c^{4} d^{3} x^{2}-45 D c^{5} d^{2} x^{2}+12 A \,c^{3} d^{4} x +9 B \,c^{4} d^{3} x -30 C \,c^{5} d^{2} x -54 D c^{6} d x +39 A \,c^{4} d^{3}+3 B \,c^{5} d^{2}-10 C \,c^{6} d -18 D c^{7}\right )}{105 \left (d x +c \right )^{2} c^{5} d^{4} \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}\) | \(243\) |
trager | \(-\frac {\left (-24 A \,d^{7} x^{4}-18 B c \,d^{6} x^{4}-10 C \,c^{2} d^{5} x^{4}+3 D c^{3} d^{4} x^{4}-72 A c \,d^{6} x^{3}-54 B \,c^{2} d^{5} x^{3}-30 C \,c^{3} d^{4} x^{3}+9 D c^{4} d^{3} x^{3}-60 A \,c^{2} d^{5} x^{2}-45 B \,c^{3} d^{4} x^{2}-25 C \,c^{4} d^{3} x^{2}-45 D c^{5} d^{2} x^{2}+12 A \,c^{3} d^{4} x +9 B \,c^{4} d^{3} x -30 C \,c^{5} d^{2} x -54 D c^{6} d x +39 A \,c^{4} d^{3}+3 B \,c^{5} d^{2}-10 C \,c^{6} d -18 D c^{7}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{105 c^{5} d^{4} \left (d x +c \right )^{4} \left (-d x +c \right )}\) | \(245\) |
default | \(\frac {D x}{d^{3} c^{2} \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {\left (C d -3 D c \right ) \left (-\frac {1}{3 c d \left (x +\frac {c}{d}\right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d \,c^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{d^{4}}+\frac {\left (B \,d^{2}-2 C c d +3 D c^{2}\right ) \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right )^{2} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}+\frac {3 d \left (-\frac {1}{3 c d \left (x +\frac {c}{d}\right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d \,c^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{d^{5}}+\frac {\left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \left (-\frac {1}{7 c d \left (x +\frac {c}{d}\right )^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}+\frac {4 d \left (-\frac {1}{5 c d \left (x +\frac {c}{d}\right )^{2} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}+\frac {3 d \left (-\frac {1}{3 c d \left (x +\frac {c}{d}\right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}-\frac {-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d}{3 d \,c^{3} \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{5 c}\right )}{7 c}\right )}{d^{6}}\) | \(540\) |
Input:
int((D*x^3+C*x^2+B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(3/2),x,method=_RETURNVER BOSE)
Output:
-1/105*(-d*x+c)*(-24*A*d^7*x^4-18*B*c*d^6*x^4-10*C*c^2*d^5*x^4+3*D*c^3*d^4 *x^4-72*A*c*d^6*x^3-54*B*c^2*d^5*x^3-30*C*c^3*d^4*x^3+9*D*c^4*d^3*x^3-60*A *c^2*d^5*x^2-45*B*c^3*d^4*x^2-25*C*c^4*d^3*x^2-45*D*c^5*d^2*x^2+12*A*c^3*d ^4*x+9*B*c^4*d^3*x-30*C*c^5*d^2*x-54*D*c^6*d*x+39*A*c^4*d^3+3*B*c^5*d^2-10 *C*c^6*d-18*D*c^7)/(d*x+c)^2/c^5/d^4/(-d^2*x^2+c^2)^(3/2)
Leaf count of result is larger than twice the leaf count of optimal. 500 vs. \(2 (217) = 434\).
Time = 0.13 (sec) , antiderivative size = 500, normalized size of antiderivative = 2.15 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {18 \, D c^{8} + 10 \, C c^{7} d - 3 \, B c^{6} d^{2} - 39 \, A c^{5} d^{3} - {\left (18 \, D c^{3} d^{5} + 10 \, C c^{2} d^{6} - 3 \, B c d^{7} - 39 \, A d^{8}\right )} x^{5} - 3 \, {\left (18 \, D c^{4} d^{4} + 10 \, C c^{3} d^{5} - 3 \, B c^{2} d^{6} - 39 \, A c d^{7}\right )} x^{4} - 2 \, {\left (18 \, D c^{5} d^{3} + 10 \, C c^{4} d^{4} - 3 \, B c^{3} d^{5} - 39 \, A c^{2} d^{6}\right )} x^{3} + 2 \, {\left (18 \, D c^{6} d^{2} + 10 \, C c^{5} d^{3} - 3 \, B c^{4} d^{4} - 39 \, A c^{3} d^{5}\right )} x^{2} + 3 \, {\left (18 \, D c^{7} d + 10 \, C c^{6} d^{2} - 3 \, B c^{5} d^{3} - 39 \, A c^{4} d^{4}\right )} x + {\left (18 \, D c^{7} + 10 \, C c^{6} d - 3 \, B c^{5} d^{2} - 39 \, A c^{4} d^{3} - {\left (3 \, D c^{3} d^{4} - 10 \, C c^{2} d^{5} - 18 \, B c d^{6} - 24 \, A d^{7}\right )} x^{4} - 3 \, {\left (3 \, D c^{4} d^{3} - 10 \, C c^{3} d^{4} - 18 \, B c^{2} d^{5} - 24 \, A c d^{6}\right )} x^{3} + 5 \, {\left (9 \, D c^{5} d^{2} + 5 \, C c^{4} d^{3} + 9 \, B c^{3} d^{4} + 12 \, A c^{2} d^{5}\right )} x^{2} + 3 \, {\left (18 \, D c^{6} d + 10 \, C c^{5} d^{2} - 3 \, B c^{4} d^{3} - 4 \, A c^{3} d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{105 \, {\left (c^{5} d^{9} x^{5} + 3 \, c^{6} d^{8} x^{4} + 2 \, c^{7} d^{7} x^{3} - 2 \, c^{8} d^{6} x^{2} - 3 \, c^{9} d^{5} x - c^{10} d^{4}\right )}} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(3/2),x, algorithm= "fricas")
Output:
-1/105*(18*D*c^8 + 10*C*c^7*d - 3*B*c^6*d^2 - 39*A*c^5*d^3 - (18*D*c^3*d^5 + 10*C*c^2*d^6 - 3*B*c*d^7 - 39*A*d^8)*x^5 - 3*(18*D*c^4*d^4 + 10*C*c^3*d ^5 - 3*B*c^2*d^6 - 39*A*c*d^7)*x^4 - 2*(18*D*c^5*d^3 + 10*C*c^4*d^4 - 3*B* c^3*d^5 - 39*A*c^2*d^6)*x^3 + 2*(18*D*c^6*d^2 + 10*C*c^5*d^3 - 3*B*c^4*d^4 - 39*A*c^3*d^5)*x^2 + 3*(18*D*c^7*d + 10*C*c^6*d^2 - 3*B*c^5*d^3 - 39*A*c ^4*d^4)*x + (18*D*c^7 + 10*C*c^6*d - 3*B*c^5*d^2 - 39*A*c^4*d^3 - (3*D*c^3 *d^4 - 10*C*c^2*d^5 - 18*B*c*d^6 - 24*A*d^7)*x^4 - 3*(3*D*c^4*d^3 - 10*C*c ^3*d^4 - 18*B*c^2*d^5 - 24*A*c*d^6)*x^3 + 5*(9*D*c^5*d^2 + 5*C*c^4*d^3 + 9 *B*c^3*d^4 + 12*A*c^2*d^5)*x^2 + 3*(18*D*c^6*d + 10*C*c^5*d^2 - 3*B*c^4*d^ 3 - 4*A*c^3*d^4)*x)*sqrt(-d^2*x^2 + c^2))/(c^5*d^9*x^5 + 3*c^6*d^8*x^4 + 2 *c^7*d^7*x^3 - 2*c^8*d^6*x^2 - 3*c^9*d^5*x - c^10*d^4)
\[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\int \frac {A + B x + C x^{2} + D x^{3}}{\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {3}{2}} \left (c + d x\right )^{3}}\, dx \] Input:
integrate((D*x**3+C*x**2+B*x+A)/(d*x+c)**3/(-d**2*x**2+c**2)**(3/2),x)
Output:
Integral((A + B*x + C*x**2 + D*x**3)/((-(-c + d*x)*(c + d*x))**(3/2)*(c + d*x)**3), x)
Leaf count of result is larger than twice the leaf count of optimal. 1454 vs. \(2 (217) = 434\).
Time = 0.07 (sec) , antiderivative size = 1454, normalized size of antiderivative = 6.24 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\text {Too large to display} \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(3/2),x, algorithm= "maxima")
Output:
1/7*D*c^3/(sqrt(-d^2*x^2 + c^2)*c*d^7*x^3 + 3*sqrt(-d^2*x^2 + c^2)*c^2*d^6 *x^2 + 3*sqrt(-d^2*x^2 + c^2)*c^3*d^5*x + sqrt(-d^2*x^2 + c^2)*c^4*d^4) + 4/35*D*c^3/(sqrt(-d^2*x^2 + c^2)*c^2*d^6*x^2 + 2*sqrt(-d^2*x^2 + c^2)*c^3* d^5*x + sqrt(-d^2*x^2 + c^2)*c^4*d^4) + 4/35*D*c^3/(sqrt(-d^2*x^2 + c^2)*c ^3*d^5*x + sqrt(-d^2*x^2 + c^2)*c^4*d^4) - 1/7*C*c^2/(sqrt(-d^2*x^2 + c^2) *c*d^6*x^3 + 3*sqrt(-d^2*x^2 + c^2)*c^2*d^5*x^2 + 3*sqrt(-d^2*x^2 + c^2)*c ^3*d^4*x + sqrt(-d^2*x^2 + c^2)*c^4*d^3) - 4/35*C*c^2/(sqrt(-d^2*x^2 + c^2 )*c^2*d^5*x^2 + 2*sqrt(-d^2*x^2 + c^2)*c^3*d^4*x + sqrt(-d^2*x^2 + c^2)*c^ 4*d^3) - 4/35*C*c^2/(sqrt(-d^2*x^2 + c^2)*c^3*d^4*x + sqrt(-d^2*x^2 + c^2) *c^4*d^3) - 3/5*D*c^2/(sqrt(-d^2*x^2 + c^2)*c*d^6*x^2 + 2*sqrt(-d^2*x^2 + c^2)*c^2*d^5*x + sqrt(-d^2*x^2 + c^2)*c^3*d^4) - 3/5*D*c^2/(sqrt(-d^2*x^2 + c^2)*c^2*d^5*x + sqrt(-d^2*x^2 + c^2)*c^3*d^4) + 1/7*B*c/(sqrt(-d^2*x^2 + c^2)*c*d^5*x^3 + 3*sqrt(-d^2*x^2 + c^2)*c^2*d^4*x^2 + 3*sqrt(-d^2*x^2 + c^2)*c^3*d^3*x + sqrt(-d^2*x^2 + c^2)*c^4*d^2) + 4/35*B*c/(sqrt(-d^2*x^2 + c^2)*c^2*d^4*x^2 + 2*sqrt(-d^2*x^2 + c^2)*c^3*d^3*x + sqrt(-d^2*x^2 + c^2 )*c^4*d^2) + 4/35*B*c/(sqrt(-d^2*x^2 + c^2)*c^3*d^3*x + sqrt(-d^2*x^2 + c^ 2)*c^4*d^2) + 2/5*C*c/(sqrt(-d^2*x^2 + c^2)*c*d^5*x^2 + 2*sqrt(-d^2*x^2 + c^2)*c^2*d^4*x + sqrt(-d^2*x^2 + c^2)*c^3*d^3) + 2/5*C*c/(sqrt(-d^2*x^2 + c^2)*c^2*d^4*x + sqrt(-d^2*x^2 + c^2)*c^3*d^3) + D*c/(sqrt(-d^2*x^2 + c^2) *c*d^5*x + sqrt(-d^2*x^2 + c^2)*c^2*d^4) - 1/7*A/(sqrt(-d^2*x^2 + c^2)*...
\[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\int { \frac {D x^{3} + C x^{2} + B x + A}{{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{3}} \,d x } \] Input:
integrate((D*x^3+C*x^2+B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(3/2),x, algorithm= "giac")
Output:
integrate((D*x^3 + C*x^2 + B*x + A)/((-d^2*x^2 + c^2)^(3/2)*(d*x + c)^3), x)
Timed out. \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\int \frac {A+B\,x+C\,x^2+x^3\,D}{{\left (c^2-d^2\,x^2\right )}^{3/2}\,{\left (c+d\,x\right )}^3} \,d x \] Input:
int((A + B*x + C*x^2 + x^3*D)/((c^2 - d^2*x^2)^(3/2)*(c + d*x)^3),x)
Output:
int((A + B*x + C*x^2 + x^3*D)/((c^2 - d^2*x^2)^(3/2)*(c + d*x)^3), x)
Time = 0.20 (sec) , antiderivative size = 466, normalized size of antiderivative = 2.00 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {-7 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{6}+28 c^{7}-21 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{5} d x -21 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4} d^{2} x^{2}-7 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d^{3} x^{3}+24 a \,d^{6} x^{4}-39 a \,c^{4} d^{2}-3 b \,c^{5} d -4 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{3} d^{2}-4 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{5} x^{3}-3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4} d -12 a \,c^{3} d^{3} x +60 a \,c^{2} d^{4} x^{2}+72 a c \,d^{5} x^{3}-9 b \,c^{4} d^{2} x +45 b \,c^{3} d^{3} x^{2}+54 b \,c^{2} d^{4} x^{3}+18 b c \,d^{5} x^{4}-12 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{3} x -12 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{4} x^{2}-9 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d^{2} x -9 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{3} x^{2}-3 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{4} x^{3}+84 c^{6} d x +70 c^{5} d^{2} x^{2}+21 c^{4} d^{3} x^{3}+7 c^{3} d^{4} x^{4}}{105 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{5} d^{3} \left (d^{3} x^{3}+3 c \,d^{2} x^{2}+3 c^{2} d x +c^{3}\right )} \] Input:
int((D*x^3+C*x^2+B*x+A)/(d*x+c)^3/(-d^2*x^2+c^2)^(3/2),x)
Output:
( - 4*sqrt(c**2 - d**2*x**2)*a*c**3*d**2 - 12*sqrt(c**2 - d**2*x**2)*a*c** 2*d**3*x - 12*sqrt(c**2 - d**2*x**2)*a*c*d**4*x**2 - 4*sqrt(c**2 - d**2*x* *2)*a*d**5*x**3 - 3*sqrt(c**2 - d**2*x**2)*b*c**4*d - 9*sqrt(c**2 - d**2*x **2)*b*c**3*d**2*x - 9*sqrt(c**2 - d**2*x**2)*b*c**2*d**3*x**2 - 3*sqrt(c* *2 - d**2*x**2)*b*c*d**4*x**3 - 7*sqrt(c**2 - d**2*x**2)*c**6 - 21*sqrt(c* *2 - d**2*x**2)*c**5*d*x - 21*sqrt(c**2 - d**2*x**2)*c**4*d**2*x**2 - 7*sq rt(c**2 - d**2*x**2)*c**3*d**3*x**3 - 39*a*c**4*d**2 - 12*a*c**3*d**3*x + 60*a*c**2*d**4*x**2 + 72*a*c*d**5*x**3 + 24*a*d**6*x**4 - 3*b*c**5*d - 9*b *c**4*d**2*x + 45*b*c**3*d**3*x**2 + 54*b*c**2*d**4*x**3 + 18*b*c*d**5*x** 4 + 28*c**7 + 84*c**6*d*x + 70*c**5*d**2*x**2 + 21*c**4*d**3*x**3 + 7*c**3 *d**4*x**4)/(105*sqrt(c**2 - d**2*x**2)*c**5*d**3*(c**3 + 3*c**2*d*x + 3*c *d**2*x**2 + d**3*x**3))