Integrand size = 39, antiderivative size = 308 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {4 c^2 \left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \sqrt {c^2-d^2 x^2}}{d^4}-\frac {c \left (26 c^2 C d-30 B c d^2+24 A d^3-23 c^3 D\right ) x \sqrt {c^2-d^2 x^2}}{16 d^3}-\frac {\left (18 c C d-6 B d^2-23 c^2 D\right ) x^3 \sqrt {c^2-d^2 x^2}}{24 d}+\frac {1}{6} d D x^5 \sqrt {c^2-d^2 x^2}-\frac {\left (5 c^2 C d-3 B c d^2+A d^3-7 c^3 D\right ) \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}+\frac {(C d-3 c D) \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4}+\frac {c^3 \left (26 c^2 C d-30 B c d^2+40 A d^3-23 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{16 d^4} \] Output:
4*c^2*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(-d^2*x^2+c^2)^(1/2)/d^4-1/16*c*(24*A* d^3-30*B*c*d^2+26*C*c^2*d-23*D*c^3)*x*(-d^2*x^2+c^2)^(1/2)/d^3-1/24*(-6*B* d^2+18*C*c*d-23*D*c^2)*x^3*(-d^2*x^2+c^2)^(1/2)/d+1/6*d*D*x^5*(-d^2*x^2+c^ 2)^(1/2)-1/3*(A*d^3-3*B*c*d^2+5*C*c^2*d-7*D*c^3)*(-d^2*x^2+c^2)^(3/2)/d^4+ 1/5*(C*d-3*D*c)*(-d^2*x^2+c^2)^(5/2)/d^4+1/16*c^3*(40*A*d^3-30*B*c*d^2+26* C*c^2*d-23*D*c^3)*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d^4
Time = 1.38 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.70 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (-544 c^5 D+c^4 d (608 C+345 D x)-2 c^3 d^2 (360 B+x (195 C+136 D x))-12 c d^4 x (30 A+x (20 B+3 x (5 C+4 D x)))+4 d^5 x^2 (20 A+x (15 B+2 x (6 C+5 D x)))+2 c^2 d^3 (440 A+x (225 B+x (152 C+115 D x)))\right )+30 c^3 \left (-26 c^2 C d+30 B c d^2-40 A d^3+23 c^3 D\right ) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{240 d^4} \] Input:
Integrate[((c^2 - d^2*x^2)^(5/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^3,x]
Output:
(Sqrt[c^2 - d^2*x^2]*(-544*c^5*D + c^4*d*(608*C + 345*D*x) - 2*c^3*d^2*(36 0*B + x*(195*C + 136*D*x)) - 12*c*d^4*x*(30*A + x*(20*B + 3*x*(5*C + 4*D*x ))) + 4*d^5*x^2*(20*A + x*(15*B + 2*x*(6*C + 5*D*x))) + 2*c^2*d^3*(440*A + x*(225*B + x*(152*C + 115*D*x)))) + 30*c^3*(-26*c^2*C*d + 30*B*c*d^2 - 40 *A*d^3 + 23*c^3*D)*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/(240*d ^4)
Time = 1.20 (sec) , antiderivative size = 286, normalized size of antiderivative = 0.93, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.282, Rules used = {2170, 25, 2170, 27, 671, 464, 469, 455, 211, 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.
\(\displaystyle \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx\) |
\(\Big \downarrow \) 2170 |
\(\displaystyle -\frac {\int -\frac {\left (c^2-d^2 x^2\right )^{5/2} \left ((6 C d-13 c D) x^2 d^4+2 \left (3 B d^2-4 c^2 D\right ) x d^3+\left (6 A d^3-c^3 D\right ) d^2\right )}{(c+d x)^3}dx}{6 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{6 d^4 (c+d x)}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left ((6 C d-13 c D) x^2 d^4+2 \left (3 B d^2-4 c^2 D\right ) x d^3+\left (6 A d^3-c^3 D\right ) d^2\right )}{(c+d x)^3}dx}{6 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{6 d^4 (c+d x)}\) |
\(\Big \downarrow \) 2170 |
\(\displaystyle \frac {-\frac {\int \frac {3 d^6 \left (-7 D c^3+4 C d c^2-10 A d^3+d \left (-17 D c^2+14 C d c-10 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^3}dx}{5 d^4}-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (6 C d-13 c D)}{5 (c+d x)^2}}{6 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{6 d^4 (c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {3}{5} d^2 \int \frac {\left (-7 D c^3+4 C d c^2-10 A d^3+d \left (-17 D c^2+14 C d c-10 B d^2\right ) x\right ) \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^3}dx-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (6 C d-13 c D)}{5 (c+d x)^2}}{6 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{6 d^4 (c+d x)}\) |
\(\Big \downarrow \) 671 |
\(\displaystyle \frac {-\frac {3}{5} d^2 \left (-\frac {\left (40 A d^3-30 B c d^2-23 c^3 D+26 c^2 C d\right ) \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2}dx}{c}-\frac {10 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^3}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (6 C d-13 c D)}{5 (c+d x)^2}}{6 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{6 d^4 (c+d x)}\) |
\(\Big \downarrow \) 464 |
\(\displaystyle \frac {-\frac {3}{5} d^2 \left (-\frac {\left (40 A d^3-30 B c d^2-23 c^3 D+26 c^2 C d\right ) \int (c-d x)^2 \sqrt {c^2-d^2 x^2}dx}{c}-\frac {10 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^3}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (6 C d-13 c D)}{5 (c+d x)^2}}{6 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{6 d^4 (c+d x)}\) |
\(\Big \downarrow \) 469 |
\(\displaystyle \frac {-\frac {3}{5} d^2 \left (-\frac {\left (40 A d^3-30 B c d^2-23 c^3 D+26 c^2 C d\right ) \left (\frac {5}{4} c \int (c-d x) \sqrt {c^2-d^2 x^2}dx+\frac {(c-d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\right )}{c}-\frac {10 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^3}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (6 C d-13 c D)}{5 (c+d x)^2}}{6 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{6 d^4 (c+d x)}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {-\frac {3}{5} d^2 \left (-\frac {\left (40 A d^3-30 B c d^2-23 c^3 D+26 c^2 C d\right ) \left (\frac {5}{4} c \left (c \int \sqrt {c^2-d^2 x^2}dx+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )+\frac {(c-d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\right )}{c}-\frac {10 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^3}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (6 C d-13 c D)}{5 (c+d x)^2}}{6 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{6 d^4 (c+d x)}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {-\frac {3}{5} d^2 \left (-\frac {\left (40 A d^3-30 B c d^2-23 c^3 D+26 c^2 C d\right ) \left (\frac {5}{4} c \left (c \left (\frac {1}{2} c^2 \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )+\frac {(c-d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\right )}{c}-\frac {10 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^3}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (6 C d-13 c D)}{5 (c+d x)^2}}{6 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{6 d^4 (c+d x)}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {-\frac {3}{5} d^2 \left (-\frac {\left (40 A d^3-30 B c d^2-23 c^3 D+26 c^2 C d\right ) \left (\frac {5}{4} c \left (c \left (\frac {1}{2} c^2 \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )+\frac {(c-d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\right )}{c}-\frac {10 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^3}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (6 C d-13 c D)}{5 (c+d x)^2}}{6 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{6 d^4 (c+d x)}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {-\frac {3}{5} d^2 \left (-\frac {\left (\frac {5}{4} c \left (c \left (\frac {c^2 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )+\frac {(c-d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\right ) \left (40 A d^3-30 B c d^2-23 c^3 D+26 c^2 C d\right )}{c}-\frac {10 \left (c^2-d^2 x^2\right )^{7/2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{c d (c+d x)^3}\right )-\frac {d \left (c^2-d^2 x^2\right )^{7/2} (6 C d-13 c D)}{5 (c+d x)^2}}{6 d^5}-\frac {D \left (c^2-d^2 x^2\right )^{7/2}}{6 d^4 (c+d x)}\) |
Input:
Int[((c^2 - d^2*x^2)^(5/2)*(A + B*x + C*x^2 + D*x^3))/(c + d*x)^3,x]
Output:
-1/6*(D*(c^2 - d^2*x^2)^(7/2))/(d^4*(c + d*x)) + (-1/5*(d*(6*C*d - 13*c*D) *(c^2 - d^2*x^2)^(7/2))/(c + d*x)^2 - (3*d^2*((-10*(c^2*C*d - B*c*d^2 + A* d^3 - c^3*D)*(c^2 - d^2*x^2)^(7/2))/(c*d*(c + d*x)^3) - ((26*c^2*C*d - 30* B*c*d^2 + 40*A*d^3 - 23*c^3*D)*(((c - d*x)*(c^2 - d^2*x^2)^(3/2))/(4*d) + (5*c*((c^2 - d^2*x^2)^(3/2)/(3*d) + c*((x*Sqrt[c^2 - d^2*x^2])/2 + (c^2*Ar cTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/(2*d))))/4))/c))/5)/(6*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)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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])
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]
Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Int[( a + b*x^2)^(n + p)/(a/c + b*(x/d))^n, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b *c^2 + a*d^2, 0] && IntegerQ[n] && RationalQ[p] && (LtQ[0, -n, p] || LtQ[p, -n, 0]) && NeQ[n, 2] && NeQ[n, -1]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[ d*(c + d*x)^(n - 1)*((a + b*x^2)^(p + 1)/(b*(n + 2*p + 1))), x] + Simp[2*c* ((n + p)/(n + 2*p + 1)) Int[(c + d*x)^(n - 1)*(a + b*x^2)^p, x], x] /; Fr eeQ[{a, b, c, d, p}, x] && EqQ[b*c^2 + a*d^2, 0] && GtQ[n, 0] && NeQ[n + 2* 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]
Leaf count of result is larger than twice the leaf count of optimal. \(880\) vs. \(2(282)=564\).
Time = 0.44 (sec) , antiderivative size = 881, normalized size of antiderivative = 2.86
method | result | size |
default | \(\frac {D \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{6}+\frac {5 c^{2} \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )}{6}\right )}{d^{3}}+\frac {\left (C d -3 D c \right ) \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{5}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 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}}}{8 d^{2}}+\frac {3 c^{2} \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \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 )}{2 \sqrt {d^{2}}}\right )}{4}\right )\right )}{d^{4}}+\frac {\left (B \,d^{2}-2 C c d +3 D c^{2}\right ) \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{3 c d \left (x +\frac {c}{d}\right )^{2}}+\frac {5 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{5}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 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}}}{8 d^{2}}+\frac {3 c^{2} \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \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 )}{2 \sqrt {d^{2}}}\right )}{4}\right )\right )}{3 c}\right )}{d^{5}}+\frac {\left (A \,d^{3}-B c \,d^{2}+C \,c^{2} d -D c^{3}\right ) \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{c d \left (x +\frac {c}{d}\right )^{3}}+\frac {4 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{3 c d \left (x +\frac {c}{d}\right )^{2}}+\frac {5 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{5}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 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}}}{8 d^{2}}+\frac {3 c^{2} \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \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 )}{2 \sqrt {d^{2}}}\right )}{4}\right )\right )}{3 c}\right )}{c}\right )}{d^{6}}\) | \(881\) |
Input:
int((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x,method=_RETURNVER BOSE)
Output:
D/d^3*(1/6*x*(-d^2*x^2+c^2)^(5/2)+5/6*c^2*(1/4*x*(-d^2*x^2+c^2)^(3/2)+3/4* c^2*(1/2*x*(-d^2*x^2+c^2)^(1/2)+1/2*c^2/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/( -d^2*x^2+c^2)^(1/2)))))+1/d^4*(C*d-3*D*c)*(1/5*(-d^2*(x+c/d)^2+2*c*d*(x+c/ d))^(5/2)+c*d*(-1/8*(-2*d^2*(x+c/d)+2*c*d)/d^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/ d))^(3/2)+3/4*c^2*(-1/4*(-2*d^2*(x+c/d)+2*c*d)/d^2*(-d^2*(x+c/d)^2+2*c*d*( x+c/d))^(1/2)+1/2*c^2/(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/d^5*(B*d^2-2*C*c*d+3*D*c^2)*(1/3/c/d/(x+c/d)^2*(-d ^2*(x+c/d)^2+2*c*d*(x+c/d))^(7/2)+5/3*d/c*(1/5*(-d^2*(x+c/d)^2+2*c*d*(x+c/ d))^(5/2)+c*d*(-1/8*(-2*d^2*(x+c/d)+2*c*d)/d^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/ d))^(3/2)+3/4*c^2*(-1/4*(-2*d^2*(x+c/d)+2*c*d)/d^2*(-d^2*(x+c/d)^2+2*c*d*( x+c/d))^(1/2)+1/2*c^2/(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/d^6*(A*d^3-B*c*d^2+C*c^2*d-D*c^3)*(1/c/d/(x+c/d)^ 3*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(7/2)+4*d/c*(1/3/c/d/(x+c/d)^2*(-d^2*(x+c /d)^2+2*c*d*(x+c/d))^(7/2)+5/3*d/c*(1/5*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(5/ 2)+c*d*(-1/8*(-2*d^2*(x+c/d)+2*c*d)/d^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d))^(3/ 2)+3/4*c^2*(-1/4*(-2*d^2*(x+c/d)+2*c*d)/d^2*(-d^2*(x+c/d)^2+2*c*d*(x+c/d)) ^(1/2)+1/2*c^2/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*(x+c/d)^2+2*c*d*(x+c /d))^(1/2)))))))
Time = 0.09 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.79 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {30 \, {\left (23 \, D c^{6} - 26 \, C c^{5} d + 30 \, B c^{4} d^{2} - 40 \, A c^{3} d^{3}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (40 \, D d^{5} x^{5} - 544 \, D c^{5} + 608 \, C c^{4} d - 720 \, B c^{3} d^{2} + 880 \, A c^{2} d^{3} - 48 \, {\left (3 \, D c d^{4} - C d^{5}\right )} x^{4} + 10 \, {\left (23 \, D c^{2} d^{3} - 18 \, C c d^{4} + 6 \, B d^{5}\right )} x^{3} - 16 \, {\left (17 \, D c^{3} d^{2} - 19 \, C c^{2} d^{3} + 15 \, B c d^{4} - 5 \, A d^{5}\right )} x^{2} + 15 \, {\left (23 \, D c^{4} d - 26 \, C c^{3} d^{2} + 30 \, B c^{2} d^{3} - 24 \, A c d^{4}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{240 \, d^{4}} \] Input:
integrate((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x, algorithm= "fricas")
Output:
1/240*(30*(23*D*c^6 - 26*C*c^5*d + 30*B*c^4*d^2 - 40*A*c^3*d^3)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + (40*D*d^5*x^5 - 544*D*c^5 + 608*C*c^4*d - 720*B*c^3*d^2 + 880*A*c^2*d^3 - 48*(3*D*c*d^4 - C*d^5)*x^4 + 10*(23*D*c^ 2*d^3 - 18*C*c*d^4 + 6*B*d^5)*x^3 - 16*(17*D*c^3*d^2 - 19*C*c^2*d^3 + 15*B *c*d^4 - 5*A*d^5)*x^2 + 15*(23*D*c^4*d - 26*C*c^3*d^2 + 30*B*c^2*d^3 - 24* A*c*d^4)*x)*sqrt(-d^2*x^2 + c^2))/d^4
\[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\int \frac {\left (- \left (- c + d x\right ) \left (c + d x\right )\right )^{\frac {5}{2}} \left (A + B x + C x^{2} + D x^{3}\right )}{\left (c + d x\right )^{3}}\, dx \] Input:
integrate((-d**2*x**2+c**2)**(5/2)*(D*x**3+C*x**2+B*x+A)/(d*x+c)**3,x)
Output:
Integral((-(-c + d*x)*(c + d*x))**(5/2)*(A + B*x + C*x**2 + D*x**3)/(c + d *x)**3, x)
Result contains complex when optimal does not.
Time = 0.20 (sec) , antiderivative size = 978, normalized size of antiderivative = 3.18 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\text {Too large to display} \] Input:
integrate((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x, algorithm= "maxima")
Output:
-1/3*(-d^2*x^2 + c^2)^(5/2)*D*c^3/(d^6*x^2 + 2*c*d^5*x + c^2*d^4) - 5/6*(- d^2*x^2 + c^2)^(3/2)*D*c^4/(d^5*x + c*d^4) + 1/3*(-d^2*x^2 + c^2)^(5/2)*C* c^2/(d^5*x^2 + 2*c*d^4*x + c^2*d^3) + 3/4*(-d^2*x^2 + c^2)^(5/2)*D*c^2/(d^ 5*x + c*d^4) + 5/6*(-d^2*x^2 + c^2)^(3/2)*C*c^3/(d^4*x + c*d^3) - 3/4*I*D* c^6*arcsin(d*x/c + 2)/d^4 + 7/8*I*C*c^5*arcsin(d*x/c + 2)/d^3 - 5/8*I*B*c^ 4*arcsin(d*x/c + 2)/d^2 - 35/16*D*c^6*arcsin(d*x/c)/d^4 + 5/2*C*c^5*arcsin (d*x/c)/d^3 - 5/2*B*c^4*arcsin(d*x/c)/d^2 + 5/2*A*c^3*arcsin(d*x/c)/d - 1/ 3*(-d^2*x^2 + c^2)^(5/2)*B*c/(d^4*x^2 + 2*c*d^3*x + c^2*d^2) - 1/2*(-d^2*x ^2 + c^2)^(5/2)*C*c/(d^4*x + c*d^3) - 5/6*(-d^2*x^2 + c^2)^(3/2)*B*c^2/(d^ 3*x + c*d^2) + 3/4*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*D*c^4*x/d^3 + 5/16*sqrt (-d^2*x^2 + c^2)*D*c^4*x/d^3 - 7/8*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*C*c^3*x /d^2 + 5/8*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*B*c^2*x/d + 1/3*(-d^2*x^2 + c^2 )^(5/2)*A/(d^3*x^2 + 2*c*d^2*x + c^2*d) + 1/4*(-d^2*x^2 + c^2)^(5/2)*B/(d^ 3*x + c*d^2) + 5/6*(-d^2*x^2 + c^2)^(3/2)*A*c/(d^2*x + c*d) + 3/2*sqrt(d^2 *x^2 + 4*c*d*x + 3*c^2)*D*c^5/d^4 - 5/2*sqrt(-d^2*x^2 + c^2)*D*c^5/d^4 - 7 /4*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*C*c^4/d^3 + 5/2*sqrt(-d^2*x^2 + c^2)*C* c^4/d^3 + 5/4*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*B*c^3/d^2 - 5/2*sqrt(-d^2*x^ 2 + c^2)*B*c^3/d^2 + 5/2*sqrt(-d^2*x^2 + c^2)*A*c^2/d - 13/24*(-d^2*x^2 + c^2)^(3/2)*D*c^2*x/d^3 + 1/4*(-d^2*x^2 + c^2)^(3/2)*C*c*x/d^2 + 5/4*(-d^2* x^2 + c^2)^(3/2)*D*c^3/d^4 - 5/6*(-d^2*x^2 + c^2)^(3/2)*C*c^2/d^3 + 5/1...
Time = 0.14 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.83 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {1}{240} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, D d x - \frac {6 \, {\left (3 \, D c d^{8} - C d^{9}\right )}}{d^{8}}\right )} x + \frac {5 \, {\left (23 \, D c^{2} d^{7} - 18 \, C c d^{8} + 6 \, B d^{9}\right )}}{d^{8}}\right )} x - \frac {8 \, {\left (17 \, D c^{3} d^{6} - 19 \, C c^{2} d^{7} + 15 \, B c d^{8} - 5 \, A d^{9}\right )}}{d^{8}}\right )} x + \frac {15 \, {\left (23 \, D c^{4} d^{5} - 26 \, C c^{3} d^{6} + 30 \, B c^{2} d^{7} - 24 \, A c d^{8}\right )}}{d^{8}}\right )} x - \frac {16 \, {\left (34 \, D c^{5} d^{4} - 38 \, C c^{4} d^{5} + 45 \, B c^{3} d^{6} - 55 \, A c^{2} d^{7}\right )}}{d^{8}}\right )} - \frac {{\left (23 \, D c^{6} - 26 \, C c^{5} d + 30 \, B c^{4} d^{2} - 40 \, A c^{3} d^{3}\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{16 \, d^{3} {\left | d \right |}} \] Input:
integrate((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x, algorithm= "giac")
Output:
1/240*sqrt(-d^2*x^2 + c^2)*((2*((4*(5*D*d*x - 6*(3*D*c*d^8 - C*d^9)/d^8)*x + 5*(23*D*c^2*d^7 - 18*C*c*d^8 + 6*B*d^9)/d^8)*x - 8*(17*D*c^3*d^6 - 19*C *c^2*d^7 + 15*B*c*d^8 - 5*A*d^9)/d^8)*x + 15*(23*D*c^4*d^5 - 26*C*c^3*d^6 + 30*B*c^2*d^7 - 24*A*c*d^8)/d^8)*x - 16*(34*D*c^5*d^4 - 38*C*c^4*d^5 + 45 *B*c^3*d^6 - 55*A*c^2*d^7)/d^8) - 1/16*(23*D*c^6 - 26*C*c^5*d + 30*B*c^4*d ^2 - 40*A*c^3*d^3)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d^3*abs(d))
Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}\,\left (A+B\,x+C\,x^2+x^3\,D\right )}{{\left (c+d\,x\right )}^3} \,d x \] Input:
int(((c^2 - d^2*x^2)^(5/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^3,x)
Output:
int(((c^2 - d^2*x^2)^(5/2)*(A + B*x + C*x^2 + x^3*D))/(c + d*x)^3, x)
Time = 0.21 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.14 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2} \left (A+B x+C x^2+D x^3\right )}{(c+d x)^3} \, dx=\frac {600 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{3} d^{2}-450 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{4} d +45 \mathit {asin} \left (\frac {d x}{c}\right ) c^{6}+880 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{2}-360 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{3} x +80 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{4} x^{2}-720 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3} d +450 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{2} x -240 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{3} x^{2}+60 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{4} x^{3}+64 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{5}-45 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4} d x +32 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d^{2} x^{2}+50 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{3} x^{3}-96 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{4} x^{4}+40 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{5} x^{5}-880 a \,c^{3} d^{2}+720 b \,c^{4} d -64 c^{6}}{240 d^{3}} \] Input:
int((-d^2*x^2+c^2)^(5/2)*(D*x^3+C*x^2+B*x+A)/(d*x+c)^3,x)
Output:
(600*asin((d*x)/c)*a*c**3*d**2 - 450*asin((d*x)/c)*b*c**4*d + 45*asin((d*x )/c)*c**6 + 880*sqrt(c**2 - d**2*x**2)*a*c**2*d**2 - 360*sqrt(c**2 - d**2* x**2)*a*c*d**3*x + 80*sqrt(c**2 - d**2*x**2)*a*d**4*x**2 - 720*sqrt(c**2 - d**2*x**2)*b*c**3*d + 450*sqrt(c**2 - d**2*x**2)*b*c**2*d**2*x - 240*sqrt (c**2 - d**2*x**2)*b*c*d**3*x**2 + 60*sqrt(c**2 - d**2*x**2)*b*d**4*x**3 + 64*sqrt(c**2 - d**2*x**2)*c**5 - 45*sqrt(c**2 - d**2*x**2)*c**4*d*x + 32* sqrt(c**2 - d**2*x**2)*c**3*d**2*x**2 + 50*sqrt(c**2 - d**2*x**2)*c**2*d** 3*x**3 - 96*sqrt(c**2 - d**2*x**2)*c*d**4*x**4 + 40*sqrt(c**2 - d**2*x**2) *d**5*x**5 - 880*a*c**3*d**2 + 720*b*c**4*d - 64*c**6)/(240*d**3)