Integrand size = 27, antiderivative size = 187 \[ \int \frac {x^3 \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\frac {c^5 x \sqrt {c^2-d^2 x^2}}{8 d^3}+\frac {c^3 x^3 \sqrt {c^2-d^2 x^2}}{12 d}-\frac {1}{3} c d x^5 \sqrt {c^2-d^2 x^2}-\frac {2 c^4 \left (c^2-d^2 x^2\right )^{3/2}}{3 d^4}+\frac {3 c^2 \left (c^2-d^2 x^2\right )^{5/2}}{5 d^4}-\frac {\left (c^2-d^2 x^2\right )^{7/2}}{7 d^4}-\frac {c^7 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d^4} \] Output:
1/8*c^5*x*(-d^2*x^2+c^2)^(1/2)/d^3+1/12*c^3*x^3*(-d^2*x^2+c^2)^(1/2)/d-1/3 *c*d*x^5*(-d^2*x^2+c^2)^(1/2)-2/3*c^4*(-d^2*x^2+c^2)^(3/2)/d^4+3/5*c^2*(-d ^2*x^2+c^2)^(5/2)/d^4-1/7*(-d^2*x^2+c^2)^(7/2)/d^4-1/8*c^7*arctan(d*x/(-d^ 2*x^2+c^2)^(1/2))/d^4
Time = 0.28 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.71 \[ \int \frac {x^3 \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\frac {d \sqrt {c^2-d^2 x^2} \left (-176 c^6+105 c^5 d x-88 c^4 d^2 x^2+70 c^3 d^3 x^3+144 c^2 d^4 x^4-280 c d^5 x^5+120 d^6 x^6\right )-105 c^7 \sqrt {-d^2} \log \left (-\sqrt {-d^2} x+\sqrt {c^2-d^2 x^2}\right )}{840 d^5} \] Input:
Integrate[(x^3*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^2,x]
Output:
(d*Sqrt[c^2 - d^2*x^2]*(-176*c^6 + 105*c^5*d*x - 88*c^4*d^2*x^2 + 70*c^3*d ^3*x^3 + 144*c^2*d^4*x^4 - 280*c*d^5*x^5 + 120*d^6*x^6) - 105*c^7*Sqrt[-d^ 2]*Log[-(Sqrt[-d^2]*x) + Sqrt[c^2 - d^2*x^2]])/(840*d^5)
Time = 0.63 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.12, number of steps used = 17, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.593, Rules used = {562, 541, 25, 27, 533, 27, 533, 25, 27, 533, 25, 27, 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.
| \(\displaystyle \int \frac {x^3 \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx\) |
| \(\Big \downarrow \) 562 |
| \(\displaystyle \int x^3 (c-d x)^2 \sqrt {c^2-d^2 x^2}dx\) |
| \(\Big \downarrow \) 541 |
| \(\displaystyle -\frac {\int -c d^2 x^3 (11 c-14 d x) \sqrt {c^2-d^2 x^2}dx}{7 d^2}-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int c d^2 x^3 (11 c-14 d x) \sqrt {c^2-d^2 x^2}dx}{7 d^2}-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} c \int x^3 (11 c-14 d x) \sqrt {c^2-d^2 x^2}dx-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 533 |
| \(\displaystyle \frac {1}{7} c \left (\frac {\int -6 c d x^2 (7 c-11 d x) \sqrt {c^2-d^2 x^2}dx}{6 d^2}+\frac {7 x^3 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} c \left (\frac {7 x^3 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}-\frac {c \int x^2 (7 c-11 d x) \sqrt {c^2-d^2 x^2}dx}{d}\right )-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 533 |
| \(\displaystyle \frac {1}{7} c \left (\frac {7 x^3 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}-\frac {c \left (\frac {\int -c d x (22 c-35 d x) \sqrt {c^2-d^2 x^2}dx}{5 d^2}+\frac {11 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}\right )}{d}\right )-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{7} c \left (\frac {7 x^3 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}-\frac {c \left (\frac {11 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}-\frac {\int c d x (22 c-35 d x) \sqrt {c^2-d^2 x^2}dx}{5 d^2}\right )}{d}\right )-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} c \left (\frac {7 x^3 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}-\frac {c \left (\frac {11 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}-\frac {c \int x (22 c-35 d x) \sqrt {c^2-d^2 x^2}dx}{5 d}\right )}{d}\right )-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 533 |
| \(\displaystyle \frac {1}{7} c \left (\frac {7 x^3 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}-\frac {c \left (\frac {11 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}-\frac {c \left (\frac {\int -c d (35 c-88 d x) \sqrt {c^2-d^2 x^2}dx}{4 d^2}+\frac {35 x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\right )}{5 d}\right )}{d}\right )-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{7} c \left (\frac {7 x^3 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}-\frac {c \left (\frac {11 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}-\frac {c \left (\frac {35 x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}-\frac {\int c d (35 c-88 d x) \sqrt {c^2-d^2 x^2}dx}{4 d^2}\right )}{5 d}\right )}{d}\right )-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {1}{7} c \left (\frac {7 x^3 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}-\frac {c \left (\frac {11 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}-\frac {c \left (\frac {35 x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}-\frac {c \int (35 c-88 d x) \sqrt {c^2-d^2 x^2}dx}{4 d}\right )}{5 d}\right )}{d}\right )-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {1}{7} c \left (\frac {7 x^3 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}-\frac {c \left (\frac {11 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}-\frac {c \left (\frac {35 x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}-\frac {c \left (35 c \int \sqrt {c^2-d^2 x^2}dx+\frac {88 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )}{4 d}\right )}{5 d}\right )}{d}\right )-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {1}{7} c \left (\frac {7 x^3 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}-\frac {c \left (\frac {11 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}-\frac {c \left (\frac {35 x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}-\frac {c \left (35 c \left (\frac {1}{2} c^2 \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {88 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )}{4 d}\right )}{5 d}\right )}{d}\right )-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {1}{7} c \left (\frac {7 x^3 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}-\frac {c \left (\frac {11 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}-\frac {c \left (\frac {35 x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}-\frac {c \left (35 c \left (\frac {1}{2} c^2 \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {88 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )}{4 d}\right )}{5 d}\right )}{d}\right )-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {1}{7} c \left (\frac {7 x^3 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}-\frac {c \left (\frac {11 x^2 \left (c^2-d^2 x^2\right )^{3/2}}{5 d}-\frac {c \left (\frac {35 x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}-\frac {c \left (35 c \left (\frac {c^2 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {88 \left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )}{4 d}\right )}{5 d}\right )}{d}\right )-\frac {1}{7} x^4 \left (c^2-d^2 x^2\right )^{3/2}\) |
Input:
Int[(x^3*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^2,x]
Output:
-1/7*(x^4*(c^2 - d^2*x^2)^(3/2)) + (c*((7*x^3*(c^2 - d^2*x^2)^(3/2))/(3*d) - (c*((11*x^2*(c^2 - d^2*x^2)^(3/2))/(5*d) - (c*((35*x*(c^2 - d^2*x^2)^(3 /2))/(4*d) - (c*((88*(c^2 - d^2*x^2)^(3/2))/(3*d) + 35*c*((x*Sqrt[c^2 - d^ 2*x^2])/2 + (c^2*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]])/(2*d))))/(4*d)))/(5*d) ))/d))/7
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[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* p + 2)) Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer Q[2*p]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[d^n*x^(m + n - 1)*((a + b*x^2)^(p + 1)/(b*(m + n + 2*p + 1))), x ] + Simp[1/(b*(m + n + 2*p + 1)) Int[x^m*(a + b*x^2)^p*ExpandToSum[b*(m + n + 2*p + 1)*(c + d*x)^n - b*d^n*(m + n + 2*p + 1)*x^n - a*d^n*(m + n - 1) *x^(n - 2), x], x], x] /; FreeQ[{a, b, c, d, m, p}, x] && IGtQ[n, 1] && IGt Q[m, -2] && GtQ[p, -1] && IntegerQ[2*p]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo l] :> Simp[c^(2*n)/a^n Int[x^m*((a + b*x^2)^(n + p)/(c - d*x)^n), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[b*c^2 + a*d^2, 0] && IGtQ[m, 0] && ILtQ[n, 0] && IGtQ[n + p + 1/2, 0]
Time = 0.23 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.64
| method | result | size |
| risch | \(-\frac {\left (-120 d^{6} x^{6}+280 c \,d^{5} x^{5}-144 c^{2} d^{4} x^{4}-70 d^{3} c^{3} x^{3}+88 c^{4} d^{2} x^{2}-105 c^{5} d x +176 c^{6}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{840 d^{4}}-\frac {c^{7} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{8 d^{3} \sqrt {d^{2}}}\) | \(119\) |
| default | \(-\frac {\left (-d^{2} x^{2}+c^{2}\right )^{\frac {7}{2}}}{7 d^{4}}-\frac {2 c \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 {3 c^{2} \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 {c^{3} \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}}\) | \(565\) |
Input:
int(x^3*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
-1/840*(-120*d^6*x^6+280*c*d^5*x^5-144*c^2*d^4*x^4-70*c^3*d^3*x^3+88*c^4*d ^2*x^2-105*c^5*d*x+176*c^6)/d^4*(-d^2*x^2+c^2)^(1/2)-1/8*c^7/d^3/(d^2)^(1/ 2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))
Time = 0.10 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.62 \[ \int \frac {x^3 \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\frac {210 \, c^{7} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (120 \, d^{6} x^{6} - 280 \, c d^{5} x^{5} + 144 \, c^{2} d^{4} x^{4} + 70 \, c^{3} d^{3} x^{3} - 88 \, c^{4} d^{2} x^{2} + 105 \, c^{5} d x - 176 \, c^{6}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{840 \, d^{4}} \] Input:
integrate(x^3*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^2,x, algorithm="fricas")
Output:
1/840*(210*c^7*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + (120*d^6*x^6 - 280*c*d^5*x^5 + 144*c^2*d^4*x^4 + 70*c^3*d^3*x^3 - 88*c^4*d^2*x^2 + 105*c^ 5*d*x - 176*c^6)*sqrt(-d^2*x^2 + c^2))/d^4
Time = 2.00 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.36 \[ \int \frac {x^3 \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=c^{2} \left (\begin {cases} \sqrt {c^{2} - d^{2} x^{2}} \left (- \frac {2 c^{4}}{15 d^{4}} - \frac {c^{2} x^{2}}{15 d^{2}} + \frac {x^{4}}{5}\right ) & \text {for}\: d^{2} \neq 0 \\\frac {x^{4} \sqrt {c^{2}}}{4} & \text {otherwise} \end {cases}\right ) - 2 c d \left (\begin {cases} \frac {c^{6} \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{16 d^{4}} + \sqrt {c^{2} - d^{2} x^{2}} \left (- \frac {c^{4} x}{16 d^{4}} - \frac {c^{2} x^{3}}{24 d^{2}} + \frac {x^{5}}{6}\right ) & \text {for}\: d^{2} \neq 0 \\\frac {x^{5} \sqrt {c^{2}}}{5} & \text {otherwise} \end {cases}\right ) + d^{2} \left (\begin {cases} \sqrt {c^{2} - d^{2} x^{2}} \left (- \frac {8 c^{6}}{105 d^{6}} - \frac {4 c^{4} x^{2}}{105 d^{4}} - \frac {c^{2} x^{4}}{35 d^{2}} + \frac {x^{6}}{7}\right ) & \text {for}\: d^{2} \neq 0 \\\frac {x^{6} \sqrt {c^{2}}}{6} & \text {otherwise} \end {cases}\right ) \] Input:
integrate(x**3*(-d**2*x**2+c**2)**(5/2)/(d*x+c)**2,x)
Output:
c**2*Piecewise((sqrt(c**2 - d**2*x**2)*(-2*c**4/(15*d**4) - c**2*x**2/(15* d**2) + x**4/5), Ne(d**2, 0)), (x**4*sqrt(c**2)/4, True)) - 2*c*d*Piecewis e((c**6*Piecewise((log(-2*d**2*x + 2*sqrt(-d**2)*sqrt(c**2 - d**2*x**2))/s qrt(-d**2), Ne(c**2, 0)), (x*log(x)/sqrt(-d**2*x**2), True))/(16*d**4) + s qrt(c**2 - d**2*x**2)*(-c**4*x/(16*d**4) - c**2*x**3/(24*d**2) + x**5/6), Ne(d**2, 0)), (x**5*sqrt(c**2)/5, True)) + d**2*Piecewise((sqrt(c**2 - d** 2*x**2)*(-8*c**6/(105*d**6) - 4*c**4*x**2/(105*d**4) - c**2*x**4/(35*d**2) + x**6/7), Ne(d**2, 0)), (x**6*sqrt(c**2)/6, True))
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.34 \[ \int \frac {x^3 \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=-\frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} c^{3}}{4 \, {\left (d^{5} x + c d^{4}\right )}} - \frac {i \, c^{7} \arcsin \left (\frac {d x}{c} + 2\right )}{2 \, d^{4}} - \frac {5 \, c^{7} \arcsin \left (\frac {d x}{c}\right )}{8 \, d^{4}} + \frac {\sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} c^{5} x}{2 \, d^{3}} - \frac {5 \, \sqrt {-d^{2} x^{2} + c^{2}} c^{5} x}{8 \, d^{3}} + \frac {\sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} c^{6}}{d^{4}} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{3} x}{3 \, d^{3}} - \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c^{4}}{12 \, d^{4}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} c x}{3 \, d^{3}} + \frac {3 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} c^{2}}{5 \, d^{4}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {7}{2}}}{7 \, d^{4}} \] Input:
integrate(x^3*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^2,x, algorithm="maxima")
Output:
-1/4*(-d^2*x^2 + c^2)^(5/2)*c^3/(d^5*x + c*d^4) - 1/2*I*c^7*arcsin(d*x/c + 2)/d^4 - 5/8*c^7*arcsin(d*x/c)/d^4 + 1/2*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)* c^5*x/d^3 - 5/8*sqrt(-d^2*x^2 + c^2)*c^5*x/d^3 + sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*c^6/d^4 + 1/3*(-d^2*x^2 + c^2)^(3/2)*c^3*x/d^3 - 5/12*(-d^2*x^2 + c ^2)^(3/2)*c^4/d^4 - 1/3*(-d^2*x^2 + c^2)^(5/2)*c*x/d^3 + 3/5*(-d^2*x^2 + c ^2)^(5/2)*c^2/d^4 - 1/7*(-d^2*x^2 + c^2)^(7/2)/d^4
Time = 0.16 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.50 \[ \int \frac {x^3 \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\frac {{\left (13440 \, c^{8} d^{8} \arctan \left (\sqrt {\frac {2 \, c}{d x + c} - 1}\right ) \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + \frac {{\left (105 \, c^{8} d^{8} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {13}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 3780 \, c^{8} d^{8} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {11}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + 189 \, c^{8} d^{8} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {9}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 4992 \, c^{8} d^{8} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 1981 \, c^{8} d^{8} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 700 \, c^{8} d^{8} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 105 \, c^{8} d^{8} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )\right )} {\left (d x + c\right )}^{7}}{c^{7}}\right )} {\left | d \right |}}{53760 \, c d^{13}} \] Input:
integrate(x^3*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^2,x, algorithm="giac")
Output:
1/53760*(13440*c^8*d^8*arctan(sqrt(2*c/(d*x + c) - 1))*sgn(1/(d*x + c))*sg n(d) + (105*c^8*d^8*(2*c/(d*x + c) - 1)^(13/2)*sgn(1/(d*x + c))*sgn(d) - 3 780*c^8*d^8*(2*c/(d*x + c) - 1)^(11/2)*sgn(1/(d*x + c))*sgn(d) + 189*c^8*d ^8*(2*c/(d*x + c) - 1)^(9/2)*sgn(1/(d*x + c))*sgn(d) - 4992*c^8*d^8*(2*c/( d*x + c) - 1)^(7/2)*sgn(1/(d*x + c))*sgn(d) - 1981*c^8*d^8*(2*c/(d*x + c) - 1)^(5/2)*sgn(1/(d*x + c))*sgn(d) - 700*c^8*d^8*(2*c/(d*x + c) - 1)^(3/2) *sgn(1/(d*x + c))*sgn(d) - 105*c^8*d^8*sqrt(2*c/(d*x + c) - 1)*sgn(1/(d*x + c))*sgn(d))*(d*x + c)^7/c^7)*abs(d)/(c*d^13)
Timed out. \[ \int \frac {x^3 \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\int \frac {x^3\,{\left (c^2-d^2\,x^2\right )}^{5/2}}{{\left (c+d\,x\right )}^2} \,d x \] Input:
int((x^3*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^2,x)
Output:
int((x^3*(c^2 - d^2*x^2)^(5/2))/(c + d*x)^2, x)
Time = 0.22 (sec) , antiderivative size = 176, normalized size of antiderivative = 0.94 \[ \int \frac {x^3 \left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\frac {-105 \mathit {asin} \left (\frac {d x}{c}\right ) c^{7}-176 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{6}+105 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{5} d x -88 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{4} d^{2} x^{2}+70 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3} d^{3} x^{3}+144 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d^{4} x^{4}-280 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{5} x^{5}+120 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{6} x^{6}+176 c^{7}}{840 d^{4}} \] Input:
int(x^3*(-d^2*x^2+c^2)^(5/2)/(d*x+c)^2,x)
Output:
( - 105*asin((d*x)/c)*c**7 - 176*sqrt(c**2 - d**2*x**2)*c**6 + 105*sqrt(c* *2 - d**2*x**2)*c**5*d*x - 88*sqrt(c**2 - d**2*x**2)*c**4*d**2*x**2 + 70*s qrt(c**2 - d**2*x**2)*c**3*d**3*x**3 + 144*sqrt(c**2 - d**2*x**2)*c**2*d** 4*x**4 - 280*sqrt(c**2 - d**2*x**2)*c*d**5*x**5 + 120*sqrt(c**2 - d**2*x** 2)*d**6*x**6 + 176*c**7)/(840*d**4)