Integrand size = 24, antiderivative size = 86 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\frac {5}{8} c^2 x \sqrt {c^2-d^2 x^2}+\frac {(8 c-3 d x) \left (c^2-d^2 x^2\right )^{3/2}}{12 d}+\frac {5 c^4 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d} \] Output:
5/8*c^2*x*(-d^2*x^2+c^2)^(1/2)+1/12*(-3*d*x+8*c)*(-d^2*x^2+c^2)^(3/2)/d+5/ 8*c^4*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d
Time = 0.32 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.07 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (16 c^3+9 c^2 d x-16 c d^2 x^2+6 d^3 x^3\right )-30 c^4 \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{24 d} \] Input:
Integrate[(c^2 - d^2*x^2)^(5/2)/(c + d*x)^2,x]
Output:
(Sqrt[c^2 - d^2*x^2]*(16*c^3 + 9*c^2*d*x - 16*c*d^2*x^2 + 6*d^3*x^3) - 30* c^4*ArcTan[(d*x)/(Sqrt[c^2] - Sqrt[c^2 - d^2*x^2])])/(24*d)
Time = 0.38 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.31, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {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.
\(\displaystyle \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx\) |
\(\Big \downarrow \) 464 |
\(\displaystyle \int (c-d x)^2 \sqrt {c^2-d^2 x^2}dx\) |
\(\Big \downarrow \) 469 |
\(\displaystyle \frac {5}{4} c \int (c-d x) \sqrt {c^2-d^2 x^2}dx+\frac {(c-d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {5}{4} c \left (c \int \sqrt {c^2-d^2 x^2}dx+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )+\frac {(c-d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {5}{4} c \left (c \left (\frac {1}{2} c^2 \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )+\frac {(c-d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {5}{4} c \left (c \left (\frac {1}{2} c^2 \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )+\frac {(c-d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {5}{4} c \left (c \left (\frac {c^2 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {\left (c^2-d^2 x^2\right )^{3/2}}{3 d}\right )+\frac {(c-d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\) |
Input:
Int[(c^2 - d^2*x^2)^(5/2)/(c + d*x)^2,x]
Output:
((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*ArcTan[(d*x)/Sqrt[c^2 - d^2*x^2]]) /(2*d))))/4
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]
Time = 0.36 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\frac {\left (6 d^{3} x^{3}-16 c \,d^{2} x^{2}+9 c^{2} d x +16 c^{3}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{24 d}+\frac {5 c^{4} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{8 \sqrt {d^{2}}}\) | \(83\) |
default | \(\frac {\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {7}{2}}}{3 c d \left (x +\frac {c}{d}\right )^{2}}+\frac {5 d \left (\frac {\left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {5}{2}}}{5}+c d \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \left (-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )\right )^{\frac {3}{2}}}{8 d^{2}}+\frac {3 c^{2} \left (-\frac {\left (-2 d^{2} \left (x +\frac {c}{d}\right )+2 c d \right ) \sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}{4 d^{2}}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} \left (x +\frac {c}{d}\right )^{2}+2 c d \left (x +\frac {c}{d}\right )}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )\right )}{3 c}}{d^{2}}\) | \(244\) |
Input:
int((-d^2*x^2+c^2)^(5/2)/(d*x+c)^2,x,method=_RETURNVERBOSE)
Output:
1/24*(6*d^3*x^3-16*c*d^2*x^2+9*c^2*d*x+16*c^3)/d*(-d^2*x^2+c^2)^(1/2)+5/8* c^4/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))
Time = 0.10 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.98 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=-\frac {30 \, c^{4} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (6 \, d^{3} x^{3} - 16 \, c d^{2} x^{2} + 9 \, c^{2} d x + 16 \, c^{3}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{24 \, d} \] Input:
integrate((-d^2*x^2+c^2)^(5/2)/(d*x+c)^2,x, algorithm="fricas")
Output:
-1/24*(30*c^4*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) - (6*d^3*x^3 - 16* c*d^2*x^2 + 9*c^2*d*x + 16*c^3)*sqrt(-d^2*x^2 + c^2))/d
Time = 1.73 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.94 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=c^{2} \left (\begin {cases} \frac {c^{2} \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {x \sqrt {c^{2} - d^{2} x^{2}}}{2} & \text {for}\: d^{2} \neq 0 \\x \sqrt {c^{2}} & \text {otherwise} \end {cases}\right ) - 2 c d \left (\begin {cases} \sqrt {c^{2} - d^{2} x^{2}} \left (- \frac {c^{2}}{3 d^{2}} + \frac {x^{2}}{3}\right ) & \text {for}\: d^{2} \neq 0 \\\frac {x^{2} \sqrt {c^{2}}}{2} & \text {otherwise} \end {cases}\right ) + d^{2} \left (\begin {cases} \frac {c^{4} \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8 d^{2}} + \sqrt {c^{2} - d^{2} x^{2}} \left (- \frac {c^{2} x}{8 d^{2}} + \frac {x^{3}}{4}\right ) & \text {for}\: d^{2} \neq 0 \\\frac {x^{3} \sqrt {c^{2}}}{3} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((-d**2*x**2+c**2)**(5/2)/(d*x+c)**2,x)
Output:
c**2*Piecewise((c**2*Piecewise((log(-2*d**2*x + 2*sqrt(-d**2)*sqrt(c**2 - d**2*x**2))/sqrt(-d**2), Ne(c**2, 0)), (x*log(x)/sqrt(-d**2*x**2), True))/ 2 + x*sqrt(c**2 - d**2*x**2)/2, Ne(d**2, 0)), (x*sqrt(c**2), True)) - 2*c* d*Piecewise((sqrt(c**2 - d**2*x**2)*(-c**2/(3*d**2) + x**2/3), Ne(d**2, 0) ), (x**2*sqrt(c**2)/2, True)) + d**2*Piecewise((c**4*Piecewise((log(-2*d** 2*x + 2*sqrt(-d**2)*sqrt(c**2 - d**2*x**2))/sqrt(-d**2), Ne(c**2, 0)), (x* log(x)/sqrt(-d**2*x**2), True))/(8*d**2) + sqrt(c**2 - d**2*x**2)*(-c**2*x /(8*d**2) + x**3/4), Ne(d**2, 0)), (x**3*sqrt(c**2)/3, True))
Result contains complex when optimal does not.
Time = 0.11 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.38 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=-\frac {5 i \, c^{4} \arcsin \left (\frac {d x}{c} + 2\right )}{8 \, d} + \frac {5}{8} \, \sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} c^{2} x + \frac {5 \, \sqrt {d^{2} x^{2} + 4 \, c d x + 3 \, c^{2}} c^{3}}{4 \, d} + \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}}}{4 \, {\left (d^{2} x + c d\right )}} + \frac {5 \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} c}{12 \, d} \] Input:
integrate((-d^2*x^2+c^2)^(5/2)/(d*x+c)^2,x, algorithm="maxima")
Output:
-5/8*I*c^4*arcsin(d*x/c + 2)/d + 5/8*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*c^2*x + 5/4*sqrt(d^2*x^2 + 4*c*d*x + 3*c^2)*c^3/d + 1/4*(-d^2*x^2 + c^2)^(5/2)/ (d^2*x + c*d) + 5/12*(-d^2*x^2 + c^2)^(3/2)*c/d
Leaf count of result is larger than twice the leaf count of optimal. 184 vs. \(2 (74) = 148\).
Time = 0.16 (sec) , antiderivative size = 184, normalized size of antiderivative = 2.14 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=-\frac {{\left (240 \, c^{5} d^{5} \arctan \left (\sqrt {\frac {2 \, c}{d x + c} - 1}\right ) \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) + \frac {{\left (15 \, c^{5} d^{5} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {7}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 73 \, c^{5} d^{5} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {5}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 55 \, c^{5} d^{5} {\left (\frac {2 \, c}{d x + c} - 1\right )}^{\frac {3}{2}} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right ) - 15 \, c^{5} d^{5} \sqrt {\frac {2 \, c}{d x + c} - 1} \mathrm {sgn}\left (\frac {1}{d x + c}\right ) \mathrm {sgn}\left (d\right )\right )} {\left (d x + c\right )}^{4}}{c^{4}}\right )} {\left | d \right |}}{192 \, c d^{7}} \] Input:
integrate((-d^2*x^2+c^2)^(5/2)/(d*x+c)^2,x, algorithm="giac")
Output:
-1/192*(240*c^5*d^5*arctan(sqrt(2*c/(d*x + c) - 1))*sgn(1/(d*x + c))*sgn(d ) + (15*c^5*d^5*(2*c/(d*x + c) - 1)^(7/2)*sgn(1/(d*x + c))*sgn(d) - 73*c^5 *d^5*(2*c/(d*x + c) - 1)^(5/2)*sgn(1/(d*x + c))*sgn(d) - 55*c^5*d^5*(2*c/( d*x + c) - 1)^(3/2)*sgn(1/(d*x + c))*sgn(d) - 15*c^5*d^5*sqrt(2*c/(d*x + c ) - 1)*sgn(1/(d*x + c))*sgn(d))*(d*x + c)^4/c^4)*abs(d)/(c*d^7)
Timed out. \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\int \frac {{\left (c^2-d^2\,x^2\right )}^{5/2}}{{\left (c+d\,x\right )}^2} \,d x \] Input:
int((c^2 - d^2*x^2)^(5/2)/(c + d*x)^2,x)
Output:
int((c^2 - d^2*x^2)^(5/2)/(c + d*x)^2, x)
Time = 0.19 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.21 \[ \int \frac {\left (c^2-d^2 x^2\right )^{5/2}}{(c+d x)^2} \, dx=\frac {15 \mathit {asin} \left (\frac {d x}{c}\right ) c^{4}+16 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{3}+9 \sqrt {-d^{2} x^{2}+c^{2}}\, c^{2} d x -16 \sqrt {-d^{2} x^{2}+c^{2}}\, c \,d^{2} x^{2}+6 \sqrt {-d^{2} x^{2}+c^{2}}\, d^{3} x^{3}-16 c^{4}}{24 d} \] Input:
int((-d^2*x^2+c^2)^(5/2)/(d*x+c)^2,x)
Output:
(15*asin((d*x)/c)*c**4 + 16*sqrt(c**2 - d**2*x**2)*c**3 + 9*sqrt(c**2 - d* *2*x**2)*c**2*d*x - 16*sqrt(c**2 - d**2*x**2)*c*d**2*x**2 + 6*sqrt(c**2 - d**2*x**2)*d**3*x**3 - 16*c**4)/(24*d)