Integrand size = 22, antiderivative size = 103 \[ \int (A+B x) \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {3}{8} A c^2 x \sqrt {c^2-d^2 x^2}+\frac {1}{4} A x \left (c^2-d^2 x^2\right )^{3/2}-\frac {B \left (c^2-d^2 x^2\right )^{5/2}}{5 d^2}+\frac {3 A c^4 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d} \] Output:
3/8*A*c^2*x*(-d^2*x^2+c^2)^(1/2)+1/4*A*x*(-d^2*x^2+c^2)^(3/2)-1/5*B*(-d^2* x^2+c^2)^(5/2)/d^2+3/8*A*c^4*arctan(d*x/(-d^2*x^2+c^2)^(1/2))/d
Time = 0.45 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.04 \[ \int (A+B x) \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {1}{40} \sqrt {c^2-d^2 x^2} \left (5 A x \left (5 c^2-2 d^2 x^2\right )-\frac {8 B \left (c^2-d^2 x^2\right )^2}{d^2}\right )-\frac {3 A c^4 \log \left (-\sqrt {-d^2} x+\sqrt {c^2-d^2 x^2}\right )}{8 \sqrt {-d^2}} \] Input:
Integrate[(A + B*x)*(c^2 - d^2*x^2)^(3/2),x]
Output:
(Sqrt[c^2 - d^2*x^2]*(5*A*x*(5*c^2 - 2*d^2*x^2) - (8*B*(c^2 - d^2*x^2)^2)/ d^2))/40 - (3*A*c^4*Log[-(Sqrt[-d^2]*x) + Sqrt[c^2 - d^2*x^2]])/(8*Sqrt[-d ^2])
Time = 0.32 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {455, 211, 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.
\(\displaystyle \int (A+B x) \left (c^2-d^2 x^2\right )^{3/2} \, dx\) |
\(\Big \downarrow \) 455 |
\(\displaystyle A \int \left (c^2-d^2 x^2\right )^{3/2}dx-\frac {B \left (c^2-d^2 x^2\right )^{5/2}}{5 d^2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle A \left (\frac {3}{4} c^2 \int \sqrt {c^2-d^2 x^2}dx+\frac {1}{4} x \left (c^2-d^2 x^2\right )^{3/2}\right )-\frac {B \left (c^2-d^2 x^2\right )^{5/2}}{5 d^2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle A \left (\frac {3}{4} c^2 \left (\frac {1}{2} c^2 \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {1}{4} x \left (c^2-d^2 x^2\right )^{3/2}\right )-\frac {B \left (c^2-d^2 x^2\right )^{5/2}}{5 d^2}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle A \left (\frac {3}{4} c^2 \left (\frac {1}{2} c^2 \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {1}{4} x \left (c^2-d^2 x^2\right )^{3/2}\right )-\frac {B \left (c^2-d^2 x^2\right )^{5/2}}{5 d^2}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle A \left (\frac {3}{4} c^2 \left (\frac {c^2 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )+\frac {1}{4} x \left (c^2-d^2 x^2\right )^{3/2}\right )-\frac {B \left (c^2-d^2 x^2\right )^{5/2}}{5 d^2}\) |
Input:
Int[(A + B*x)*(c^2 - d^2*x^2)^(3/2),x]
Output:
-1/5*(B*(c^2 - d^2*x^2)^(5/2))/d^2 + A*((x*(c^2 - d^2*x^2)^(3/2))/4 + (3*c ^2*((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]
Time = 0.30 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.94
method | result | size |
default | \(A \left (\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4}+\frac {3 c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4}\right )-\frac {B \left (-d^{2} x^{2}+c^{2}\right )^{\frac {5}{2}}}{5 d^{2}}\) | \(97\) |
risch | \(\frac {\left (-8 B \,d^{4} x^{4}-10 A \,d^{4} x^{3}+16 x^{2} c^{2} B \,d^{2}+25 A \,c^{2} d^{2} x -8 B \,c^{4}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{40 d^{2}}+\frac {3 A \,c^{4} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{8 \sqrt {d^{2}}}\) | \(101\) |
Input:
int((B*x+A)*(-d^2*x^2+c^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
A*(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))))-1/5*B*(-d^2*x^2+c ^2)^(5/2)/d^2
Time = 0.09 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.99 \[ \int (A+B x) \left (c^2-d^2 x^2\right )^{3/2} \, dx=-\frac {30 \, A c^{4} d \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + {\left (8 \, B d^{4} x^{4} + 10 \, A d^{4} x^{3} - 16 \, B c^{2} d^{2} x^{2} - 25 \, A c^{2} d^{2} x + 8 \, B c^{4}\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{40 \, d^{2}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(3/2),x, algorithm="fricas")
Output:
-1/40*(30*A*c^4*d*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) + (8*B*d^4*x^4 + 10*A*d^4*x^3 - 16*B*c^2*d^2*x^2 - 25*A*c^2*d^2*x + 8*B*c^4)*sqrt(-d^2*x ^2 + c^2))/d^2
Time = 0.47 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.47 \[ \int (A+B x) \left (c^2-d^2 x^2\right )^{3/2} \, dx=\begin {cases} \frac {3 A c^{4} \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right )}{8} + \sqrt {c^{2} - d^{2} x^{2}} \cdot \left (\frac {5 A c^{2} x}{8} - \frac {A d^{2} x^{3}}{4} - \frac {B c^{4}}{5 d^{2}} + \frac {2 B c^{2} x^{2}}{5} - \frac {B d^{2} x^{4}}{5}\right ) & \text {for}\: d^{2} \neq 0 \\\left (A x + \frac {B x^{2}}{2}\right ) \left (c^{2}\right )^{\frac {3}{2}} & \text {otherwise} \end {cases} \] Input:
integrate((B*x+A)*(-d**2*x**2+c**2)**(3/2),x)
Output:
Piecewise((3*A*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 + sqrt(c**2 - d**2*x**2)*(5*A*c**2*x/8 - A*d**2*x**3/4 - B*c**4/(5*d**2) + 2*B*c**2*x**2/5 - B*d**2*x**4/5), Ne(d**2, 0)), ((A*x + B*x**2/2)*(c**2) **(3/2), True))
Time = 0.11 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.74 \[ \int (A+B x) \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {3 \, A c^{4} \arcsin \left (\frac {d x}{c}\right )}{8 \, d} + \frac {3}{8} \, \sqrt {-d^{2} x^{2} + c^{2}} A c^{2} x + \frac {1}{4} \, {\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A x - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {5}{2}} B}{5 \, d^{2}} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(3/2),x, algorithm="maxima")
Output:
3/8*A*c^4*arcsin(d*x/c)/d + 3/8*sqrt(-d^2*x^2 + c^2)*A*c^2*x + 1/4*(-d^2*x ^2 + c^2)^(3/2)*A*x - 1/5*(-d^2*x^2 + c^2)^(5/2)*B/d^2
Time = 0.14 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.83 \[ \int (A+B x) \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {3 \, A c^{4} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{8 \, {\left | d \right |}} - \frac {1}{40} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left (\frac {8 \, B c^{4}}{d^{2}} - {\left (25 \, A c^{2} + 2 \, {\left (8 \, B c^{2} - {\left (4 \, B d^{2} x + 5 \, A d^{2}\right )} x\right )} x\right )} x\right )} \] Input:
integrate((B*x+A)*(-d^2*x^2+c^2)^(3/2),x, algorithm="giac")
Output:
3/8*A*c^4*arcsin(d*x/c)*sgn(c)*sgn(d)/abs(d) - 1/40*sqrt(-d^2*x^2 + c^2)*( 8*B*c^4/d^2 - (25*A*c^2 + 2*(8*B*c^2 - (4*B*d^2*x + 5*A*d^2)*x)*x)*x)
Time = 10.29 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.66 \[ \int (A+B x) \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {A\,x\,{\left (c^2-d^2\,x^2\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ \frac {d^2\,x^2}{c^2}\right )}{{\left (1-\frac {d^2\,x^2}{c^2}\right )}^{3/2}}-\frac {B\,{\left (c^2-d^2\,x^2\right )}^{5/2}}{5\,d^2} \] Input:
int((c^2 - d^2*x^2)^(3/2)*(A + B*x),x)
Output:
(A*x*(c^2 - d^2*x^2)^(3/2)*hypergeom([-3/2, 1/2], 3/2, (d^2*x^2)/c^2))/(1 - (d^2*x^2)/c^2)^(3/2) - (B*(c^2 - d^2*x^2)^(5/2))/(5*d^2)
Time = 0.21 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.33 \[ \int (A+B x) \left (c^2-d^2 x^2\right )^{3/2} \, dx=\frac {15 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{4} d +25 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d^{2} x -10 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{4} x^{3}-8 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{4}+16 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d^{2} x^{2}-8 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{4} x^{4}+8 b \,c^{5}}{40 d^{2}} \] Input:
int((B*x+A)*(-d^2*x^2+c^2)^(3/2),x)
Output:
(15*asin((d*x)/c)*a*c**4*d + 25*sqrt(c**2 - d**2*x**2)*a*c**2*d**2*x - 10* sqrt(c**2 - d**2*x**2)*a*d**4*x**3 - 8*sqrt(c**2 - d**2*x**2)*b*c**4 + 16* sqrt(c**2 - d**2*x**2)*b*c**2*d**2*x**2 - 8*sqrt(c**2 - d**2*x**2)*b*d**4* x**4 + 8*b*c**5)/(40*d**2)