Integrand size = 27, antiderivative size = 132 \[ \int (A+B x) (c+d x) \sqrt {c^2-d^2 x^2} \, dx=\frac {c (B c+4 A d) x \sqrt {c^2-d^2 x^2}}{8 d}-\frac {(B c+4 A d) \left (c^2-d^2 x^2\right )^{3/2}}{12 d^2}-\frac {B (c+d x) \left (c^2-d^2 x^2\right )^{3/2}}{4 d^2}+\frac {c^3 (B c+4 A d) \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{8 d^2} \] Output:
1/8*c*(4*A*d+B*c)*x*(-d^2*x^2+c^2)^(1/2)/d-1/12*(4*A*d+B*c)*(-d^2*x^2+c^2) ^(3/2)/d^2-1/4*B*(d*x+c)*(-d^2*x^2+c^2)^(3/2)/d^2+1/8*c^3*(4*A*d+B*c)*arct an(d*x/(-d^2*x^2+c^2)^(1/2))/d^2
Time = 0.72 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.95 \[ \int (A+B x) (c+d x) \sqrt {c^2-d^2 x^2} \, dx=\frac {\sqrt {c^2-d^2 x^2} \left (4 A d \left (-2 c^2+3 c d x+2 d^2 x^2\right )+B \left (-8 c^3-3 c^2 d x+8 c d^2 x^2+6 d^3 x^3\right )\right )-6 c^3 (B c+4 A d) \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{24 d^2} \] Input:
Integrate[(A + B*x)*(c + d*x)*Sqrt[c^2 - d^2*x^2],x]
Output:
(Sqrt[c^2 - d^2*x^2]*(4*A*d*(-2*c^2 + 3*c*d*x + 2*d^2*x^2) + B*(-8*c^3 - 3 *c^2*d*x + 8*c*d^2*x^2 + 6*d^3*x^3)) - 6*c^3*(B*c + 4*A*d)*ArcTan[(d*x)/(S qrt[c^2] - Sqrt[c^2 - d^2*x^2])])/(24*d^2)
Time = 0.34 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.94, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {676, 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.
\(\displaystyle \int (A+B x) (c+d x) \sqrt {c^2-d^2 x^2} \, dx\) |
\(\Big \downarrow \) 676 |
\(\displaystyle \frac {c (4 A d+B c) \int \sqrt {c^2-d^2 x^2}dx}{4 d}-\frac {\left (c^2-d^2 x^2\right )^{3/2} (A d+B c)}{3 d^2}-\frac {B x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {c (4 A d+B c) \left (\frac {1}{2} c^2 \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )}{4 d}-\frac {\left (c^2-d^2 x^2\right )^{3/2} (A d+B c)}{3 d^2}-\frac {B x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {c (4 A d+B c) \left (\frac {1}{2} c^2 \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )}{4 d}-\frac {\left (c^2-d^2 x^2\right )^{3/2} (A d+B c)}{3 d^2}-\frac {B x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {c (4 A d+B c) \left (\frac {c^2 \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{2 d}+\frac {1}{2} x \sqrt {c^2-d^2 x^2}\right )}{4 d}-\frac {\left (c^2-d^2 x^2\right )^{3/2} (A d+B c)}{3 d^2}-\frac {B x \left (c^2-d^2 x^2\right )^{3/2}}{4 d}\) |
Input:
Int[(A + B*x)*(c + d*x)*Sqrt[c^2 - d^2*x^2],x]
Output:
-1/3*((B*c + A*d)*(c^2 - d^2*x^2)^(3/2))/d^2 - (B*x*(c^2 - d^2*x^2)^(3/2)) /(4*d) + (c*(B*c + 4*A*d)*((x*Sqrt[c^2 - d^2*x^2])/2 + (c^2*ArcTan[(d*x)/S qrt[c^2 - d^2*x^2]])/(2*d)))/(4*d)
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[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x _Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p + 3))/(c*(2*p + 3)) Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g , p}, x] && !LeQ[p, -1]
Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.92
method | result | size |
risch | \(-\frac {\left (-6 B \,d^{3} x^{3}-8 A \,d^{3} x^{2}-8 B c \,d^{2} x^{2}-12 A c \,d^{2} x +3 B \,c^{2} d x +8 A \,c^{2} d +8 B \,c^{3}\right ) \sqrt {-d^{2} x^{2}+c^{2}}}{24 d^{2}}+\frac {c^{3} \left (4 A d +B c \right ) \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{8 d \sqrt {d^{2}}}\) | \(122\) |
default | \(A c \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )-\frac {\left (A d +B c \right ) \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{3 d^{2}}+B d \left (-\frac {x \left (-d^{2} x^{2}+c^{2}\right )^{\frac {3}{2}}}{4 d^{2}}+\frac {c^{2} \left (\frac {x \sqrt {-d^{2} x^{2}+c^{2}}}{2}+\frac {c^{2} \arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{2 \sqrt {d^{2}}}\right )}{4 d^{2}}\right )\) | \(163\) |
Input:
int((B*x+A)*(d*x+c)*(-d^2*x^2+c^2)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/24/d^2*(-6*B*d^3*x^3-8*A*d^3*x^2-8*B*c*d^2*x^2-12*A*c*d^2*x+3*B*c^2*d*x +8*A*c^2*d+8*B*c^3)*(-d^2*x^2+c^2)^(1/2)+1/8*c^3/d*(4*A*d+B*c)/(d^2)^(1/2) *arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2)^(1/2))
Time = 0.08 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.92 \[ \int (A+B x) (c+d x) \sqrt {c^2-d^2 x^2} \, dx=-\frac {6 \, {\left (B c^{4} + 4 \, A c^{3} d\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) - {\left (6 \, B d^{3} x^{3} - 8 \, B c^{3} - 8 \, A c^{2} d + 8 \, {\left (B c d^{2} + A d^{3}\right )} x^{2} - 3 \, {\left (B c^{2} d - 4 \, A c d^{2}\right )} x\right )} \sqrt {-d^{2} x^{2} + c^{2}}}{24 \, d^{2}} \] Input:
integrate((B*x+A)*(d*x+c)*(-d^2*x^2+c^2)^(1/2),x, algorithm="fricas")
Output:
-1/24*(6*(B*c^4 + 4*A*c^3*d)*arctan(-(c - sqrt(-d^2*x^2 + c^2))/(d*x)) - ( 6*B*d^3*x^3 - 8*B*c^3 - 8*A*c^2*d + 8*(B*c*d^2 + A*d^3)*x^2 - 3*(B*c^2*d - 4*A*c*d^2)*x)*sqrt(-d^2*x^2 + c^2))/d^2
Time = 0.62 (sec) , antiderivative size = 226, normalized size of antiderivative = 1.71 \[ \int (A+B x) (c+d x) \sqrt {c^2-d^2 x^2} \, dx=\begin {cases} \sqrt {c^{2} - d^{2} x^{2}} \left (\frac {B d x^{3}}{4} - \frac {x^{2} \left (- A d^{3} - B c d^{2}\right )}{3 d^{2}} - \frac {x \left (- A c d^{2} + \frac {B c^{2} d}{4}\right )}{2 d^{2}} - \frac {A c^{2} d + B c^{3} + \frac {2 c^{2} \left (- A d^{3} - B c d^{2}\right )}{3 d^{2}}}{d^{2}}\right ) + \left (A c^{3} + \frac {c^{2} \left (- A c d^{2} + \frac {B c^{2} d}{4}\right )}{2 d^{2}}\right ) \left (\begin {cases} \frac {\log {\left (- 2 d^{2} x + 2 \sqrt {- d^{2}} \sqrt {c^{2} - d^{2} x^{2}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: c^{2} \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {- d^{2} x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: d^{2} \neq 0 \\\left (A c x + \frac {B d x^{3}}{3} + \frac {x^{2} \left (A d + B c\right )}{2}\right ) \sqrt {c^{2}} & \text {otherwise} \end {cases} \] Input:
integrate((B*x+A)*(d*x+c)*(-d**2*x**2+c**2)**(1/2),x)
Output:
Piecewise((sqrt(c**2 - d**2*x**2)*(B*d*x**3/4 - x**2*(-A*d**3 - B*c*d**2)/ (3*d**2) - x*(-A*c*d**2 + B*c**2*d/4)/(2*d**2) - (A*c**2*d + B*c**3 + 2*c* *2*(-A*d**3 - B*c*d**2)/(3*d**2))/d**2) + (A*c**3 + c**2*(-A*c*d**2 + B*c* *2*d/4)/(2*d**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)), Ne (d**2, 0)), ((A*c*x + B*d*x**3/3 + x**2*(A*d + B*c)/2)*sqrt(c**2), True))
Time = 0.11 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.05 \[ \int (A+B x) (c+d x) \sqrt {c^2-d^2 x^2} \, dx=\frac {B c^{4} \arcsin \left (\frac {d x}{c}\right )}{8 \, d^{2}} + \frac {A c^{3} \arcsin \left (\frac {d x}{c}\right )}{2 \, d} + \frac {1}{2} \, \sqrt {-d^{2} x^{2} + c^{2}} A c x + \frac {\sqrt {-d^{2} x^{2} + c^{2}} B c^{2} x}{8 \, d} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B x}{4 \, d} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} B c}{3 \, d^{2}} - \frac {{\left (-d^{2} x^{2} + c^{2}\right )}^{\frac {3}{2}} A}{3 \, d} \] Input:
integrate((B*x+A)*(d*x+c)*(-d^2*x^2+c^2)^(1/2),x, algorithm="maxima")
Output:
1/8*B*c^4*arcsin(d*x/c)/d^2 + 1/2*A*c^3*arcsin(d*x/c)/d + 1/2*sqrt(-d^2*x^ 2 + c^2)*A*c*x + 1/8*sqrt(-d^2*x^2 + c^2)*B*c^2*x/d - 1/4*(-d^2*x^2 + c^2) ^(3/2)*B*x/d - 1/3*(-d^2*x^2 + c^2)^(3/2)*B*c/d^2 - 1/3*(-d^2*x^2 + c^2)^( 3/2)*A/d
Time = 0.14 (sec) , antiderivative size = 123, normalized size of antiderivative = 0.93 \[ \int (A+B x) (c+d x) \sqrt {c^2-d^2 x^2} \, dx=\frac {{\left (B c^{4} + 4 \, A c^{3} d\right )} \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{8 \, d {\left | d \right |}} + \frac {1}{24} \, \sqrt {-d^{2} x^{2} + c^{2}} {\left ({\left (2 \, {\left (3 \, B d x + \frac {4 \, {\left (B c d^{4} + A d^{5}\right )}}{d^{4}}\right )} x - \frac {3 \, {\left (B c^{2} d^{3} - 4 \, A c d^{4}\right )}}{d^{4}}\right )} x - \frac {8 \, {\left (B c^{3} d^{2} + A c^{2} d^{3}\right )}}{d^{4}}\right )} \] Input:
integrate((B*x+A)*(d*x+c)*(-d^2*x^2+c^2)^(1/2),x, algorithm="giac")
Output:
1/8*(B*c^4 + 4*A*c^3*d)*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) + 1/24*sqrt (-d^2*x^2 + c^2)*((2*(3*B*d*x + 4*(B*c*d^4 + A*d^5)/d^4)*x - 3*(B*c^2*d^3 - 4*A*c*d^4)/d^4)*x - 8*(B*c^3*d^2 + A*c^2*d^3)/d^4)
Timed out. \[ \int (A+B x) (c+d x) \sqrt {c^2-d^2 x^2} \, dx=\int \sqrt {c^2-d^2\,x^2}\,\left (A+B\,x\right )\,\left (c+d\,x\right ) \,d x \] Input:
int((c^2 - d^2*x^2)^(1/2)*(A + B*x)*(c + d*x),x)
Output:
int((c^2 - d^2*x^2)^(1/2)*(A + B*x)*(c + d*x), x)
Time = 0.21 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.47 \[ \int (A+B x) (c+d x) \sqrt {c^2-d^2 x^2} \, dx=\frac {12 \mathit {asin} \left (\frac {d x}{c}\right ) a \,c^{3} d +3 \mathit {asin} \left (\frac {d x}{c}\right ) b \,c^{4}-8 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,c^{2} d +12 \sqrt {-d^{2} x^{2}+c^{2}}\, a c \,d^{2} x +8 \sqrt {-d^{2} x^{2}+c^{2}}\, a \,d^{3} x^{2}-8 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{3}-3 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,c^{2} d x +8 \sqrt {-d^{2} x^{2}+c^{2}}\, b c \,d^{2} x^{2}+6 \sqrt {-d^{2} x^{2}+c^{2}}\, b \,d^{3} x^{3}+8 a \,c^{3} d +8 b \,c^{4}}{24 d^{2}} \] Input:
int((B*x+A)*(d*x+c)*(-d^2*x^2+c^2)^(1/2),x)
Output:
(12*asin((d*x)/c)*a*c**3*d + 3*asin((d*x)/c)*b*c**4 - 8*sqrt(c**2 - d**2*x **2)*a*c**2*d + 12*sqrt(c**2 - d**2*x**2)*a*c*d**2*x + 8*sqrt(c**2 - d**2* x**2)*a*d**3*x**2 - 8*sqrt(c**2 - d**2*x**2)*b*c**3 - 3*sqrt(c**2 - d**2*x **2)*b*c**2*d*x + 8*sqrt(c**2 - d**2*x**2)*b*c*d**2*x**2 + 6*sqrt(c**2 - d **2*x**2)*b*d**3*x**3 + 8*a*c**3*d + 8*b*c**4)/(24*d**2)