Integrand size = 27, antiderivative size = 62 \[ \int \frac {(A+B x) (c+d x)}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {(B c+A d) (c+d x)}{c d^2 \sqrt {c^2-d^2 x^2}}-\frac {B \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^2} \] Output:
(A*d+B*c)*(d*x+c)/c/d^2/(-d^2*x^2+c^2)^(1/2)-B*arctan(d*x/(-d^2*x^2+c^2)^( 1/2))/d^2
Time = 0.44 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.24 \[ \int \frac {(A+B x) (c+d x)}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {(B c+A d) \sqrt {c^2-d^2 x^2}}{c d^2 (c-d x)}+\frac {2 B \arctan \left (\frac {d x}{\sqrt {c^2}-\sqrt {c^2-d^2 x^2}}\right )}{d^2} \] Input:
Integrate[((A + B*x)*(c + d*x))/(c^2 - d^2*x^2)^(3/2),x]
Output:
((B*c + A*d)*Sqrt[c^2 - d^2*x^2])/(c*d^2*(c - d*x)) + (2*B*ArcTan[(d*x)/(S qrt[c^2] - Sqrt[c^2 - d^2*x^2])])/d^2
Time = 0.31 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {665, 27, 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.
\(\displaystyle \int \frac {(A+B x) (c+d x)}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 665 |
\(\displaystyle \frac {(c+d x) (A d+B c)}{c d^2 \sqrt {c^2-d^2 x^2}}-\frac {\int \frac {B}{\sqrt {c^2-d^2 x^2}}dx}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(c+d x) (A d+B c)}{c d^2 \sqrt {c^2-d^2 x^2}}-\frac {B \int \frac {1}{\sqrt {c^2-d^2 x^2}}dx}{d}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {(c+d x) (A d+B c)}{c d^2 \sqrt {c^2-d^2 x^2}}-\frac {B \int \frac {1}{\frac {d^2 x^2}{c^2-d^2 x^2}+1}d\frac {x}{\sqrt {c^2-d^2 x^2}}}{d}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {(c+d x) (A d+B c)}{c d^2 \sqrt {c^2-d^2 x^2}}-\frac {B \arctan \left (\frac {d x}{\sqrt {c^2-d^2 x^2}}\right )}{d^2}\) |
Input:
Int[((A + B*x)*(c + d*x))/(c^2 - d^2*x^2)^(3/2),x]
Output:
((B*c + A*d)*(c + d*x))/(c*d^2*Sqrt[c^2 - d^2*x^2]) - (B*ArcTan[(d*x)/Sqrt [c^2 - d^2*x^2]])/d^2
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)^(-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_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_) + (c_.)*( x_)^2)^(3/2), x_Symbol] :> Simp[(-2^(m - 1))*d^(m - 2)*(e*f + d*g)^n*((d + e*x)/(c*e^(n - 1)*Sqrt[a + c*x^2])), x] + Simp[1/(c*e^(n - 2)) Int[Expand ToSum[(2^(m - 1)*d^(m - 1)*(e*f + d*g)^n - e^n*(d + e*x)^(m - 1)*(f + g*x)^ n)/(d - e*x), x]/Sqrt[a + c*x^2], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && EqQ[c*d^2 + a*e^2, 0] && IGtQ[m, 0] && IGtQ[n, 0]
Time = 0.30 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.65
method | result | size |
default | \(\frac {A x}{c \sqrt {-d^{2} x^{2}+c^{2}}}+\frac {A d +B c}{d^{2} \sqrt {-d^{2} x^{2}+c^{2}}}+B d \left (\frac {x}{\sqrt {-d^{2} x^{2}+c^{2}}\, d^{2}}-\frac {\arctan \left (\frac {\sqrt {d^{2}}\, x}{\sqrt {-d^{2} x^{2}+c^{2}}}\right )}{d^{2} \sqrt {d^{2}}}\right )\) | \(102\) |
Input:
int((B*x+A)*(d*x+c)/(-d^2*x^2+c^2)^(3/2),x,method=_RETURNVERBOSE)
Output:
A/c*x/(-d^2*x^2+c^2)^(1/2)+(A*d+B*c)/d^2/(-d^2*x^2+c^2)^(1/2)+B*d*(1/(-d^2 *x^2+c^2)^(1/2)/d^2*x-1/d^2/(d^2)^(1/2)*arctan((d^2)^(1/2)*x/(-d^2*x^2+c^2 )^(1/2)))
Time = 0.09 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.69 \[ \int \frac {(A+B x) (c+d x)}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {B c^{2} + A c d - {\left (B c d + A d^{2}\right )} x - 2 \, {\left (B c d x - B c^{2}\right )} \arctan \left (-\frac {c - \sqrt {-d^{2} x^{2} + c^{2}}}{d x}\right ) + \sqrt {-d^{2} x^{2} + c^{2}} {\left (B c + A d\right )}}{c d^{3} x - c^{2} d^{2}} \] Input:
integrate((B*x+A)*(d*x+c)/(-d^2*x^2+c^2)^(3/2),x, algorithm="fricas")
Output:
-(B*c^2 + A*c*d - (B*c*d + A*d^2)*x - 2*(B*c*d*x - B*c^2)*arctan(-(c - sqr t(-d^2*x^2 + c^2))/(d*x)) + sqrt(-d^2*x^2 + c^2)*(B*c + A*d))/(c*d^3*x - c ^2*d^2)
Result contains complex when optimal does not.
Time = 4.64 (sec) , antiderivative size = 206, normalized size of antiderivative = 3.32 \[ \int \frac {(A+B x) (c+d x)}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=A c \left (\begin {cases} - \frac {i x}{c^{3} \sqrt {-1 + \frac {d^{2} x^{2}}{c^{2}}}} & \text {for}\: \left |{\frac {d^{2} x^{2}}{c^{2}}}\right | > 1 \\\frac {x}{c^{3} \sqrt {1 - \frac {d^{2} x^{2}}{c^{2}}}} & \text {otherwise} \end {cases}\right ) + A d \left (\begin {cases} \frac {1}{d^{2} \sqrt {c^{2} - d^{2} x^{2}}} & \text {for}\: d \neq 0 \\\frac {x^{2}}{2 \left (c^{2}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + B c \left (\begin {cases} \frac {1}{d^{2} \sqrt {c^{2} - d^{2} x^{2}}} & \text {for}\: d \neq 0 \\\frac {x^{2}}{2 \left (c^{2}\right )^{\frac {3}{2}}} & \text {otherwise} \end {cases}\right ) + B d \left (\begin {cases} \frac {i \operatorname {acosh}{\left (\frac {d x}{c} \right )}}{d^{3}} - \frac {i x}{c d^{2} \sqrt {-1 + \frac {d^{2} x^{2}}{c^{2}}}} & \text {for}\: \left |{\frac {d^{2} x^{2}}{c^{2}}}\right | > 1 \\- \frac {\operatorname {asin}{\left (\frac {d x}{c} \right )}}{d^{3}} + \frac {x}{c d^{2} \sqrt {1 - \frac {d^{2} x^{2}}{c^{2}}}} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((B*x+A)*(d*x+c)/(-d**2*x**2+c**2)**(3/2),x)
Output:
A*c*Piecewise((-I*x/(c**3*sqrt(-1 + d**2*x**2/c**2)), Abs(d**2*x**2/c**2) > 1), (x/(c**3*sqrt(1 - d**2*x**2/c**2)), True)) + A*d*Piecewise((1/(d**2* sqrt(c**2 - d**2*x**2)), Ne(d, 0)), (x**2/(2*(c**2)**(3/2)), True)) + B*c* Piecewise((1/(d**2*sqrt(c**2 - d**2*x**2)), Ne(d, 0)), (x**2/(2*(c**2)**(3 /2)), True)) + B*d*Piecewise((I*acosh(d*x/c)/d**3 - I*x/(c*d**2*sqrt(-1 + d**2*x**2/c**2)), Abs(d**2*x**2/c**2) > 1), (-asin(d*x/c)/d**3 + x/(c*d**2 *sqrt(1 - d**2*x**2/c**2)), True))
Time = 0.12 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.50 \[ \int \frac {(A+B x) (c+d x)}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {A x}{\sqrt {-d^{2} x^{2} + c^{2}} c} + \frac {B x}{\sqrt {-d^{2} x^{2} + c^{2}} d} - \frac {B \arcsin \left (\frac {d x}{c}\right )}{d^{2}} + \frac {B c}{\sqrt {-d^{2} x^{2} + c^{2}} d^{2}} + \frac {A}{\sqrt {-d^{2} x^{2} + c^{2}} d} \] Input:
integrate((B*x+A)*(d*x+c)/(-d^2*x^2+c^2)^(3/2),x, algorithm="maxima")
Output:
A*x/(sqrt(-d^2*x^2 + c^2)*c) + B*x/(sqrt(-d^2*x^2 + c^2)*d) - B*arcsin(d*x /c)/d^2 + B*c/(sqrt(-d^2*x^2 + c^2)*d^2) + A/(sqrt(-d^2*x^2 + c^2)*d)
Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.18 \[ \int \frac {(A+B x) (c+d x)}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=-\frac {B \arcsin \left (\frac {d x}{c}\right ) \mathrm {sgn}\left (c\right ) \mathrm {sgn}\left (d\right )}{d {\left | d \right |}} + \frac {2 \, {\left (B c + A d\right )}}{c d {\left (\frac {c d + \sqrt {-d^{2} x^{2} + c^{2}} {\left | d \right |}}{d^{2} x} - 1\right )} {\left | d \right |}} \] Input:
integrate((B*x+A)*(d*x+c)/(-d^2*x^2+c^2)^(3/2),x, algorithm="giac")
Output:
-B*arcsin(d*x/c)*sgn(c)*sgn(d)/(d*abs(d)) + 2*(B*c + A*d)/(c*d*((c*d + sqr t(-d^2*x^2 + c^2)*abs(d))/(d^2*x) - 1)*abs(d))
Time = 9.79 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.71 \[ \int \frac {(A+B x) (c+d x)}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {B\,d\,\ln \left (x\,\sqrt {-d^2}+\sqrt {c^2-d^2\,x^2}\right )}{{\left (-d^2\right )}^{3/2}}+\frac {B\,c}{d^2\,\sqrt {c^2-d^2\,x^2}}+\frac {B\,x}{d\,\sqrt {c^2-d^2\,x^2}}+\frac {A\,\sqrt {c^2-d^2\,x^2}}{c\,d\,\left (c-d\,x\right )} \] Input:
int(((A + B*x)*(c + d*x))/(c^2 - d^2*x^2)^(3/2),x)
Output:
(B*d*log(x*(-d^2)^(1/2) + (c^2 - d^2*x^2)^(1/2)))/(-d^2)^(3/2) + (B*c)/(d^ 2*(c^2 - d^2*x^2)^(1/2)) + (B*x)/(d*(c^2 - d^2*x^2)^(1/2)) + (A*(c^2 - d^2 *x^2)^(1/2))/(c*d*(c - d*x))
Time = 0.21 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.31 \[ \int \frac {(A+B x) (c+d x)}{\left (c^2-d^2 x^2\right )^{3/2}} \, dx=\frac {-\mathit {asin} \left (\frac {d x}{c}\right ) \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) b c +\mathit {asin} \left (\frac {d x}{c}\right ) b c -2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) a d -2 \tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right ) b c}{c \,d^{2} \left (\tan \left (\frac {\mathit {asin} \left (\frac {d x}{c}\right )}{2}\right )-1\right )} \] Input:
int((B*x+A)*(d*x+c)/(-d^2*x^2+c^2)^(3/2),x)
Output:
( - asin((d*x)/c)*tan(asin((d*x)/c)/2)*b*c + asin((d*x)/c)*b*c - 2*tan(asi n((d*x)/c)/2)*a*d - 2*tan(asin((d*x)/c)/2)*b*c)/(c*d**2*(tan(asin((d*x)/c) /2) - 1))