\(\int \sqrt {a+a x} \sqrt {c-c x} \, dx\) [1139]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F]
   Maxima [A] (verification not implemented)
   Giac [B] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 20, antiderivative size = 67 \[ \int \sqrt {a+a x} \sqrt {c-c x} \, dx=\frac {1}{2} x \sqrt {a+a x} \sqrt {c-c x}+\sqrt {a} \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+a x}}{\sqrt {a} \sqrt {c-c x}}\right ) \]

[Out]

arctan(c^(1/2)*(a*x+a)^(1/2)/a^(1/2)/(-c*x+c)^(1/2))*a^(1/2)*c^(1/2)+1/2*x*(a*x+a)^(1/2)*(-c*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {38, 65, 223, 209} \[ \int \sqrt {a+a x} \sqrt {c-c x} \, dx=\sqrt {a} \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a x+a}}{\sqrt {a} \sqrt {c-c x}}\right )+\frac {1}{2} x \sqrt {a x+a} \sqrt {c-c x} \]

[In]

Int[Sqrt[a + a*x]*Sqrt[c - c*x],x]

[Out]

(x*Sqrt[a + a*x]*Sqrt[c - c*x])/2 + Sqrt[a]*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + a*x])/(Sqrt[a]*Sqrt[c - c*x])]

Rule 38

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[x*(a + b*x)^m*((c + d*x)^m/(2*m + 1))
, x] + Dist[2*a*c*(m/(2*m + 1)), Int[(a + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] &&
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0]

Rule 65

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
 FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 223

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} x \sqrt {a+a x} \sqrt {c-c x}+\frac {1}{2} (a c) \int \frac {1}{\sqrt {a+a x} \sqrt {c-c x}} \, dx \\ & = \frac {1}{2} x \sqrt {a+a x} \sqrt {c-c x}+c \text {Subst}\left (\int \frac {1}{\sqrt {2 c-\frac {c x^2}{a}}} \, dx,x,\sqrt {a+a x}\right ) \\ & = \frac {1}{2} x \sqrt {a+a x} \sqrt {c-c x}+c \text {Subst}\left (\int \frac {1}{1+\frac {c x^2}{a}} \, dx,x,\frac {\sqrt {a+a x}}{\sqrt {c-c x}}\right ) \\ & = \frac {1}{2} x \sqrt {a+a x} \sqrt {c-c x}+\sqrt {a} \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+a x}}{\sqrt {a} \sqrt {c-c x}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.08 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.03 \[ \int \sqrt {a+a x} \sqrt {c-c x} \, dx=\frac {\sqrt {a (1+x)} \left (x \sqrt {1+x} \sqrt {c-c x}-2 \sqrt {c} \arcsin \left (\frac {\sqrt {c-c x}}{\sqrt {2} \sqrt {c}}\right )\right )}{2 \sqrt {1+x}} \]

[In]

Integrate[Sqrt[a + a*x]*Sqrt[c - c*x],x]

[Out]

(Sqrt[a*(1 + x)]*(x*Sqrt[1 + x]*Sqrt[c - c*x] - 2*Sqrt[c]*ArcSin[Sqrt[c - c*x]/(Sqrt[2]*Sqrt[c])]))/(2*Sqrt[1
+ x])

Maple [A] (verified)

Time = 0.18 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.27

method result size
risch \(-\frac {x \left (-1+x \right ) \left (1+x \right ) a c}{2 \sqrt {a \left (1+x \right )}\, \sqrt {-c \left (-1+x \right )}}+\frac {\arctan \left (\frac {\sqrt {a c}\, x}{\sqrt {-a c \,x^{2}+a c}}\right ) a c \sqrt {-a \left (1+x \right ) c \left (-1+x \right )}}{2 \sqrt {a c}\, \sqrt {a \left (1+x \right )}\, \sqrt {-c \left (-1+x \right )}}\) \(85\)
default \(-\frac {\sqrt {a x +a}\, \left (-c x +c \right )^{\frac {3}{2}}}{2 c}+\frac {a \left (\frac {\sqrt {-c x +c}\, \sqrt {a x +a}}{a}+\frac {c \sqrt {\left (-c x +c \right ) \left (a x +a \right )}\, \arctan \left (\frac {\sqrt {a c}\, x}{\sqrt {-a c \,x^{2}+a c}}\right )}{\sqrt {-c x +c}\, \sqrt {a x +a}\, \sqrt {a c}}\right )}{2}\) \(102\)

[In]

int((a*x+a)^(1/2)*(-c*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*x*(-1+x)*(1+x)*a*c/(a*(1+x))^(1/2)/(-c*(-1+x))^(1/2)+1/2/(a*c)^(1/2)*arctan((a*c)^(1/2)*x/(-a*c*x^2+a*c)^
(1/2))*a*c*(-a*(1+x)*c*(-1+x))^(1/2)/(a*(1+x))^(1/2)/(-c*(-1+x))^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.90 \[ \int \sqrt {a+a x} \sqrt {c-c x} \, dx=\left [\frac {1}{2} \, \sqrt {a x + a} \sqrt {-c x + c} x + \frac {1}{4} \, \sqrt {-a c} \log \left (2 \, a c x^{2} + 2 \, \sqrt {-a c} \sqrt {a x + a} \sqrt {-c x + c} x - a c\right ), \frac {1}{2} \, \sqrt {a x + a} \sqrt {-c x + c} x - \frac {1}{2} \, \sqrt {a c} \arctan \left (\frac {\sqrt {a c} \sqrt {a x + a} \sqrt {-c x + c} x}{a c x^{2} - a c}\right )\right ] \]

[In]

integrate((a*x+a)^(1/2)*(-c*x+c)^(1/2),x, algorithm="fricas")

[Out]

[1/2*sqrt(a*x + a)*sqrt(-c*x + c)*x + 1/4*sqrt(-a*c)*log(2*a*c*x^2 + 2*sqrt(-a*c)*sqrt(a*x + a)*sqrt(-c*x + c)
*x - a*c), 1/2*sqrt(a*x + a)*sqrt(-c*x + c)*x - 1/2*sqrt(a*c)*arctan(sqrt(a*c)*sqrt(a*x + a)*sqrt(-c*x + c)*x/
(a*c*x^2 - a*c))]

Sympy [F]

\[ \int \sqrt {a+a x} \sqrt {c-c x} \, dx=\int \sqrt {a \left (x + 1\right )} \sqrt {- c \left (x - 1\right )}\, dx \]

[In]

integrate((a*x+a)**(1/2)*(-c*x+c)**(1/2),x)

[Out]

Integral(sqrt(a*(x + 1))*sqrt(-c*(x - 1)), x)

Maxima [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.42 \[ \int \sqrt {a+a x} \sqrt {c-c x} \, dx=\frac {a c \arcsin \left (x\right )}{2 \, \sqrt {a c}} + \frac {1}{2} \, \sqrt {-a c x^{2} + a c} x \]

[In]

integrate((a*x+a)^(1/2)*(-c*x+c)^(1/2),x, algorithm="maxima")

[Out]

1/2*a*c*arcsin(x)/sqrt(a*c) + 1/2*sqrt(-a*c*x^2 + a*c)*x

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (49) = 98\).

Time = 0.38 (sec) , antiderivative size = 173, normalized size of antiderivative = 2.58 \[ \int \sqrt {a+a x} \sqrt {c-c x} \, dx=-\frac {{\left (\frac {2 \, a^{2} c \log \left ({\left | -\sqrt {-a c} \sqrt {a x + a} + \sqrt {-{\left (a x + a\right )} a c + 2 \, a^{2} c} \right |}\right )}{\sqrt {-a c}} - \sqrt {-{\left (a x + a\right )} a c + 2 \, a^{2} c} \sqrt {a x + a}\right )} {\left | a \right |}}{a^{2}} + \frac {{\left (\frac {2 \, a^{3} c \log \left ({\left | -\sqrt {-a c} \sqrt {a x + a} + \sqrt {-{\left (a x + a\right )} a c + 2 \, a^{2} c} \right |}\right )}{\sqrt {-a c}} + \sqrt {-{\left (a x + a\right )} a c + 2 \, a^{2} c} \sqrt {a x + a} {\left (a x - 2 \, a\right )}\right )} {\left | a \right |}}{2 \, a^{3}} \]

[In]

integrate((a*x+a)^(1/2)*(-c*x+c)^(1/2),x, algorithm="giac")

[Out]

-(2*a^2*c*log(abs(-sqrt(-a*c)*sqrt(a*x + a) + sqrt(-(a*x + a)*a*c + 2*a^2*c)))/sqrt(-a*c) - sqrt(-(a*x + a)*a*
c + 2*a^2*c)*sqrt(a*x + a))*abs(a)/a^2 + 1/2*(2*a^3*c*log(abs(-sqrt(-a*c)*sqrt(a*x + a) + sqrt(-(a*x + a)*a*c
+ 2*a^2*c)))/sqrt(-a*c) + sqrt(-(a*x + a)*a*c + 2*a^2*c)*sqrt(a*x + a)*(a*x - 2*a))*abs(a)/a^3

Mupad [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.88 \[ \int \sqrt {a+a x} \sqrt {c-c x} \, dx=\frac {x\,\sqrt {a+a\,x}\,\sqrt {c-c\,x}}{2}-\frac {\sqrt {a}\,\sqrt {-c}\,\ln \left (\sqrt {-c}\,\sqrt {a\,\left (x+1\right )}\,\sqrt {-c\,\left (x-1\right )}-\sqrt {a}\,c\,x\right )}{2} \]

[In]

int((a + a*x)^(1/2)*(c - c*x)^(1/2),x)

[Out]

(x*(a + a*x)^(1/2)*(c - c*x)^(1/2))/2 - (a^(1/2)*(-c)^(1/2)*log((-c)^(1/2)*(a*(x + 1))^(1/2)*(-c*(x - 1))^(1/2
) - a^(1/2)*c*x))/2