3.11.78 \(\int \sqrt {a+b x} \sqrt {a c-b c x} \, dx\)

Optimal. Leaf size=68 \[ \frac {a^2 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c (a-b x)}}\right )}{b}+\frac {1}{2} x \sqrt {a+b x} \sqrt {a c-b c x} \]

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Rubi [A]  time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {38, 63, 217, 203} \begin {gather*} \frac {a^2 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c (a-b x)}}\right )}{b}+\frac {1}{2} x \sqrt {a+b x} \sqrt {a c-b c x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x]*Sqrt[a*c - b*c*x],x]

[Out]

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

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 63

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 203

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

Rule 217

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*} \int \sqrt {a+b x} \sqrt {a c-b c x} \, dx &=\frac {1}{2} x \sqrt {a+b x} \sqrt {a c-b c x}+\frac {1}{2} \left (a^2 c\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx\\ &=\frac {1}{2} x \sqrt {a+b x} \sqrt {a c-b c x}+\frac {\left (a^2 c\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {2 a c-c x^2}} \, dx,x,\sqrt {a+b x}\right )}{b}\\ &=\frac {1}{2} x \sqrt {a+b x} \sqrt {a c-b c x}+\frac {\left (a^2 c\right ) \operatorname {Subst}\left (\int \frac {1}{1+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c (a-b x)}}\right )}{b}\\ &=\frac {1}{2} x \sqrt {a+b x} \sqrt {a c-b c x}+\frac {a^2 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {c (a-b x)}}\right )}{b}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 95, normalized size = 1.40 \begin {gather*} \frac {c \left (-2 a^{5/2} \sqrt {a-b x} \sqrt {\frac {b x}{a}+1} \sin ^{-1}\left (\frac {\sqrt {a-b x}}{\sqrt {2} \sqrt {a}}\right )+a^2 b x-b^3 x^3\right )}{2 b \sqrt {a+b x} \sqrt {c (a-b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c*(a^2*b*x - b^3*x^3 - 2*a^(5/2)*Sqrt[a - b*x]*Sqrt[1 + (b*x)/a]*ArcSin[Sqrt[a - b*x]/(Sqrt[2]*Sqrt[a])]))/(2
*b*Sqrt[c*(a - b*x)]*Sqrt[a + b*x])

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IntegrateAlgebraic [A]  time = 0.16, size = 114, normalized size = 1.68 \begin {gather*} \frac {a^2 c \sqrt {a c-b c x} \left (c-\frac {a c-b c x}{a+b x}\right )}{b \sqrt {a+b x} \left (\frac {a c-b c x}{a+b x}+c\right )^2}-\frac {a^2 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {a c-b c x}}{\sqrt {c} \sqrt {a+b x}}\right )}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[Sqrt[a + b*x]*Sqrt[a*c - b*c*x],x]

[Out]

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

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fricas [A]  time = 1.40, size = 159, normalized size = 2.34 \begin {gather*} \left [\frac {a^{2} \sqrt {-c} \log \left (2 \, b^{2} c x^{2} + 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {-c} x - a^{2} c\right ) + 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} b x}{4 \, b}, -\frac {a^{2} \sqrt {c} \arctan \left (\frac {\sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {c} x}{b^{2} c x^{2} - a^{2} c}\right ) - \sqrt {-b c x + a c} \sqrt {b x + a} b x}{2 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [B]  time = 0.01, size = 127, normalized size = 1.87 \begin {gather*} \frac {\sqrt {\left (b x +a \right ) \left (-b c x +a c \right )}\, a^{2} c \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-b^{2} c \,x^{2}+a^{2} c}}\right )}{2 \sqrt {-b c x +a c}\, \sqrt {b x +a}\, \sqrt {b^{2} c}}+\frac {\sqrt {-b c x +a c}\, \sqrt {b x +a}\, a}{2 b}-\frac {\sqrt {b x +a}\, \left (-b c x +a c \right )^{\frac {3}{2}}}{2 b c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b/c*(b*x+a)^(1/2)*(-b*c*x+a*c)^(3/2)+1/2*a/b*(-b*c*x+a*c)^(1/2)*(b*x+a)^(1/2)+1/2*a^2*c*((b*x+a)*(-b*c*x+
a*c))^(1/2)/(-b*c*x+a*c)^(1/2)/(b*x+a)^(1/2)/(b^2*c)^(1/2)*arctan((b^2*c)^(1/2)/(-b^2*c*x^2+a^2*c)^(1/2)*x)

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maxima [A]  time = 3.09, size = 39, normalized size = 0.57 \begin {gather*} \frac {a^{2} \sqrt {c} \arcsin \left (\frac {b x}{a}\right )}{2 \, b} + \frac {1}{2} \, \sqrt {-b^{2} c x^{2} + a^{2} c} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.20, size = 72, normalized size = 1.06 \begin {gather*} \frac {x\,\sqrt {a\,c-b\,c\,x}\,\sqrt {a+b\,x}}{2}-\frac {a^2\,\sqrt {b}\,c^2\,\ln \left (\sqrt {-b\,c}\,\sqrt {c\,\left (a-b\,x\right )}\,\sqrt {a+b\,x}-b^{3/2}\,c\,x\right )}{2\,{\left (-b\,c\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {- c \left (- a + b x\right )} \sqrt {a + b x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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