3.30 \(\int \frac {A+B x+C x^2}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx\)

Optimal. Leaf size=177 \[ \frac {\left (a^2 C+2 A b^2\right ) \sqrt {a^2 c-b^2 c x^2} \tan ^{-1}\left (\frac {b \sqrt {c} x}{\sqrt {a^2 c-b^2 c x^2}}\right )}{2 b^3 \sqrt {c} \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {B \left (a^2-b^2 x^2\right )}{b^2 \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {C x \left (a^2-b^2 x^2\right )}{2 b^2 \sqrt {a+b x} \sqrt {a c-b c x}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.152, Rules used = {901, 1815, 641, 217, 203} \[ \frac {\left (a^2 C+2 A b^2\right ) \sqrt {a^2 c-b^2 c x^2} \tan ^{-1}\left (\frac {b \sqrt {c} x}{\sqrt {a^2 c-b^2 c x^2}}\right )}{2 b^3 \sqrt {c} \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {B \left (a^2-b^2 x^2\right )}{b^2 \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {C x \left (a^2-b^2 x^2\right )}{2 b^2 \sqrt {a+b x} \sqrt {a c-b c x}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*x + C*x^2)/(Sqrt[a + b*x]*Sqrt[a*c - b*c*x]),x]

[Out]

-((B*(a^2 - b^2*x^2))/(b^2*Sqrt[a + b*x]*Sqrt[a*c - b*c*x])) - (C*x*(a^2 - b^2*x^2))/(2*b^2*Sqrt[a + b*x]*Sqrt
[a*c - b*c*x]) + ((2*A*b^2 + a^2*C)*Sqrt[a^2*c - b^2*c*x^2]*ArcTan[(b*Sqrt[c]*x)/Sqrt[a^2*c - b^2*c*x^2]])/(2*
b^3*Sqrt[c]*Sqrt[a + b*x]*Sqrt[a*c - b*c*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]

Rule 641

Int[((d_) + (e_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(e*(a + c*x^2)^(p + 1))/(2*c*(p + 1)),
x] + Dist[d, Int[(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, p}, x] && NeQ[p, -1]

Rule 901

Int[((d_) + (e_.)*(x_))^(m_)*((f_) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :>
Dist[((d + e*x)^FracPart[m]*(f + g*x)^FracPart[m])/(d*f + e*g*x^2)^FracPart[m], Int[(d*f + e*g*x^2)^m*(a + b*x
+ c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && EqQ[m - n, 0] && EqQ[e*f + d*g, 0]

Rule 1815

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Si
mp[(e*x^(q - 1)*(a + b*x^2)^(p + 1))/(b*(q + 2*p + 1)), x] + Dist[1/(b*(q + 2*p + 1)), Int[(a + b*x^2)^p*Expan
dToSum[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x]
&& PolyQ[Pq, x] &&  !LeQ[p, -1]

Rubi steps

\begin {align*} \int \frac {A+B x+C x^2}{\sqrt {a+b x} \sqrt {a c-b c x}} \, dx &=\frac {\sqrt {a^2 c-b^2 c x^2} \int \frac {A+B x+C x^2}{\sqrt {a^2 c-b^2 c x^2}} \, dx}{\sqrt {a+b x} \sqrt {a c-b c x}}\\ &=-\frac {C x \left (a^2-b^2 x^2\right )}{2 b^2 \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {\sqrt {a^2 c-b^2 c x^2} \int \frac {-c \left (2 A b^2+a^2 C\right )-2 b^2 B c x}{\sqrt {a^2 c-b^2 c x^2}} \, dx}{2 b^2 c \sqrt {a+b x} \sqrt {a c-b c x}}\\ &=-\frac {B \left (a^2-b^2 x^2\right )}{b^2 \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {C x \left (a^2-b^2 x^2\right )}{2 b^2 \sqrt {a+b x} \sqrt {a c-b c x}}+\frac {\left (\left (2 A b^2+a^2 C\right ) \sqrt {a^2 c-b^2 c x^2}\right ) \int \frac {1}{\sqrt {a^2 c-b^2 c x^2}} \, dx}{2 b^2 \sqrt {a+b x} \sqrt {a c-b c x}}\\ &=-\frac {B \left (a^2-b^2 x^2\right )}{b^2 \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {C x \left (a^2-b^2 x^2\right )}{2 b^2 \sqrt {a+b x} \sqrt {a c-b c x}}+\frac {\left (\left (2 A b^2+a^2 C\right ) \sqrt {a^2 c-b^2 c x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{1+b^2 c x^2} \, dx,x,\frac {x}{\sqrt {a^2 c-b^2 c x^2}}\right )}{2 b^2 \sqrt {a+b x} \sqrt {a c-b c x}}\\ &=-\frac {B \left (a^2-b^2 x^2\right )}{b^2 \sqrt {a+b x} \sqrt {a c-b c x}}-\frac {C x \left (a^2-b^2 x^2\right )}{2 b^2 \sqrt {a+b x} \sqrt {a c-b c x}}+\frac {\left (2 A b^2+a^2 C\right ) \sqrt {a^2 c-b^2 c x^2} \tan ^{-1}\left (\frac {b \sqrt {c} x}{\sqrt {a^2 c-b^2 c x^2}}\right )}{2 b^3 \sqrt {c} \sqrt {a+b x} \sqrt {a c-b c x}}\\ \end {align*}

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Mathematica [A]  time = 0.44, size = 169, normalized size = 0.95 \[ -\frac {\sqrt {a-b x} \left (\sqrt {\frac {b x}{a}+1} \left (4 \tan ^{-1}\left (\frac {\sqrt {a-b x}}{\sqrt {a+b x}}\right ) \left (a (a C-b B)+A b^2\right )+b \sqrt {a-b x} \sqrt {a+b x} (2 B+C x)\right )-2 \sqrt {a} \sqrt {a+b x} (a C-2 b B) \sin ^{-1}\left (\frac {\sqrt {a-b x}}{\sqrt {2} \sqrt {a}}\right )\right )}{2 b^3 \sqrt {\frac {b x}{a}+1} \sqrt {c (a-b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*x + C*x^2)/(Sqrt[a + b*x]*Sqrt[a*c - b*c*x]),x]

[Out]

-1/2*(Sqrt[a - b*x]*(-2*Sqrt[a]*(-2*b*B + a*C)*Sqrt[a + b*x]*ArcSin[Sqrt[a - b*x]/(Sqrt[2]*Sqrt[a])] + Sqrt[1
+ (b*x)/a]*(b*Sqrt[a - b*x]*Sqrt[a + b*x]*(2*B + C*x) + 4*(A*b^2 + a*(-(b*B) + a*C))*ArcTan[Sqrt[a - b*x]/Sqrt
[a + b*x]])))/(b^3*Sqrt[c*(a - b*x)]*Sqrt[1 + (b*x)/a])

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fricas [A]  time = 0.90, size = 196, normalized size = 1.11 \[ \left [-\frac {{\left (C a^{2} + 2 \, A b^{2}\right )} \sqrt {-c} \log \left (2 \, b^{2} c x^{2} - 2 \, \sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {-c} x - a^{2} c\right ) + 2 \, {\left (C b x + 2 \, B b\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{4 \, b^{3} c}, -\frac {{\left (C a^{2} + 2 \, A b^{2}\right )} \sqrt {c} \arctan \left (\frac {\sqrt {-b c x + a c} \sqrt {b x + a} b \sqrt {c} x}{b^{2} c x^{2} - a^{2} c}\right ) + {\left (C b x + 2 \, B b\right )} \sqrt {-b c x + a c} \sqrt {b x + a}}{2 \, b^{3} c}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/4*((C*a^2 + 2*A*b^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*(C*b*x + 2*B*b)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/(b^3*c), -1/2*((C*a^2 + 2*A*b^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)) + (C*b*x + 2*B*b)*sqrt(-b*c*x + a*c)*sqrt(b*x + a))/(
b^3*c)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.02, size = 180, normalized size = 1.02 \[ \frac {\sqrt {b x +a}\, \sqrt {-\left (b x -a \right ) c}\, \left (2 A \,b^{2} c \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}}\right )+C \,a^{2} c \arctan \left (\frac {\sqrt {b^{2} c}\, x}{\sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}}\right )-\sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, C x -2 \sqrt {b^{2} c}\, \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, B \right )}{2 \sqrt {-\left (b^{2} x^{2}-a^{2}\right ) c}\, \sqrt {b^{2} c}\, b^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*x+a)^(1/2)*(-(b*x-a)*c)^(1/2)/b^2*(2*A*arctan((b^2*c)^(1/2)/(-(b^2*x^2-a^2)*c)^(1/2)*x)*b^2*c+C*arctan(
(b^2*c)^(1/2)/(-(b^2*x^2-a^2)*c)^(1/2)*x)*a^2*c-C*(b^2*c)^(1/2)*(-(b^2*x^2-a^2)*c)^(1/2)*x-2*B*(b^2*c)^(1/2)*(
-(b^2*x^2-a^2)*c)^(1/2))/(-(b^2*x^2-a^2)*c)^(1/2)/c/(b^2*c)^(1/2)

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maxima [A]  time = 2.50, size = 88, normalized size = 0.50 \[ \frac {C a^{2} \arcsin \left (\frac {b x}{a}\right )}{2 \, b^{3} \sqrt {c}} + \frac {A \arcsin \left (\frac {b x}{a}\right )}{b \sqrt {c}} - \frac {\sqrt {-b^{2} c x^{2} + a^{2} c} C x}{2 \, b^{2} c} - \frac {\sqrt {-b^{2} c x^{2} + a^{2} c} B}{b^{2} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*C*a^2*arcsin(b*x/a)/(b^3*sqrt(c)) + A*arcsin(b*x/a)/(b*sqrt(c)) - 1/2*sqrt(-b^2*c*x^2 + a^2*c)*C*x/(b^2*c)
- sqrt(-b^2*c*x^2 + a^2*c)*B/(b^2*c)

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mupad [B]  time = 14.95, size = 489, normalized size = 2.76 \[ -\frac {\frac {2\,C\,a^2\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^7}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^7}-\frac {2\,C\,a^2\,c^3\,\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}{\sqrt {a+b\,x}-\sqrt {a}}-\frac {14\,C\,a^2\,c\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^5}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^5}+\frac {14\,C\,a^2\,c^2\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^3}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^3}}{b^3\,c^4+\frac {b^3\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^8}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^8}+\frac {4\,b^3\,c^3\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^2}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^2}+\frac {6\,b^3\,c^2\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^4}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^4}+\frac {4\,b^3\,c\,{\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}^6}{{\left (\sqrt {a+b\,x}-\sqrt {a}\right )}^6}}-\frac {4\,A\,\mathrm {atan}\left (\frac {b\,\left (\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}\right )}{\sqrt {b^2\,c}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}\right )}{\sqrt {b^2\,c}}-\frac {2\,C\,a^2\,\mathrm {atan}\left (\frac {\sqrt {a\,c-b\,c\,x}-\sqrt {a\,c}}{\sqrt {c}\,\left (\sqrt {a+b\,x}-\sqrt {a}\right )}\right )}{b^3\,\sqrt {c}}-\frac {B\,\sqrt {a\,c-b\,c\,x}\,\sqrt {a+b\,x}}{b^2\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((2*C*a^2*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^7)/((a + b*x)^(1/2) - a^(1/2))^7 - (2*C*a^2*c^3*((a*c - b*c*x)
^(1/2) - (a*c)^(1/2)))/((a + b*x)^(1/2) - a^(1/2)) - (14*C*a^2*c*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^5)/((a +
b*x)^(1/2) - a^(1/2))^5 + (14*C*a^2*c^2*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^3)/((a + b*x)^(1/2) - a^(1/2))^3)/
(b^3*c^4 + (b^3*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^8)/((a + b*x)^(1/2) - a^(1/2))^8 + (4*b^3*c^3*((a*c - b*c*
x)^(1/2) - (a*c)^(1/2))^2)/((a + b*x)^(1/2) - a^(1/2))^2 + (6*b^3*c^2*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^4)/(
(a + b*x)^(1/2) - a^(1/2))^4 + (4*b^3*c*((a*c - b*c*x)^(1/2) - (a*c)^(1/2))^6)/((a + b*x)^(1/2) - a^(1/2))^6)
- (4*A*atan((b*((a*c - b*c*x)^(1/2) - (a*c)^(1/2)))/((b^2*c)^(1/2)*((a + b*x)^(1/2) - a^(1/2)))))/(b^2*c)^(1/2
) - (2*C*a^2*atan(((a*c - b*c*x)^(1/2) - (a*c)^(1/2))/(c^(1/2)*((a + b*x)^(1/2) - a^(1/2)))))/(b^3*c^(1/2)) -
(B*(a*c - b*c*x)^(1/2)*(a + b*x)^(1/2))/(b^2*c)

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sympy [C]  time = 56.83, size = 338, normalized size = 1.91 \[ - \frac {i A {G_{6, 6}^{6, 2}\left (\begin {matrix} \frac {1}{4}, \frac {3}{4} & \frac {1}{2}, \frac {1}{2}, 1, 1 \\0, \frac {1}{4}, \frac {1}{2}, \frac {3}{4}, 1, 0 & \end {matrix} \middle | {\frac {a^{2}}{b^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} b \sqrt {c}} + \frac {A {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 1 & \\- \frac {1}{4}, \frac {1}{4} & - \frac {1}{2}, 0, 0, 0 \end {matrix} \middle | {\frac {a^{2} e^{- 2 i \pi }}{b^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} b \sqrt {c}} - \frac {i B a {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {1}{4}, \frac {1}{4} & 0, 0, \frac {1}{2}, 1 \\- \frac {1}{2}, - \frac {1}{4}, 0, \frac {1}{4}, \frac {1}{2}, 0 & \end {matrix} \middle | {\frac {a^{2}}{b^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} b^{2} \sqrt {c}} - \frac {B a {G_{6, 6}^{2, 6}\left (\begin {matrix} -1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 1 & \\- \frac {3}{4}, - \frac {1}{4} & -1, - \frac {1}{2}, - \frac {1}{2}, 0 \end {matrix} \middle | {\frac {a^{2} e^{- 2 i \pi }}{b^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} b^{2} \sqrt {c}} - \frac {i C a^{2} {G_{6, 6}^{6, 2}\left (\begin {matrix} - \frac {3}{4}, - \frac {1}{4} & - \frac {1}{2}, - \frac {1}{2}, 0, 1 \\-1, - \frac {3}{4}, - \frac {1}{2}, - \frac {1}{4}, 0, 0 & \end {matrix} \middle | {\frac {a^{2}}{b^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} b^{3} \sqrt {c}} + \frac {C a^{2} {G_{6, 6}^{2, 6}\left (\begin {matrix} - \frac {3}{2}, - \frac {5}{4}, -1, - \frac {3}{4}, - \frac {1}{2}, 1 & \\- \frac {5}{4}, - \frac {3}{4} & - \frac {3}{2}, -1, -1, 0 \end {matrix} \middle | {\frac {a^{2} e^{- 2 i \pi }}{b^{2} x^{2}}} \right )}}{4 \pi ^{\frac {3}{2}} b^{3} \sqrt {c}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*A*meijerg(((1/4, 3/4), (1/2, 1/2, 1, 1)), ((0, 1/4, 1/2, 3/4, 1, 0), ()), a**2/(b**2*x**2))/(4*pi**(3/2)*b*
sqrt(c)) + A*meijerg(((-1/2, -1/4, 0, 1/4, 1/2, 1), ()), ((-1/4, 1/4), (-1/2, 0, 0, 0)), a**2*exp_polar(-2*I*p
i)/(b**2*x**2))/(4*pi**(3/2)*b*sqrt(c)) - I*B*a*meijerg(((-1/4, 1/4), (0, 0, 1/2, 1)), ((-1/2, -1/4, 0, 1/4, 1
/2, 0), ()), a**2/(b**2*x**2))/(4*pi**(3/2)*b**2*sqrt(c)) - B*a*meijerg(((-1, -3/4, -1/2, -1/4, 0, 1), ()), ((
-3/4, -1/4), (-1, -1/2, -1/2, 0)), a**2*exp_polar(-2*I*pi)/(b**2*x**2))/(4*pi**(3/2)*b**2*sqrt(c)) - I*C*a**2*
meijerg(((-3/4, -1/4), (-1/2, -1/2, 0, 1)), ((-1, -3/4, -1/2, -1/4, 0, 0), ()), a**2/(b**2*x**2))/(4*pi**(3/2)
*b**3*sqrt(c)) + C*a**2*meijerg(((-3/2, -5/4, -1, -3/4, -1/2, 1), ()), ((-5/4, -3/4), (-3/2, -1, -1, 0)), a**2
*exp_polar(-2*I*pi)/(b**2*x**2))/(4*pi**(3/2)*b**3*sqrt(c))

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