3.837 \(\int \frac{a+b x+c x^2}{\sqrt{d+e x} \sqrt{f+g x}} \, dx\)

Optimal. Leaf size=164 \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{g} \sqrt{d+e x}}{\sqrt{e} \sqrt{f+g x}}\right ) \left (4 e g (2 a e g-b (d g+e f))+c \left (3 d^2 g^2+2 d e f g+3 e^2 f^2\right )\right )}{4 e^{5/2} g^{5/2}}-\frac{\sqrt{d+e x} \sqrt{f+g x} (-4 b e g+5 c d g+3 c e f)}{4 e^2 g^2}+\frac{c (d+e x)^{3/2} \sqrt{f+g x}}{2 e^2 g} \]

[Out]

-((3*c*e*f + 5*c*d*g - 4*b*e*g)*Sqrt[d + e*x]*Sqrt[f + g*x])/(4*e^2*g^2) + (c*(d + e*x)^(3/2)*Sqrt[f + g*x])/(
2*e^2*g) + ((c*(3*e^2*f^2 + 2*d*e*f*g + 3*d^2*g^2) + 4*e*g*(2*a*e*g - b*(e*f + d*g)))*ArcTanh[(Sqrt[g]*Sqrt[d
+ e*x])/(Sqrt[e]*Sqrt[f + g*x])])/(4*e^(5/2)*g^(5/2))

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Rubi [A]  time = 0.142798, antiderivative size = 164, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {951, 80, 63, 217, 206} \[ \frac{\tanh ^{-1}\left (\frac{\sqrt{g} \sqrt{d+e x}}{\sqrt{e} \sqrt{f+g x}}\right ) \left (4 e g (2 a e g-b (d g+e f))+c \left (3 d^2 g^2+2 d e f g+3 e^2 f^2\right )\right )}{4 e^{5/2} g^{5/2}}-\frac{\sqrt{d+e x} \sqrt{f+g x} (-4 b e g+5 c d g+3 c e f)}{4 e^2 g^2}+\frac{c (d+e x)^{3/2} \sqrt{f+g x}}{2 e^2 g} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)/(Sqrt[d + e*x]*Sqrt[f + g*x]),x]

[Out]

-((3*c*e*f + 5*c*d*g - 4*b*e*g)*Sqrt[d + e*x]*Sqrt[f + g*x])/(4*e^2*g^2) + (c*(d + e*x)^(3/2)*Sqrt[f + g*x])/(
2*e^2*g) + ((c*(3*e^2*f^2 + 2*d*e*f*g + 3*d^2*g^2) + 4*e*g*(2*a*e*g - b*(e*f + d*g)))*ArcTanh[(Sqrt[g]*Sqrt[d
+ e*x])/(Sqrt[e]*Sqrt[f + g*x])])/(4*e^(5/2)*g^(5/2))

Rule 951

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Simp[(c^p*(d + e*x)^(m + 2*p)*(f + g*x)^(n + 1))/(g*e^(2*p)*(m + n + 2*p + 1)), x] + Dist[1/(g*e^(2*p)*(m +
n + 2*p + 1)), Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x + c*x^2)^p - c^p*
(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x
] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*
p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 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 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 206

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

Rubi steps

\begin{align*} \int \frac{a+b x+c x^2}{\sqrt{d+e x} \sqrt{f+g x}} \, dx &=\frac{c (d+e x)^{3/2} \sqrt{f+g x}}{2 e^2 g}+\frac{\int \frac{\frac{1}{2} \left (4 a e^2 g-c d (3 e f+d g)\right )-\frac{1}{2} e (3 c e f+5 c d g-4 b e g) x}{\sqrt{d+e x} \sqrt{f+g x}} \, dx}{2 e^2 g}\\ &=-\frac{(3 c e f+5 c d g-4 b e g) \sqrt{d+e x} \sqrt{f+g x}}{4 e^2 g^2}+\frac{c (d+e x)^{3/2} \sqrt{f+g x}}{2 e^2 g}+\frac{\left (c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )+4 e g (2 a e g-b (e f+d g))\right ) \int \frac{1}{\sqrt{d+e x} \sqrt{f+g x}} \, dx}{8 e^2 g^2}\\ &=-\frac{(3 c e f+5 c d g-4 b e g) \sqrt{d+e x} \sqrt{f+g x}}{4 e^2 g^2}+\frac{c (d+e x)^{3/2} \sqrt{f+g x}}{2 e^2 g}+\frac{\left (c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )+4 e g (2 a e g-b (e f+d g))\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{f-\frac{d g}{e}+\frac{g x^2}{e}}} \, dx,x,\sqrt{d+e x}\right )}{4 e^3 g^2}\\ &=-\frac{(3 c e f+5 c d g-4 b e g) \sqrt{d+e x} \sqrt{f+g x}}{4 e^2 g^2}+\frac{c (d+e x)^{3/2} \sqrt{f+g x}}{2 e^2 g}+\frac{\left (c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )+4 e g (2 a e g-b (e f+d g))\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{g x^2}{e}} \, dx,x,\frac{\sqrt{d+e x}}{\sqrt{f+g x}}\right )}{4 e^3 g^2}\\ &=-\frac{(3 c e f+5 c d g-4 b e g) \sqrt{d+e x} \sqrt{f+g x}}{4 e^2 g^2}+\frac{c (d+e x)^{3/2} \sqrt{f+g x}}{2 e^2 g}+\frac{\left (c \left (3 e^2 f^2+2 d e f g+3 d^2 g^2\right )+4 e g (2 a e g-b (e f+d g))\right ) \tanh ^{-1}\left (\frac{\sqrt{g} \sqrt{d+e x}}{\sqrt{e} \sqrt{f+g x}}\right )}{4 e^{5/2} g^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.71818, size = 173, normalized size = 1.05 \[ \frac{\sqrt{e f-d g} \sqrt{\frac{e (f+g x)}{e f-d g}} \sinh ^{-1}\left (\frac{\sqrt{g} \sqrt{d+e x}}{\sqrt{e f-d g}}\right ) \left (4 e g (2 a e g-b (d g+e f))+c \left (3 d^2 g^2+2 d e f g+3 e^2 f^2\right )\right )+e \sqrt{g} \sqrt{d+e x} (f+g x) (4 b e g+c (-3 d g-3 e f+2 e g x))}{4 e^3 g^{5/2} \sqrt{f+g x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)/(Sqrt[d + e*x]*Sqrt[f + g*x]),x]

[Out]

(e*Sqrt[g]*Sqrt[d + e*x]*(f + g*x)*(4*b*e*g + c*(-3*e*f - 3*d*g + 2*e*g*x)) + Sqrt[e*f - d*g]*(c*(3*e^2*f^2 +
2*d*e*f*g + 3*d^2*g^2) + 4*e*g*(2*a*e*g - b*(e*f + d*g)))*Sqrt[(e*(f + g*x))/(e*f - d*g)]*ArcSinh[(Sqrt[g]*Sqr
t[d + e*x])/Sqrt[e*f - d*g]])/(4*e^3*g^(5/2)*Sqrt[f + g*x])

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Maple [B]  time = 0.336, size = 425, normalized size = 2.6 \begin{align*}{\frac{1}{8\,{e}^{2}{g}^{2}} \left ( 8\,\ln \left ( 1/2\,{\frac{2\,egx+2\,\sqrt{ \left ( gx+f \right ) \left ( ex+d \right ) }\sqrt{eg}+dg+ef}{\sqrt{eg}}} \right ) a{e}^{2}{g}^{2}-4\,\ln \left ( 1/2\,{\frac{2\,egx+2\,\sqrt{ \left ( gx+f \right ) \left ( ex+d \right ) }\sqrt{eg}+dg+ef}{\sqrt{eg}}} \right ) bde{g}^{2}-4\,\ln \left ( 1/2\,{\frac{2\,egx+2\,\sqrt{ \left ( gx+f \right ) \left ( ex+d \right ) }\sqrt{eg}+dg+ef}{\sqrt{eg}}} \right ) b{e}^{2}fg+3\,\ln \left ( 1/2\,{\frac{2\,egx+2\,\sqrt{ \left ( gx+f \right ) \left ( ex+d \right ) }\sqrt{eg}+dg+ef}{\sqrt{eg}}} \right ) c{d}^{2}{g}^{2}+2\,\ln \left ( 1/2\,{\frac{2\,egx+2\,\sqrt{ \left ( gx+f \right ) \left ( ex+d \right ) }\sqrt{eg}+dg+ef}{\sqrt{eg}}} \right ) cdefg+3\,\ln \left ( 1/2\,{\frac{2\,egx+2\,\sqrt{ \left ( gx+f \right ) \left ( ex+d \right ) }\sqrt{eg}+dg+ef}{\sqrt{eg}}} \right ) c{e}^{2}{f}^{2}+4\,\sqrt{eg}\sqrt{ \left ( gx+f \right ) \left ( ex+d \right ) }xceg+8\,\sqrt{eg}\sqrt{ \left ( gx+f \right ) \left ( ex+d \right ) }beg-6\,\sqrt{eg}\sqrt{ \left ( gx+f \right ) \left ( ex+d \right ) }cdg-6\,\sqrt{eg}\sqrt{ \left ( gx+f \right ) \left ( ex+d \right ) }cef \right ) \sqrt{ex+d}\sqrt{gx+f}{\frac{1}{\sqrt{ \left ( gx+f \right ) \left ( ex+d \right ) }}}{\frac{1}{\sqrt{eg}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)/(e*x+d)^(1/2)/(g*x+f)^(1/2),x)

[Out]

1/8*(8*ln(1/2*(2*e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*a*e^2*g^2-4*ln(1/2*(2*e*g*x
+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*b*d*e*g^2-4*ln(1/2*(2*e*g*x+2*((g*x+f)*(e*x+d))^(
1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*b*e^2*f*g+3*ln(1/2*(2*e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e
*f)/(e*g)^(1/2))*c*d^2*g^2+2*ln(1/2*(2*e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*c*d*e
*f*g+3*ln(1/2*(2*e*g*x+2*((g*x+f)*(e*x+d))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*c*e^2*f^2+4*(e*g)^(1/2)*((g
*x+f)*(e*x+d))^(1/2)*x*c*e*g+8*(e*g)^(1/2)*((g*x+f)*(e*x+d))^(1/2)*b*e*g-6*(e*g)^(1/2)*((g*x+f)*(e*x+d))^(1/2)
*c*d*g-6*(e*g)^(1/2)*((g*x+f)*(e*x+d))^(1/2)*c*e*f)*(e*x+d)^(1/2)*(g*x+f)^(1/2)/(e*g)^(1/2)/g^2/e^2/((g*x+f)*(
e*x+d))^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(e*x+d)^(1/2)/(g*x+f)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.99842, size = 879, normalized size = 5.36 \begin{align*} \left [\frac{{\left (3 \, c e^{2} f^{2} + 2 \,{\left (c d e - 2 \, b e^{2}\right )} f g +{\left (3 \, c d^{2} - 4 \, b d e + 8 \, a e^{2}\right )} g^{2}\right )} \sqrt{e g} \log \left (8 \, e^{2} g^{2} x^{2} + e^{2} f^{2} + 6 \, d e f g + d^{2} g^{2} + 4 \,{\left (2 \, e g x + e f + d g\right )} \sqrt{e g} \sqrt{e x + d} \sqrt{g x + f} + 8 \,{\left (e^{2} f g + d e g^{2}\right )} x\right ) + 4 \,{\left (2 \, c e^{2} g^{2} x - 3 \, c e^{2} f g -{\left (3 \, c d e - 4 \, b e^{2}\right )} g^{2}\right )} \sqrt{e x + d} \sqrt{g x + f}}{16 \, e^{3} g^{3}}, -\frac{{\left (3 \, c e^{2} f^{2} + 2 \,{\left (c d e - 2 \, b e^{2}\right )} f g +{\left (3 \, c d^{2} - 4 \, b d e + 8 \, a e^{2}\right )} g^{2}\right )} \sqrt{-e g} \arctan \left (\frac{{\left (2 \, e g x + e f + d g\right )} \sqrt{-e g} \sqrt{e x + d} \sqrt{g x + f}}{2 \,{\left (e^{2} g^{2} x^{2} + d e f g +{\left (e^{2} f g + d e g^{2}\right )} x\right )}}\right ) - 2 \,{\left (2 \, c e^{2} g^{2} x - 3 \, c e^{2} f g -{\left (3 \, c d e - 4 \, b e^{2}\right )} g^{2}\right )} \sqrt{e x + d} \sqrt{g x + f}}{8 \, e^{3} g^{3}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(e*x+d)^(1/2)/(g*x+f)^(1/2),x, algorithm="fricas")

[Out]

[1/16*((3*c*e^2*f^2 + 2*(c*d*e - 2*b*e^2)*f*g + (3*c*d^2 - 4*b*d*e + 8*a*e^2)*g^2)*sqrt(e*g)*log(8*e^2*g^2*x^2
+ e^2*f^2 + 6*d*e*f*g + d^2*g^2 + 4*(2*e*g*x + e*f + d*g)*sqrt(e*g)*sqrt(e*x + d)*sqrt(g*x + f) + 8*(e^2*f*g
+ d*e*g^2)*x) + 4*(2*c*e^2*g^2*x - 3*c*e^2*f*g - (3*c*d*e - 4*b*e^2)*g^2)*sqrt(e*x + d)*sqrt(g*x + f))/(e^3*g^
3), -1/8*((3*c*e^2*f^2 + 2*(c*d*e - 2*b*e^2)*f*g + (3*c*d^2 - 4*b*d*e + 8*a*e^2)*g^2)*sqrt(-e*g)*arctan(1/2*(2
*e*g*x + e*f + d*g)*sqrt(-e*g)*sqrt(e*x + d)*sqrt(g*x + f)/(e^2*g^2*x^2 + d*e*f*g + (e^2*f*g + d*e*g^2)*x)) -
2*(2*c*e^2*g^2*x - 3*c*e^2*f*g - (3*c*d*e - 4*b*e^2)*g^2)*sqrt(e*x + d)*sqrt(g*x + f))/(e^3*g^3)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b x + c x^{2}}{\sqrt{d + e x} \sqrt{f + g x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x**2+b*x+a)/(e*x+d)**(1/2)/(g*x+f)**(1/2),x)

[Out]

Integral((a + b*x + c*x**2)/(sqrt(d + e*x)*sqrt(f + g*x)), x)

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Giac [A]  time = 1.18298, size = 242, normalized size = 1.48 \begin{align*} \frac{1}{4} \, \sqrt{{\left (x e + d\right )} g e - d g e + f e^{2}} \sqrt{x e + d}{\left (\frac{2 \,{\left (x e + d\right )} c e^{\left (-3\right )}}{g} - \frac{{\left (5 \, c d g^{2} e^{5} + 3 \, c f g e^{6} - 4 \, b g^{2} e^{6}\right )} e^{\left (-8\right )}}{g^{3}}\right )} - \frac{{\left (3 \, c d^{2} g^{2} + 2 \, c d f g e - 4 \, b d g^{2} e + 3 \, c f^{2} e^{2} - 4 \, b f g e^{2} + 8 \, a g^{2} e^{2}\right )} e^{\left (-\frac{5}{2}\right )} \log \left ({\left | -\sqrt{x e + d} \sqrt{g} e^{\frac{1}{2}} + \sqrt{{\left (x e + d\right )} g e - d g e + f e^{2}} \right |}\right )}{4 \, g^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(e*x+d)^(1/2)/(g*x+f)^(1/2),x, algorithm="giac")

[Out]

1/4*sqrt((x*e + d)*g*e - d*g*e + f*e^2)*sqrt(x*e + d)*(2*(x*e + d)*c*e^(-3)/g - (5*c*d*g^2*e^5 + 3*c*f*g*e^6 -
4*b*g^2*e^6)*e^(-8)/g^3) - 1/4*(3*c*d^2*g^2 + 2*c*d*f*g*e - 4*b*d*g^2*e + 3*c*f^2*e^2 - 4*b*f*g*e^2 + 8*a*g^2
*e^2)*e^(-5/2)*log(abs(-sqrt(x*e + d)*sqrt(g)*e^(1/2) + sqrt((x*e + d)*g*e - d*g*e + f*e^2)))/g^(5/2)