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

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

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Rubi [A]  time = 0.18, antiderivative size = 164, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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} \end {gather*}

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 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 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 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])

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 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])

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*}

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Mathematica [A]  time = 0.77, size = 173, normalized size = 1.05 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.39, size = 229, normalized size = 1.40 \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {g} \sqrt {d+e x}}\right ) \left (8 a e^2 g^2-4 b d e g^2-4 b e^2 f g+3 c d^2 g^2+2 c d e f g+3 c e^2 f^2\right )}{4 e^{5/2} g^{5/2}}+\frac {\sqrt {f+g x} (e f-d g) \left (\frac {4 b e^2 g (f+g x)}{d+e x}-4 b e g^2-\frac {3 c e^2 f (f+g x)}{d+e x}-\frac {5 c d e g (f+g x)}{d+e x}+3 c d g^2+5 c e f g\right )}{4 e^2 g^2 \sqrt {d+e x} \left (\frac {e (f+g x)}{d+e x}-g\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((e*f - d*g)*Sqrt[f + g*x]*(5*c*e*f*g + 3*c*d*g^2 - 4*b*e*g^2 - (3*c*e^2*f*(f + g*x))/(d + e*x) - (5*c*d*e*g*(
f + g*x))/(d + e*x) + (4*b*e^2*g*(f + g*x))/(d + e*x)))/(4*e^2*g^2*Sqrt[d + e*x]*(-g + (e*(f + g*x))/(d + e*x)
)^2) + ((3*c*e^2*f^2 + 2*c*d*e*f*g - 4*b*e^2*f*g + 3*c*d^2*g^2 - 4*b*d*e*g^2 + 8*a*e^2*g^2)*ArcTanh[(Sqrt[e]*S
qrt[f + g*x])/(Sqrt[g]*Sqrt[d + e*x])])/(4*e^(5/2)*g^(5/2))

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fricas [A]  time = 0.50, size = 380, normalized size = 2.32 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.26, size = 179, normalized size = 1.09 \begin {gather*} \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 {gather*}

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)

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maple [B]  time = 0.03, size = 425, normalized size = 2.59 \begin {gather*} \frac {\left (8 a \,e^{2} g^{2} \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )-4 b d e \,g^{2} \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )-4 b \,e^{2} f g \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+3 c \,d^{2} g^{2} \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+2 c d e f g \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+3 c \,e^{2} f^{2} \ln \left (\frac {2 e g x +d g +e f +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}}{2 \sqrt {e g}}\right )+4 \sqrt {e g}\, \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, c e g x +8 \sqrt {e g}\, \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, b e g -6 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, c d g -6 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\, c e f \right ) \sqrt {e x +d}\, \sqrt {g x +f}}{8 \sqrt {e g}\, \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, e^{2} g^{2}} \end {gather*}

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

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 >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(d*g-e*f>0)', see `assume?` for
more details)Is d*g-e*f zero or nonzero?

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mupad [B]  time = 22.38, size = 833, normalized size = 5.08 \begin {gather*} \frac {\frac {\left (2\,b\,d\,g+2\,b\,e\,f\right )\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{g^3\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}+\frac {\left (2\,b\,d\,g+2\,b\,e\,f\right )\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3}{e\,g^2\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^3}-\frac {8\,b\,\sqrt {d}\,\sqrt {f}\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{g^2\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^2}}{\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^4}+\frac {e^2}{g^2}-\frac {2\,e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{g\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^2}}-\frac {\frac {\left (\sqrt {d+e\,x}-\sqrt {d}\right )\,\left (\frac {3\,c\,d^2\,e\,g^2}{2}+c\,d\,e^2\,f\,g+\frac {3\,c\,e^3\,f^2}{2}\right )}{g^6\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}-\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^3\,\left (\frac {11\,c\,d^2\,g^2}{2}+25\,c\,d\,e\,f\,g+\frac {11\,c\,e^2\,f^2}{2}\right )}{g^5\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^3}+\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^7\,\left (\frac {3\,c\,d^2\,g^2}{2}+c\,d\,e\,f\,g+\frac {3\,c\,e^2\,f^2}{2}\right )}{e^2\,g^3\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^7}-\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^5\,\left (\frac {11\,c\,d^2\,g^2}{2}+25\,c\,d\,e\,f\,g+\frac {11\,c\,e^2\,f^2}{2}\right )}{e\,g^4\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^5}+\frac {\sqrt {d}\,\sqrt {f}\,\left (32\,c\,d\,g+32\,c\,e\,f\right )\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{g^4\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^4}}{\frac {{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^8}{{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^8}+\frac {e^4}{g^4}-\frac {4\,e\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^6}{g\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^6}-\frac {4\,e^3\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^2}{g^3\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^2}+\frac {6\,e^2\,{\left (\sqrt {d+e\,x}-\sqrt {d}\right )}^4}{g^2\,{\left (\sqrt {f+g\,x}-\sqrt {f}\right )}^4}}-\frac {4\,a\,\mathrm {atan}\left (\frac {e\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}{\sqrt {-e\,g}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}\right )}{\sqrt {-e\,g}}-\frac {2\,b\,\mathrm {atanh}\left (\frac {\sqrt {g}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{\sqrt {e}\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}\right )\,\left (d\,g+e\,f\right )}{e^{3/2}\,g^{3/2}}+\frac {c\,\mathrm {atanh}\left (\frac {\sqrt {g}\,\left (\sqrt {d+e\,x}-\sqrt {d}\right )}{\sqrt {e}\,\left (\sqrt {f+g\,x}-\sqrt {f}\right )}\right )\,\left (3\,d^2\,g^2+2\,d\,e\,f\,g+3\,e^2\,f^2\right )}{2\,e^{5/2}\,g^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((2*b*d*g + 2*b*e*f)*((d + e*x)^(1/2) - d^(1/2)))/(g^3*((f + g*x)^(1/2) - f^(1/2))) + ((2*b*d*g + 2*b*e*f)*((
d + e*x)^(1/2) - d^(1/2))^3)/(e*g^2*((f + g*x)^(1/2) - f^(1/2))^3) - (8*b*d^(1/2)*f^(1/2)*((d + e*x)^(1/2) - d
^(1/2))^2)/(g^2*((f + g*x)^(1/2) - f^(1/2))^2))/(((d + e*x)^(1/2) - d^(1/2))^4/((f + g*x)^(1/2) - f^(1/2))^4 +
e^2/g^2 - (2*e*((d + e*x)^(1/2) - d^(1/2))^2)/(g*((f + g*x)^(1/2) - f^(1/2))^2)) - ((((d + e*x)^(1/2) - d^(1/
2))*((3*c*e^3*f^2)/2 + (3*c*d^2*e*g^2)/2 + c*d*e^2*f*g))/(g^6*((f + g*x)^(1/2) - f^(1/2))) - (((d + e*x)^(1/2)
- d^(1/2))^3*((11*c*d^2*g^2)/2 + (11*c*e^2*f^2)/2 + 25*c*d*e*f*g))/(g^5*((f + g*x)^(1/2) - f^(1/2))^3) + (((d
+ e*x)^(1/2) - d^(1/2))^7*((3*c*d^2*g^2)/2 + (3*c*e^2*f^2)/2 + c*d*e*f*g))/(e^2*g^3*((f + g*x)^(1/2) - f^(1/2
))^7) - (((d + e*x)^(1/2) - d^(1/2))^5*((11*c*d^2*g^2)/2 + (11*c*e^2*f^2)/2 + 25*c*d*e*f*g))/(e*g^4*((f + g*x)
^(1/2) - f^(1/2))^5) + (d^(1/2)*f^(1/2)*(32*c*d*g + 32*c*e*f)*((d + e*x)^(1/2) - d^(1/2))^4)/(g^4*((f + g*x)^(
1/2) - f^(1/2))^4))/(((d + e*x)^(1/2) - d^(1/2))^8/((f + g*x)^(1/2) - f^(1/2))^8 + e^4/g^4 - (4*e*((d + e*x)^(
1/2) - d^(1/2))^6)/(g*((f + g*x)^(1/2) - f^(1/2))^6) - (4*e^3*((d + e*x)^(1/2) - d^(1/2))^2)/(g^3*((f + g*x)^(
1/2) - f^(1/2))^2) + (6*e^2*((d + e*x)^(1/2) - d^(1/2))^4)/(g^2*((f + g*x)^(1/2) - f^(1/2))^4)) - (4*a*atan((e
*((f + g*x)^(1/2) - f^(1/2)))/((-e*g)^(1/2)*((d + e*x)^(1/2) - d^(1/2)))))/(-e*g)^(1/2) - (2*b*atanh((g^(1/2)*
((d + e*x)^(1/2) - d^(1/2)))/(e^(1/2)*((f + g*x)^(1/2) - f^(1/2))))*(d*g + e*f))/(e^(3/2)*g^(3/2)) + (c*atanh(
(g^(1/2)*((d + e*x)^(1/2) - d^(1/2)))/(e^(1/2)*((f + g*x)^(1/2) - f^(1/2))))*(3*d^2*g^2 + 3*e^2*f^2 + 2*d*e*f*
g))/(2*e^(5/2)*g^(5/2))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {a + b x + c x^{2}}{\sqrt {d + e x} \sqrt {f + g x}}\, dx \end {gather*}

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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