3.9.38 \(\int \frac {a+b x+c x^2}{(d+e x)^{3/2} \sqrt {f+g x}} \, dx\) [838]

Optimal. Leaf size=129 \[ -\frac {2 \left (a+\frac {d (c d-b e)}{e^2}\right ) \sqrt {f+g x}}{(e f-d g) \sqrt {d+e x}}+\frac {c \sqrt {d+e x} \sqrt {f+g x}}{e^2 g}-\frac {(c e f+3 c d g-2 b e g) \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{e^{5/2} g^{3/2}} \]

[Out]

-(-2*b*e*g+3*c*d*g+c*e*f)*arctanh(g^(1/2)*(e*x+d)^(1/2)/e^(1/2)/(g*x+f)^(1/2))/e^(5/2)/g^(3/2)-2*(a+d*(-b*e+c*
d)/e^2)*(g*x+f)^(1/2)/(-d*g+e*f)/(e*x+d)^(1/2)+c*(e*x+d)^(1/2)*(g*x+f)^(1/2)/e^2/g

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Rubi [A]
time = 0.09, antiderivative size = 129, 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 = {963, 81, 65, 223, 212} \begin {gather*} -\frac {2 \sqrt {f+g x} \left (a+\frac {d (c d-b e)}{e^2}\right )}{\sqrt {d+e x} (e f-d g)}-\frac {(-2 b e g+3 c d g+c e f) \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{e^{5/2} g^{3/2}}+\frac {c \sqrt {d+e x} \sqrt {f+g x}}{e^2 g} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x + c*x^2)/((d + e*x)^(3/2)*Sqrt[f + g*x]),x]

[Out]

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

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 81

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 212

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

Rule 963

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{Qx = PolynomialQuotient[(a + b*x + c*x^2)^p, d + e*x, x], R = PolynomialRemainder[(a + b*x + c*x^2)^p,
d + e*x, x]}, Simp[R*(d + e*x)^(m + 1)*((f + g*x)^(n + 1)/((m + 1)*(e*f - d*g))), x] + Dist[1/((m + 1)*(e*f -
d*g)), Int[(d + e*x)^(m + 1)*(f + g*x)^n*ExpandToSum[(m + 1)*(e*f - d*g)*Qx - g*R*(m + n + 2), 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] && LtQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.34, size = 131, normalized size = 1.02 \begin {gather*} \frac {\sqrt {f+g x} \left (2 e (b d-a e) g+c \left (-3 d^2 g+e^2 f x+d e (f-g x)\right )\right )}{e^2 g (e f-d g) \sqrt {d+e x}}+\frac {(2 b e g-c (e f+3 d g)) \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {g} \sqrt {d+e x}}\right )}{e^{5/2} g^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x + c*x^2)/((d + e*x)^(3/2)*Sqrt[f + g*x]),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(696\) vs. \(2(109)=218\).
time = 0.08, size = 697, normalized size = 5.40

method	result	size
default	\(\frac {\sqrt {g x +f}\, \left (2 \ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) b d \,e^{2} g^{2} x -2 \ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) b \,e^{3} f g x -3 \ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) c \,d^{2} e \,g^{2} x +2 \ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) c d \,e^{2} f g x +\ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) c \,e^{3} f^{2} x +2 \ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) b \,d^{2} e \,g^{2}-2 \ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) b d \,e^{2} f g -3 \ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) c \,d^{3} g^{2}+2 \ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) c \,d^{2} e f g +\ln \left (\frac {2 e g x +2 \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+d g +e f}{2 \sqrt {e g}}\right ) c d \,e^{2} f^{2}+2 c d e g x \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}-2 c \,e^{2} f x \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+4 a \,e^{2} g \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}-4 b d e g \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}+6 c \,d^{2} g \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}-2 c d e f \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, \sqrt {e g}\right )}{2 g \sqrt {e g}\, \left (d g -e f \right ) \sqrt {\left (e x +d \right ) \left (g x +f \right )}\, e^{2} \sqrt {e x +d}}\)	\(697\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x^2+b*x+a)/(e*x+d)^(3/2)/(g*x+f)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/2*(g*x+f)^(1/2)*(2*ln(1/2*(2*e*g*x+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*b*d*e^2*g^2*x
-2*ln(1/2*(2*e*g*x+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*b*e^3*f*g*x-3*ln(1/2*(2*e*g*x+2
*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*c*d^2*e*g^2*x+2*ln(1/2*(2*e*g*x+2*((e*x+d)*(g*x+f))
^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*c*d*e^2*f*g*x+ln(1/2*(2*e*g*x+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+d
*g+e*f)/(e*g)^(1/2))*c*e^3*f^2*x+2*ln(1/2*(2*e*g*x+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))
*b*d^2*e*g^2-2*ln(1/2*(2*e*g*x+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*b*d*e^2*f*g-3*ln(1/
2*(2*e*g*x+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*c*d^3*g^2+2*ln(1/2*(2*e*g*x+2*((e*x+d)*
(g*x+f))^(1/2)*(e*g)^(1/2)+d*g+e*f)/(e*g)^(1/2))*c*d^2*e*f*g+ln(1/2*(2*e*g*x+2*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(
1/2)+d*g+e*f)/(e*g)^(1/2))*c*d*e^2*f^2+2*c*d*e*g*x*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)-2*c*e^2*f*x*((e*x+d)*(g
*x+f))^(1/2)*(e*g)^(1/2)+4*a*e^2*g*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)-4*b*d*e*g*((e*x+d)*(g*x+f))^(1/2)*(e*g)
^(1/2)+6*c*d^2*g*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2)-2*c*d*e*f*((e*x+d)*(g*x+f))^(1/2)*(e*g)^(1/2))/g/(e*g)^(1
/2)/(d*g-e*f)/((e*x+d)*(g*x+f))^(1/2)/e^2/(e*x+d)^(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)^(3/2)/(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?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 277 vs. \(2 (112) = 224\).
time = 5.18, size = 564, normalized size = 4.37 \begin {gather*} \left [-\frac {{\left (3 \, c d^{3} g^{2} - {\left (c f^{2} - 2 \, b f g\right )} x e^{3} - {\left (c d f^{2} - 2 \, b d f g + 2 \, {\left (c d f g + b d g^{2}\right )} x\right )} e^{2} + {\left (3 \, c d^{2} g^{2} x - 2 \, c d^{2} f g - 2 \, b d^{2} g^{2}\right )} e\right )} \sqrt {g} e^{\frac {1}{2}} \log \left (d^{2} g^{2} + 4 \, {\left (d g + {\left (2 \, g x + f\right )} e\right )} \sqrt {g x + f} \sqrt {x e + d} \sqrt {g} e^{\frac {1}{2}} + {\left (8 \, g^{2} x^{2} + 8 \, f g x + f^{2}\right )} e^{2} + 2 \, {\left (4 \, d g^{2} x + 3 \, d f g\right )} e\right ) - 4 \, {\left (3 \, c d^{2} g^{2} e - {\left (c f g x - 2 \, a g^{2}\right )} e^{3} + {\left (c d g^{2} x - c d f g - 2 \, b d g^{2}\right )} e^{2}\right )} \sqrt {g x + f} \sqrt {x e + d}}{4 \, {\left (d^{2} g^{3} e^{3} - f g^{2} x e^{5} + {\left (d g^{3} x - d f g^{2}\right )} e^{4}\right )}}, \frac {{\left (3 \, c d^{3} g^{2} - {\left (c f^{2} - 2 \, b f g\right )} x e^{3} - {\left (c d f^{2} - 2 \, b d f g + 2 \, {\left (c d f g + b d g^{2}\right )} x\right )} e^{2} + {\left (3 \, c d^{2} g^{2} x - 2 \, c d^{2} f g - 2 \, b d^{2} g^{2}\right )} e\right )} \sqrt {-g e} \arctan \left (\frac {{\left (d g + {\left (2 \, g x + f\right )} e\right )} \sqrt {g x + f} \sqrt {-g e} \sqrt {x e + d}}{2 \, {\left ({\left (g^{2} x^{2} + f g x\right )} e^{2} + {\left (d g^{2} x + d f g\right )} e\right )}}\right ) + 2 \, {\left (3 \, c d^{2} g^{2} e - {\left (c f g x - 2 \, a g^{2}\right )} e^{3} + {\left (c d g^{2} x - c d f g - 2 \, b d g^{2}\right )} e^{2}\right )} \sqrt {g x + f} \sqrt {x e + d}}{2 \, {\left (d^{2} g^{3} e^{3} - f g^{2} x e^{5} + {\left (d g^{3} x - d f g^{2}\right )} e^{4}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(e*x+d)^(3/2)/(g*x+f)^(1/2),x, algorithm="fricas")

[Out]

[-1/4*((3*c*d^3*g^2 - (c*f^2 - 2*b*f*g)*x*e^3 - (c*d*f^2 - 2*b*d*f*g + 2*(c*d*f*g + b*d*g^2)*x)*e^2 + (3*c*d^2
*g^2*x - 2*c*d^2*f*g - 2*b*d^2*g^2)*e)*sqrt(g)*e^(1/2)*log(d^2*g^2 + 4*(d*g + (2*g*x + f)*e)*sqrt(g*x + f)*sqr
t(x*e + d)*sqrt(g)*e^(1/2) + (8*g^2*x^2 + 8*f*g*x + f^2)*e^2 + 2*(4*d*g^2*x + 3*d*f*g)*e) - 4*(3*c*d^2*g^2*e -
(c*f*g*x - 2*a*g^2)*e^3 + (c*d*g^2*x - c*d*f*g - 2*b*d*g^2)*e^2)*sqrt(g*x + f)*sqrt(x*e + d))/(d^2*g^3*e^3 -
f*g^2*x*e^5 + (d*g^3*x - d*f*g^2)*e^4), 1/2*((3*c*d^3*g^2 - (c*f^2 - 2*b*f*g)*x*e^3 - (c*d*f^2 - 2*b*d*f*g + 2
*(c*d*f*g + b*d*g^2)*x)*e^2 + (3*c*d^2*g^2*x - 2*c*d^2*f*g - 2*b*d^2*g^2)*e)*sqrt(-g*e)*arctan(1/2*(d*g + (2*g
*x + f)*e)*sqrt(g*x + f)*sqrt(-g*e)*sqrt(x*e + d)/((g^2*x^2 + f*g*x)*e^2 + (d*g^2*x + d*f*g)*e)) + 2*(3*c*d^2*
g^2*e - (c*f*g*x - 2*a*g^2)*e^3 + (c*d*g^2*x - c*d*f*g - 2*b*d*g^2)*e^2)*sqrt(g*x + f)*sqrt(x*e + d))/(d^2*g^3
*e^3 - f*g^2*x*e^5 + (d*g^3*x - d*f*g^2)*e^4)]

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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}}{\left (d + e x\right )^{\frac {3}{2}} \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)**(3/2)/(g*x+f)**(1/2),x)

[Out]

Integral((a + b*x + c*x**2)/((d + e*x)**(3/2)*sqrt(f + g*x)), x)

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Giac [A]
time = 4.10, size = 218, normalized size = 1.69 \begin {gather*} \frac {{\left (3 \, c d g + c f e - 2 \, b g e\right )} e^{\left (-\frac {5}{2}\right )} \log \left ({\left | -\sqrt {g x + f} \sqrt {g} e^{\frac {1}{2}} + \sqrt {d g^{2} + {\left (g x + f\right )} g e - f g e} \right |}\right )}{\sqrt {g} {\left | g \right |}} + \frac {\sqrt {g x + f} {\left (\frac {{\left (c d g^{3} e^{2} - c f g^{2} e^{3}\right )} {\left (g x + f\right )}}{d g^{3} {\left | g \right |} e^{3} - f g^{2} {\left | g \right |} e^{4}} + \frac {3 \, c d^{2} g^{4} e - 2 \, c d f g^{3} e^{2} - 2 \, b d g^{4} e^{2} + c f^{2} g^{2} e^{3} + 2 \, a g^{4} e^{3}}{d g^{3} {\left | g \right |} e^{3} - f g^{2} {\left | g \right |} e^{4}}\right )}}{\sqrt {d g^{2} + {\left (g x + f\right )} g e - f g e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x^2+b*x+a)/(e*x+d)^(3/2)/(g*x+f)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {c\,x^2+b\,x+a}{\sqrt {f+g\,x}\,{\left (d+e\,x\right )}^{3/2}} \,d x \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)^(3/2)),x)

[Out]

int((a + b*x + c*x^2)/((f + g*x)^(1/2)*(d + e*x)^(3/2)), x)

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