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

Optimal. Leaf size=241 \[ -\frac{\sqrt{d+e x} \sqrt{a e+c d x} (c d f-a e g)^2 \tanh ^{-1}\left (\frac{\sqrt{g} \sqrt{a e+c d x}}{\sqrt{c} \sqrt{d} \sqrt{f+g x}}\right )}{4 c^{3/2} d^{3/2} g^{3/2} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{(f+g x)^{3/2} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{2 g \sqrt{d+e x}}+\frac{\sqrt{f+g x} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac{a e}{c d}-\frac{f}{g}\right )}{4 \sqrt{d+e x}} \]

[Out]

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

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Rubi [A]  time = 0.352413, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 48, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {864, 870, 891, 63, 217, 206} \[ -\frac{\sqrt{d+e x} \sqrt{a e+c d x} (c d f-a e g)^2 \tanh ^{-1}\left (\frac{\sqrt{g} \sqrt{a e+c d x}}{\sqrt{c} \sqrt{d} \sqrt{f+g x}}\right )}{4 c^{3/2} d^{3/2} g^{3/2} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}+\frac{(f+g x)^{3/2} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{2 g \sqrt{d+e x}}+\frac{\sqrt{f+g x} \sqrt{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \left (\frac{a e}{c d}-\frac{f}{g}\right )}{4 \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 864

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

Rule 870

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

Rule 891

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :>
Dist[(a + b*x + c*x^2)^FracPart[p]/((d + e*x)^FracPart[p]*(a/d + (c*x)/e)^FracPart[p]), Int[(d + e*x)^(m + p)*
(f + g*x)^n*(a/d + (c*x)/e)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2
- 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&  !IntegerQ[p] &&  !IGtQ[m, 0] &&  !IGtQ[n, 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{\sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{\sqrt{d+e x}} \, dx &=\frac{(f+g x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 g \sqrt{d+e x}}-\frac{(c d f-a e g) \int \frac{\sqrt{d+e x} \sqrt{f+g x}}{\sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{4 g}\\ &=\frac{\left (\frac{a e}{c d}-\frac{f}{g}\right ) \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 \sqrt{d+e x}}+\frac{(f+g x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 g \sqrt{d+e x}}-\frac{(c d f-a e g)^2 \int \frac{\sqrt{d+e x}}{\sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}} \, dx}{8 c d g}\\ &=\frac{\left (\frac{a e}{c d}-\frac{f}{g}\right ) \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 \sqrt{d+e x}}+\frac{(f+g x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 g \sqrt{d+e x}}-\frac{\left ((c d f-a e g)^2 \sqrt{a e+c d x} \sqrt{d+e x}\right ) \int \frac{1}{\sqrt{a e+c d x} \sqrt{f+g x}} \, dx}{8 c d g \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac{\left (\frac{a e}{c d}-\frac{f}{g}\right ) \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 \sqrt{d+e x}}+\frac{(f+g x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 g \sqrt{d+e x}}-\frac{\left ((c d f-a e g)^2 \sqrt{a e+c d x} \sqrt{d+e x}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{f-\frac{a e g}{c d}+\frac{g x^2}{c d}}} \, dx,x,\sqrt{a e+c d x}\right )}{4 c^2 d^2 g \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac{\left (\frac{a e}{c d}-\frac{f}{g}\right ) \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 \sqrt{d+e x}}+\frac{(f+g x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 g \sqrt{d+e x}}-\frac{\left ((c d f-a e g)^2 \sqrt{a e+c d x} \sqrt{d+e x}\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{g x^2}{c d}} \, dx,x,\frac{\sqrt{a e+c d x}}{\sqrt{f+g x}}\right )}{4 c^2 d^2 g \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ &=\frac{\left (\frac{a e}{c d}-\frac{f}{g}\right ) \sqrt{f+g x} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{4 \sqrt{d+e x}}+\frac{(f+g x)^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{2 g \sqrt{d+e x}}-\frac{(c d f-a e g)^2 \sqrt{a e+c d x} \sqrt{d+e x} \tanh ^{-1}\left (\frac{\sqrt{g} \sqrt{a e+c d x}}{\sqrt{c} \sqrt{d} \sqrt{f+g x}}\right )}{4 c^{3/2} d^{3/2} g^{3/2} \sqrt{a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\\ \end{align*}

Mathematica [A]  time = 0.609267, size = 215, normalized size = 0.89 \[ \frac{\sqrt{c} \sqrt{d} \sqrt{d+e x} \left (\sqrt{c} \sqrt{d} \sqrt{g} \sqrt{c d} (f+g x) (a e+c d x) (a e g+c d (f+2 g x))-\sqrt{a e+c d x} (c d f-a e g)^{5/2} \sqrt{\frac{c d (f+g x)}{c d f-a e g}} \sinh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{g} \sqrt{a e+c d x}}{\sqrt{c d} \sqrt{c d f-a e g}}\right )\right )}{4 g^{3/2} (c d)^{5/2} \sqrt{f+g x} \sqrt{(d+e x) (a e+c d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.323, size = 385, normalized size = 1.6 \begin{align*} -{\frac{1}{8\,cdg}\sqrt{gx+f}\sqrt{cde{x}^{2}+a{e}^{2}x+c{d}^{2}x+ade} \left ( \ln \left ({\frac{1}{2} \left ( 2\,xcdg+aeg+cdf+2\,\sqrt{cdg{x}^{2}+aegx+cdfx+aef}\sqrt{cdg} \right ){\frac{1}{\sqrt{cdg}}}} \right ){a}^{2}{e}^{2}{g}^{2}-2\,\ln \left ( 1/2\,{\frac{2\,xcdg+aeg+cdf+2\,\sqrt{cdg{x}^{2}+aegx+cdfx+aef}\sqrt{cdg}}{\sqrt{cdg}}} \right ) acdefg+\ln \left ({\frac{1}{2} \left ( 2\,xcdg+aeg+cdf+2\,\sqrt{cdg{x}^{2}+aegx+cdfx+aef}\sqrt{cdg} \right ){\frac{1}{\sqrt{cdg}}}} \right ){c}^{2}{d}^{2}{f}^{2}-4\,\sqrt{cdg}\sqrt{cdg{x}^{2}+aegx+cdfx+aef}xcdg-2\,\sqrt{cdg}\sqrt{cdg{x}^{2}+aegx+cdfx+aef}aeg-2\,\sqrt{cdg}\sqrt{cdg{x}^{2}+aegx+cdfx+aef}cdf \right ){\frac{1}{\sqrt{ex+d}}}{\frac{1}{\sqrt{cdg{x}^{2}+aegx+cdfx+aef}}}{\frac{1}{\sqrt{cdg}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{g x + f}}{\sqrt{e x + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(g*x + f)/sqrt(e*x + d), x)

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Fricas [A]  time = 7.31236, size = 1427, normalized size = 5.92 \begin{align*} \left [\frac{4 \,{\left (2 \, c^{2} d^{2} g^{2} x + c^{2} d^{2} f g + a c d e g^{2}\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d} \sqrt{g x + f} +{\left (c^{2} d^{3} f^{2} - 2 \, a c d^{2} e f g + a^{2} d e^{2} g^{2} +{\left (c^{2} d^{2} e f^{2} - 2 \, a c d e^{2} f g + a^{2} e^{3} g^{2}\right )} x\right )} \sqrt{c d g} \log \left (-\frac{8 \, c^{2} d^{2} e g^{2} x^{3} + c^{2} d^{3} f^{2} + 6 \, a c d^{2} e f g + a^{2} d e^{2} g^{2} - 4 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x}{\left (2 \, c d g x + c d f + a e g\right )} \sqrt{c d g} \sqrt{e x + d} \sqrt{g x + f} + 8 \,{\left (c^{2} d^{2} e f g +{\left (c^{2} d^{3} + a c d e^{2}\right )} g^{2}\right )} x^{2} +{\left (c^{2} d^{2} e f^{2} + 2 \,{\left (4 \, c^{2} d^{3} + 3 \, a c d e^{2}\right )} f g +{\left (8 \, a c d^{2} e + a^{2} e^{3}\right )} g^{2}\right )} x}{e x + d}\right )}{16 \,{\left (c^{2} d^{2} e g^{2} x + c^{2} d^{3} g^{2}\right )}}, \frac{2 \,{\left (2 \, c^{2} d^{2} g^{2} x + c^{2} d^{2} f g + a c d e g^{2}\right )} \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{e x + d} \sqrt{g x + f} +{\left (c^{2} d^{3} f^{2} - 2 \, a c d^{2} e f g + a^{2} d e^{2} g^{2} +{\left (c^{2} d^{2} e f^{2} - 2 \, a c d e^{2} f g + a^{2} e^{3} g^{2}\right )} x\right )} \sqrt{-c d g} \arctan \left (\frac{2 \, \sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{-c d g} \sqrt{e x + d} \sqrt{g x + f}}{2 \, c d e g x^{2} + c d^{2} f + a d e g +{\left (c d e f +{\left (2 \, c d^{2} + a e^{2}\right )} g\right )} x}\right )}{8 \,{\left (c^{2} d^{2} e g^{2} x + c^{2} d^{3} g^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c d e x^{2} + a d e +{\left (c d^{2} + a e^{2}\right )} x} \sqrt{g x + f}}{\sqrt{e x + d}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*sqrt(g*x + f)/sqrt(e*x + d), x)