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

Optimal. Leaf size=342 \[ \frac {\left (-\sqrt {-a} \sqrt {c} (d g+e f)-a e g+c d f\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {\sqrt {c} f-\sqrt {-a} g}}{\sqrt {f+g x} \sqrt {\sqrt {c} d-\sqrt {-a} e}}\right )}{\sqrt {-a} c \sqrt {\sqrt {c} d-\sqrt {-a} e} \sqrt {\sqrt {c} f-\sqrt {-a} g}}-\frac {\left (\sqrt {-a} \sqrt {c} (d g+e f)-a e g+c d f\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {\sqrt {-a} g+\sqrt {c} f}}{\sqrt {f+g x} \sqrt {\sqrt {-a} e+\sqrt {c} d}}\right )}{\sqrt {-a} c \sqrt {\sqrt {-a} e+\sqrt {c} d} \sqrt {\sqrt {-a} g+\sqrt {c} f}}+\frac {2 \sqrt {e} \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{c} \]

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Rubi [A]  time = 2.09, antiderivative size = 342, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {906, 63, 217, 206, 6725, 93, 208} \begin {gather*} \frac {\left (-\sqrt {-a} \sqrt {c} (d g+e f)-a e g+c d f\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {\sqrt {c} f-\sqrt {-a} g}}{\sqrt {f+g x} \sqrt {\sqrt {c} d-\sqrt {-a} e}}\right )}{\sqrt {-a} c \sqrt {\sqrt {c} d-\sqrt {-a} e} \sqrt {\sqrt {c} f-\sqrt {-a} g}}-\frac {\left (\sqrt {-a} \sqrt {c} (d g+e f)-a e g+c d f\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {\sqrt {-a} g+\sqrt {c} f}}{\sqrt {f+g x} \sqrt {\sqrt {-a} e+\sqrt {c} d}}\right )}{\sqrt {-a} c \sqrt {\sqrt {-a} e+\sqrt {c} d} \sqrt {\sqrt {-a} g+\sqrt {c} f}}+\frac {2 \sqrt {e} \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {g} \sqrt {d+e x}}{\sqrt {e} \sqrt {f+g x}}\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[e]*Sqrt[g]*ArcTanh[(Sqrt[g]*Sqrt[d + e*x])/(Sqrt[e]*Sqrt[f + g*x])])/c + ((c*d*f - a*e*g - Sqrt[-a]*Sq
rt[c]*(e*f + d*g))*ArcTanh[(Sqrt[Sqrt[c]*f - Sqrt[-a]*g]*Sqrt[d + e*x])/(Sqrt[Sqrt[c]*d - Sqrt[-a]*e]*Sqrt[f +
g*x])])/(Sqrt[-a]*c*Sqrt[Sqrt[c]*d - Sqrt[-a]*e]*Sqrt[Sqrt[c]*f - Sqrt[-a]*g]) - ((c*d*f - a*e*g + Sqrt[-a]*S
qrt[c]*(e*f + d*g))*ArcTanh[(Sqrt[Sqrt[c]*f + Sqrt[-a]*g]*Sqrt[d + e*x])/(Sqrt[Sqrt[c]*d + Sqrt[-a]*e]*Sqrt[f
+ g*x])])/(Sqrt[-a]*c*Sqrt[Sqrt[c]*d + Sqrt[-a]*e]*Sqrt[Sqrt[c]*f + Sqrt[-a]*g])

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 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

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 208

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

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 906

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

Rule 6725

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionExpand[u/(a + b*x^n), x]}, Int[v, x]
/; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [A]  time = 1.27, size = 339, normalized size = 0.99 \begin {gather*} \frac {\frac {\frac {\left (\sqrt {-a} \sqrt {c} d-a e\right ) \sqrt {\sqrt {-a} g+\sqrt {c} f} \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {\sqrt {-a} g+\sqrt {c} f}}{\sqrt {f+g x} \sqrt {\sqrt {-a} e+\sqrt {c} d}}\right )}{\sqrt {\sqrt {-a} e+\sqrt {c} d}}-\frac {\left (\sqrt {-a} \sqrt {c} d+a e\right ) \sqrt {\sqrt {-a} g-\sqrt {c} f} \tanh ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {\sqrt {-a} g-\sqrt {c} f}}{\sqrt {f+g x} \sqrt {\sqrt {-a} e-\sqrt {c} d}}\right )}{\sqrt {\sqrt {-a} e-\sqrt {c} d}}}{a}+\frac {2 \sqrt {g} \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 )}{\sqrt {f+g x}}}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((2*Sqrt[g]*Sqrt[e*f - d*g]*Sqrt[(e*(f + g*x))/(e*f - d*g)]*ArcSinh[(Sqrt[g]*Sqrt[d + e*x])/Sqrt[e*f - d*g]])/
Sqrt[f + g*x] + (-(((Sqrt[-a]*Sqrt[c]*d + a*e)*Sqrt[-(Sqrt[c]*f) + Sqrt[-a]*g]*ArcTanh[(Sqrt[-(Sqrt[c]*f) + Sq
rt[-a]*g]*Sqrt[d + e*x])/(Sqrt[-(Sqrt[c]*d) + Sqrt[-a]*e]*Sqrt[f + g*x])])/Sqrt[-(Sqrt[c]*d) + Sqrt[-a]*e]) +
((Sqrt[-a]*Sqrt[c]*d - a*e)*Sqrt[Sqrt[c]*f + Sqrt[-a]*g]*ArcTanh[(Sqrt[Sqrt[c]*f + Sqrt[-a]*g]*Sqrt[d + e*x])/
(Sqrt[Sqrt[c]*d + Sqrt[-a]*e]*Sqrt[f + g*x])])/Sqrt[Sqrt[c]*d + Sqrt[-a]*e])/a)/c

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IntegrateAlgebraic [C]  time = 1.08, size = 378, normalized size = 1.11 \begin {gather*} \frac {\sqrt {a e^2+c d^2} \left (\sqrt {a} g-i \sqrt {c} f\right ) \tan ^{-1}\left (\frac {\sqrt {f+g x} \sqrt {a e^2+c d^2}}{\sqrt {d+e x} \sqrt {-i \sqrt {a} \sqrt {c} d g+i \sqrt {a} \sqrt {c} e f-a e g-c d f}}\right )}{\sqrt {a} c \sqrt {-\left (\left (\sqrt {c} d-i \sqrt {a} e\right ) \left (\sqrt {c} f+i \sqrt {a} g\right )\right )}}+\frac {\sqrt {a e^2+c d^2} \left (\sqrt {a} g+i \sqrt {c} f\right ) \tan ^{-1}\left (\frac {\sqrt {f+g x} \sqrt {a e^2+c d^2}}{\sqrt {d+e x} \sqrt {i \sqrt {a} \sqrt {c} d g-i \sqrt {a} \sqrt {c} e f-a e g-c d f}}\right )}{\sqrt {a} c \sqrt {-\left (\left (\sqrt {c} d+i \sqrt {a} e\right ) \left (\sqrt {c} f-i \sqrt {a} g\right )\right )}}+\frac {2 \sqrt {e} \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {e} \sqrt {f+g x}}{\sqrt {g} \sqrt {d+e x}}\right )}{c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[c*d^2 + a*e^2]*((-I)*Sqrt[c]*f + Sqrt[a]*g)*ArcTan[(Sqrt[c*d^2 + a*e^2]*Sqrt[f + g*x])/(Sqrt[-(c*d*f) +
I*Sqrt[a]*Sqrt[c]*e*f - I*Sqrt[a]*Sqrt[c]*d*g - a*e*g]*Sqrt[d + e*x])])/(Sqrt[a]*c*Sqrt[-((Sqrt[c]*d - I*Sqrt[
a]*e)*(Sqrt[c]*f + I*Sqrt[a]*g))]) + (Sqrt[c*d^2 + a*e^2]*(I*Sqrt[c]*f + Sqrt[a]*g)*ArcTan[(Sqrt[c*d^2 + a*e^2
]*Sqrt[f + g*x])/(Sqrt[-(c*d*f) - I*Sqrt[a]*Sqrt[c]*e*f + I*Sqrt[a]*Sqrt[c]*d*g - a*e*g]*Sqrt[d + e*x])])/(Sqr
t[a]*c*Sqrt[-((Sqrt[c]*d + I*Sqrt[a]*e)*(Sqrt[c]*f - I*Sqrt[a]*g))]) + (2*Sqrt[e]*Sqrt[g]*ArcTanh[(Sqrt[e]*Sqr
t[f + g*x])/(Sqrt[g]*Sqrt[d + e*x])])/c

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \mathit {sage}_{0} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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maple [B]  time = 0.03, size = 1569, normalized size = 4.59

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F(-1)]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \text {Hanged} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

\text{Hanged}

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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