\(\int \frac {1}{\sqrt {d+e x} (a d e+(c d^2+a e^2) x+c d e x^2)} \, dx\) [2005]

Optimal result

Integrand size = 37, antiderivative size = 91 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {2}{\left (c d^2-a e^2\right ) \sqrt {d+e x}}-\frac {2 \sqrt {c} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {640, 53, 65, 214} \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}-\frac {2 \sqrt {c} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{3/2}} \]

[In]

Int[1/(Sqrt[d + e*x]*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)),x]

[Out]

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

Rule 53

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n + 1
)/((b*c - a*d)*(m + 1))), x] - Dist[d*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

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 214

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

Rule 640

Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)^(m + p)*(a
/d + (c/e)*x)^p, x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && EqQ[c*d^2 - b*d*e + a*e^2, 0] &&
IntegerQ[p]

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

Mathematica [A] (verified)

Time = 0.11 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {2}{\left (c d^2-a e^2\right ) \sqrt {d+e x}}-\frac {2 \sqrt {c} \sqrt {d} \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{\left (-c d^2+a e^2\right )^{3/2}} \]

[In]

Integrate[1/(Sqrt[d + e*x]*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)),x]

[Out]

2/((c*d^2 - a*e^2)*Sqrt[d + e*x]) - (2*Sqrt[c]*Sqrt[d]*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) +
a*e^2]])/(-(c*d^2) + a*e^2)^(3/2)

Maple [A] (verified)

Time = 2.94 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(-\frac {2 c d \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right ) \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}-\frac {2}{\left (e^{2} a -c \,d^{2}\right ) \sqrt {e x +d}}\)	\(88\)
default	\(-\frac {2 c d \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right ) \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}-\frac {2}{\left (e^{2} a -c \,d^{2}\right ) \sqrt {e x +d}}\)	\(88\)
pseudoelliptic	\(-\frac {2 \left (c d \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right ) \sqrt {e x +d}+\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\right )}{\left (e^{2} a -c \,d^{2}\right ) \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\, \sqrt {e x +d}}\)	\(97\)

[In]

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

[Out]

-2*c*d/(a*e^2-c*d^2)/((a*e^2-c*d^2)*c*d)^(1/2)*arctan(c*d*(e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2))-2/(a*e^2-c*
d^2)/(e*x+d)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.36 (sec) , antiderivative size = 261, normalized size of antiderivative = 2.87 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\left [-\frac {{\left (e x + d\right )} \sqrt {\frac {c d}{c d^{2} - a e^{2}}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} + 2 \, {\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {\frac {c d}{c d^{2} - a e^{2}}}}{c d x + a e}\right ) - 2 \, \sqrt {e x + d}}{c d^{3} - a d e^{2} + {\left (c d^{2} e - a e^{3}\right )} x}, -\frac {2 \, {\left ({\left (e x + d\right )} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}} \arctan \left (-\frac {{\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}}}{c d e x + c d^{2}}\right ) - \sqrt {e x + d}\right )}}{c d^{3} - a d e^{2} + {\left (c d^{2} e - a e^{3}\right )} x}\right ] \]

[In]

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

[Out]

[-((e*x + d)*sqrt(c*d/(c*d^2 - a*e^2))*log((c*d*e*x + 2*c*d^2 - a*e^2 + 2*(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(c
*d/(c*d^2 - a*e^2)))/(c*d*x + a*e)) - 2*sqrt(e*x + d))/(c*d^3 - a*d*e^2 + (c*d^2*e - a*e^3)*x), -2*((e*x + d)*
sqrt(-c*d/(c*d^2 - a*e^2))*arctan(-(c*d^2 - a*e^2)*sqrt(e*x + d)*sqrt(-c*d/(c*d^2 - a*e^2))/(c*d*e*x + c*d^2))
- sqrt(e*x + d))/(c*d^3 - a*d*e^2 + (c*d^2*e - a*e^3)*x)]

Sympy [A] (verification not implemented)

Time = 2.06 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.01 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\begin {cases} \frac {2 \left (- \frac {e}{\sqrt {d + e x} \left (a e^{2} - c d^{2}\right )} - \frac {e \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e^{2} - c d^{2}}{c d}}} \right )}}{\sqrt {\frac {a e^{2} - c d^{2}}{c d}} \left (a e^{2} - c d^{2}\right )}\right )}{e} & \text {for}\: e \neq 0 \\\frac {\log {\left (x \right )}}{c d^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((2*(-e/(sqrt(d + e*x)*(a*e**2 - c*d**2)) - e*atan(sqrt(d + e*x)/sqrt((a*e**2 - c*d**2)/(c*d)))/(sqrt
((a*e**2 - c*d**2)/(c*d))*(a*e**2 - c*d**2)))/e, Ne(e, 0)), (log(x)/(c*d**(5/2)), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^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(a*e^2-c*d^2>0)', see `assume?`
for more de

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.98 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=\frac {2 \, c d \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{\sqrt {-c^{2} d^{3} + a c d e^{2}} {\left (c d^{2} - a e^{2}\right )}} + \frac {2}{{\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d}} \]

[In]

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

[Out]

2*c*d*arctan(sqrt(e*x + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))/(sqrt(-c^2*d^3 + a*c*d*e^2)*(c*d^2 - a*e^2)) + 2/((
c*d^2 - a*e^2)*sqrt(e*x + d))

Mupad [B] (verification not implemented)

Time = 10.05 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\sqrt {d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )} \, dx=-\frac {2}{\left (a\,e^2-c\,d^2\right )\,\sqrt {d+e\,x}}-\frac {2\,\sqrt {c}\,\sqrt {d}\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}}{\sqrt {a\,e^2-c\,d^2}}\right )}{{\left (a\,e^2-c\,d^2\right )}^{3/2}} \]

[In]

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

[Out]

- 2/((a*e^2 - c*d^2)*(d + e*x)^(1/2)) - (2*c^(1/2)*d^(1/2)*atan((c^(1/2)*d^(1/2)*(d + e*x)^(1/2))/(a*e^2 - c*d
^2)^(1/2)))/(a*e^2 - c*d^2)^(3/2)