\(\int \frac {\sqrt {d+e x}}{(a d e+(c d^2+a e^2) x+c d e x^2)^2} \, dx\) [2015]

Optimal result

Integrand size = 37, antiderivative size = 128 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {3 e}{\left (c d^2-a e^2\right )^2 \sqrt {d+e x}}-\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x) \sqrt {d+e x}}+\frac {3 \sqrt {c} \sqrt {d} e \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{5/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {640, 44, 53, 65, 214} \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {3 \sqrt {c} \sqrt {d} e \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{5/2}}-\frac {3 e}{\sqrt {d+e x} \left (c d^2-a e^2\right )^2}-\frac {1}{\sqrt {d+e x} \left (c d^2-a e^2\right ) (a e+c d x)} \]

[In]

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

[Out]

(-3*e)/((c*d^2 - a*e^2)^2*Sqrt[d + e*x]) - 1/((c*d^2 - a*e^2)*(a*e + c*d*x)*Sqrt[d + e*x]) + (3*Sqrt[c]*Sqrt[d
]*e*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(c*d^2 - a*e^2)^(5/2)

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

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

Mathematica [A] (verified)

Time = 0.32 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {2 a e^2+c d (d+3 e x)}{\left (c d^2-a e^2\right )^2 (a e+c d x) \sqrt {d+e x}}-\frac {3 \sqrt {c} \sqrt {d} e \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{\left (-c d^2+a e^2\right )^{5/2}} \]

[In]

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

[Out]

-((2*a*e^2 + c*d*(d + 3*e*x))/((c*d^2 - a*e^2)^2*(a*e + c*d*x)*Sqrt[d + e*x])) - (3*Sqrt[c]*Sqrt[d]*e*ArcTan[(
Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/(-(c*d^2) + a*e^2)^(5/2)

Maple [A] (verified)

Time = 7.64 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.78

method	result	size
pseudoelliptic	\(\frac {e \left (-\frac {2}{\sqrt {e x +d}}-\frac {c d \sqrt {e x +d}}{\left (c d x +a e \right ) e}-\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right ) c d}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{2}}\)	\(100\)
derivativedivides	\(2 e \left (-\frac {c d \left (\frac {\sqrt {e x +d}}{2 c d \left (e x +d \right )+2 e^{2} a -2 c \,d^{2}}+\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{2 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{2}}-\frac {1}{\left (e^{2} a -c \,d^{2}\right )^{2} \sqrt {e x +d}}\right )\)	\(125\)
default	\(2 e \left (-\frac {c d \left (\frac {\sqrt {e x +d}}{2 c d \left (e x +d \right )+2 e^{2} a -2 c \,d^{2}}+\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{2 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{2}}-\frac {1}{\left (e^{2} a -c \,d^{2}\right )^{2} \sqrt {e x +d}}\right )\)	\(125\)

[In]

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

[Out]

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 243 vs. \(2 (110) = 220\).

Time = 0.39 (sec) , antiderivative size = 502, normalized size of antiderivative = 3.92 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\left [\frac {3 \, {\left (c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt {\frac {c d}{c d^{2} - a e^{2}}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} + 2 \, {\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {\frac {c d}{c d^{2} - a e^{2}}}}{c d x + a e}\right ) - 2 \, {\left (3 \, c d e x + c d^{2} + 2 \, a e^{2}\right )} \sqrt {e x + d}}{2 \, {\left (a c^{2} d^{5} e - 2 \, a^{2} c d^{3} e^{3} + a^{3} d e^{5} + {\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x^{2} + {\left (c^{3} d^{6} - a c^{2} d^{4} e^{2} - a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} x\right )}}, \frac {3 \, {\left (c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}} \arctan \left (-\frac {{\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}}}{c d e x + c d^{2}}\right ) - {\left (3 \, c d e x + c d^{2} + 2 \, a e^{2}\right )} \sqrt {e x + d}}{a c^{2} d^{5} e - 2 \, a^{2} c d^{3} e^{3} + a^{3} d e^{5} + {\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x^{2} + {\left (c^{3} d^{6} - a c^{2} d^{4} e^{2} - a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} x}\right ] \]

[In]

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

[Out]

[1/2*(3*(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)*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*(3*c*d*e*x + c*d^2 + 2*a*e^2)
*sqrt(e*x + d))/(a*c^2*d^5*e - 2*a^2*c*d^3*e^3 + a^3*d*e^5 + (c^3*d^5*e - 2*a*c^2*d^3*e^3 + a^2*c*d*e^5)*x^2 +
(c^3*d^6 - a*c^2*d^4*e^2 - a^2*c*d^2*e^4 + a^3*e^6)*x), (3*(c*d*e^2*x^2 + a*d*e^2 + (c*d^2*e + a*e^3)*x)*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)) - (
3*c*d*e*x + c*d^2 + 2*a*e^2)*sqrt(e*x + d))/(a*c^2*d^5*e - 2*a^2*c*d^3*e^3 + a^3*d*e^5 + (c^3*d^5*e - 2*a*c^2*
d^3*e^3 + a^2*c*d*e^5)*x^2 + (c^3*d^6 - a*c^2*d^4*e^2 - a^2*c*d^2*e^4 + a^3*e^6)*x)]

Sympy [F]

\[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {3}{2}} \left (a e + c d x\right )^{2}}\, dx \]

[In]

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

[Out]

Integral(1/((d + e*x)**(3/2)*(a*e + c*d*x)**2), x)

Maxima [F(-2)]

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

[In]

integrate((e*x+d)^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^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 = 0.27 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.34 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {3 \, c d e \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} - \frac {3 \, {\left (e x + d\right )} c d e - 2 \, c d^{2} e + 2 \, a e^{3}}{{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} {\left ({\left (e x + d\right )}^{\frac {3}{2}} c d - \sqrt {e x + d} c d^{2} + \sqrt {e x + d} a e^{2}\right )}} \]

[In]

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

[Out]

-3*c*d*e*arctan(sqrt(e*x + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))/((c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*sqrt(-c^2*d
^3 + a*c*d*e^2)) - (3*(e*x + d)*c*d*e - 2*c*d^2*e + 2*a*e^3)/((c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*((e*x + d)^(
3/2)*c*d - sqrt(e*x + d)*c*d^2 + sqrt(e*x + d)*a*e^2))

Mupad [B] (verification not implemented)

Time = 9.79 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {\frac {2\,e}{a\,e^2-c\,d^2}+\frac {3\,c\,d\,e\,\left (d+e\,x\right )}{{\left (a\,e^2-c\,d^2\right )}^2}}{\left (a\,e^2-c\,d^2\right )\,\sqrt {d+e\,x}+c\,d\,{\left (d+e\,x\right )}^{3/2}}-\frac {3\,\sqrt {c}\,\sqrt {d}\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{{\left (a\,e^2-c\,d^2\right )}^{5/2}}\right )}{{\left (a\,e^2-c\,d^2\right )}^{5/2}} \]

[In]

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

[Out]

- ((2*e)/(a*e^2 - c*d^2) + (3*c*d*e*(d + e*x))/(a*e^2 - c*d^2)^2)/((a*e^2 - c*d^2)*(d + e*x)^(1/2) + c*d*(d +
e*x)^(3/2)) - (3*c^(1/2)*d^(1/2)*e*atan((c^(1/2)*d^(1/2)*(d + e*x)^(1/2)*(a^2*e^4 + c^2*d^4 - 2*a*c*d^2*e^2))/
(a*e^2 - c*d^2)^(5/2)))/(a*e^2 - c*d^2)^(5/2)