Integrand size = 37, antiderivative size = 128 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {3 e}{\left (c d^2-a e^2\right )^2 \sqrt {d+e x}}-\frac {1}{\left (c d^2-a e^2\right ) (a e+c d x) \sqrt {d+e x}}+\frac {3 \sqrt {c} \sqrt {d} e \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{5/2}} \]
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)
Time = 0.32 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.92 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {2 a e^2+c d (d+3 e x)}{\left (c d^2-a e^2\right )^2 (a e+c d x) \sqrt {d+e x}}-\frac {3 \sqrt {c} \sqrt {d} e \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{\left (-c d^2+a e^2\right )^{5/2}} \]
-((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)
Time = 0.25 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.14, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {1121, 52, 61, 73, 221}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {\sqrt {d+e x}}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 1121 |
\(\displaystyle \int \frac {1}{(d+e x)^{3/2} (a e+c d x)^2}dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {3 e \int \frac {1}{(a e+c d x) (d+e x)^{3/2}}dx}{2 \left (c d^2-a e^2\right )}-\frac {1}{\sqrt {d+e x} \left (c d^2-a e^2\right ) (a e+c d x)}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle -\frac {3 e \left (\frac {c d \int \frac {1}{(a e+c d x) \sqrt {d+e x}}dx}{c d^2-a e^2}+\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{\sqrt {d+e x} \left (c d^2-a e^2\right ) (a e+c d x)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {3 e \left (\frac {2 c d \int \frac {1}{-\frac {c d^2}{e}+\frac {c (d+e x) d}{e}+a e}d\sqrt {d+e x}}{e \left (c d^2-a e^2\right )}+\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{\sqrt {d+e x} \left (c d^2-a e^2\right ) (a e+c d x)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {3 e \left (\frac {2}{\sqrt {d+e x} \left (c d^2-a e^2\right )}-\frac {2 \sqrt {c} \sqrt {d} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\left (c d^2-a e^2\right )^{3/2}}\right )}{2 \left (c d^2-a e^2\right )}-\frac {1}{\sqrt {d+e x} \left (c d^2-a e^2\right ) (a e+c d x)}\) |
-(1/((c*d^2 - a*e^2)*(a*e + c*d*x)*Sqrt[d + e*x])) - (3*e*(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)))/(2*(c*d^2 - a*e^2))
3.21.15.3.1 Defintions of rubi rules used
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] - Simp[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] && ILtQ[m, -1] && FractionQ[n] && LtQ[n, 0]
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] - Simp[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] && 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]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[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] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
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]
Int[((d_) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_ Symbol] :> Int[ExpandIntegrand[(d + e*x)^(m + p)*(a/d + (c/e)*x)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && (Int egerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && LtQ[c, 0]))
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 | \(2 e \left (-\frac {c d \left (\frac {\sqrt {e x +d}}{2 c d \left (e x +d \right )+2 e^{2} a -2 c \,d^{2}}+\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{2 \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\left (e^{2} a -c \,d^{2}\right )^{2}}-\frac {1}{\left (e^{2} a -c \,d^{2}\right )^{2} \sqrt {e x +d}}\right )\) | \(125\) |
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)
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 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\left [\frac {3 \, {\left (c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt {\frac {c d}{c d^{2} - a e^{2}}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} + 2 \, {\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {\frac {c d}{c d^{2} - a e^{2}}}}{c d x + a e}\right ) - 2 \, {\left (3 \, c d e x + c d^{2} + 2 \, a e^{2}\right )} \sqrt {e x + d}}{2 \, {\left (a c^{2} d^{5} e - 2 \, a^{2} c d^{3} e^{3} + a^{3} d e^{5} + {\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x^{2} + {\left (c^{3} d^{6} - a c^{2} d^{4} e^{2} - a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} x\right )}}, \frac {3 \, {\left (c d e^{2} x^{2} + a d e^{2} + {\left (c d^{2} e + a e^{3}\right )} x\right )} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}} \arctan \left (-\frac {{\left (c d^{2} - a e^{2}\right )} \sqrt {e x + d} \sqrt {-\frac {c d}{c d^{2} - a e^{2}}}}{c d e x + c d^{2}}\right ) - {\left (3 \, c d e x + c d^{2} + 2 \, a e^{2}\right )} \sqrt {e x + d}}{a c^{2} d^{5} e - 2 \, a^{2} c d^{3} e^{3} + a^{3} d e^{5} + {\left (c^{3} d^{5} e - 2 \, a c^{2} d^{3} e^{3} + a^{2} c d e^{5}\right )} x^{2} + {\left (c^{3} d^{6} - a c^{2} d^{4} e^{2} - a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} x}\right ] \]
[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)*sqr t(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)*s qrt(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)]
\[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\int \frac {1}{\left (d + e x\right )^{\frac {3}{2}} \left (a e + c d x\right )^{2}}\, dx \]
Exception generated. \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(a*e^2-c*d^2>0)', see `assume?` f or more de
Time = 0.27 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.34 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {3 \, c d e \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} - \frac {3 \, {\left (e x + d\right )} c d e - 2 \, c d^{2} e + 2 \, a e^{3}}{{\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} {\left ({\left (e x + d\right )}^{\frac {3}{2}} c d - \sqrt {e x + d} c d^{2} + \sqrt {e x + d} a e^{2}\right )}} \]
-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))
Time = 9.79 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.21 \[ \int \frac {\sqrt {d+e x}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {\frac {2\,e}{a\,e^2-c\,d^2}+\frac {3\,c\,d\,e\,\left (d+e\,x\right )}{{\left (a\,e^2-c\,d^2\right )}^2}}{\left (a\,e^2-c\,d^2\right )\,\sqrt {d+e\,x}+c\,d\,{\left (d+e\,x\right )}^{3/2}}-\frac {3\,\sqrt {c}\,\sqrt {d}\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}\,\left (a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{{\left (a\,e^2-c\,d^2\right )}^{5/2}}\right )}{{\left (a\,e^2-c\,d^2\right )}^{5/2}} \]