Integrand size = 37, antiderivative size = 94 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {\sqrt {d+e x}}{c d (a e+c d x)}-\frac {e \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{3/2} d^{3/2} \sqrt {c d^2-a e^2}} \] Output:
-(e*x+d)^(1/2)/c/d/(c*d*x+a*e)-e*arctanh(c^(1/2)*d^(1/2)*(e*x+d)^(1/2)/(-a *e^2+c*d^2)^(1/2))/c^(3/2)/d^(3/2)/(-a*e^2+c*d^2)^(1/2)
Time = 0.21 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.99 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=-\frac {\sqrt {d+e x}}{a c d e+c^2 d^2 x}+\frac {e \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{c^{3/2} d^{3/2} \sqrt {-c d^2+a e^2}} \] Input:
Integrate[(d + e*x)^(5/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]
Output:
-(Sqrt[d + e*x]/(a*c*d*e + c^2*d^2*x)) + (e*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/(c^(3/2)*d^(3/2)*Sqrt[-(c*d^2) + a*e^2])
Time = 0.34 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {1121, 51, 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.
\(\displaystyle \int \frac {(d+e x)^{5/2}}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^2} \, dx\) |
\(\Big \downarrow \) 1121 |
\(\displaystyle \int \frac {\sqrt {d+e x}}{(a e+c d x)^2}dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {e \int \frac {1}{(a e+c d x) \sqrt {d+e x}}dx}{2 c d}-\frac {\sqrt {d+e x}}{c d (a e+c d x)}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\int \frac {1}{-\frac {c d^2}{e}+\frac {c (d+e x) d}{e}+a e}d\sqrt {d+e x}}{c d}-\frac {\sqrt {d+e x}}{c d (a e+c d x)}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {e \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{3/2} d^{3/2} \sqrt {c d^2-a e^2}}-\frac {\sqrt {d+e x}}{c d (a e+c d x)}\) |
Input:
Int[(d + e*x)^(5/2)/(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^2,x]
Output:
-(Sqrt[d + e*x]/(c*d*(a*e + c*d*x))) - (e*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(c^(3/2)*d^(3/2)*Sqrt[c*d^2 - a*e^2])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d, n}, x ] && ILtQ[m, -1] && FractionQ[n] && GtQ[n, 0]
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 = 4.04 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.80
method | result | size |
pseudoelliptic | \(\frac {-\frac {\sqrt {e x +d}}{c d x +a e}+\frac {e \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}}{c d}\) | \(75\) |
derivativedivides | \(2 e \left (-\frac {\sqrt {e x +d}}{2 c d \left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )}+\frac {\arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 c d \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )\) | \(95\) |
default | \(2 e \left (-\frac {\sqrt {e x +d}}{2 c d \left (c d \left (e x +d \right )+a \,e^{2}-c \,d^{2}\right )}+\frac {\arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )}{2 c d \sqrt {c d \left (a \,e^{2}-c \,d^{2}\right )}}\right )\) | \(95\) |
Input:
int((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*x^2*e)^2,x,method=_RETURNVERB OSE)
Output:
1/c/d*(-(e*x+d)^(1/2)/(c*d*x+a*e)+e/(c*d*(a*e^2-c*d^2))^(1/2)*arctan(c*d*( e*x+d)^(1/2)/(c*d*(a*e^2-c*d^2))^(1/2)))
Time = 0.09 (sec) , antiderivative size = 309, normalized size of antiderivative = 3.29 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\left [\frac {\sqrt {c^{2} d^{3} - a c d e^{2}} {\left (c d e x + a e^{2}\right )} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {c^{2} d^{3} - a c d e^{2}} \sqrt {e x + d}}{c d x + a e}\right ) - 2 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} \sqrt {e x + d}}{2 \, {\left (a c^{3} d^{4} e - a^{2} c^{2} d^{2} e^{3} + {\left (c^{4} d^{5} - a c^{3} d^{3} e^{2}\right )} x\right )}}, \frac {\sqrt {-c^{2} d^{3} + a c d e^{2}} {\left (c d e x + a e^{2}\right )} \arctan \left (\frac {\sqrt {-c^{2} d^{3} + a c d e^{2}} \sqrt {e x + d}}{c d e x + c d^{2}}\right ) - {\left (c^{2} d^{3} - a c d e^{2}\right )} \sqrt {e x + d}}{a c^{3} d^{4} e - a^{2} c^{2} d^{2} e^{3} + {\left (c^{4} d^{5} - a c^{3} d^{3} e^{2}\right )} x}\right ] \] Input:
integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm=" fricas")
Output:
[1/2*(sqrt(c^2*d^3 - a*c*d*e^2)*(c*d*e*x + a*e^2)*log((c*d*e*x + 2*c*d^2 - a*e^2 - 2*sqrt(c^2*d^3 - a*c*d*e^2)*sqrt(e*x + d))/(c*d*x + a*e)) - 2*(c^ 2*d^3 - a*c*d*e^2)*sqrt(e*x + d))/(a*c^3*d^4*e - a^2*c^2*d^2*e^3 + (c^4*d^ 5 - a*c^3*d^3*e^2)*x), (sqrt(-c^2*d^3 + a*c*d*e^2)*(c*d*e*x + a*e^2)*arcta n(sqrt(-c^2*d^3 + a*c*d*e^2)*sqrt(e*x + d)/(c*d*e*x + c*d^2)) - (c^2*d^3 - a*c*d*e^2)*sqrt(e*x + d))/(a*c^3*d^4*e - a^2*c^2*d^2*e^3 + (c^4*d^5 - a*c ^3*d^3*e^2)*x)]
\[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\int \frac {\sqrt {d + e x}}{\left (a e + c d x\right )^{2}}\, dx \] Input:
integrate((e*x+d)**(5/2)/(a*d*e+(a*e**2+c*d**2)*x+c*d*e*x**2)**2,x)
Output:
Integral(sqrt(d + e*x)/(a*e + c*d*x)**2, x)
Exception generated. \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm=" maxima")
Output:
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 = 0.13 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {e \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{\sqrt {-c^{2} d^{3} + a c d e^{2}} c d} - \frac {\sqrt {e x + d} e}{{\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )} c d} \] Input:
integrate((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x, algorithm=" giac")
Output:
e*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) - sqrt(e*x + d)*e/(((e*x + d)*c*d - c*d^2 + a*e^2)*c*d)
Time = 5.19 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {e\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}}{\sqrt {a\,e^2-c\,d^2}}\right )}{c^{3/2}\,d^{3/2}\,\sqrt {a\,e^2-c\,d^2}}-\frac {e\,\sqrt {d+e\,x}}{x\,c^2\,d^2\,e+a\,c\,d\,e^2} \] Input:
int((d + e*x)^(5/2)/(x*(a*e^2 + c*d^2) + a*d*e + c*d*e*x^2)^2,x)
Output:
(e*atan((c^(1/2)*d^(1/2)*(d + e*x)^(1/2))/(a*e^2 - c*d^2)^(1/2)))/(c^(3/2) *d^(3/2)*(a*e^2 - c*d^2)^(1/2)) - (e*(d + e*x)^(1/2))/(a*c*d*e^2 + c^2*d^2 *e*x)
Time = 0.22 (sec) , antiderivative size = 180, normalized size of antiderivative = 1.91 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^2} \, dx=\frac {\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) a \,e^{2}+\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}\, \mathit {atan} \left (\frac {\sqrt {e x +d}\, c d}{\sqrt {d}\, \sqrt {c}\, \sqrt {a \,e^{2}-c \,d^{2}}}\right ) c d e x -\sqrt {e x +d}\, a c d \,e^{2}+\sqrt {e x +d}\, c^{2} d^{3}}{c^{2} d^{2} \left (a c d \,e^{2} x -c^{2} d^{3} x +a^{2} e^{3}-a c \,d^{2} e \right )} \] Input:
int((e*x+d)^(5/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^2,x)
Output:
(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d)*s qrt(c)*sqrt(a*e**2 - c*d**2)))*a*e**2 + sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d* *2)*atan((sqrt(d + e*x)*c*d)/(sqrt(d)*sqrt(c)*sqrt(a*e**2 - c*d**2)))*c*d* e*x - sqrt(d + e*x)*a*c*d*e**2 + sqrt(d + e*x)*c**2*d**3)/(c**2*d**2*(a**2 *e**3 - a*c*d**2*e + a*c*d*e**2*x - c**2*d**3*x))