Integrand size = 37, antiderivative size = 114 \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 \left (c d^2-a e^2\right ) \sqrt {d+e x}}{c^2 d^2}+\frac {2 (d+e x)^{3/2}}{3 c d}-\frac {2 \left (c d^2-a e^2\right )^{3/2} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{c^{5/2} d^{5/2}} \]
2/3*(e*x+d)^(3/2)/c/d-2*(-a*e^2+c*d^2)^(3/2)*arctanh(c^(1/2)*d^(1/2)*(e*x+ d)^(1/2)/(-a*e^2+c*d^2)^(1/2))/c^(5/2)/d^(5/2)+2*(-a*e^2+c*d^2)*(e*x+d)^(1 /2)/c^2/d^2
Time = 0.17 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.89 \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 \sqrt {d+e x} \left (-3 a e^2+c d (4 d+e x)\right )}{3 c^2 d^2}+\frac {2 \left (-c d^2+a e^2\right )^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{c^{5/2} d^{5/2}} \]
(2*Sqrt[d + e*x]*(-3*a*e^2 + c*d*(4*d + e*x)))/(3*c^2*d^2) + (2*(-(c*d^2) + a*e^2)^(3/2)*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^ 2]])/(c^(5/2)*d^(5/2))
Time = 0.24 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.17, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {1121, 60, 60, 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.
\(\displaystyle \int \frac {(d+e x)^{5/2}}{x \left (a e^2+c d^2\right )+a d e+c d e x^2} \, dx\) |
\(\Big \downarrow \) 1121 |
\(\displaystyle \int \frac {(d+e x)^{3/2}}{a e+c d x}dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {\left (d^2-\frac {a e^2}{c}\right ) \int \frac {\sqrt {d+e x}}{a e+c d x}dx}{d}+\frac {2 (d+e x)^{3/2}}{3 c d}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {\left (d^2-\frac {a e^2}{c}\right ) \int \frac {1}{(a e+c d x) \sqrt {d+e x}}dx}{d}+\frac {2 \sqrt {d+e x}}{c d}\right )}{d}+\frac {2 (d+e x)^{3/2}}{3 c d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {2 \left (d^2-\frac {a e^2}{c}\right ) \int \frac {1}{-\frac {c d^2}{e}+\frac {c (d+e x) d}{e}+a e}d\sqrt {d+e x}}{d e}+\frac {2 \sqrt {d+e x}}{c d}\right )}{d}+\frac {2 (d+e x)^{3/2}}{3 c d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\left (d^2-\frac {a e^2}{c}\right ) \left (\frac {2 \sqrt {d+e x}}{c d}-\frac {2 \left (d^2-\frac {a e^2}{c}\right ) \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c} d^{3/2} \sqrt {c d^2-a e^2}}\right )}{d}+\frac {2 (d+e x)^{3/2}}{3 c d}\) |
(2*(d + e*x)^(3/2))/(3*c*d) + ((d^2 - (a*e^2)/c)*((2*Sqrt[d + e*x])/(c*d) - (2*(d^2 - (a*e^2)/c)*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(Sqrt[c]*d^(3/2)*Sqrt[c*d^2 - a*e^2])))/d
3.21.2.3.1 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/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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 = 2.86 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.98
method | result | size |
pseudoelliptic | \(-\frac {2 \left (-\left (e^{2} a -c \,d^{2}\right )^{2} \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )+\sqrt {e x +d}\, \left (-\frac {4 \left (\frac {e x}{4}+d \right ) d c}{3}+e^{2} a \right ) \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\right )}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}\, c^{2} d^{2}}\) | \(112\) |
risch | \(-\frac {2 \left (-x c d e +3 e^{2} a -4 c \,d^{2}\right ) \sqrt {e x +d}}{3 c^{2} d^{2}}+\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{c^{2} d^{2} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\) | \(114\) |
derivativedivides | \(-\frac {2 \left (-\frac {c d \left (e x +d \right )^{\frac {3}{2}}}{3}+a \,e^{2} \sqrt {e x +d}-c \,d^{2} \sqrt {e x +d}\right )}{c^{2} d^{2}}+\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{c^{2} d^{2} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\) | \(125\) |
default | \(-\frac {2 \left (-\frac {c d \left (e x +d \right )^{\frac {3}{2}}}{3}+a \,e^{2} \sqrt {e x +d}-c \,d^{2} \sqrt {e x +d}\right )}{c^{2} d^{2}}+\frac {2 \left (a^{2} e^{4}-2 a c \,d^{2} e^{2}+c^{2} d^{4}\right ) \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{c^{2} d^{2} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\) | \(125\) |
-2*(-(a*e^2-c*d^2)^2*arctan(c*d*(e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2))+( e*x+d)^(1/2)*(-4/3*(1/4*e*x+d)*d*c+e^2*a)*((a*e^2-c*d^2)*c*d)^(1/2))/((a*e ^2-c*d^2)*c*d)^(1/2)/c^2/d^2
Time = 0.36 (sec) , antiderivative size = 254, normalized size of antiderivative = 2.23 \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\left [\frac {3 \, {\left (c d^{2} - a e^{2}\right )} \sqrt {\frac {c d^{2} - a e^{2}}{c d}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {e x + d} c d \sqrt {\frac {c d^{2} - a e^{2}}{c d}}}{c d x + a e}\right ) + 2 \, {\left (c d e x + 4 \, c d^{2} - 3 \, a e^{2}\right )} \sqrt {e x + d}}{3 \, c^{2} d^{2}}, -\frac {2 \, {\left (3 \, {\left (c d^{2} - a e^{2}\right )} \sqrt {-\frac {c d^{2} - a e^{2}}{c d}} \arctan \left (-\frac {\sqrt {e x + d} c d \sqrt {-\frac {c d^{2} - a e^{2}}{c d}}}{c d^{2} - a e^{2}}\right ) - {\left (c d e x + 4 \, c d^{2} - 3 \, a e^{2}\right )} \sqrt {e x + d}\right )}}{3 \, c^{2} d^{2}}\right ] \]
[1/3*(3*(c*d^2 - a*e^2)*sqrt((c*d^2 - a*e^2)/(c*d))*log((c*d*e*x + 2*c*d^2 - a*e^2 - 2*sqrt(e*x + d)*c*d*sqrt((c*d^2 - a*e^2)/(c*d)))/(c*d*x + a*e)) + 2*(c*d*e*x + 4*c*d^2 - 3*a*e^2)*sqrt(e*x + d))/(c^2*d^2), -2/3*(3*(c*d^ 2 - a*e^2)*sqrt(-(c*d^2 - a*e^2)/(c*d))*arctan(-sqrt(e*x + d)*c*d*sqrt(-(c *d^2 - a*e^2)/(c*d))/(c*d^2 - a*e^2)) - (c*d*e*x + 4*c*d^2 - 3*a*e^2)*sqrt (e*x + d))/(c^2*d^2)]
Time = 5.40 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.07 \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\begin {cases} \frac {2 \left (\frac {e \left (d + e x\right )^{\frac {3}{2}}}{3 c d} + \frac {\sqrt {d + e x} \left (- a e^{3} + c d^{2} e\right )}{c^{2} d^{2}} + \frac {e \left (a e^{2} - c d^{2}\right )^{2} \operatorname {atan}{\left (\frac {\sqrt {d + e x}}{\sqrt {\frac {a e^{2} - c d^{2}}{c d}}} \right )}}{c^{3} d^{3} \sqrt {\frac {a e^{2} - c d^{2}}{c d}}}\right )}{e} & \text {for}\: e \neq 0 \\\frac {\sqrt {d} \log {\left (x \right )}}{c} & \text {otherwise} \end {cases} \]
Piecewise((2*(e*(d + e*x)**(3/2)/(3*c*d) + sqrt(d + e*x)*(-a*e**3 + c*d**2 *e)/(c**2*d**2) + e*(a*e**2 - c*d**2)**2*atan(sqrt(d + e*x)/sqrt((a*e**2 - c*d**2)/(c*d)))/(c**3*d**3*sqrt((a*e**2 - c*d**2)/(c*d))))/e, Ne(e, 0)), (sqrt(d)*log(x)/c, True))
Exception generated. \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\text {Exception raised: ValueError} \]
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 = 0.28 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.18 \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{\sqrt {-c^{2} d^{3} + a c d e^{2}} c^{2} d^{2}} + \frac {2 \, {\left ({\left (e x + d\right )}^{\frac {3}{2}} c^{2} d^{2} + 3 \, \sqrt {e x + d} c^{2} d^{3} - 3 \, \sqrt {e x + d} a c d e^{2}\right )}}{3 \, c^{3} d^{3}} \]
2*(c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*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^2*d^2) + 2/3*((e*x + d)^(3/ 2)*c^2*d^2 + 3*sqrt(e*x + d)*c^2*d^3 - 3*sqrt(e*x + d)*a*c*d*e^2)/(c^3*d^3 )
Time = 0.09 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.06 \[ \int \frac {(d+e x)^{5/2}}{a d e+\left (c d^2+a e^2\right ) x+c d e x^2} \, dx=\frac {2\,{\left (d+e\,x\right )}^{3/2}}{3\,c\,d}+\frac {2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,{\left (a\,e^2-c\,d^2\right )}^{3/2}\,\sqrt {d+e\,x}}{a^2\,e^4-2\,a\,c\,d^2\,e^2+c^2\,d^4}\right )\,{\left (a\,e^2-c\,d^2\right )}^{3/2}}{c^{5/2}\,d^{5/2}}-\frac {2\,\left (a\,e^2-c\,d^2\right )\,\sqrt {d+e\,x}}{c^2\,d^2} \]