Integrand size = 37, antiderivative size = 146 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {\sqrt {d+e x}}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2}+\frac {3 e \sqrt {d+e x}}{4 \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac {3 e^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 \sqrt {c} \sqrt {d} \left (c d^2-a e^2\right )^{5/2}} \]
-3/4*e^2*arctanh(c^(1/2)*d^(1/2)*(e*x+d)^(1/2)/(-a*e^2+c*d^2)^(1/2))/(-a*e ^2+c*d^2)^(5/2)/c^(1/2)/d^(1/2)-1/2*(e*x+d)^(1/2)/(-a*e^2+c*d^2)/(c*d*x+a* e)^2+3/4*e*(e*x+d)^(1/2)/(-a*e^2+c*d^2)^2/(c*d*x+a*e)
Time = 0.28 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {1}{4} \left (\frac {\sqrt {d+e x} \left (5 a e^2+c d (-2 d+3 e x)\right )}{\left (c d^2-a e^2\right )^2 (a e+c d x)^2}+\frac {3 e^2 \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{\sqrt {c} \sqrt {d} \left (-c d^2+a e^2\right )^{5/2}}\right ) \]
((Sqrt[d + e*x]*(5*a*e^2 + c*d*(-2*d + 3*e*x)))/((c*d^2 - a*e^2)^2*(a*e + c*d*x)^2) + (3*e^2*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^2]])/(Sqrt[c]*Sqrt[d]*(-(c*d^2) + a*e^2)^(5/2)))/4
Time = 0.25 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.08, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {1121, 52, 52, 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.
\(\displaystyle \int \frac {(d+e x)^{5/2}}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 1121 |
\(\displaystyle \int \frac {1}{\sqrt {d+e x} (a e+c d x)^3}dx\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {3 e \int \frac {1}{(a e+c d x)^2 \sqrt {d+e x}}dx}{4 \left (c d^2-a e^2\right )}-\frac {\sqrt {d+e x}}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle -\frac {3 e \left (-\frac {e \int \frac {1}{(a e+c d x) \sqrt {d+e x}}dx}{2 \left (c d^2-a e^2\right )}-\frac {\sqrt {d+e x}}{\left (c d^2-a e^2\right ) (a e+c d x)}\right )}{4 \left (c d^2-a e^2\right )}-\frac {\sqrt {d+e x}}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle -\frac {3 e \left (-\frac {\int \frac {1}{-\frac {c d^2}{e}+\frac {c (d+e x) d}{e}+a e}d\sqrt {d+e x}}{c d^2-a e^2}-\frac {\sqrt {d+e x}}{\left (c d^2-a e^2\right ) (a e+c d x)}\right )}{4 \left (c d^2-a e^2\right )}-\frac {\sqrt {d+e x}}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle -\frac {3 e \left (\frac {e \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c} \sqrt {d} \left (c d^2-a e^2\right )^{3/2}}-\frac {\sqrt {d+e x}}{\left (c d^2-a e^2\right ) (a e+c d x)}\right )}{4 \left (c d^2-a e^2\right )}-\frac {\sqrt {d+e x}}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2}\) |
-1/2*Sqrt[d + e*x]/((c*d^2 - a*e^2)*(a*e + c*d*x)^2) - (3*e*(-(Sqrt[d + e* x]/((c*d^2 - a*e^2)*(a*e + c*d*x))) + (e*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[c*d^2 - a*e^2]])/(Sqrt[c]*Sqrt[d]*(c*d^2 - a*e^2)^(3/2))))/(4* (c*d^2 - a*e^2))
3.21.23.3.1 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] :> 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 = 10.96 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.88
method | result | size |
pseudoelliptic | \(\frac {\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right ) \left (c d x +a e \right )^{2} e^{2}}{4}+\frac {5 \left (-\frac {2 d \left (-\frac {3 e x}{2}+d \right ) c}{5}+e^{2} a \right ) \sqrt {e x +d}\, \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}{4}}{\left (e^{2} a -c \,d^{2}\right )^{2} \left (c d x +a e \right )^{2} \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\) | \(129\) |
derivativedivides | \(2 e^{2} \left (\frac {\sqrt {e x +d}}{4 \left (e^{2} a -c \,d^{2}\right ) \left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {\frac {3 \sqrt {e x +d}}{8 \left (e^{2} a -c \,d^{2}\right ) \left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )}+\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \left (e^{2} a -c \,d^{2}\right ) \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}}{e^{2} a -c \,d^{2}}\right )\) | \(175\) |
default | \(2 e^{2} \left (\frac {\sqrt {e x +d}}{4 \left (e^{2} a -c \,d^{2}\right ) \left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {\frac {3 \sqrt {e x +d}}{8 \left (e^{2} a -c \,d^{2}\right ) \left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )}+\frac {3 \arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \left (e^{2} a -c \,d^{2}\right ) \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}}{e^{2} a -c \,d^{2}}\right )\) | \(175\) |
5/4*(3/5*arctan(c*d*(e*x+d)^(1/2)/((a*e^2-c*d^2)*c*d)^(1/2))*(c*d*x+a*e)^2 *e^2+(-2/5*d*(-3/2*e*x+d)*c+e^2*a)*(e*x+d)^(1/2)*((a*e^2-c*d^2)*c*d)^(1/2) )/((a*e^2-c*d^2)*c*d)^(1/2)/(a*e^2-c*d^2)^2/(c*d*x+a*e)^2
Leaf count of result is larger than twice the leaf count of optimal. 318 vs. \(2 (122) = 244\).
Time = 0.33 (sec) , antiderivative size = 654, normalized size of antiderivative = 4.48 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\left [\frac {3 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {c^{2} d^{3} - a c d e^{2}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} - 2 \, \sqrt {c^{2} d^{3} - a c d e^{2}} \sqrt {e x + d}}{c d x + a e}\right ) - 2 \, {\left (2 \, c^{3} d^{5} - 7 \, a c^{2} d^{3} e^{2} + 5 \, a^{2} c d e^{4} - 3 \, {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} c^{4} d^{7} e^{2} - 3 \, a^{3} c^{3} d^{5} e^{4} + 3 \, a^{4} c^{2} d^{3} e^{6} - a^{5} c d e^{8} + {\left (c^{6} d^{9} - 3 \, a c^{5} d^{7} e^{2} + 3 \, a^{2} c^{4} d^{5} e^{4} - a^{3} c^{3} d^{3} e^{6}\right )} x^{2} + 2 \, {\left (a c^{5} d^{8} e - 3 \, a^{2} c^{4} d^{6} e^{3} + 3 \, a^{3} c^{3} d^{4} e^{5} - a^{4} c^{2} d^{2} e^{7}\right )} x\right )}}, \frac {3 \, {\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}} \arctan \left (\frac {\sqrt {-c^{2} d^{3} + a c d e^{2}} \sqrt {e x + d}}{c d e x + c d^{2}}\right ) - {\left (2 \, c^{3} d^{5} - 7 \, a c^{2} d^{3} e^{2} + 5 \, a^{2} c d e^{4} - 3 \, {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} c^{4} d^{7} e^{2} - 3 \, a^{3} c^{3} d^{5} e^{4} + 3 \, a^{4} c^{2} d^{3} e^{6} - a^{5} c d e^{8} + {\left (c^{6} d^{9} - 3 \, a c^{5} d^{7} e^{2} + 3 \, a^{2} c^{4} d^{5} e^{4} - a^{3} c^{3} d^{3} e^{6}\right )} x^{2} + 2 \, {\left (a c^{5} d^{8} e - 3 \, a^{2} c^{4} d^{6} e^{3} + 3 \, a^{3} c^{3} d^{4} e^{5} - a^{4} c^{2} d^{2} e^{7}\right )} x\right )}}\right ] \]
[1/8*(3*(c^2*d^2*e^2*x^2 + 2*a*c*d*e^3*x + a^2*e^4)*sqrt(c^2*d^3 - a*c*d*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*(2*c^3*d^5 - 7*a*c^2*d^3*e^2 + 5*a^2*c*d*e^4 - 3* (c^3*d^4*e - a*c^2*d^2*e^3)*x)*sqrt(e*x + d))/(a^2*c^4*d^7*e^2 - 3*a^3*c^3 *d^5*e^4 + 3*a^4*c^2*d^3*e^6 - a^5*c*d*e^8 + (c^6*d^9 - 3*a*c^5*d^7*e^2 + 3*a^2*c^4*d^5*e^4 - a^3*c^3*d^3*e^6)*x^2 + 2*(a*c^5*d^8*e - 3*a^2*c^4*d^6* e^3 + 3*a^3*c^3*d^4*e^5 - a^4*c^2*d^2*e^7)*x), 1/4*(3*(c^2*d^2*e^2*x^2 + 2 *a*c*d*e^3*x + a^2*e^4)*sqrt(-c^2*d^3 + a*c*d*e^2)*arctan(sqrt(-c^2*d^3 + a*c*d*e^2)*sqrt(e*x + d)/(c*d*e*x + c*d^2)) - (2*c^3*d^5 - 7*a*c^2*d^3*e^2 + 5*a^2*c*d*e^4 - 3*(c^3*d^4*e - a*c^2*d^2*e^3)*x)*sqrt(e*x + d))/(a^2*c^ 4*d^7*e^2 - 3*a^3*c^3*d^5*e^4 + 3*a^4*c^2*d^3*e^6 - a^5*c*d*e^8 + (c^6*d^9 - 3*a*c^5*d^7*e^2 + 3*a^2*c^4*d^5*e^4 - a^3*c^3*d^3*e^6)*x^2 + 2*(a*c^5*d ^8*e - 3*a^2*c^4*d^6*e^3 + 3*a^3*c^3*d^4*e^5 - a^4*c^2*d^2*e^7)*x)]
Timed out. \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]
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 = 0.29 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.20 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {3 \, e^{2} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{4 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} + \frac {3 \, {\left (e x + d\right )}^{\frac {3}{2}} c d e^{2} - 5 \, \sqrt {e x + d} c d^{2} e^{2} + 5 \, \sqrt {e x + d} a e^{4}}{4 \, {\left (c^{2} d^{4} - 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} {\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )}^{2}} \]
3/4*e^2*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)) + 1/4*(3*(e*x + d)^(3/ 2)*c*d*e^2 - 5*sqrt(e*x + d)*c*d^2*e^2 + 5*sqrt(e*x + d)*a*e^4)/((c^2*d^4 - 2*a*c*d^2*e^2 + a^2*e^4)*((e*x + d)*c*d - c*d^2 + a*e^2)^2)
Time = 10.00 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.21 \[ \int \frac {(d+e x)^{5/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {\frac {5\,e^2\,\sqrt {d+e\,x}}{4\,\left (a\,e^2-c\,d^2\right )}+\frac {3\,c\,d\,e^2\,{\left (d+e\,x\right )}^{3/2}}{4\,{\left (a\,e^2-c\,d^2\right )}^2}}{a^2\,e^4+c^2\,d^4-\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,\left (d+e\,x\right )+c^2\,d^2\,{\left (d+e\,x\right )}^2-2\,a\,c\,d^2\,e^2}+\frac {3\,e^2\,\mathrm {atan}\left (\frac {c\,d\,\sqrt {d+e\,x}}{\sqrt {c\,d}\,\sqrt {a\,e^2-c\,d^2}}\right )}{4\,\sqrt {c\,d}\,{\left (a\,e^2-c\,d^2\right )}^{5/2}} \]
((5*e^2*(d + e*x)^(1/2))/(4*(a*e^2 - c*d^2)) + (3*c*d*e^2*(d + e*x)^(3/2)) /(4*(a*e^2 - c*d^2)^2))/(a^2*e^4 + c^2*d^4 - (2*c^2*d^3 - 2*a*c*d*e^2)*(d + e*x) + c^2*d^2*(d + e*x)^2 - 2*a*c*d^2*e^2) + (3*e^2*atan((c*d*(d + e*x) ^(1/2))/((c*d)^(1/2)*(a*e^2 - c*d^2)^(1/2))))/(4*(c*d)^(1/2)*(a*e^2 - c*d^ 2)^(5/2))