Integrand size = 37, antiderivative size = 144 \[ \int \frac {(d+e x)^{7/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {\sqrt {d+e x}}{2 c d (a e+c d x)^2}-\frac {e \sqrt {d+e x}}{4 c d \left (c d^2-a e^2\right ) (a e+c d x)}+\frac {e^2 \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{4 c^{3/2} d^{3/2} \left (c d^2-a e^2\right )^{3/2}} \]
1/4*e^2*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)^(3/2)-1/2*(e*x+d)^(1/2)/c/d/(c*d*x+a*e)^2-1/4*e*( e*x+d)^(1/2)/c/d/(-a*e^2+c*d^2)/(c*d*x+a*e)
Time = 0.44 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x)^{7/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {\sqrt {d+e x} \left (a e^2-c d (2 d+e x)\right )}{4 c d \left (c d^2-a e^2\right ) (a e+c d x)^2}+\frac {e^2 \arctan \left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {-c d^2+a e^2}}\right )}{4 c^{3/2} d^{3/2} \left (-c d^2+a e^2\right )^{3/2}} \]
(Sqrt[d + e*x]*(a*e^2 - c*d*(2*d + e*x)))/(4*c*d*(c*d^2 - a*e^2)*(a*e + c* d*x)^2) + (e^2*ArcTan[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/Sqrt[-(c*d^2) + a*e^ 2]])/(4*c^(3/2)*d^(3/2)*(-(c*d^2) + a*e^2)^(3/2))
Time = 0.25 (sec) , antiderivative size = 142, normalized size of antiderivative = 0.99, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {1121, 51, 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.
\(\displaystyle \int \frac {(d+e x)^{7/2}}{\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 1121 |
\(\displaystyle \int \frac {\sqrt {d+e x}}{(a e+c d x)^3}dx\) |
\(\Big \downarrow \) 51 |
\(\displaystyle \frac {e \int \frac {1}{(a e+c d x)^2 \sqrt {d+e x}}dx}{4 c d}-\frac {\sqrt {d+e x}}{2 c d (a e+c d x)^2}\) |
\(\Big \downarrow \) 52 |
\(\displaystyle \frac {e \left (-\frac {e \int \frac {1}{(a e+c d x) \sqrt {d+e x}}dx}{2 \left (c d^2-a e^2\right )}-\frac {\sqrt {d+e x}}{\left (c d^2-a e^2\right ) (a e+c d x)}\right )}{4 c d}-\frac {\sqrt {d+e x}}{2 c d (a e+c d x)^2}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {e \left (-\frac {\int \frac {1}{-\frac {c d^2}{e}+\frac {c (d+e x) d}{e}+a e}d\sqrt {d+e x}}{c d^2-a e^2}-\frac {\sqrt {d+e x}}{\left (c d^2-a e^2\right ) (a e+c d x)}\right )}{4 c d}-\frac {\sqrt {d+e x}}{2 c d (a e+c d x)^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {e \left (\frac {e \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {c d^2-a e^2}}\right )}{\sqrt {c} \sqrt {d} \left (c d^2-a e^2\right )^{3/2}}-\frac {\sqrt {d+e x}}{\left (c d^2-a e^2\right ) (a e+c d x)}\right )}{4 c d}-\frac {\sqrt {d+e x}}{2 c d (a e+c d x)^2}\) |
-1/2*Sqrt[d + e*x]/(c*d*(a*e + c*d*x)^2) + (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)
3.21.22.3.1 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 + 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] :> 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 = 9.73 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.78
method | result | size |
pseudoelliptic | \(\frac {e^{2} \left (-\frac {\sqrt {e x +d}\, \left (-x c d e +e^{2} a -2 c \,d^{2}\right )}{\left (c d x +a e \right )^{2} e^{2}}+\frac {\arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{4 \left (e^{2} a -c \,d^{2}\right ) c d}\) | \(113\) |
derivativedivides | \(2 e^{2} \left (\frac {\frac {\left (e x +d \right )^{\frac {3}{2}}}{8 e^{2} a -8 c \,d^{2}}-\frac {\sqrt {e x +d}}{8 c d}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {\arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \left (e^{2} a -c \,d^{2}\right ) c d \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )\) | \(136\) |
default | \(2 e^{2} \left (\frac {\frac {\left (e x +d \right )^{\frac {3}{2}}}{8 e^{2} a -8 c \,d^{2}}-\frac {\sqrt {e x +d}}{8 c d}}{\left (c d \left (e x +d \right )+e^{2} a -c \,d^{2}\right )^{2}}+\frac {\arctan \left (\frac {c d \sqrt {e x +d}}{\sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )}{8 \left (e^{2} a -c \,d^{2}\right ) c d \sqrt {\left (e^{2} a -c \,d^{2}\right ) c d}}\right )\) | \(136\) |
1/4*e^2/(a*e^2-c*d^2)/c/d*(-(e*x+d)^(1/2)*(-c*d*e*x+a*e^2-2*c*d^2)/(c*d*x+ a*e)^2/e^2+1/((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)))
Leaf count of result is larger than twice the leaf count of optimal. 273 vs. \(2 (120) = 240\).
Time = 0.35 (sec) , antiderivative size = 565, normalized size of antiderivative = 3.92 \[ \int \frac {(d+e x)^{7/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\left [\frac {{\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {c^{2} d^{3} - a c d e^{2}} \log \left (\frac {c d e x + 2 \, c d^{2} - a e^{2} + 2 \, \sqrt {c^{2} d^{3} - a c d e^{2}} \sqrt {e x + d}}{c d x + a e}\right ) - 2 \, {\left (2 \, c^{3} d^{5} - 3 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4} + {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{8 \, {\left (a^{2} c^{4} d^{6} e^{2} - 2 \, a^{3} c^{3} d^{4} e^{4} + a^{4} c^{2} d^{2} e^{6} + {\left (c^{6} d^{8} - 2 \, a c^{5} d^{6} e^{2} + a^{2} c^{4} d^{4} e^{4}\right )} x^{2} + 2 \, {\left (a c^{5} d^{7} e - 2 \, a^{2} c^{4} d^{5} e^{3} + a^{3} c^{3} d^{3} e^{5}\right )} x\right )}}, -\frac {{\left (c^{2} d^{2} e^{2} x^{2} + 2 \, a c d e^{3} x + a^{2} e^{4}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}} \arctan \left (\frac {\sqrt {-c^{2} d^{3} + a c d e^{2}} \sqrt {e x + d}}{c d e x + c d^{2}}\right ) + {\left (2 \, c^{3} d^{5} - 3 \, a c^{2} d^{3} e^{2} + a^{2} c d e^{4} + {\left (c^{3} d^{4} e - a c^{2} d^{2} e^{3}\right )} x\right )} \sqrt {e x + d}}{4 \, {\left (a^{2} c^{4} d^{6} e^{2} - 2 \, a^{3} c^{3} d^{4} e^{4} + a^{4} c^{2} d^{2} e^{6} + {\left (c^{6} d^{8} - 2 \, a c^{5} d^{6} e^{2} + a^{2} c^{4} d^{4} e^{4}\right )} x^{2} + 2 \, {\left (a c^{5} d^{7} e - 2 \, a^{2} c^{4} d^{5} e^{3} + a^{3} c^{3} d^{3} e^{5}\right )} x\right )}}\right ] \]
[1/8*((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 - 3*a*c^2*d^3*e^2 + a^2*c*d*e^4 + (c^3*d ^4*e - a*c^2*d^2*e^3)*x)*sqrt(e*x + d))/(a^2*c^4*d^6*e^2 - 2*a^3*c^3*d^4*e ^4 + a^4*c^2*d^2*e^6 + (c^6*d^8 - 2*a*c^5*d^6*e^2 + a^2*c^4*d^4*e^4)*x^2 + 2*(a*c^5*d^7*e - 2*a^2*c^4*d^5*e^3 + a^3*c^3*d^3*e^5)*x), -1/4*((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 - 3*a* c^2*d^3*e^2 + a^2*c*d*e^4 + (c^3*d^4*e - a*c^2*d^2*e^3)*x)*sqrt(e*x + d))/ (a^2*c^4*d^6*e^2 - 2*a^3*c^3*d^4*e^4 + a^4*c^2*d^2*e^6 + (c^6*d^8 - 2*a*c^ 5*d^6*e^2 + a^2*c^4*d^4*e^4)*x^2 + 2*(a*c^5*d^7*e - 2*a^2*c^4*d^5*e^3 + a^ 3*c^3*d^3*e^5)*x)]
Timed out. \[ \int \frac {(d+e x)^{7/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(d+e x)^{7/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\text {Exception raised: ValueError} \]
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 = 0.28 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.08 \[ \int \frac {(d+e x)^{7/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=-\frac {e^{2} \arctan \left (\frac {\sqrt {e x + d} c d}{\sqrt {-c^{2} d^{3} + a c d e^{2}}}\right )}{4 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} \sqrt {-c^{2} d^{3} + a c d e^{2}}} - \frac {{\left (e x + d\right )}^{\frac {3}{2}} c d e^{2} + \sqrt {e x + d} c d^{2} e^{2} - \sqrt {e x + d} a e^{4}}{4 \, {\left (c^{2} d^{3} - a c d e^{2}\right )} {\left ({\left (e x + d\right )} c d - c d^{2} + a e^{2}\right )}^{2}} \]
-1/4*e^2*arctan(sqrt(e*x + d)*c*d/sqrt(-c^2*d^3 + a*c*d*e^2))/((c^2*d^3 - a*c*d*e^2)*sqrt(-c^2*d^3 + a*c*d*e^2)) - 1/4*((e*x + d)^(3/2)*c*d*e^2 + sq rt(e*x + d)*c*d^2*e^2 - sqrt(e*x + d)*a*e^4)/((c^2*d^3 - a*c*d*e^2)*((e*x + d)*c*d - c*d^2 + a*e^2)^2)
Time = 0.11 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.15 \[ \int \frac {(d+e x)^{7/2}}{\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx=\frac {\frac {e^2\,{\left (d+e\,x\right )}^{3/2}}{4\,\left (a\,e^2-c\,d^2\right )}-\frac {e^2\,\sqrt {d+e\,x}}{4\,c\,d}}{a^2\,e^4+c^2\,d^4-\left (2\,c^2\,d^3-2\,a\,c\,d\,e^2\right )\,\left (d+e\,x\right )+c^2\,d^2\,{\left (d+e\,x\right )}^2-2\,a\,c\,d^2\,e^2}+\frac {e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,\sqrt {d}\,\sqrt {d+e\,x}}{\sqrt {a\,e^2-c\,d^2}}\right )}{4\,c^{3/2}\,d^{3/2}\,{\left (a\,e^2-c\,d^2\right )}^{3/2}} \]