Integrand size = 37, antiderivative size = 244 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^3} \, dx=\frac {5 \left (c d^2-a e^2\right ) \left (c d^2+a e^2+2 c d e x\right ) \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}{8 e^3}-\frac {5 c d \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{3/2}}{3 e^2}+\frac {2 \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{e (d+e x)^2}-\frac {5 \left (c d^2-a e^2\right )^3 \text {arctanh}\left (\frac {c d^2+a e^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {a d e+\left (c d^2+a e^2\right ) x+c d e x^2}}\right )}{16 \sqrt {c} \sqrt {d} e^{7/2}} \]
-5/3*c*d*(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(3/2)/e^2+2*(a*d*e+(a*e^2+c*d^2 )*x+c*d*e*x^2)^(5/2)/e/(e*x+d)^2-5/16*(-a*e^2+c*d^2)^3*arctanh(1/2*(2*c*d* e*x+a*e^2+c*d^2)/c^(1/2)/d^(1/2)/e^(1/2)/(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2) ^(1/2))/e^(7/2)/c^(1/2)/d^(1/2)+5/8*(-a*e^2+c*d^2)*(2*c*d*e*x+a*e^2+c*d^2) *(a*d*e+(a*e^2+c*d^2)*x+c*d*e*x^2)^(1/2)/e^3
Time = 0.42 (sec) , antiderivative size = 183, normalized size of antiderivative = 0.75 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^3} \, dx=\frac {\sqrt {e} (a e+c d x) (d+e x) \left (33 a^2 e^4+2 a c d e^2 (-20 d+13 e x)+c^2 d^2 \left (15 d^2-10 d e x+8 e^2 x^2\right )\right )-\frac {15 \left (c d^2-a e^2\right )^3 \sqrt {a e+c d x} \sqrt {d+e x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {d} \sqrt {d+e x}}{\sqrt {e} \sqrt {a e+c d x}}\right )}{\sqrt {c} \sqrt {d}}}{24 e^{7/2} \sqrt {(a e+c d x) (d+e x)}} \]
(Sqrt[e]*(a*e + c*d*x)*(d + e*x)*(33*a^2*e^4 + 2*a*c*d*e^2*(-20*d + 13*e*x ) + c^2*d^2*(15*d^2 - 10*d*e*x + 8*e^2*x^2)) - (15*(c*d^2 - a*e^2)^3*Sqrt[ a*e + c*d*x]*Sqrt[d + e*x]*ArcTanh[(Sqrt[c]*Sqrt[d]*Sqrt[d + e*x])/(Sqrt[e ]*Sqrt[a*e + c*d*x])])/(Sqrt[c]*Sqrt[d]))/(24*e^(7/2)*Sqrt[(a*e + c*d*x)*( d + e*x)])
Time = 0.38 (sec) , antiderivative size = 264, 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 = {1130, 1131, 1087, 1092, 219}
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.
| \(\displaystyle \int \frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{(d+e x)^3} \, dx\) |
| \(\Big \downarrow \) 1130 |
| \(\displaystyle \frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{e (d+e x)^2}-\frac {5 c d \int \frac {\left (c d e x^2+\left (c d^2+a e^2\right ) x+a d e\right )^{3/2}}{d+e x}dx}{e}\) |
| \(\Big \downarrow \) 1131 |
| \(\displaystyle \frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{e (d+e x)^2}-\frac {5 c d \left (\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e}-\frac {\left (c d^2-a e^2\right ) \int \sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}dx}{2 e}\right )}{e}\) |
| \(\Big \downarrow \) 1087 |
| \(\displaystyle \frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{e (d+e x)^2}-\frac {5 c d \left (\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e}-\frac {\left (c d^2-a e^2\right ) \left (\frac {\left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 c d e}-\frac {\left (c d^2-a e^2\right )^2 \int \frac {1}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}dx}{8 c d e}\right )}{2 e}\right )}{e}\) |
| \(\Big \downarrow \) 1092 |
| \(\displaystyle \frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{e (d+e x)^2}-\frac {5 c d \left (\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e}-\frac {\left (c d^2-a e^2\right ) \left (\frac {\left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 c d e}-\frac {\left (c d^2-a e^2\right )^2 \int \frac {1}{4 c d e-\frac {\left (c d^2+2 c e x d+a e^2\right )^2}{c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}d\frac {c d^2+2 c e x d+a e^2}{\sqrt {c d e x^2+\left (c d^2+a e^2\right ) x+a d e}}}{4 c d e}\right )}{2 e}\right )}{e}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {2 \left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{5/2}}{e (d+e x)^2}-\frac {5 c d \left (\frac {\left (x \left (a e^2+c d^2\right )+a d e+c d e x^2\right )^{3/2}}{3 e}-\frac {\left (c d^2-a e^2\right ) \left (\frac {\left (a e^2+c d^2+2 c d e x\right ) \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}{4 c d e}-\frac {\left (c d^2-a e^2\right )^2 \text {arctanh}\left (\frac {a e^2+c d^2+2 c d e x}{2 \sqrt {c} \sqrt {d} \sqrt {e} \sqrt {x \left (a e^2+c d^2\right )+a d e+c d e x^2}}\right )}{8 c^{3/2} d^{3/2} e^{3/2}}\right )}{2 e}\right )}{e}\) |
(2*(a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(5/2))/(e*(d + e*x)^2) - (5*c*d *((a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2)^(3/2)/(3*e) - ((c*d^2 - a*e^2)*( ((c*d^2 + a*e^2 + 2*c*d*e*x)*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])/ (4*c*d*e) - ((c*d^2 - a*e^2)^2*ArcTanh[(c*d^2 + a*e^2 + 2*c*d*e*x)/(2*Sqrt [c]*Sqrt[d]*Sqrt[e]*Sqrt[a*d*e + (c*d^2 + a*e^2)*x + c*d*e*x^2])])/(8*c^(3 /2)*d^(3/2)*e^(3/2))))/(2*e)))/e
3.20.38.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(b + 2*c*x) *((a + b*x + c*x^2)^p/(2*c*(2*p + 1))), x] - Simp[p*((b^2 - 4*a*c)/(2*c*(2* p + 1))) Int[(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[3*p])
Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[I nt[1/(4*c - x^2), x], x, (b + 2*c*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a , b, c}, x]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + p + 1))), x] - Simp[c*(p/(e^2*(m + p + 1))) Int[(d + e*x)^(m + 2)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && (LtQ[m, -2] || EqQ[m + 2*p + 1, 0]) && NeQ[m + p + 1, 0] & & IntegerQ[2*p]
Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[(d + e*x)^(m + 1)*((a + b*x + c*x^2)^p/(e*(m + 2*p + 1))), x ] - Simp[p*((2*c*d - b*e)/(e^2*(m + 2*p + 1))) Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c*d^2 - b *d*e + a*e^2, 0] && GtQ[p, 0] && (LeQ[-2, m, 0] || EqQ[m + p + 1, 0]) && Ne Q[m + 2*p + 1, 0] && IntegerQ[2*p]
Leaf count of result is larger than twice the leaf count of optimal. \(488\) vs. \(2(216)=432\).
Time = 2.88 (sec) , antiderivative size = 489, normalized size of antiderivative = 2.00
| method | result | size |
| default | \(\frac {\frac {2 \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{3}}-\frac {8 c d e \left (\frac {2 \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {7}{2}}}{3 \left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )^{2}}-\frac {10 c d e \left (\frac {\left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {5}{2}}}{5}+\frac {\left (e^{2} a -c \,d^{2}\right ) \left (\frac {\left (2 c d e \left (x +\frac {d}{e}\right )+e^{2} a -c \,d^{2}\right ) \left (c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )\right )^{\frac {3}{2}}}{8 c d e}-\frac {3 \left (e^{2} a -c \,d^{2}\right )^{2} \left (\frac {\left (2 c d e \left (x +\frac {d}{e}\right )+e^{2} a -c \,d^{2}\right ) \sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}}{4 c d e}-\frac {\left (e^{2} a -c \,d^{2}\right )^{2} \ln \left (\frac {\frac {e^{2} a}{2}-\frac {c \,d^{2}}{2}+c d e \left (x +\frac {d}{e}\right )}{\sqrt {c d e}}+\sqrt {c d e \left (x +\frac {d}{e}\right )^{2}+\left (e^{2} a -c \,d^{2}\right ) \left (x +\frac {d}{e}\right )}\right )}{8 c d e \sqrt {c d e}}\right )}{16 c d e}\right )}{2}\right )}{3 \left (e^{2} a -c \,d^{2}\right )}\right )}{e^{2} a -c \,d^{2}}}{e^{3}}\) | \(489\) |
1/e^3*(2/(a*e^2-c*d^2)/(x+d/e)^3*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^( 7/2)-8*c*d*e/(a*e^2-c*d^2)*(2/3/(a*e^2-c*d^2)/(x+d/e)^2*(c*d*e*(x+d/e)^2+( a*e^2-c*d^2)*(x+d/e))^(7/2)-10/3*c*d*e/(a*e^2-c*d^2)*(1/5*(c*d*e*(x+d/e)^2 +(a*e^2-c*d^2)*(x+d/e))^(5/2)+1/2*(a*e^2-c*d^2)*(1/8*(2*c*d*e*(x+d/e)+e^2* a-c*d^2)/c/d/e*(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^(3/2)-3/16*(a*e^2-c *d^2)^2/c/d/e*(1/4*(2*c*d*e*(x+d/e)+e^2*a-c*d^2)/c/d/e*(c*d*e*(x+d/e)^2+(a *e^2-c*d^2)*(x+d/e))^(1/2)-1/8*(a*e^2-c*d^2)^2/c/d/e*ln((1/2*e^2*a-1/2*c*d ^2+c*d*e*(x+d/e))/(c*d*e)^(1/2)+(c*d*e*(x+d/e)^2+(a*e^2-c*d^2)*(x+d/e))^(1 /2))/(c*d*e)^(1/2))))))
Time = 0.32 (sec) , antiderivative size = 534, normalized size of antiderivative = 2.19 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^3} \, dx=\left [\frac {15 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt {c d e} \log \left (8 \, c^{2} d^{2} e^{2} x^{2} + c^{2} d^{4} + 6 \, a c d^{2} e^{2} + a^{2} e^{4} - 4 \, \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {c d e} + 8 \, {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right ) + 4 \, {\left (8 \, c^{3} d^{3} e^{3} x^{2} + 15 \, c^{3} d^{5} e - 40 \, a c^{2} d^{3} e^{3} + 33 \, a^{2} c d e^{5} - 2 \, {\left (5 \, c^{3} d^{4} e^{2} - 13 \, a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{96 \, c d e^{4}}, \frac {15 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \sqrt {-c d e} \arctan \left (\frac {\sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x} {\left (2 \, c d e x + c d^{2} + a e^{2}\right )} \sqrt {-c d e}}{2 \, {\left (c^{2} d^{2} e^{2} x^{2} + a c d^{2} e^{2} + {\left (c^{2} d^{3} e + a c d e^{3}\right )} x\right )}}\right ) + 2 \, {\left (8 \, c^{3} d^{3} e^{3} x^{2} + 15 \, c^{3} d^{5} e - 40 \, a c^{2} d^{3} e^{3} + 33 \, a^{2} c d e^{5} - 2 \, {\left (5 \, c^{3} d^{4} e^{2} - 13 \, a c^{2} d^{2} e^{4}\right )} x\right )} \sqrt {c d e x^{2} + a d e + {\left (c d^{2} + a e^{2}\right )} x}}{48 \, c d e^{4}}\right ] \]
[1/96*(15*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(c*d *e)*log(8*c^2*d^2*e^2*x^2 + c^2*d^4 + 6*a*c*d^2*e^2 + a^2*e^4 - 4*sqrt(c*d *e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt(c*d*e ) + 8*(c^2*d^3*e + a*c*d*e^3)*x) + 4*(8*c^3*d^3*e^3*x^2 + 15*c^3*d^5*e - 4 0*a*c^2*d^3*e^3 + 33*a^2*c*d*e^5 - 2*(5*c^3*d^4*e^2 - 13*a*c^2*d^2*e^4)*x) *sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x))/(c*d*e^4), 1/48*(15*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*sqrt(-c*d*e)*arctan(1/2*sq rt(c*d*e*x^2 + a*d*e + (c*d^2 + a*e^2)*x)*(2*c*d*e*x + c*d^2 + a*e^2)*sqrt (-c*d*e)/(c^2*d^2*e^2*x^2 + a*c*d^2*e^2 + (c^2*d^3*e + a*c*d*e^3)*x)) + 2* (8*c^3*d^3*e^3*x^2 + 15*c^3*d^5*e - 40*a*c^2*d^3*e^3 + 33*a^2*c*d*e^5 - 2* (5*c^3*d^4*e^2 - 13*a*c^2*d^2*e^4)*x)*sqrt(c*d*e*x^2 + a*d*e + (c*d^2 + a* e^2)*x))/(c*d*e^4)]
Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^3} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^3} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e*(a*e^2-c*d^2)>0)', see `assume ?` for mor
Time = 0.35 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.96 \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^3} \, dx=\frac {1}{24} \, \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e} {\left (2 \, {\left (\frac {4 \, c^{2} d^{2} x}{e} - \frac {5 \, c^{4} d^{5} e - 13 \, a c^{3} d^{3} e^{3}}{c^{2} d^{2} e^{3}}\right )} x + \frac {15 \, c^{4} d^{6} - 40 \, a c^{3} d^{4} e^{2} + 33 \, a^{2} c^{2} d^{2} e^{4}}{c^{2} d^{2} e^{3}}\right )} + \frac {5 \, {\left (c^{3} d^{6} - 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} - a^{3} e^{6}\right )} \log \left ({\left | -c d^{2} - a e^{2} - 2 \, \sqrt {c d e} {\left (\sqrt {c d e} x - \sqrt {c d e x^{2} + c d^{2} x + a e^{2} x + a d e}\right )} \right |}\right )}{16 \, \sqrt {c d e} e^{3}} \]
1/24*sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e)*(2*(4*c^2*d^2*x/e - (5*c^ 4*d^5*e - 13*a*c^3*d^3*e^3)/(c^2*d^2*e^3))*x + (15*c^4*d^6 - 40*a*c^3*d^4* e^2 + 33*a^2*c^2*d^2*e^4)/(c^2*d^2*e^3)) + 5/16*(c^3*d^6 - 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4 - a^3*e^6)*log(abs(-c*d^2 - a*e^2 - 2*sqrt(c*d*e)*(sqrt (c*d*e)*x - sqrt(c*d*e*x^2 + c*d^2*x + a*e^2*x + a*d*e))))/(sqrt(c*d*e)*e^ 3)
Timed out. \[ \int \frac {\left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^{5/2}}{(d+e x)^3} \, dx=\int \frac {{\left (c\,d\,e\,x^2+\left (c\,d^2+a\,e^2\right )\,x+a\,d\,e\right )}^{5/2}}{{\left (d+e\,x\right )}^3} \,d x \]