Integrand size = 26, antiderivative size = 106 \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=-\frac {2 e (e x)^{3/2}}{3 b^2 c}-\frac {a^{3/2} e^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{7/2} c}+\frac {a^{3/2} e^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{7/2} c} \]
-2/3*e*(e*x)^(3/2)/b^2/c-a^(3/2)*e^(5/2)*arctan(b^(1/2)*(e*x)^(1/2)/a^(1/2 )/e^(1/2))/b^(7/2)/c+a^(3/2)*e^(5/2)*arctanh(b^(1/2)*(e*x)^(1/2)/a^(1/2)/e ^(1/2))/b^(7/2)/c
Time = 0.06 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.80 \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=-\frac {(e x)^{5/2} \left (2 b^{3/2} x^{3/2}+3 a^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )-3 a^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{3 b^{7/2} c x^{5/2}} \]
-1/3*((e*x)^(5/2)*(2*b^(3/2)*x^(3/2) + 3*a^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/ Sqrt[a]] - 3*a^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[x])/Sqrt[a]]))/(b^(7/2)*c*x^(5/ 2))
Time = 0.24 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.13, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {82, 262, 27, 266, 27, 827, 218, 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.
\(\displaystyle \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx\) |
\(\Big \downarrow \) 82 |
\(\displaystyle \int \frac {(e x)^{5/2}}{a^2 c-b^2 c x^2}dx\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {a^2 e^2 \int \frac {\sqrt {e x}}{c \left (a^2-b^2 x^2\right )}dx}{b^2}-\frac {2 e (e x)^{3/2}}{3 b^2 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 e^2 \int \frac {\sqrt {e x}}{a^2-b^2 x^2}dx}{b^2 c}-\frac {2 e (e x)^{3/2}}{3 b^2 c}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 a^2 e \int \frac {e^3 x}{a^2 e^2-b^2 e^2 x^2}d\sqrt {e x}}{b^2 c}-\frac {2 e (e x)^{3/2}}{3 b^2 c}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {2 a^2 e^3 \int \frac {e x}{a^2 e^2-b^2 e^2 x^2}d\sqrt {e x}}{b^2 c}-\frac {2 e (e x)^{3/2}}{3 b^2 c}\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {2 a^2 e^3 \left (\frac {\int \frac {1}{a e-b e x}d\sqrt {e x}}{2 b}-\frac {\int \frac {1}{a e+b x e}d\sqrt {e x}}{2 b}\right )}{b^2 c}-\frac {2 e (e x)^{3/2}}{3 b^2 c}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 a^2 e^3 \left (\frac {\int \frac {1}{a e-b e x}d\sqrt {e x}}{2 b}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{2 \sqrt {a} b^{3/2} \sqrt {e}}\right )}{b^2 c}-\frac {2 e (e x)^{3/2}}{3 b^2 c}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 a^2 e^3 \left (\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{2 \sqrt {a} b^{3/2} \sqrt {e}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{2 \sqrt {a} b^{3/2} \sqrt {e}}\right )}{b^2 c}-\frac {2 e (e x)^{3/2}}{3 b^2 c}\) |
(-2*e*(e*x)^(3/2))/(3*b^2*c) + (2*a^2*e^3*(-1/2*ArcTan[(Sqrt[b]*Sqrt[e*x]) /(Sqrt[a]*Sqrt[e])]/(Sqrt[a]*b^(3/2)*Sqrt[e]) + ArcTanh[(Sqrt[b]*Sqrt[e*x] )/(Sqrt[a]*Sqrt[e])]/(2*Sqrt[a]*b^(3/2)*Sqrt[e])))/(b^2*c)
3.1.50.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_) )^(p_.), x_] :> Int[(a*c + b*d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && IntegerQ[m]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ [a/b, 0]
Time = 0.68 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.66
method | result | size |
pseudoelliptic | \(-\frac {2 e^{2} \left (b x \sqrt {e x}\, \sqrt {a e b}-\frac {3 e \left (\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )-\arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )\right ) a^{2}}{2}\right )}{3 \sqrt {a e b}\, c \,b^{3}}\) | \(70\) |
derivativedivides | \(-\frac {2 e \left (\frac {\left (e x \right )^{\frac {3}{2}}}{3 b^{2}}-\frac {a^{2} e^{2} \operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 b^{3} \sqrt {a e b}}+\frac {a^{2} e^{2} \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 b^{3} \sqrt {a e b}}\right )}{c}\) | \(80\) |
default | \(\frac {2 e \left (-\frac {\left (e x \right )^{\frac {3}{2}}}{3 b^{2}}+\frac {a^{2} e^{2} \operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 b^{3} \sqrt {a e b}}-\frac {a^{2} e^{2} \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 b^{3} \sqrt {a e b}}\right )}{c}\) | \(80\) |
risch | \(-\frac {2 x^{2} e^{3}}{3 b^{2} \sqrt {e x}\, c}+\frac {\left (\frac {a^{2} \operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{b^{3} \sqrt {a e b}}-\frac {a^{2} \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{b^{3} \sqrt {a e b}}\right ) e^{3}}{c}\) | \(84\) |
-2/3/(a*e*b)^(1/2)*e^2*(b*x*(e*x)^(1/2)*(a*e*b)^(1/2)-3/2*e*(arctanh(b*(e* x)^(1/2)/(a*e*b)^(1/2))-arctan(b*(e*x)^(1/2)/(a*e*b)^(1/2)))*a^2)/c/b^3
Time = 0.24 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.04 \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=\left [-\frac {4 \, \sqrt {e x} b e^{2} x + 6 \, a \sqrt {\frac {a e}{b}} e^{2} \arctan \left (\frac {\sqrt {e x} b \sqrt {\frac {a e}{b}}}{a e}\right ) - 3 \, a \sqrt {\frac {a e}{b}} e^{2} \log \left (\frac {b e x + 2 \, \sqrt {e x} b \sqrt {\frac {a e}{b}} + a e}{b x - a}\right )}{6 \, b^{3} c}, -\frac {4 \, \sqrt {e x} b e^{2} x + 6 \, a \sqrt {-\frac {a e}{b}} e^{2} \arctan \left (\frac {\sqrt {e x} b \sqrt {-\frac {a e}{b}}}{a e}\right ) - 3 \, a \sqrt {-\frac {a e}{b}} e^{2} \log \left (\frac {b e x - 2 \, \sqrt {e x} b \sqrt {-\frac {a e}{b}} - a e}{b x + a}\right )}{6 \, b^{3} c}\right ] \]
[-1/6*(4*sqrt(e*x)*b*e^2*x + 6*a*sqrt(a*e/b)*e^2*arctan(sqrt(e*x)*b*sqrt(a *e/b)/(a*e)) - 3*a*sqrt(a*e/b)*e^2*log((b*e*x + 2*sqrt(e*x)*b*sqrt(a*e/b) + a*e)/(b*x - a)))/(b^3*c), -1/6*(4*sqrt(e*x)*b*e^2*x + 6*a*sqrt(-a*e/b)*e ^2*arctan(sqrt(e*x)*b*sqrt(-a*e/b)/(a*e)) - 3*a*sqrt(-a*e/b)*e^2*log((b*e* x - 2*sqrt(e*x)*b*sqrt(-a*e/b) - a*e)/(b*x + a)))/(b^3*c)]
Leaf count of result is larger than twice the leaf count of optimal. 211 vs. \(2 (97) = 194\).
Time = 2.44 (sec) , antiderivative size = 211, normalized size of antiderivative = 1.99 \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=\begin {cases} \frac {a^{\frac {3}{2}} e^{\frac {5}{2}} \operatorname {acoth}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{b^{\frac {7}{2}} c} + \frac {a^{\frac {3}{2}} e^{\frac {5}{2}} \operatorname {atan}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{b^{\frac {7}{2}} c} - \frac {2 e^{\frac {5}{2}} x^{\frac {3}{2}}}{3 b^{2} c} - \frac {e^{\frac {5}{2}} x^{\frac {5}{2}}}{5 a b c} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {a^{\frac {3}{2}} e^{\frac {5}{2}} \operatorname {atan}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{b^{\frac {7}{2}} c} + \frac {a^{\frac {3}{2}} e^{\frac {5}{2}} \operatorname {atanh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{b^{\frac {7}{2}} c} - \frac {2 e^{\frac {5}{2}} x^{\frac {3}{2}}}{3 b^{2} c} - \frac {e^{\frac {5}{2}} x^{\frac {5}{2}}}{5 a b c} & \text {otherwise} \end {cases} \]
Piecewise((a**(3/2)*e**(5/2)*acoth(sqrt(a)/(sqrt(b)*sqrt(x)))/(b**(7/2)*c) + a**(3/2)*e**(5/2)*atan(sqrt(a)/(sqrt(b)*sqrt(x)))/(b**(7/2)*c) - 2*e**( 5/2)*x**(3/2)/(3*b**2*c) - e**(5/2)*x**(5/2)/(5*a*b*c), Abs(a/(b*x)) > 1), (a**(3/2)*e**(5/2)*atan(sqrt(a)/(sqrt(b)*sqrt(x)))/(b**(7/2)*c) + a**(3/2 )*e**(5/2)*atanh(sqrt(a)/(sqrt(b)*sqrt(x)))/(b**(7/2)*c) - 2*e**(5/2)*x**( 3/2)/(3*b**2*c) - e**(5/2)*x**(5/2)/(5*a*b*c), True))
Exception generated. \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=\text {Exception raised: ValueError} \]
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>0)', see `assume?` for more de tails)Is e
Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.81 \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=-\frac {1}{3} \, e^{2} {\left (\frac {3 \, a^{2} e \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} b^{3} c} + \frac {3 \, a^{2} e \arctan \left (\frac {\sqrt {e x} b}{\sqrt {-a b e}}\right )}{\sqrt {-a b e} b^{3} c} + \frac {2 \, \sqrt {e x} x}{b^{2} c}\right )} \]
-1/3*e^2*(3*a^2*e*arctan(sqrt(e*x)*b/sqrt(a*b*e))/(sqrt(a*b*e)*b^3*c) + 3* a^2*e*arctan(sqrt(e*x)*b/sqrt(-a*b*e))/(sqrt(-a*b*e)*b^3*c) + 2*sqrt(e*x)* x/(b^2*c))
Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.70 \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=\frac {a^{3/2}\,e^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{b^{7/2}\,c}-\frac {a^{3/2}\,e^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{b^{7/2}\,c}-\frac {2\,e\,{\left (e\,x\right )}^{3/2}}{3\,b^2\,c} \]