\(\int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx\) [50]

Optimal result

Integrand size = 26, antiderivative size = 106 \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=-\frac {2 e (e x)^{3/2}}{3 b^2 c}-\frac {a^{3/2} e^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{7/2} c}+\frac {a^{3/2} e^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{7/2} c} \]

[Out]

-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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {74, 327, 335, 304, 211, 214} \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=-\frac {a^{3/2} e^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{7/2} c}+\frac {a^{3/2} e^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{7/2} c}-\frac {2 e (e x)^{3/2}}{3 b^2 c} \]

[In]

Int[(e*x)^(5/2)/((a + b*x)*(a*c - b*c*x)),x]

[Out]

(-2*e*(e*x)^(3/2))/(3*b^2*c) - (a^(3/2)*e^(5/2)*ArcTan[(Sqrt[b]*Sqrt[e*x])/(Sqrt[a]*Sqrt[e])])/(b^(7/2)*c) + (
a^(3/2)*e^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[e*x])/(Sqrt[a]*Sqrt[e])])/(b^(7/2)*c)

Rule 74

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> 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] && Integer
Q[m] && (NeQ[m, -1] || (EqQ[e, 0] && (EqQ[p, 1] ||  !IntegerQ[p])))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

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]

Rule 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[a/b, 0]

Rule 327

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n
)^(p + 1)/(b*(m + n*p + 1))), x] - Dist[a*c^n*((m - n + 1)/(b*(m + n*p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(e x)^{5/2}}{a^2 c-b^2 c x^2} \, dx \\ & = -\frac {2 e (e x)^{3/2}}{3 b^2 c}+\frac {\left (a^2 e^2\right ) \int \frac {\sqrt {e x}}{a^2 c-b^2 c x^2} \, dx}{b^2} \\ & = -\frac {2 e (e x)^{3/2}}{3 b^2 c}+\frac {\left (2 a^2 e\right ) \text {Subst}\left (\int \frac {x^2}{a^2 c-\frac {b^2 c x^4}{e^2}} \, dx,x,\sqrt {e x}\right )}{b^2} \\ & = -\frac {2 e (e x)^{3/2}}{3 b^2 c}+\frac {\left (a^2 e^3\right ) \text {Subst}\left (\int \frac {1}{a e-b x^2} \, dx,x,\sqrt {e x}\right )}{b^3 c}-\frac {\left (a^2 e^3\right ) \text {Subst}\left (\int \frac {1}{a e+b x^2} \, dx,x,\sqrt {e x}\right )}{b^3 c} \\ & = -\frac {2 e (e x)^{3/2}}{3 b^2 c}-\frac {a^{3/2} e^{5/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{7/2} c}+\frac {a^{3/2} e^{5/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{7/2} c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.80 \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=-\frac {(e x)^{5/2} \left (2 b^{3/2} x^{3/2}+3 a^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )-3 a^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{3 b^{7/2} c x^{5/2}} \]

[In]

Integrate[(e*x)^(5/2)/((a + b*x)*(a*c - b*c*x)),x]

[Out]

-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))

Maple [A] (verified)

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	\(-\frac {2 x^{2} e^{3}}{3 b^{2} \sqrt {e x}\, c}+\frac {\left (\frac {a^{2} \operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{b^{3} \sqrt {a e b}}-\frac {a^{2} \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{b^{3} \sqrt {a e b}}\right ) e^{3}}{c}\)	\(84\)

[In]

int((e*x)^(5/2)/(b*x+a)/(-b*c*x+a*c),x,method=_RETURNVERBOSE)

[Out]

-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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 216, normalized size of antiderivative = 2.04 \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=\left [-\frac {4 \, \sqrt {e x} b e^{2} x + 6 \, a \sqrt {\frac {a e}{b}} e^{2} \arctan \left (\frac {\sqrt {e x} b \sqrt {\frac {a e}{b}}}{a e}\right ) - 3 \, a \sqrt {\frac {a e}{b}} e^{2} \log \left (\frac {b e x + 2 \, \sqrt {e x} b \sqrt {\frac {a e}{b}} + a e}{b x - a}\right )}{6 \, b^{3} c}, -\frac {4 \, \sqrt {e x} b e^{2} x + 6 \, a \sqrt {-\frac {a e}{b}} e^{2} \arctan \left (\frac {\sqrt {e x} b \sqrt {-\frac {a e}{b}}}{a e}\right ) - 3 \, a \sqrt {-\frac {a e}{b}} e^{2} \log \left (\frac {b e x - 2 \, \sqrt {e x} b \sqrt {-\frac {a e}{b}} - a e}{b x + a}\right )}{6 \, b^{3} c}\right ] \]

[In]

integrate((e*x)^(5/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="fricas")

[Out]

[-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*l
og((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)]

Sympy [B] (verification not implemented)

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 \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=\begin {cases} \frac {a^{\frac {3}{2}} e^{\frac {5}{2}} \operatorname {acoth}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{b^{\frac {7}{2}} c} + \frac {a^{\frac {3}{2}} e^{\frac {5}{2}} \operatorname {atan}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{b^{\frac {7}{2}} c} - \frac {2 e^{\frac {5}{2}} x^{\frac {3}{2}}}{3 b^{2} c} - \frac {e^{\frac {5}{2}} x^{\frac {5}{2}}}{5 a b c} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {a^{\frac {3}{2}} e^{\frac {5}{2}} \operatorname {atan}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{b^{\frac {7}{2}} c} + \frac {a^{\frac {3}{2}} e^{\frac {5}{2}} \operatorname {atanh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{b^{\frac {7}{2}} c} - \frac {2 e^{\frac {5}{2}} x^{\frac {3}{2}}}{3 b^{2} c} - \frac {e^{\frac {5}{2}} x^{\frac {5}{2}}}{5 a b c} & \text {otherwise} \end {cases} \]

[In]

integrate((e*x)**(5/2)/(b*x+a)/(-b*c*x+a*c),x)

[Out]

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)/(s
qrt(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))

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((e*x)^(5/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more
details)Is e

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.81 \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=-\frac {1}{3} \, e^{2} {\left (\frac {3 \, a^{2} e \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} b^{3} c} + \frac {3 \, a^{2} e \arctan \left (\frac {\sqrt {e x} b}{\sqrt {-a b e}}\right )}{\sqrt {-a b e} b^{3} c} + \frac {2 \, \sqrt {e x} x}{b^{2} c}\right )} \]

[In]

integrate((e*x)^(5/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="giac")

[Out]

-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))

Mupad [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.70 \[ \int \frac {(e x)^{5/2}}{(a+b x) (a c-b c x)} \, dx=\frac {a^{3/2}\,e^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{b^{7/2}\,c}-\frac {a^{3/2}\,e^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{b^{7/2}\,c}-\frac {2\,e\,{\left (e\,x\right )}^{3/2}}{3\,b^2\,c} \]

[In]

int((e*x)^(5/2)/((a*c - b*c*x)*(a + b*x)),x)

[Out]

(a^(3/2)*e^(5/2)*atanh((b^(1/2)*(e*x)^(1/2))/(a^(1/2)*e^(1/2))))/(b^(7/2)*c) - (a^(3/2)*e^(5/2)*atan((b^(1/2)*
(e*x)^(1/2))/(a^(1/2)*e^(1/2))))/(b^(7/2)*c) - (2*e*(e*x)^(3/2))/(3*b^2*c)