\(\int \frac {(e x)^{3/2}}{(a+b x) (a c-b c x)} \, dx\) [51]

Optimal result

Integrand size = 26, antiderivative size = 103 \[ \int \frac {(e x)^{3/2}}{(a+b x) (a c-b c x)} \, dx=-\frac {2 e \sqrt {e x}}{b^2 c}+\frac {\sqrt {a} e^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{5/2} c}+\frac {\sqrt {a} e^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{5/2} c} \]

[Out]

e^(3/2)*arctan(b^(1/2)*(e*x)^(1/2)/a^(1/2)/e^(1/2))*a^(1/2)/b^(5/2)/c+e^(3/2)*arctanh(b^(1/2)*(e*x)^(1/2)/a^(1
/2)/e^(1/2))*a^(1/2)/b^(5/2)/c-2*e*(e*x)^(1/2)/b^2/c

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 103, 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, 218, 214, 211} \[ \int \frac {(e x)^{3/2}}{(a+b x) (a c-b c x)} \, dx=\frac {\sqrt {a} e^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{5/2} c}+\frac {\sqrt {a} e^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{5/2} c}-\frac {2 e \sqrt {e x}}{b^2 c} \]

[In]

Int[(e*x)^(3/2)/((a + b*x)*(a*c - b*c*x)),x]

[Out]

(-2*e*Sqrt[e*x])/(b^2*c) + (Sqrt[a]*e^(3/2)*ArcTan[(Sqrt[b]*Sqrt[e*x])/(Sqrt[a]*Sqrt[e])])/(b^(5/2)*c) + (Sqrt
[a]*e^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[e*x])/(Sqrt[a]*Sqrt[e])])/(b^(5/2)*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 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[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)^{3/2}}{a^2 c-b^2 c x^2} \, dx \\ & = -\frac {2 e \sqrt {e x}}{b^2 c}+\frac {\left (a^2 e^2\right ) \int \frac {1}{\sqrt {e x} \left (a^2 c-b^2 c x^2\right )} \, dx}{b^2} \\ & = -\frac {2 e \sqrt {e x}}{b^2 c}+\frac {\left (2 a^2 e\right ) \text {Subst}\left (\int \frac {1}{a^2 c-\frac {b^2 c x^4}{e^2}} \, dx,x,\sqrt {e x}\right )}{b^2} \\ & = -\frac {2 e \sqrt {e x}}{b^2 c}+\frac {\left (a e^2\right ) \text {Subst}\left (\int \frac {1}{a e-b x^2} \, dx,x,\sqrt {e x}\right )}{b^2 c}+\frac {\left (a e^2\right ) \text {Subst}\left (\int \frac {1}{a e+b x^2} \, dx,x,\sqrt {e x}\right )}{b^2 c} \\ & = -\frac {2 e \sqrt {e x}}{b^2 c}+\frac {\sqrt {a} e^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{5/2} c}+\frac {\sqrt {a} e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{5/2} c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.78 \[ \int \frac {(e x)^{3/2}}{(a+b x) (a c-b c x)} \, dx=\frac {(e x)^{3/2} \left (-2 \sqrt {b} \sqrt {x}+\sqrt {a} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+\sqrt {a} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{b^{5/2} c x^{3/2}} \]

[In]

Integrate[(e*x)^(3/2)/((a + b*x)*(a*c - b*c*x)),x]

[Out]

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

Maple [A] (verified)

Time = 0.46 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.61

method	result	size
pseudoelliptic	\(-\frac {e \left (2 \sqrt {e x}\, \sqrt {a e b}-\left (\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )+\arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )\right ) e a \right )}{\sqrt {a e b}\, c \,b^{2}}\)	\(63\)
derivativedivides	\(-\frac {2 e \left (\frac {\sqrt {e x}}{b^{2}}-\frac {a e \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 b^{2} \sqrt {a e b}}-\frac {a e \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 b^{2} \sqrt {a e b}}\right )}{c}\)	\(71\)
default	\(\frac {2 e \left (-\frac {\sqrt {e x}}{b^{2}}+\frac {a e \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 b^{2} \sqrt {a e b}}+\frac {a e \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 b^{2} \sqrt {a e b}}\right )}{c}\)	\(72\)
risch	\(-\frac {2 x \,e^{2}}{b^{2} \sqrt {e x}\, c}+\frac {\left (\frac {a \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{b^{2} \sqrt {a e b}}+\frac {a \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{b^{2} \sqrt {a e b}}\right ) e^{2}}{c}\)	\(77\)

[In]

int((e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x,method=_RETURNVERBOSE)

[Out]

-e*(2*(e*x)^(1/2)*(a*e*b)^(1/2)-(arctanh(b*(e*x)^(1/2)/(a*e*b)^(1/2))+arctan(b*(e*x)^(1/2)/(a*e*b)^(1/2)))*e*a
)/(a*e*b)^(1/2)/c/b^2

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.89 \[ \int \frac {(e x)^{3/2}}{(a+b x) (a c-b c x)} \, dx=\left [\frac {2 \, \sqrt {\frac {a e}{b}} e \arctan \left (\frac {\sqrt {e x} b \sqrt {\frac {a e}{b}}}{a e}\right ) + \sqrt {\frac {a e}{b}} e \log \left (\frac {b e x + 2 \, \sqrt {e x} b \sqrt {\frac {a e}{b}} + a e}{b x - a}\right ) - 4 \, \sqrt {e x} e}{2 \, b^{2} c}, -\frac {2 \, \sqrt {-\frac {a e}{b}} e \arctan \left (\frac {\sqrt {e x} b \sqrt {-\frac {a e}{b}}}{a e}\right ) - \sqrt {-\frac {a e}{b}} e \log \left (\frac {b e x + 2 \, \sqrt {e x} b \sqrt {-\frac {a e}{b}} - a e}{b x + a}\right ) + 4 \, \sqrt {e x} e}{2 \, b^{2} c}\right ] \]

[In]

integrate((e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="fricas")

[Out]

[1/2*(2*sqrt(a*e/b)*e*arctan(sqrt(e*x)*b*sqrt(a*e/b)/(a*e)) + sqrt(a*e/b)*e*log((b*e*x + 2*sqrt(e*x)*b*sqrt(a*
e/b) + a*e)/(b*x - a)) - 4*sqrt(e*x)*e)/(b^2*c), -1/2*(2*sqrt(-a*e/b)*e*arctan(sqrt(e*x)*b*sqrt(-a*e/b)/(a*e))
- sqrt(-a*e/b)*e*log((b*e*x + 2*sqrt(e*x)*b*sqrt(-a*e/b) - a*e)/(b*x + a)) + 4*sqrt(e*x)*e)/(b^2*c)]

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 207 vs. \(2 (95) = 190\).

Time = 1.56 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.01 \[ \int \frac {(e x)^{3/2}}{(a+b x) (a c-b c x)} \, dx=\begin {cases} \frac {\sqrt {a} e^{\frac {3}{2}} \operatorname {acoth}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{b^{\frac {5}{2}} c} - \frac {\sqrt {a} e^{\frac {3}{2}} \operatorname {atan}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{b^{\frac {5}{2}} c} - \frac {2 e^{\frac {3}{2}} \sqrt {x}}{b^{2} c} - \frac {e^{\frac {3}{2}} x^{\frac {3}{2}}}{3 a b c} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\- \frac {\sqrt {a} e^{\frac {3}{2}} \operatorname {atan}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{b^{\frac {5}{2}} c} + \frac {\sqrt {a} e^{\frac {3}{2}} \operatorname {atanh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{b^{\frac {5}{2}} c} - \frac {2 e^{\frac {3}{2}} \sqrt {x}}{b^{2} c} - \frac {e^{\frac {3}{2}} x^{\frac {3}{2}}}{3 a b c} & \text {otherwise} \end {cases} \]

[In]

integrate((e*x)**(3/2)/(b*x+a)/(-b*c*x+a*c),x)

[Out]

Piecewise((sqrt(a)*e**(3/2)*acoth(sqrt(a)/(sqrt(b)*sqrt(x)))/(b**(5/2)*c) - sqrt(a)*e**(3/2)*atan(sqrt(a)/(sqr
t(b)*sqrt(x)))/(b**(5/2)*c) - 2*e**(3/2)*sqrt(x)/(b**2*c) - e**(3/2)*x**(3/2)/(3*a*b*c), Abs(a/(b*x)) > 1), (-
sqrt(a)*e**(3/2)*atan(sqrt(a)/(sqrt(b)*sqrt(x)))/(b**(5/2)*c) + sqrt(a)*e**(3/2)*atanh(sqrt(a)/(sqrt(b)*sqrt(x
)))/(b**(5/2)*c) - 2*e**(3/2)*sqrt(x)/(b**2*c) - e**(3/2)*x**(3/2)/(3*a*b*c), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {(e x)^{3/2}}{(a+b x) (a c-b c x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((e*x)^(3/2)/(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 = 77, normalized size of antiderivative = 0.75 \[ \int \frac {(e x)^{3/2}}{(a+b x) (a c-b c x)} \, dx=e {\left (\frac {a e \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} b^{2} c} - \frac {a e \arctan \left (\frac {\sqrt {e x} b}{\sqrt {-a b e}}\right )}{\sqrt {-a b e} b^{2} c} - \frac {2 \, \sqrt {e x}}{b^{2} c}\right )} \]

[In]

integrate((e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="giac")

[Out]

e*(a*e*arctan(sqrt(e*x)*b/sqrt(a*b*e))/(sqrt(a*b*e)*b^2*c) - a*e*arctan(sqrt(e*x)*b/sqrt(-a*b*e))/(sqrt(-a*b*e
)*b^2*c) - 2*sqrt(e*x)/(b^2*c))

Mupad [B] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.71 \[ \int \frac {(e x)^{3/2}}{(a+b x) (a c-b c x)} \, dx=\frac {\sqrt {a}\,e^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{b^{5/2}\,c}-\frac {2\,e\,\sqrt {e\,x}}{b^2\,c}+\frac {\sqrt {a}\,e^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{b^{5/2}\,c} \]

[In]

int((e*x)^(3/2)/((a*c - b*c*x)*(a + b*x)),x)

[Out]

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