\(\int \frac {1}{(e x)^{3/2} (a+b x) (a c-b c x)} \, dx\) [54]

Optimal result

Integrand size = 26, antiderivative size = 106 \[ \int \frac {1}{(e x)^{3/2} (a+b x) (a c-b c x)} \, dx=-\frac {2}{a^2 c e \sqrt {e x}}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{5/2} c e^{3/2}}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{5/2} c e^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.05 (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, 331, 335, 304, 211, 214} \[ \int \frac {1}{(e x)^{3/2} (a+b x) (a c-b c x)} \, dx=-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{5/2} c e^{3/2}}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{5/2} c e^{3/2}}-\frac {2}{a^2 c e \sqrt {e x}} \]

[In]

Int[1/((e*x)^(3/2)*(a + b*x)*(a*c - b*c*x)),x]

[Out]

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

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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 {1}{(e x)^{3/2} \left (a^2 c-b^2 c x^2\right )} \, dx \\ & = -\frac {2}{a^2 c e \sqrt {e x}}+\frac {b^2 \int \frac {\sqrt {e x}}{a^2 c-b^2 c x^2} \, dx}{a^2 e^2} \\ & = -\frac {2}{a^2 c e \sqrt {e x}}+\frac {\left (2 b^2\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 )}{a^2 e^3} \\ & = -\frac {2}{a^2 c e \sqrt {e x}}+\frac {b \text {Subst}\left (\int \frac {1}{a e-b x^2} \, dx,x,\sqrt {e x}\right )}{a^2 c e}-\frac {b \text {Subst}\left (\int \frac {1}{a e+b x^2} \, dx,x,\sqrt {e x}\right )}{a^2 c e} \\ & = -\frac {2}{a^2 c e \sqrt {e x}}-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{5/2} c e^{3/2}}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{5/2} c e^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.77 \[ \int \frac {1}{(e x)^{3/2} (a+b x) (a c-b c x)} \, dx=\frac {x \left (-2 \sqrt {a}-\sqrt {b} \sqrt {x} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+\sqrt {b} \sqrt {x} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{a^{5/2} c (e x)^{3/2}} \]

[In]

Integrate[1/((e*x)^(3/2)*(a + b*x)*(a*c - b*c*x)),x]

[Out]

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

Maple [A] (verified)

Time = 0.93 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.70

method	result	size
pseudoelliptic	\(\frac {b \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right ) \sqrt {e x}-b \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right ) \sqrt {e x}-2 \sqrt {a e b}}{e c \,a^{2} \sqrt {a e b}\, \sqrt {e x}}\)	\(74\)
risch	\(-\frac {2}{a^{2} c e \sqrt {e x}}+\frac {\frac {b \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{a^{2} \sqrt {a e b}}-\frac {b \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{a^{2} \sqrt {a e b}}}{e c}\)	\(77\)
derivativedivides	\(-\frac {2 e \left (-\frac {b \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a^{2} e^{2} \sqrt {a e b}}+\frac {b \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a^{2} e^{2} \sqrt {a e b}}+\frac {1}{a^{2} e^{2} \sqrt {e x}}\right )}{c}\)	\(78\)
default	\(\frac {2 e \left (\frac {b \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a^{2} e^{2} \sqrt {a e b}}-\frac {b \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a^{2} e^{2} \sqrt {a e b}}-\frac {1}{a^{2} e^{2} \sqrt {e x}}\right )}{c}\)	\(79\)

[In]

int(1/(e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.08 \[ \int \frac {1}{(e x)^{3/2} (a+b x) (a c-b c x)} \, dx=\left [\frac {2 \, e x \sqrt {\frac {b}{a e}} \arctan \left (\frac {\sqrt {e x} a \sqrt {\frac {b}{a e}}}{b x}\right ) + e x \sqrt {\frac {b}{a e}} \log \left (\frac {b x + 2 \, \sqrt {e x} a \sqrt {\frac {b}{a e}} + a}{b x - a}\right ) - 4 \, \sqrt {e x}}{2 \, a^{2} c e^{2} x}, -\frac {2 \, e x \sqrt {-\frac {b}{a e}} \arctan \left (\frac {\sqrt {e x} a \sqrt {-\frac {b}{a e}}}{b x}\right ) - e x \sqrt {-\frac {b}{a e}} \log \left (\frac {b x - 2 \, \sqrt {e x} a \sqrt {-\frac {b}{a e}} - a}{b x + a}\right ) + 4 \, \sqrt {e x}}{2 \, a^{2} c e^{2} x}\right ] \]

[In]

integrate(1/(e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="fricas")

[Out]

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

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.60 (sec) , antiderivative size = 369, normalized size of antiderivative = 3.48 \[ \int \frac {1}{(e x)^{3/2} (a+b x) (a c-b c x)} \, dx=\begin {cases} \frac {1}{3 a b c e^{\frac {3}{2}} x^{\frac {3}{2}}} - \frac {2}{a^{2} c e^{\frac {3}{2}} \sqrt {x}} + \frac {\sqrt {b} \operatorname {acoth}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{a^{\frac {5}{2}} c e^{\frac {3}{2}}} - \frac {\sqrt {b} \operatorname {atan}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{a^{\frac {5}{2}} c e^{\frac {3}{2}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\- \frac {i \left (1 + i\right )}{6 a b c e^{\frac {3}{2}} x^{\frac {3}{2}}} + \frac {1 + i}{6 a b c e^{\frac {3}{2}} x^{\frac {3}{2}}} + \frac {-6 - 6 i}{6 a^{2} c e^{\frac {3}{2}} \sqrt {x}} - \frac {i \left (-6 - 6 i\right )}{6 a^{2} c e^{\frac {3}{2}} \sqrt {x}} + \frac {\sqrt {b} \left (-3 - 3 i\right ) \operatorname {atan}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{6 a^{\frac {5}{2}} c e^{\frac {3}{2}}} - \frac {i \sqrt {b} \left (-3 - 3 i\right ) \operatorname {atan}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{6 a^{\frac {5}{2}} c e^{\frac {3}{2}}} - \frac {i \sqrt {b} \left (3 + 3 i\right ) \operatorname {atanh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{6 a^{\frac {5}{2}} c e^{\frac {3}{2}}} + \frac {\sqrt {b} \left (3 + 3 i\right ) \operatorname {atanh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{6 a^{\frac {5}{2}} c e^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(e*x)**(3/2)/(b*x+a)/(-b*c*x+a*c),x)

[Out]

Piecewise((1/(3*a*b*c*e**(3/2)*x**(3/2)) - 2/(a**2*c*e**(3/2)*sqrt(x)) + sqrt(b)*acoth(sqrt(b)*sqrt(x)/sqrt(a)
)/(a**(5/2)*c*e**(3/2)) - sqrt(b)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(a**(5/2)*c*e**(3/2)), Abs(b*x/a) > 1), (-I*(1
+ I)/(6*a*b*c*e**(3/2)*x**(3/2)) + (1 + I)/(6*a*b*c*e**(3/2)*x**(3/2)) + (-6 - 6*I)/(6*a**2*c*e**(3/2)*sqrt(x
)) - I*(-6 - 6*I)/(6*a**2*c*e**(3/2)*sqrt(x)) + sqrt(b)*(-3 - 3*I)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**(5/2)*c
*e**(3/2)) - I*sqrt(b)*(-3 - 3*I)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**(5/2)*c*e**(3/2)) - I*sqrt(b)*(3 + 3*I)*
atanh(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**(5/2)*c*e**(3/2)) + sqrt(b)*(3 + 3*I)*atanh(sqrt(b)*sqrt(x)/sqrt(a))/(6*a
**(5/2)*c*e**(3/2)), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(e x)^{3/2} (a+b x) (a c-b c x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(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 = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.73 \[ \int \frac {1}{(e x)^{3/2} (a+b x) (a c-b c x)} \, dx=-\frac {\frac {b \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} a^{2} c} + \frac {b \arctan \left (\frac {\sqrt {e x} b}{\sqrt {-a b e}}\right )}{\sqrt {-a b e} a^{2} c} + \frac {2}{\sqrt {e x} a^{2} c}}{e} \]

[In]

integrate(1/(e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="giac")

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.47 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.72 \[ \int \frac {1}{(e x)^{3/2} (a+b x) (a c-b c x)} \, dx=\frac {\sqrt {b}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{a^{5/2}\,c\,e^{3/2}}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{a^{5/2}\,c\,e^{3/2}}-\frac {2}{a^2\,c\,e\,\sqrt {e\,x}} \]

[In]

int(1/((a*c - b*c*x)*(e*x)^(3/2)*(a + b*x)),x)

[Out]

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