\(\int \frac {1}{(e x)^{5/2} (a+b x) (a c-b c x)} \, dx\) [55]

Optimal result

Integrand size = 26, antiderivative size = 107 \[ \int \frac {1}{(e x)^{5/2} (a+b x) (a c-b c x)} \, dx=-\frac {2}{3 a^2 c e (e x)^{3/2}}+\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{7/2} c e^{5/2}}+\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{7/2} c e^{5/2}} \]

[Out]

-2/3/a^2/c/e/(e*x)^(3/2)+b^(3/2)*arctan(b^(1/2)*(e*x)^(1/2)/a^(1/2)/e^(1/2))/a^(7/2)/c/e^(5/2)+b^(3/2)*arctanh
(b^(1/2)*(e*x)^(1/2)/a^(1/2)/e^(1/2))/a^(7/2)/c/e^(5/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 107, 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, 218, 214, 211} \[ \int \frac {1}{(e x)^{5/2} (a+b x) (a c-b c x)} \, dx=\frac {b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{7/2} c e^{5/2}}+\frac {b^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{7/2} c e^{5/2}}-\frac {2}{3 a^2 c e (e x)^{3/2}} \]

[In]

Int[1/((e*x)^(5/2)*(a + b*x)*(a*c - b*c*x)),x]

[Out]

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

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

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.80 \[ \int \frac {1}{(e x)^{5/2} (a+b x) (a c-b c x)} \, dx=\frac {x \left (-2 a^{3/2}+3 b^{3/2} x^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+3 b^{3/2} x^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{3 a^{7/2} c (e x)^{5/2}} \]

[In]

Integrate[1/((e*x)^(5/2)*(a + b*x)*(a*c - b*c*x)),x]

[Out]

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

Maple [A] (verified)

Time = 0.44 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.78

method	result	size
derivativedivides	\(-\frac {2 e \left (\frac {1}{3 a^{2} e^{2} \left (e x \right )^{\frac {3}{2}}}-\frac {b^{2} \operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a^{3} e^{3} \sqrt {a e b}}-\frac {b^{2} \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a^{3} e^{3} \sqrt {a e b}}\right )}{c}\)	\(83\)
default	\(\frac {2 e \left (-\frac {1}{3 a^{2} e^{2} \left (e x \right )^{\frac {3}{2}}}+\frac {b^{2} \operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a^{3} e^{3} \sqrt {a e b}}+\frac {b^{2} \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a^{3} e^{3} \sqrt {a e b}}\right )}{c}\)	\(83\)
risch	\(-\frac {2}{3 a^{2} x \sqrt {e x}\, e^{2} c}+\frac {\frac {b^{2} \operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{a^{3} \sqrt {a e b}}+\frac {b^{2} \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{a^{3} \sqrt {a e b}}}{e^{2} c}\)	\(83\)
pseudoelliptic	\(-\frac {2 \left (-\frac {3 \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right ) b^{2} x \sqrt {e x}}{2}-\frac {3 \,\operatorname {arctanh}\left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right ) b^{2} x \sqrt {e x}}{2}+a \sqrt {a e b}\right )}{3 \sqrt {e x}\, \sqrt {a e b}\, e^{2} c \,a^{3} x}\)	\(85\)

[In]

int(1/(e*x)^(5/2)/(b*x+a)/(-b*c*x+a*c),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 235, normalized size of antiderivative = 2.20 \[ \int \frac {1}{(e x)^{5/2} (a+b x) (a c-b c x)} \, dx=\left [-\frac {6 \, b e x^{2} \sqrt {\frac {b}{a e}} \arctan \left (\frac {\sqrt {e x} a \sqrt {\frac {b}{a e}}}{b x}\right ) - 3 \, b e x^{2} \sqrt {\frac {b}{a e}} \log \left (\frac {b x + 2 \, \sqrt {e x} a \sqrt {\frac {b}{a e}} + a}{b x - a}\right ) + 4 \, \sqrt {e x} a}{6 \, a^{3} c e^{3} x^{2}}, -\frac {6 \, b e x^{2} \sqrt {-\frac {b}{a e}} \arctan \left (\frac {\sqrt {e x} a \sqrt {-\frac {b}{a e}}}{b x}\right ) - 3 \, b e x^{2} \sqrt {-\frac {b}{a e}} \log \left (\frac {b x + 2 \, \sqrt {e x} a \sqrt {-\frac {b}{a e}} - a}{b x + a}\right ) + 4 \, \sqrt {e x} a}{6 \, a^{3} c e^{3} x^{2}}\right ] \]

[In]

integrate(1/(e*x)^(5/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="fricas")

[Out]

[-1/6*(6*b*e*x^2*sqrt(b/(a*e))*arctan(sqrt(e*x)*a*sqrt(b/(a*e))/(b*x)) - 3*b*e*x^2*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)/(a^3*c*e^3*x^2), -1/6*(6*b*e*x^2*sqrt(-b/(a*e))*arc
tan(sqrt(e*x)*a*sqrt(-b/(a*e))/(b*x)) - 3*b*e*x^2*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)/(a^3*c*e^3*x^2)]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.45 (sec) , antiderivative size = 371, normalized size of antiderivative = 3.47 \[ \int \frac {1}{(e x)^{5/2} (a+b x) (a c-b c x)} \, dx=\begin {cases} \frac {1}{5 a b c e^{\frac {5}{2}} x^{\frac {5}{2}}} - \frac {2}{3 a^{2} c e^{\frac {5}{2}} x^{\frac {3}{2}}} + \frac {b^{\frac {3}{2}} \operatorname {acoth}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{a^{\frac {7}{2}} c e^{\frac {5}{2}}} + \frac {b^{\frac {3}{2}} \operatorname {atan}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{a^{\frac {7}{2}} c e^{\frac {5}{2}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\- \frac {i \left (3 + 3 i\right )}{30 a b c e^{\frac {5}{2}} x^{\frac {5}{2}}} + \frac {3 + 3 i}{30 a b c e^{\frac {5}{2}} x^{\frac {5}{2}}} + \frac {-10 - 10 i}{30 a^{2} c e^{\frac {5}{2}} x^{\frac {3}{2}}} - \frac {i \left (-10 - 10 i\right )}{30 a^{2} c e^{\frac {5}{2}} x^{\frac {3}{2}}} - \frac {i b^{\frac {3}{2}} \cdot \left (15 + 15 i\right ) \operatorname {atan}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{30 a^{\frac {7}{2}} c e^{\frac {5}{2}}} + \frac {b^{\frac {3}{2}} \cdot \left (15 + 15 i\right ) \operatorname {atan}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{30 a^{\frac {7}{2}} c e^{\frac {5}{2}}} - \frac {i b^{\frac {3}{2}} \cdot \left (15 + 15 i\right ) \operatorname {atanh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{30 a^{\frac {7}{2}} c e^{\frac {5}{2}}} + \frac {b^{\frac {3}{2}} \cdot \left (15 + 15 i\right ) \operatorname {atanh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{30 a^{\frac {7}{2}} c e^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate(1/(e*x)**(5/2)/(b*x+a)/(-b*c*x+a*c),x)

[Out]

Piecewise((1/(5*a*b*c*e**(5/2)*x**(5/2)) - 2/(3*a**2*c*e**(5/2)*x**(3/2)) + b**(3/2)*acoth(sqrt(b)*sqrt(x)/sqr
t(a))/(a**(7/2)*c*e**(5/2)) + b**(3/2)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(a**(7/2)*c*e**(5/2)), Abs(b*x/a) > 1), (
-I*(3 + 3*I)/(30*a*b*c*e**(5/2)*x**(5/2)) + (3 + 3*I)/(30*a*b*c*e**(5/2)*x**(5/2)) + (-10 - 10*I)/(30*a**2*c*e
**(5/2)*x**(3/2)) - I*(-10 - 10*I)/(30*a**2*c*e**(5/2)*x**(3/2)) - I*b**(3/2)*(15 + 15*I)*atan(sqrt(b)*sqrt(x)
/sqrt(a))/(30*a**(7/2)*c*e**(5/2)) + b**(3/2)*(15 + 15*I)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(30*a**(7/2)*c*e**(5/2
)) - I*b**(3/2)*(15 + 15*I)*atanh(sqrt(b)*sqrt(x)/sqrt(a))/(30*a**(7/2)*c*e**(5/2)) + b**(3/2)*(15 + 15*I)*ata
nh(sqrt(b)*sqrt(x)/sqrt(a))/(30*a**(7/2)*c*e**(5/2)), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{(e x)^{5/2} (a+b x) (a c-b c x)} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate(1/(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 = 0.27 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.83 \[ \int \frac {1}{(e x)^{5/2} (a+b x) (a c-b c x)} \, dx=\frac {b^{2} \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} a^{3} c e^{2}} - \frac {b^{2} \arctan \left (\frac {\sqrt {e x} b}{\sqrt {-a b e}}\right )}{\sqrt {-a b e} a^{3} c e^{2}} - \frac {2}{3 \, \sqrt {e x} a^{2} c e^{2} x} \]

[In]

integrate(1/(e*x)^(5/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="giac")

[Out]

b^2*arctan(sqrt(e*x)*b/sqrt(a*b*e))/(sqrt(a*b*e)*a^3*c*e^2) - b^2*arctan(sqrt(e*x)*b/sqrt(-a*b*e))/(sqrt(-a*b*
e)*a^3*c*e^2) - 2/3/(sqrt(e*x)*a^2*c*e^2*x)

Mupad [B] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(e x)^{5/2} (a+b x) (a c-b c x)} \, dx=\frac {b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{a^{7/2}\,c\,e^{5/2}}-\frac {2}{3\,a^2\,c\,e\,{\left (e\,x\right )}^{3/2}}+\frac {b^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{a^{7/2}\,c\,e^{5/2}} \]

[In]

int(1/((a*c - b*c*x)*(e*x)^(5/2)*(a + b*x)),x)

[Out]

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