3.54 \(\int \frac{1}{(e x)^{3/2} (a+b x) (a c-b c x)} \, dx\)
Optimal. Leaf size=106 \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{a^{5/2} c e^{3/2}}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{a^{5/2} c e^{3/2}}-\frac{2}{a^2 c e \sqrt{e x}} \]
[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))
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Rubi [A] time = 0.0690417, antiderivative size = 106, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used =
{73, 325, 329, 298, 205, 208} \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{a^{5/2} c e^{3/2}}+\frac{\sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{a^{5/2} c e^{3/2}}-\frac{2}{a^2 c e \sqrt{e x}} \]
Antiderivative was successfully verified.
[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 73
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]
Rule 325
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 329
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]
Rule 298
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 205
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{1}{(e x)^{3/2} (a+b x) (a c-b c x)} \, dx &=\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 ) \operatorname{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 \operatorname{Subst}\left (\int \frac{1}{a e-b x^2} \, dx,x,\sqrt{e x}\right )}{a^2 c e}-\frac{b \operatorname{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 [C] time = 0.011987, size = 34, normalized size = 0.32 \[ -\frac{2 x \, _2F_1\left (-\frac{1}{4},1;\frac{3}{4};\frac{b^2 x^2}{a^2}\right )}{a^2 c (e x)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[1/((e*x)^(3/2)*(a + b*x)*(a*c - b*c*x)),x]
[Out]
(-2*x*Hypergeometric2F1[-1/4, 1, 3/4, (b^2*x^2)/a^2])/(a^2*c*(e*x)^(3/2))
________________________________________________________________________________________
Maple [A] time = 0.013, size = 81, normalized size = 0.8 \begin{align*} -{\frac{b}{ce{a}^{2}}\arctan \left ({b\sqrt{ex}{\frac{1}{\sqrt{aeb}}}} \right ){\frac{1}{\sqrt{aeb}}}}+{\frac{b}{ce{a}^{2}}{\it Artanh} \left ({b\sqrt{ex}{\frac{1}{\sqrt{aeb}}}} \right ){\frac{1}{\sqrt{aeb}}}}-2\,{\frac{1}{ce{a}^{2}\sqrt{ex}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x)
[Out]
-1/c/e/a^2*b/(a*e*b)^(1/2)*arctan(b*(e*x)^(1/2)/(a*e*b)^(1/2))+1/c/e/a^2*b/(a*e*b)^(1/2)*arctanh(b*(e*x)^(1/2)
/(a*e*b)^(1/2))-2/a^2/c/e/(e*x)^(1/2)
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 2.29261, size = 471, normalized size = 4.44 \begin{align*} \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 ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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)]
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Sympy [B] time = 3.622, size = 1331, normalized size = 12.56 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x)**(3/2)/(b*x+a)/(-b*c*x+a*c),x)
[Out]
Piecewise((6*a**(13/2)*b**3*x**(5/2)*acoth(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a
**8*b**(7/2)*c*e**(3/2)*x**(7/2)) - 6*a**(13/2)*b**3*x**(5/2)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**9*b**(5/2)*c
*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) + 3*pi*a**(13/2)*b**3*x**(5/2)/(6*a**9*b**(5/2)*c*e*
*(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) - 6*a**(11/2)*b**4*x**(7/2)*acoth(sqrt(b)*sqrt(x)/sqrt(
a))/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) + 6*a**(11/2)*b**4*x**(7/2)*at
an(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) - 3*pi
*a**(11/2)*b**4*x**(7/2)/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) + 2*a**8*
b**(3/2)*x/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) - 14*a**7*b**(5/2)*x**2
/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) + 12*a**6*b**(7/2)*x**3/(6*a**9*b
**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)), Abs(b*x)/Abs(a) > 1), (-6*a**(13/2)*b**3*x
**(5/2)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/
2)) + 6*a**(13/2)*b**3*x**(5/2)*atanh(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b
**(7/2)*c*e**(3/2)*x**(7/2)) + 3*pi*a**(13/2)*b**3*x**(5/2)/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(
7/2)*c*e**(3/2)*x**(7/2)) + 3*I*pi*a**(13/2)*b**3*x**(5/2)/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7
/2)*c*e**(3/2)*x**(7/2)) + 6*a**(11/2)*b**4*x**(7/2)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**9*b**(5/2)*c*e**(3/2)
*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) - 6*a**(11/2)*b**4*x**(7/2)*atanh(sqrt(b)*sqrt(x)/sqrt(a))/(6
*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) - 3*pi*a**(11/2)*b**4*x**(7/2)/(6*a*
*9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) - 3*I*pi*a**(11/2)*b**4*x**(7/2)/(6*a**
9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) + 2*a**8*b**(3/2)*x/(6*a**9*b**(5/2)*c*e
**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) - 14*a**7*b**(5/2)*x**2/(6*a**9*b**(5/2)*c*e**(3/2)*x*
*(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) + 12*a**6*b**(7/2)*x**3/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6
*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)), True))
________________________________________________________________________________________
Giac [A] time = 1.21967, size = 103, normalized size = 0.97 \begin{align*} -{\left (\frac{b \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right ) e^{\left (-\frac{1}{2}\right )}}{\sqrt{a b} a^{2} c} + \frac{b \arctan \left (\frac{b \sqrt{x} e^{\frac{1}{2}}}{\sqrt{-a b e}}\right )}{\sqrt{-a b e} a^{2} c} + \frac{2 \, e^{\left (-\frac{1}{2}\right )}}{a^{2} c \sqrt{x}}\right )} e^{\left (-1\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/(e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="giac")
[Out]
-(b*arctan(b*sqrt(x)/sqrt(a*b))*e^(-1/2)/(sqrt(a*b)*a^2*c) + b*arctan(b*sqrt(x)*e^(1/2)/sqrt(-a*b*e))/(sqrt(-a
*b*e)*a^2*c) + 2*e^(-1/2)/(a^2*c*sqrt(x)))*e^(-1)