3.51 \(\int \frac{(e x)^{3/2}}{(a+b x) (a c-b c x)} \, dx\)
Optimal. Leaf size=103 \[ \frac{\sqrt{a} e^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{b^{5/2} c}+\frac{\sqrt{a} e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{b^{5/2} c}-\frac{2 e \sqrt{e x}}{b^2 c} \]
[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)
________________________________________________________________________________________
Rubi [A] time = 0.0730234, antiderivative size = 103, 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, 321, 329, 212, 208, 205} \[ \frac{\sqrt{a} e^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{b^{5/2} c}+\frac{\sqrt{a} e^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{b^{5/2} c}-\frac{2 e \sqrt{e x}}{b^2 c} \]
Antiderivative was successfully verified.
[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 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 321
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 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 212
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] &&
!GtQ[a/b, 0]
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]
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]
Rubi steps
\begin{align*} \int \frac{(e x)^{3/2}}{(a+b x) (a c-b c x)} \, dx &=\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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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] time = 0.031875, size = 80, normalized size = 0.78 \[ \frac{(e x)^{3/2} \left (\sqrt{a} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )+\sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )-2 \sqrt{b} \sqrt{x}\right )}{b^{5/2} c x^{3/2}} \]
Antiderivative was successfully verified.
[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] time = 0.011, size = 78, normalized size = 0.8 \begin{align*} -2\,{\frac{e\sqrt{ex}}{{b}^{2}c}}+{\frac{{e}^{2}a}{{b}^{2}c}\arctan \left ({b\sqrt{ex}{\frac{1}{\sqrt{aeb}}}} \right ){\frac{1}{\sqrt{aeb}}}}+{\frac{{e}^{2}a}{{b}^{2}c}{\it Artanh} \left ({b\sqrt{ex}{\frac{1}{\sqrt{aeb}}}} \right ){\frac{1}{\sqrt{aeb}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x)
[Out]
-2*e*(e*x)^(1/2)/b^2/c+1/c*e^2/b^2*a/(a*e*b)^(1/2)*arctan(b*(e*x)^(1/2)/(a*e*b)^(1/2))+1/c*e^2/b^2*a/(a*e*b)^(
1/2)*arctanh(b*(e*x)^(1/2)/(a*e*b)^(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((e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
Fricas [A] time = 2.19164, size = 439, normalized size = 4.26 \begin{align*} \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 ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[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] time = 3.14552, size = 1056, normalized size = 10.25 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x)**(3/2)/(b*x+a)/(-b*c*x+a*c),x)
[Out]
Piecewise((-12*a**(5/2)*b**(13/2)*e**(3/2)*x**7/(6*a**(5/2)*b**(17/2)*c*x**(13/2) - 6*a**(3/2)*b**(19/2)*c*x**
(15/2)) + 10*a**(3/2)*b**(15/2)*e**(3/2)*x**8/(6*a**(5/2)*b**(17/2)*c*x**(13/2) - 6*a**(3/2)*b**(19/2)*c*x**(1
5/2)) + 2*sqrt(a)*b**(17/2)*e**(3/2)*x**9/(6*a**(5/2)*b**(17/2)*c*x**(13/2) - 6*a**(3/2)*b**(19/2)*c*x**(15/2)
) + 6*a**3*b**6*e**(3/2)*x**(13/2)*acoth(sqrt(a)/(sqrt(b)*sqrt(x)))/(6*a**(5/2)*b**(17/2)*c*x**(13/2) - 6*a**(
3/2)*b**(19/2)*c*x**(15/2)) - 6*a**3*b**6*e**(3/2)*x**(13/2)*atan(sqrt(a)/(sqrt(b)*sqrt(x)))/(6*a**(5/2)*b**(1
7/2)*c*x**(13/2) - 6*a**(3/2)*b**(19/2)*c*x**(15/2)) + 3*I*pi*a**3*b**6*e**(3/2)*x**(13/2)/(6*a**(5/2)*b**(17/
2)*c*x**(13/2) - 6*a**(3/2)*b**(19/2)*c*x**(15/2)) - 6*a**2*b**7*e**(3/2)*x**(15/2)*acoth(sqrt(a)/(sqrt(b)*sqr
t(x)))/(6*a**(5/2)*b**(17/2)*c*x**(13/2) - 6*a**(3/2)*b**(19/2)*c*x**(15/2)) + 6*a**2*b**7*e**(3/2)*x**(15/2)*
atan(sqrt(a)/(sqrt(b)*sqrt(x)))/(6*a**(5/2)*b**(17/2)*c*x**(13/2) - 6*a**(3/2)*b**(19/2)*c*x**(15/2)) - 3*I*pi
*a**2*b**7*e**(3/2)*x**(15/2)/(6*a**(5/2)*b**(17/2)*c*x**(13/2) - 6*a**(3/2)*b**(19/2)*c*x**(15/2)), Abs(a)/(A
bs(b)*Abs(x)) > 1), (-6*a**(5/2)*b**(13/2)*e**(3/2)*x**7/(3*a**(5/2)*b**(17/2)*c*x**(13/2) - 3*a**(3/2)*b**(19
/2)*c*x**(15/2)) + 5*a**(3/2)*b**(15/2)*e**(3/2)*x**8/(3*a**(5/2)*b**(17/2)*c*x**(13/2) - 3*a**(3/2)*b**(19/2)
*c*x**(15/2)) + sqrt(a)*b**(17/2)*e**(3/2)*x**9/(3*a**(5/2)*b**(17/2)*c*x**(13/2) - 3*a**(3/2)*b**(19/2)*c*x**
(15/2)) - 3*a**3*b**6*e**(3/2)*x**(13/2)*atan(sqrt(a)/(sqrt(b)*sqrt(x)))/(3*a**(5/2)*b**(17/2)*c*x**(13/2) - 3
*a**(3/2)*b**(19/2)*c*x**(15/2)) + 3*a**3*b**6*e**(3/2)*x**(13/2)*atanh(sqrt(a)/(sqrt(b)*sqrt(x)))/(3*a**(5/2)
*b**(17/2)*c*x**(13/2) - 3*a**(3/2)*b**(19/2)*c*x**(15/2)) + 3*a**2*b**7*e**(3/2)*x**(15/2)*atan(sqrt(a)/(sqrt
(b)*sqrt(x)))/(3*a**(5/2)*b**(17/2)*c*x**(13/2) - 3*a**(3/2)*b**(19/2)*c*x**(15/2)) - 3*a**2*b**7*e**(3/2)*x**
(15/2)*atanh(sqrt(a)/(sqrt(b)*sqrt(x)))/(3*a**(5/2)*b**(17/2)*c*x**(13/2) - 3*a**(3/2)*b**(19/2)*c*x**(15/2)),
True))
________________________________________________________________________________________
Giac [A] time = 1.20217, size = 101, normalized size = 0.98 \begin{align*} -\frac{a \arctan \left (\frac{b \sqrt{x} e^{\frac{1}{2}}}{\sqrt{-a b e}}\right ) e^{2}}{\sqrt{-a b e} b^{2} c} + \frac{a \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right ) e^{\frac{3}{2}}}{\sqrt{a b} b^{2} c} - \frac{2 \, \sqrt{x} e^{\frac{3}{2}}}{b^{2} c} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="giac")
[Out]
-a*arctan(b*sqrt(x)*e^(1/2)/sqrt(-a*b*e))*e^2/(sqrt(-a*b*e)*b^2*c) + a*arctan(b*sqrt(x)/sqrt(a*b))*e^(3/2)/(sq
rt(a*b)*b^2*c) - 2*sqrt(x)*e^(3/2)/(b^2*c)