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

Optimal. Leaf size=106 \[ -\frac{a^{3/2} e^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{b^{7/2} c}+\frac{a^{3/2} e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{b^{7/2} c}-\frac{2 e (e x)^{3/2}}{3 b^2 c} \]

[Out]

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

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Rubi [A]  time = 0.0812784, 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, 321, 329, 298, 205, 208} \[ -\frac{a^{3/2} e^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{b^{7/2} c}+\frac{a^{3/2} e^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{b^{7/2} c}-\frac{2 e (e x)^{3/2}}{3 b^2 c} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [A]  time = 0.0449516, size = 85, normalized size = 0.8 \[ -\frac{(e x)^{5/2} \left (3 a^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )-3 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )+2 b^{3/2} x^{3/2}\right )}{3 b^{7/2} c x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.016, size = 83, normalized size = 0.8 \begin{align*} -{\frac{2\,e}{3\,{b}^{2}c} \left ( ex \right ) ^{{\frac{3}{2}}}}-{\frac{{e}^{3}{a}^{2}}{c{b}^{3}}\arctan \left ({b\sqrt{ex}{\frac{1}{\sqrt{aeb}}}} \right ){\frac{1}{\sqrt{aeb}}}}+{\frac{{e}^{3}{a}^{2}}{c{b}^{3}}{\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)^(5/2)/(b*x+a)/(-b*c*x+a*c),x)

[Out]

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

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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)^(5/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.24757, size = 483, normalized size = 4.56 \begin{align*} \left [-\frac{4 \, \sqrt{e x} b e^{2} x + 6 \, a \sqrt{\frac{a e}{b}} e^{2} \arctan \left (\frac{\sqrt{e x} b \sqrt{\frac{a e}{b}}}{a e}\right ) - 3 \, a \sqrt{\frac{a e}{b}} e^{2} \log \left (\frac{b e x + 2 \, \sqrt{e x} b \sqrt{\frac{a e}{b}} + a e}{b x - a}\right )}{6 \, b^{3} c}, -\frac{4 \, \sqrt{e x} b e^{2} x + 6 \, a \sqrt{-\frac{a e}{b}} e^{2} \arctan \left (\frac{\sqrt{e x} b \sqrt{-\frac{a e}{b}}}{a e}\right ) - 3 \, a \sqrt{-\frac{a e}{b}} e^{2} \log \left (\frac{b e x - 2 \, \sqrt{e x} b \sqrt{-\frac{a e}{b}} - a e}{b x + a}\right )}{6 \, b^{3} c}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/6*(4*sqrt(e*x)*b*e^2*x + 6*a*sqrt(a*e/b)*e^2*arctan(sqrt(e*x)*b*sqrt(a*e/b)/(a*e)) - 3*a*sqrt(a*e/b)*e^2*l
og((b*e*x + 2*sqrt(e*x)*b*sqrt(a*e/b) + a*e)/(b*x - a)))/(b^3*c), -1/6*(4*sqrt(e*x)*b*e^2*x + 6*a*sqrt(-a*e/b)
*e^2*arctan(sqrt(e*x)*b*sqrt(-a*e/b)/(a*e)) - 3*a*sqrt(-a*e/b)*e^2*log((b*e*x - 2*sqrt(e*x)*b*sqrt(-a*e/b) - a
*e)/(b*x + a)))/(b^3*c)]

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Sympy [B]  time = 5.66928, size = 898, normalized size = 8.47 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-30*a**(9/2)*b**(39/2)*e**(5/2)*x**21*acoth(sqrt(a)/(sqrt(b)*sqrt(x)))/(-30*a**3*b**23*c*x**21 + 30
*a**2*b**24*c*x**22) - 30*a**(9/2)*b**(39/2)*e**(5/2)*x**21*atan(sqrt(a)/(sqrt(b)*sqrt(x)))/(-30*a**3*b**23*c*
x**21 + 30*a**2*b**24*c*x**22) - 15*I*pi*a**(9/2)*b**(39/2)*e**(5/2)*x**21/(-30*a**3*b**23*c*x**21 + 30*a**2*b
**24*c*x**22) + 30*a**(7/2)*b**(41/2)*e**(5/2)*x**22*acoth(sqrt(a)/(sqrt(b)*sqrt(x)))/(-30*a**3*b**23*c*x**21
+ 30*a**2*b**24*c*x**22) + 30*a**(7/2)*b**(41/2)*e**(5/2)*x**22*atan(sqrt(a)/(sqrt(b)*sqrt(x)))/(-30*a**3*b**2
3*c*x**21 + 30*a**2*b**24*c*x**22) + 15*I*pi*a**(7/2)*b**(41/2)*e**(5/2)*x**22/(-30*a**3*b**23*c*x**21 + 30*a*
*2*b**24*c*x**22) + 20*a**3*b**21*e**(5/2)*x**(45/2)/(-30*a**3*b**23*c*x**21 + 30*a**2*b**24*c*x**22) - 14*a**
2*b**22*e**(5/2)*x**(47/2)/(-30*a**3*b**23*c*x**21 + 30*a**2*b**24*c*x**22) - 6*a*b**23*e**(5/2)*x**(49/2)/(-3
0*a**3*b**23*c*x**21 + 30*a**2*b**24*c*x**22), Abs(a)/(Abs(b)*Abs(x)) > 1), (-15*a**(9/2)*b**(39/2)*e**(5/2)*x
**21*atan(sqrt(a)/(sqrt(b)*sqrt(x)))/(-15*a**3*b**23*c*x**21 + 15*a**2*b**24*c*x**22) - 15*a**(9/2)*b**(39/2)*
e**(5/2)*x**21*atanh(sqrt(a)/(sqrt(b)*sqrt(x)))/(-15*a**3*b**23*c*x**21 + 15*a**2*b**24*c*x**22) + 15*a**(7/2)
*b**(41/2)*e**(5/2)*x**22*atan(sqrt(a)/(sqrt(b)*sqrt(x)))/(-15*a**3*b**23*c*x**21 + 15*a**2*b**24*c*x**22) + 1
5*a**(7/2)*b**(41/2)*e**(5/2)*x**22*atanh(sqrt(a)/(sqrt(b)*sqrt(x)))/(-15*a**3*b**23*c*x**21 + 15*a**2*b**24*c
*x**22) + 10*a**3*b**21*e**(5/2)*x**(45/2)/(-15*a**3*b**23*c*x**21 + 15*a**2*b**24*c*x**22) - 7*a**2*b**22*e**
(5/2)*x**(47/2)/(-15*a**3*b**23*c*x**21 + 15*a**2*b**24*c*x**22) - 3*a*b**23*e**(5/2)*x**(49/2)/(-15*a**3*b**2
3*c*x**21 + 15*a**2*b**24*c*x**22), True))

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Giac [A]  time = 1.24198, size = 113, normalized size = 1.07 \begin{align*} -\frac{1}{3} \,{\left (\frac{3 \, a^{2} \arctan \left (\frac{b \sqrt{x} e^{\frac{1}{2}}}{\sqrt{-a b e}}\right ) e^{2}}{\sqrt{-a b e} b^{3} c} + \frac{2 \, x^{\frac{3}{2}} e^{\frac{3}{2}}}{b^{2} c} + \frac{3 \, a^{2} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right ) e^{\frac{3}{2}}}{\sqrt{a b} b^{3} c}\right )} e \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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