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

Optimal. Leaf size=107 \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{a^{7/2} c e^{5/2}}+\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{a^{7/2} c e^{5/2}}-\frac{2}{3 a^2 c e (e x)^{3/2}} \]

[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))

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Rubi [A]  time = 0.0685933, antiderivative size = 107, 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, 212, 208, 205} \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{a^{7/2} c e^{5/2}}+\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e x}}{\sqrt{a} \sqrt{e}}\right )}{a^{7/2} c e^{5/2}}-\frac{2}{3 a^2 c e (e x)^{3/2}} \]

Antiderivative was successfully verified.

[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 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 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{1}{(e x)^{5/2} (a+b x) (a c-b c x)} \, dx &=\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 ) \operatorname{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 \operatorname{Subst}\left (\int \frac{1}{a e-b x^2} \, dx,x,\sqrt{e x}\right )}{a^3 c e^2}+\frac{b^2 \operatorname{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 [C]  time = 0.0149444, size = 36, normalized size = 0.34 \[ -\frac{2 x \, _2F_1\left (-\frac{3}{4},1;\frac{1}{4};\frac{b^2 x^2}{a^2}\right )}{3 a^2 c (e x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*x*Hypergeometric2F1[-3/4, 1, 1/4, (b^2*x^2)/a^2])/(3*a^2*c*(e*x)^(5/2))

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Maple [A]  time = 0.013, size = 84, normalized size = 0.8 \begin{align*} -{\frac{2}{3\,{a}^{2}ce} \left ( ex \right ) ^{-{\frac{3}{2}}}}+{\frac{{b}^{2}}{c{e}^{2}{a}^{3}}\arctan \left ({b\sqrt{ex}{\frac{1}{\sqrt{aeb}}}} \right ){\frac{1}{\sqrt{aeb}}}}+{\frac{{b}^{2}}{c{e}^{2}{a}^{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(1/(e*x)^(5/2)/(b*x+a)/(-b*c*x+a*c),x)

[Out]

-2/3/a^2/c/e/(e*x)^(3/2)+1/c/e^2/a^3*b^2/(a*e*b)^(1/2)*arctan(b*(e*x)^(1/2)/(a*e*b)^(1/2))+1/c/e^2/a^3*b^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(1/(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 = 2.16859, size = 510, normalized size = 4.77 \begin{align*} \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 ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [B]  time = 6.31684, size = 1253, normalized size = 11.71 \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)**(5/2)/(b*x+a)/(-b*c*x+a*c),x)

[Out]

Piecewise((-6*a**(45/2)*b**4*x**(5/2)/(-30*a**(47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6)
+ 26*a**(43/2)*b**5*x**(7/2)/(-30*a**(47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6) - 20*a*
*(41/2)*b**6*x**(9/2)/(-30*a**(47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6) - 30*a**20*b**(
13/2)*x**5*acoth(sqrt(b)*sqrt(x)/sqrt(a))/(-30*a**(47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x
**6) - 30*a**20*b**(13/2)*x**5*atan(sqrt(b)*sqrt(x)/sqrt(a))/(-30*a**(47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2
)*b**6*c*e**(5/2)*x**6) + 15*pi*a**20*b**(13/2)*x**5/(-30*a**(47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c
*e**(5/2)*x**6) + 30*a**19*b**(15/2)*x**6*acoth(sqrt(b)*sqrt(x)/sqrt(a))/(-30*a**(47/2)*b**5*c*e**(5/2)*x**5 +
30*a**(45/2)*b**6*c*e**(5/2)*x**6) + 30*a**19*b**(15/2)*x**6*atan(sqrt(b)*sqrt(x)/sqrt(a))/(-30*a**(47/2)*b**
5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6) - 15*pi*a**19*b**(15/2)*x**6/(-30*a**(47/2)*b**5*c*e**(
5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6), Abs(b*x)/Abs(a) > 1), (-6*a**(45/2)*b**4*x**(5/2)/(-30*a**(47/
2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6) + 26*a**(43/2)*b**5*x**(7/2)/(-30*a**(47/2)*b**5*
c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6) - 20*a**(41/2)*b**6*x**(9/2)/(-30*a**(47/2)*b**5*c*e**(5/
2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6) - 30*a**20*b**(13/2)*x**5*atan(sqrt(b)*sqrt(x)/sqrt(a))/(-30*a**(
47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6) - 30*a**20*b**(13/2)*x**5*atanh(sqrt(b)*sqrt(x
)/sqrt(a))/(-30*a**(47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6) + 15*pi*a**20*b**(13/2)*x*
*5/(-30*a**(47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6) - 15*I*pi*a**20*b**(13/2)*x**5/(-3
0*a**(47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6) + 30*a**19*b**(15/2)*x**6*atan(sqrt(b)*s
qrt(x)/sqrt(a))/(-30*a**(47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6) + 30*a**19*b**(15/2)*
x**6*atanh(sqrt(b)*sqrt(x)/sqrt(a))/(-30*a**(47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6) -
15*pi*a**19*b**(15/2)*x**6/(-30*a**(47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6) + 15*I*pi
*a**19*b**(15/2)*x**6/(-30*a**(47/2)*b**5*c*e**(5/2)*x**5 + 30*a**(45/2)*b**6*c*e**(5/2)*x**6), True))

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Giac [A]  time = 1.17703, size = 107, normalized size = 1. \begin{align*} -\frac{b^{2} \arctan \left (\frac{b \sqrt{x} e^{\frac{1}{2}}}{\sqrt{-a b e}}\right ) e^{\left (-2\right )}}{\sqrt{-a b e} a^{3} c} + \frac{b^{2} \arctan \left (\frac{b \sqrt{x}}{\sqrt{a b}}\right ) e^{\left (-\frac{5}{2}\right )}}{\sqrt{a b} a^{3} c} - \frac{2 \, e^{\left (-\frac{5}{2}\right )}}{3 \, a^{2} c x^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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