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]

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

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Rubi [A]  time = 0.08, antiderivative size = 106, normalized size of antiderivative = 1.00, 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 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]

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 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]

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*}

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Mathematica [A]  time = 0.06, size = 85, normalized size = 0.80 \[ -\frac {(e x)^{5/2} \left (3 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )-3 a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+2 b^{3/2} x^{3/2}\right )}{3 b^{7/2} c x^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((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
]*Sqrt[x])/Sqrt[a]]))/(b^(7/2)*c*x^(5/2))

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fricas [A]  time = 0.80, size = 216, normalized size = 2.04 \[ \left [-\frac {4 \, \sqrt {e x} b e^{2} x + 6 \, a \sqrt {\frac {a e}{b}} e^{2} \arctan \left (\frac {\sqrt {e x} b \sqrt {\frac {a e}{b}}}{a e}\right ) - 3 \, a \sqrt {\frac {a e}{b}} e^{2} \log \left (\frac {b e x + 2 \, \sqrt {e x} b \sqrt {\frac {a e}{b}} + a e}{b x - a}\right )}{6 \, b^{3} c}, -\frac {4 \, \sqrt {e x} b e^{2} x + 6 \, a \sqrt {-\frac {a e}{b}} e^{2} \arctan \left (\frac {\sqrt {e x} b \sqrt {-\frac {a e}{b}}}{a e}\right ) - 3 \, a \sqrt {-\frac {a e}{b}} e^{2} \log \left (\frac {b e x - 2 \, \sqrt {e x} b \sqrt {-\frac {a e}{b}} - a e}{b x + a}\right )}{6 \, b^{3} c}\right ] \]

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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giac [A]  time = 0.98, size = 84, normalized size = 0.79 \[ -\frac {1}{3} \, {\left (\frac {3 \, a^{2} \arctan \left (\frac {b \sqrt {x} e^{\frac {1}{2}}}{\sqrt {-a b e}}\right ) e}{\sqrt {-a b e} b^{3} c} + \frac {2 \, x^{\frac {3}{2}} e^{\frac {1}{2}}}{b^{2} c} + \frac {3 \, a^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) e^{\frac {1}{2}}}{\sqrt {a b} b^{3} c}\right )} e^{2} \]

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

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maple [A]  time = 0.02, size = 83, normalized size = 0.78 \[ \frac {a^{2} e^{3} \arctanh \left (\frac {\sqrt {e x}\, b}{\sqrt {a b e}}\right )}{\sqrt {a b e}\, b^{3} c}-\frac {a^{2} e^{3} \arctan \left (\frac {\sqrt {e x}\, b}{\sqrt {a b e}}\right )}{\sqrt {a b e}\, b^{3} c}-\frac {2 \left (e x \right )^{\frac {3}{2}} e}{3 b^{2} c} \]

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

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maxima [A]  time = 2.38, size = 110, normalized size = 1.04 \[ -\frac {\frac {6 \, a^{2} e^{4} \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} b^{3} c} + \frac {3 \, a^{2} e^{4} \log \left (\frac {\sqrt {e x} b - \sqrt {a b e}}{\sqrt {e x} b + \sqrt {a b e}}\right )}{\sqrt {a b e} b^{3} c} + \frac {4 \, \left (e x\right )^{\frac {3}{2}} e^{2}}{b^{2} c}}{6 \, e} \]

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]

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

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mupad [B]  time = 0.20, size = 74, normalized size = 0.70 \[ \frac {a^{3/2}\,e^{5/2}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{b^{7/2}\,c}-\frac {a^{3/2}\,e^{5/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{b^{7/2}\,c}-\frac {2\,e\,{\left (e\,x\right )}^{3/2}}{3\,b^2\,c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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