3.1.51 \(\int \frac {(e x)^{3/2}}{(a+b x) (a c-b c x)} \, dx\)

Optimal. Leaf size=103 \[ \frac {\sqrt {a} e^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{5/2} c}+\frac {\sqrt {a} e^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{b^{5/2} c}-\frac {2 e \sqrt {e x}}{b^2 c} \]

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Rubi [A]  time = 0.07, antiderivative size = 103, 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, 212, 208, 205} \begin {gather*} \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} \end {gather*}

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 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 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 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)^{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*}

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Mathematica [A]  time = 0.04, size = 80, normalized size = 0.78 \begin {gather*} \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}} \end {gather*}

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))

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IntegrateAlgebraic [A]  time = 0.06, size = 103, normalized size = 1.00 \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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)

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fricas [A]  time = 1.54, size = 195, normalized size = 1.89 \begin {gather*} \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 {gather*}

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)]

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giac [A]  time = 1.03, size = 79, normalized size = 0.77 \begin {gather*} -{\left (\frac {a \arctan \left (\frac {b \sqrt {x} e^{\frac {1}{2}}}{\sqrt {-a b e}}\right ) e}{\sqrt {-a b e} b^{2} c} - \frac {a \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) e^{\frac {1}{2}}}{\sqrt {a b} b^{2} c} + \frac {2 \, \sqrt {x} e^{\frac {1}{2}}}{b^{2} c}\right )} e \end {gather*}

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

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maple [A]  time = 0.01, size = 78, normalized size = 0.76 \begin {gather*} \frac {a \,e^{2} \arctanh \left (\frac {\sqrt {e x}\, b}{\sqrt {a b e}}\right )}{\sqrt {a b e}\, b^{2} c}+\frac {a \,e^{2} \arctan \left (\frac {\sqrt {e x}\, b}{\sqrt {a b e}}\right )}{\sqrt {a b e}\, b^{2} c}-\frac {2 \sqrt {e x}\, e}{b^{2} c} \end {gather*}

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

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maxima [A]  time = 2.31, size = 106, normalized size = 1.03 \begin {gather*} \frac {\frac {2 \, a e^{3} \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} b^{2} c} - \frac {a e^{3} \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^{2} c} - \frac {4 \, \sqrt {e x} e^{2}}{b^{2} c}}{2 \, e} \end {gather*}

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]

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

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mupad [B]  time = 0.15, size = 73, normalized size = 0.71 \begin {gather*} \frac {\sqrt {a}\,e^{3/2}\,\mathrm {atan}\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}+\frac {\sqrt {a}\,e^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{b^{5/2}\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x)^(3/2)/((a*c - b*c*x)*(a + b*x)),x)

[Out]

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

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sympy [B]  time = 4.71, size = 1052, normalized size = 10.21

result too large to display

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/(b*
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)*ata
nh(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))

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