3.1.54 \(\int \frac {1}{(e x)^{3/2} (a+b x) (a c-b c x)} \, dx\)

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

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Rubi [A]  time = 0.07, 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, 325, 329, 298, 205, 208} \begin {gather*} -\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{5/2} c e^{3/2}}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{5/2} c e^{3/2}}-\frac {2}{a^2 c e \sqrt {e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((e*x)^(3/2)*(a + b*x)*(a*c - b*c*x)),x]

[Out]

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

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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.02, size = 34, normalized size = 0.32 \begin {gather*} -\frac {2 x \, _2F_1\left (-\frac {1}{4},1;\frac {3}{4};\frac {b^2 x^2}{a^2}\right )}{a^2 c (e x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((e*x)^(3/2)*(a + b*x)*(a*c - b*c*x)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.07, size = 106, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{5/2} c e^{3/2}}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {e x}}{\sqrt {a} \sqrt {e}}\right )}{a^{5/2} c e^{3/2}}-\frac {2}{a^2 c e \sqrt {e x}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((e*x)^(3/2)*(a + b*x)*(a*c - b*c*x)),x]

[Out]

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

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fricas [A]  time = 1.37, size = 220, normalized size = 2.08 \begin {gather*} \left [\frac {2 \, e x \sqrt {\frac {b}{a e}} \arctan \left (\frac {\sqrt {e x} a \sqrt {\frac {b}{a e}}}{b x}\right ) + e x \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}}{2 \, a^{2} c e^{2} x}, -\frac {2 \, e x \sqrt {-\frac {b}{a e}} \arctan \left (\frac {\sqrt {e x} a \sqrt {-\frac {b}{a e}}}{b x}\right ) - e x \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}}{2 \, a^{2} c e^{2} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.92, size = 76, normalized size = 0.72 \begin {gather*} -{\left (\frac {b \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) e^{\left (-\frac {1}{2}\right )}}{\sqrt {a b} a^{2} c} + \frac {b \arctan \left (\frac {b \sqrt {x} e^{\frac {1}{2}}}{\sqrt {-a b e}}\right )}{\sqrt {-a b e} a^{2} c} + \frac {2 \, e^{\left (-\frac {1}{2}\right )}}{a^{2} c \sqrt {x}}\right )} e^{\left (-1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 81, normalized size = 0.76 \begin {gather*} \frac {b \arctanh \left (\frac {\sqrt {e x}\, b}{\sqrt {a b e}}\right )}{\sqrt {a b e}\, a^{2} c e}-\frac {b \arctan \left (\frac {\sqrt {e x}\, b}{\sqrt {a b e}}\right )}{\sqrt {a b e}\, a^{2} c e}-\frac {2}{\sqrt {e x}\, a^{2} c e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x)

[Out]

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

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maxima [A]  time = 2.46, size = 96, normalized size = 0.91 \begin {gather*} -\frac {\frac {2 \, b \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} a^{2} c} + \frac {b \log \left (\frac {\sqrt {e x} b - \sqrt {a b e}}{\sqrt {e x} b + \sqrt {a b e}}\right )}{\sqrt {a b e} a^{2} c} + \frac {4}{\sqrt {e x} a^{2} c}}{2 \, e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x)^(3/2)/(b*x+a)/(-b*c*x+a*c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.40, size = 76, normalized size = 0.72 \begin {gather*} \frac {\sqrt {b}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{a^{5/2}\,c\,e^{3/2}}-\frac {\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{a^{5/2}\,c\,e^{3/2}}-\frac {2}{a^2\,c\,e\,\sqrt {e\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*c - b*c*x)*(e*x)^(3/2)*(a + b*x)),x)

[Out]

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

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sympy [B]  time = 4.79, size = 1287, normalized size = 12.14

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x)**(3/2)/(b*x+a)/(-b*c*x+a*c),x)

[Out]

Piecewise((6*a**(13/2)*b**3*x**(5/2)*acoth(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a
**8*b**(7/2)*c*e**(3/2)*x**(7/2)) - 6*a**(13/2)*b**3*x**(5/2)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**9*b**(5/2)*c
*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) + 3*pi*a**(13/2)*b**3*x**(5/2)/(6*a**9*b**(5/2)*c*e*
*(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) - 6*a**(11/2)*b**4*x**(7/2)*acoth(sqrt(b)*sqrt(x)/sqrt(
a))/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) + 6*a**(11/2)*b**4*x**(7/2)*at
an(sqrt(b)*sqrt(x)/sqrt(a))/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) - 3*pi
*a**(11/2)*b**4*x**(7/2)/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) + 2*a**8*
b**(3/2)*x/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) - 14*a**7*b**(5/2)*x**2
/(6*a**9*b**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)) + 12*a**6*b**(7/2)*x**3/(6*a**9*b
**(5/2)*c*e**(3/2)*x**(5/2) - 6*a**8*b**(7/2)*c*e**(3/2)*x**(7/2)), Abs(b*x/a) > 1), (a**(13/2)*b**3*x**(5/2)*
(3 - 3*I)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(a**9*b**(5/2)*c*e**(3/2)*x**(5/2)*(-3 + 3*I) + a**8*b**(7/2)*c*e**(3/
2)*x**(7/2)*(3 - 3*I)) + a**(13/2)*b**3*x**(5/2)*(-3 + 3*I)*atanh(sqrt(b)*sqrt(x)/sqrt(a))/(a**9*b**(5/2)*c*e*
*(3/2)*x**(5/2)*(-3 + 3*I) + a**8*b**(7/2)*c*e**(3/2)*x**(7/2)*(3 - 3*I)) - 3*pi*a**(13/2)*b**3*x**(5/2)/(a**9
*b**(5/2)*c*e**(3/2)*x**(5/2)*(-3 + 3*I) + a**8*b**(7/2)*c*e**(3/2)*x**(7/2)*(3 - 3*I)) + a**(11/2)*b**4*x**(7
/2)*(-3 + 3*I)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(a**9*b**(5/2)*c*e**(3/2)*x**(5/2)*(-3 + 3*I) + a**8*b**(7/2)*c*e
**(3/2)*x**(7/2)*(3 - 3*I)) + a**(11/2)*b**4*x**(7/2)*(3 - 3*I)*atanh(sqrt(b)*sqrt(x)/sqrt(a))/(a**9*b**(5/2)*
c*e**(3/2)*x**(5/2)*(-3 + 3*I) + a**8*b**(7/2)*c*e**(3/2)*x**(7/2)*(3 - 3*I)) + 3*pi*a**(11/2)*b**4*x**(7/2)/(
a**9*b**(5/2)*c*e**(3/2)*x**(5/2)*(-3 + 3*I) + a**8*b**(7/2)*c*e**(3/2)*x**(7/2)*(3 - 3*I)) + a**8*b**(3/2)*x*
(-1 + I)/(a**9*b**(5/2)*c*e**(3/2)*x**(5/2)*(-3 + 3*I) + a**8*b**(7/2)*c*e**(3/2)*x**(7/2)*(3 - 3*I)) + a**7*b
**(5/2)*x**2*(7 - 7*I)/(a**9*b**(5/2)*c*e**(3/2)*x**(5/2)*(-3 + 3*I) + a**8*b**(7/2)*c*e**(3/2)*x**(7/2)*(3 -
3*I)) + a**6*b**(7/2)*x**3*(-6 + 6*I)/(a**9*b**(5/2)*c*e**(3/2)*x**(5/2)*(-3 + 3*I) + a**8*b**(7/2)*c*e**(3/2)
*x**(7/2)*(3 - 3*I)), True))

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