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

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Rubi [A]  time = 0.07, antiderivative size = 107, 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, 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 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 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)^{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*}

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Mathematica [C]  time = 0.02, 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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fricas [A]  time = 0.73, size = 235, normalized size = 2.20 \[ \left [-\frac {6 \, b e x^{2} \sqrt {\frac {b}{a e}} \arctan \left (\frac {\sqrt {e x} a \sqrt {\frac {b}{a e}}}{b x}\right ) - 3 \, b e x^{2} \sqrt {\frac {b}{a e}} \log \left (\frac {b x + 2 \, \sqrt {e x} a \sqrt {\frac {b}{a e}} + a}{b x - a}\right ) + 4 \, \sqrt {e x} a}{6 \, a^{3} c e^{3} x^{2}}, -\frac {6 \, b e x^{2} \sqrt {-\frac {b}{a e}} \arctan \left (\frac {\sqrt {e x} a \sqrt {-\frac {b}{a e}}}{b x}\right ) - 3 \, b e x^{2} \sqrt {-\frac {b}{a e}} \log \left (\frac {b x + 2 \, \sqrt {e x} a \sqrt {-\frac {b}{a e}} - a}{b x + a}\right ) + 4 \, \sqrt {e x} a}{6 \, a^{3} c e^{3} x^{2}}\right ] \]

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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giac [A]  time = 1.07, size = 79, normalized size = 0.74 \[ -\frac {b^{2} \arctan \left (\frac {b \sqrt {x} e^{\frac {1}{2}}}{\sqrt {-a b e}}\right ) e^{\left (-2\right )}}{\sqrt {-a b e} a^{3} c} + \frac {b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right ) e^{\left (-\frac {5}{2}\right )}}{\sqrt {a b} a^{3} c} - \frac {2 \, e^{\left (-\frac {5}{2}\right )}}{3 \, a^{2} c x^{\frac {3}{2}}} \]

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))

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maple [A]  time = 0.02, size = 84, normalized size = 0.79 \[ \frac {b^{2} \arctanh \left (\frac {\sqrt {e x}\, b}{\sqrt {a b e}}\right )}{\sqrt {a b e}\, a^{3} c \,e^{2}}+\frac {b^{2} \arctan \left (\frac {\sqrt {e x}\, b}{\sqrt {a b e}}\right )}{\sqrt {a b e}\, a^{3} c \,e^{2}}-\frac {2}{3 \left (e x \right )^{\frac {3}{2}} a^{2} c e} \]

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]

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

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maxima [A]  time = 2.45, size = 107, normalized size = 1.00 \[ \frac {\frac {6 \, b^{2} \arctan \left (\frac {\sqrt {e x} b}{\sqrt {a b e}}\right )}{\sqrt {a b e} a^{3} c e} - \frac {3 \, b^{2} \log \left (\frac {\sqrt {e x} b - \sqrt {a b e}}{\sqrt {e x} b + \sqrt {a b e}}\right )}{\sqrt {a b e} a^{3} c e} - \frac {4}{\left (e x\right )^{\frac {3}{2}} a^{2} c}}{6 \, e} \]

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]

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

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mupad [B]  time = 0.16, size = 75, normalized size = 0.70 \[ \frac {b^{3/2}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{a^{7/2}\,c\,e^{5/2}}-\frac {2}{3\,a^2\,c\,e\,{\left (e\,x\right )}^{3/2}}+\frac {b^{3/2}\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {e\,x}}{\sqrt {a}\,\sqrt {e}}\right )}{a^{7/2}\,c\,e^{5/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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