3.1.55 \(\int \frac {1}{(e x)^{5/2} (a+b x) (a c-b c x)} \, dx\) [55]

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

[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 = 0.04, 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 = {74, 331, 335, 218, 214, 211} \begin {gather*} \frac {b^{3/2} \text {ArcTan}\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}} \end {gather*}

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 74

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] && (NeQ[m, -1] || (EqQ[e, 0] && (EqQ[p, 1] ||  !IntegerQ[p])))

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 218

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] &&  !Gt
Q[a/b, 0]

Rule 331

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 335

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 ) \text {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 \text {Subst}\left (\int \frac {1}{a e-b x^2} \, dx,x,\sqrt {e x}\right )}{a^3 c e^2}+\frac {b^2 \text {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 [A]
time = 0.05, size = 86, normalized size = 0.80 \begin {gather*} \frac {x \left (-2 a^{3/2}+3 b^{3/2} x^{3/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )+3 b^{3/2} x^{3/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )\right )}{3 a^{7/2} c (e x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(x*(-2*a^(3/2) + 3*b^(3/2)*x^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a]] + 3*b^(3/2)*x^(3/2)*ArcTanh[(Sqrt[b]*Sqrt
[x])/Sqrt[a]]))/(3*a^(7/2)*c*(e*x)^(5/2))

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Maple [A]
time = 0.08, size = 83, normalized size = 0.78

method	result	size
derivativedivides	\(-\frac {2 e \left (\frac {1}{3 a^{2} e^{2} \left (e x \right )^{\frac {3}{2}}}-\frac {b^{2} \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a^{3} e^{3} \sqrt {a e b}}-\frac {b^{2} \arctanh \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a^{3} e^{3} \sqrt {a e b}}\right )}{c}\)	\(83\)
default	\(-\frac {2 e \left (\frac {1}{3 a^{2} e^{2} \left (e x \right )^{\frac {3}{2}}}-\frac {b^{2} \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a^{3} e^{3} \sqrt {a e b}}-\frac {b^{2} \arctanh \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{2 a^{3} e^{3} \sqrt {a e b}}\right )}{c}\)	\(83\)
risch	\(-\frac {2}{3 a^{2} x \sqrt {e x}\, e^{2} c}+\frac {\frac {b^{2} \arctan \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{a^{3} \sqrt {a e b}}+\frac {b^{2} \arctanh \left (\frac {b \sqrt {e x}}{\sqrt {a e b}}\right )}{a^{3} \sqrt {a e b}}}{e^{2} c}\)	\(83\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.48, size = 87, normalized size = 0.81 \begin {gather*} \frac {1}{6} \, {\left (\frac {6 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3} c} - \frac {3 \, b^{2} \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{\sqrt {a b} a^{3} c} - \frac {4}{a^{2} c x^{\frac {3}{2}}}\right )} e^{\left (-\frac {5}{2}\right )} \end {gather*}

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(b*sqrt(x)/sqrt(a*b))/(sqrt(a*b)*a^3*c) - 3*b^2*log((b*sqrt(x) - sqrt(a*b))/(b*sqrt(x) + sqrt
(a*b)))/(sqrt(a*b)*a^3*c) - 4/(a^2*c*x^(3/2)))*e^(-5/2)

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Fricas [A]
time = 1.66, size = 187, normalized size = 1.75 \begin {gather*} \left [-\frac {{\left (6 \, b x^{2} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}}}{b \sqrt {x}}\right ) - 3 \, b x^{2} \sqrt {\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {\frac {b}{a}} + a}{b x - a}\right ) + 4 \, a \sqrt {x}\right )} e^{\left (-\frac {5}{2}\right )}}{6 \, a^{3} c x^{2}}, -\frac {{\left (6 \, b x^{2} \sqrt {-\frac {b}{a}} \arctan \left (\frac {a \sqrt {-\frac {b}{a}}}{b \sqrt {x}}\right ) - 3 \, b x^{2} \sqrt {-\frac {b}{a}} \log \left (\frac {b x + 2 \, a \sqrt {x} \sqrt {-\frac {b}{a}} - a}{b x + a}\right ) + 4 \, a \sqrt {x}\right )} e^{\left (-\frac {5}{2}\right )}}{6 \, a^{3} c x^{2}}\right ] \end {gather*}

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*x^2*sqrt(b/a)*arctan(a*sqrt(b/a)/(b*sqrt(x))) - 3*b*x^2*sqrt(b/a)*log((b*x + 2*a*sqrt(x)*sqrt(b/a)
+ a)/(b*x - a)) + 4*a*sqrt(x))*e^(-5/2)/(a^3*c*x^2), -1/6*(6*b*x^2*sqrt(-b/a)*arctan(a*sqrt(-b/a)/(b*sqrt(x)))
- 3*b*x^2*sqrt(-b/a)*log((b*x + 2*a*sqrt(x)*sqrt(-b/a) - a)/(b*x + a)) + 4*a*sqrt(x))*e^(-5/2)/(a^3*c*x^2)]

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Sympy [C] Result contains complex when optimal does not.
time = 2.62, size = 371, normalized size = 3.47 \begin {gather*} \begin {cases} \frac {1}{5 a b c e^{\frac {5}{2}} x^{\frac {5}{2}}} - \frac {2}{3 a^{2} c e^{\frac {5}{2}} x^{\frac {3}{2}}} + \frac {b^{\frac {3}{2}} \operatorname {acoth}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{a^{\frac {7}{2}} c e^{\frac {5}{2}}} + \frac {b^{\frac {3}{2}} \operatorname {atan}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{a^{\frac {7}{2}} c e^{\frac {5}{2}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\- \frac {i \left (3 + 3 i\right )}{30 a b c e^{\frac {5}{2}} x^{\frac {5}{2}}} + \frac {3 + 3 i}{30 a b c e^{\frac {5}{2}} x^{\frac {5}{2}}} + \frac {-10 - 10 i}{30 a^{2} c e^{\frac {5}{2}} x^{\frac {3}{2}}} - \frac {i \left (-10 - 10 i\right )}{30 a^{2} c e^{\frac {5}{2}} x^{\frac {3}{2}}} - \frac {i b^{\frac {3}{2}} \cdot \left (15 + 15 i\right ) \operatorname {atan}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{30 a^{\frac {7}{2}} c e^{\frac {5}{2}}} + \frac {b^{\frac {3}{2}} \cdot \left (15 + 15 i\right ) \operatorname {atan}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{30 a^{\frac {7}{2}} c e^{\frac {5}{2}}} - \frac {i b^{\frac {3}{2}} \cdot \left (15 + 15 i\right ) \operatorname {atanh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{30 a^{\frac {7}{2}} c e^{\frac {5}{2}}} + \frac {b^{\frac {3}{2}} \cdot \left (15 + 15 i\right ) \operatorname {atanh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{30 a^{\frac {7}{2}} c e^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

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((1/(5*a*b*c*e**(5/2)*x**(5/2)) - 2/(3*a**2*c*e**(5/2)*x**(3/2)) + b**(3/2)*acoth(sqrt(b)*sqrt(x)/sqr
t(a))/(a**(7/2)*c*e**(5/2)) + b**(3/2)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(a**(7/2)*c*e**(5/2)), Abs(b*x/a) > 1), (
-I*(3 + 3*I)/(30*a*b*c*e**(5/2)*x**(5/2)) + (3 + 3*I)/(30*a*b*c*e**(5/2)*x**(5/2)) + (-10 - 10*I)/(30*a**2*c*e
**(5/2)*x**(3/2)) - I*(-10 - 10*I)/(30*a**2*c*e**(5/2)*x**(3/2)) - I*b**(3/2)*(15 + 15*I)*atan(sqrt(b)*sqrt(x)
/sqrt(a))/(30*a**(7/2)*c*e**(5/2)) + b**(3/2)*(15 + 15*I)*atan(sqrt(b)*sqrt(x)/sqrt(a))/(30*a**(7/2)*c*e**(5/2
)) - I*b**(3/2)*(15 + 15*I)*atanh(sqrt(b)*sqrt(x)/sqrt(a))/(30*a**(7/2)*c*e**(5/2)) + b**(3/2)*(15 + 15*I)*ata
nh(sqrt(b)*sqrt(x)/sqrt(a))/(30*a**(7/2)*c*e**(5/2)), True))

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Giac [A]
time = 1.47, size = 72, normalized size = 0.67 \begin {gather*} \frac {1}{3} \, {\left (\frac {3 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3} c} - \frac {3 \, b^{2} \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{\sqrt {-a b} a^{3} c} - \frac {2}{a^{2} c x^{\frac {3}{2}}}\right )} e^{\left (-\frac {5}{2}\right )} \end {gather*}

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]

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

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Mupad [B]
time = 0.16, size = 75, normalized size = 0.70 \begin {gather*} \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}} \end {gather*}

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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