3.1.28 \(\int \frac {1+a x}{x^3 \sqrt {a x} \sqrt {1-a x}} \, dx\)

Optimal. Leaf size=73 \[ -\frac {12 a^2 \sqrt {1-a x}}{5 \sqrt {a x}}-\frac {6 a^2 \sqrt {1-a x}}{5 (a x)^{3/2}}-\frac {2 a^2 \sqrt {1-a x}}{5 (a x)^{5/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {16, 78, 45, 37} \begin {gather*} -\frac {12 a^2 \sqrt {1-a x}}{5 \sqrt {a x}}-\frac {6 a^2 \sqrt {1-a x}}{5 (a x)^{3/2}}-\frac {2 a^2 \sqrt {1-a x}}{5 (a x)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + a*x)/(x^3*Sqrt[a*x]*Sqrt[1 - a*x]),x]

[Out]

(-2*a^2*Sqrt[1 - a*x])/(5*(a*x)^(5/2)) - (6*a^2*Sqrt[1 - a*x])/(5*(a*x)^(3/2)) - (12*a^2*Sqrt[1 - a*x])/(5*Sqr
t[a*x])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*Simplify[m + n + 2])/((b*c - a*d)*(m + 1)), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rubi steps

\begin {align*} \int \frac {1+a x}{x^3 \sqrt {a x} \sqrt {1-a x}} \, dx &=a^3 \int \frac {1+a x}{(a x)^{7/2} \sqrt {1-a x}} \, dx\\ &=-\frac {2 a^2 \sqrt {1-a x}}{5 (a x)^{5/2}}+\frac {1}{5} \left (9 a^3\right ) \int \frac {1}{(a x)^{5/2} \sqrt {1-a x}} \, dx\\ &=-\frac {2 a^2 \sqrt {1-a x}}{5 (a x)^{5/2}}-\frac {6 a^2 \sqrt {1-a x}}{5 (a x)^{3/2}}+\frac {1}{5} \left (6 a^3\right ) \int \frac {1}{(a x)^{3/2} \sqrt {1-a x}} \, dx\\ &=-\frac {2 a^2 \sqrt {1-a x}}{5 (a x)^{5/2}}-\frac {6 a^2 \sqrt {1-a x}}{5 (a x)^{3/2}}-\frac {12 a^2 \sqrt {1-a x}}{5 \sqrt {a x}}\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 37, normalized size = 0.51 \begin {gather*} -\frac {2 \sqrt {-a x (a x-1)} \left (6 a^2 x^2+3 a x+1\right )}{5 a x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + a*x)/(x^3*Sqrt[a*x]*Sqrt[1 - a*x]),x]

[Out]

(-2*Sqrt[-(a*x*(-1 + a*x))]*(1 + 3*a*x + 6*a^2*x^2))/(5*a*x^3)

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IntegrateAlgebraic [A]  time = 0.04, size = 55, normalized size = 0.75 \begin {gather*} -\frac {2 a^2 \sqrt {1-a x} \left (\frac {(1-a x)^2}{a^2 x^2}+\frac {5 (1-a x)}{a x}+10\right )}{5 \sqrt {a x}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(1 + a*x)/(x^3*Sqrt[a*x]*Sqrt[1 - a*x]),x]

[Out]

(-2*a^2*Sqrt[1 - a*x]*(10 + (5*(1 - a*x))/(a*x) + (1 - a*x)^2/(a^2*x^2)))/(5*Sqrt[a*x])

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fricas [A]  time = 1.33, size = 35, normalized size = 0.48 \begin {gather*} -\frac {2 \, {\left (6 \, a^{2} x^{2} + 3 \, a x + 1\right )} \sqrt {a x} \sqrt {-a x + 1}}{5 \, a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(6*a^2*x^2 + 3*a*x + 1)*sqrt(a*x)*sqrt(-a*x + 1)/(a*x^3)

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giac [B]  time = 1.24, size = 130, normalized size = 1.78 \begin {gather*} -\frac {\frac {a^{3} {\left (\sqrt {-a x + 1} - 1\right )}^{5}}{\left (a x\right )^{\frac {5}{2}}} + \frac {15 \, a^{3} {\left (\sqrt {-a x + 1} - 1\right )}^{3}}{\left (a x\right )^{\frac {3}{2}}} + \frac {110 \, a^{3} {\left (\sqrt {-a x + 1} - 1\right )}}{\sqrt {a x}} - \frac {{\left (a^{3} + \frac {15 \, a^{2} {\left (\sqrt {-a x + 1} - 1\right )}^{2}}{x} + \frac {110 \, a {\left (\sqrt {-a x + 1} - 1\right )}^{4}}{x^{2}}\right )} \left (a x\right )^{\frac {5}{2}}}{{\left (\sqrt {-a x + 1} - 1\right )}^{5}}}{80 \, a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/80*(a^3*(sqrt(-a*x + 1) - 1)^5/(a*x)^(5/2) + 15*a^3*(sqrt(-a*x + 1) - 1)^3/(a*x)^(3/2) + 110*a^3*(sqrt(-a*x
+ 1) - 1)/sqrt(a*x) - (a^3 + 15*a^2*(sqrt(-a*x + 1) - 1)^2/x + 110*a*(sqrt(-a*x + 1) - 1)^4/x^2)*(a*x)^(5/2)/
(sqrt(-a*x + 1) - 1)^5)/a

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maple [A]  time = 0.00, size = 33, normalized size = 0.45 \begin {gather*} -\frac {2 \left (6 a^{2} x^{2}+3 a x +1\right ) \sqrt {-a x +1}}{5 \sqrt {a x}\, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*(6*a^2*x^2+3*a*x+1)/x^2/(a*x)^(1/2)*(-a*x+1)^(1/2)

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maxima [A]  time = 0.96, size = 62, normalized size = 0.85 \begin {gather*} -\frac {12 \, \sqrt {-a^{2} x^{2} + a x} a}{5 \, x} - \frac {6 \, \sqrt {-a^{2} x^{2} + a x}}{5 \, x^{2}} - \frac {2 \, \sqrt {-a^{2} x^{2} + a x}}{5 \, a x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-12/5*sqrt(-a^2*x^2 + a*x)*a/x - 6/5*sqrt(-a^2*x^2 + a*x)/x^2 - 2/5*sqrt(-a^2*x^2 + a*x)/(a*x^3)

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mupad [B]  time = 2.73, size = 32, normalized size = 0.44 \begin {gather*} -\frac {\sqrt {1-a\,x}\,\left (\frac {12\,a^2\,x^2}{5}+\frac {6\,a\,x}{5}+\frac {2}{5}\right )}{x^2\,\sqrt {a\,x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((1 - a*x)^(1/2)*((6*a*x)/5 + (12*a^2*x^2)/5 + 2/5))/(x^2*(a*x)^(1/2))

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sympy [C]  time = 17.84, size = 189, normalized size = 2.59 \begin {gather*} a \left (\begin {cases} - \frac {4 a \sqrt {-1 + \frac {1}{a x}}}{3} - \frac {2 \sqrt {-1 + \frac {1}{a x}}}{3 x} & \text {for}\: \frac {1}{\left |{a x}\right |} > 1 \\- \frac {4 i a \sqrt {1 - \frac {1}{a x}}}{3} - \frac {2 i \sqrt {1 - \frac {1}{a x}}}{3 x} & \text {otherwise} \end {cases}\right ) + \begin {cases} - \frac {16 a^{2} \sqrt {-1 + \frac {1}{a x}}}{15} - \frac {8 a \sqrt {-1 + \frac {1}{a x}}}{15 x} - \frac {2 \sqrt {-1 + \frac {1}{a x}}}{5 x^{2}} & \text {for}\: \frac {1}{\left |{a x}\right |} > 1 \\- \frac {16 i a^{2} \sqrt {1 - \frac {1}{a x}}}{15} - \frac {8 i a \sqrt {1 - \frac {1}{a x}}}{15 x} - \frac {2 i \sqrt {1 - \frac {1}{a x}}}{5 x^{2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*Piecewise((-4*a*sqrt(-1 + 1/(a*x))/3 - 2*sqrt(-1 + 1/(a*x))/(3*x), 1/Abs(a*x) > 1), (-4*I*a*sqrt(1 - 1/(a*x)
)/3 - 2*I*sqrt(1 - 1/(a*x))/(3*x), True)) + Piecewise((-16*a**2*sqrt(-1 + 1/(a*x))/15 - 8*a*sqrt(-1 + 1/(a*x))
/(15*x) - 2*sqrt(-1 + 1/(a*x))/(5*x**2), 1/Abs(a*x) > 1), (-16*I*a**2*sqrt(1 - 1/(a*x))/15 - 8*I*a*sqrt(1 - 1/
(a*x))/(15*x) - 2*I*sqrt(1 - 1/(a*x))/(5*x**2), True))

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