3.17.92 \(\int \frac {\sqrt {1-2 x} (2+3 x)}{3+5 x} \, dx\)

Optimal. Leaf size=56 \[ -\frac {1}{5} (1-2 x)^{3/2}+\frac {2}{25} \sqrt {1-2 x}-\frac {2}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 56, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {80, 50, 63, 206} \begin {gather*} -\frac {1}{5} (1-2 x)^{3/2}+\frac {2}{25} \sqrt {1-2 x}-\frac {2}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[1 - 2*x]*(2 + 3*x))/(3 + 5*x),x]

[Out]

(2*Sqrt[1 - 2*x])/25 - (1 - 2*x)^(3/2)/5 - (2*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/25

Rule 50

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 80

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {\sqrt {1-2 x} (2+3 x)}{3+5 x} \, dx &=-\frac {1}{5} (1-2 x)^{3/2}+\frac {1}{5} \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx\\ &=\frac {2}{25} \sqrt {1-2 x}-\frac {1}{5} (1-2 x)^{3/2}+\frac {11}{25} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {2}{25} \sqrt {1-2 x}-\frac {1}{5} (1-2 x)^{3/2}-\frac {11}{25} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {2}{25} \sqrt {1-2 x}-\frac {1}{5} (1-2 x)^{3/2}-\frac {2}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 46, normalized size = 0.82 \begin {gather*} \frac {1}{125} \left (5 \sqrt {1-2 x} (10 x-3)-2 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[1 - 2*x]*(2 + 3*x))/(3 + 5*x),x]

[Out]

(5*Sqrt[1 - 2*x]*(-3 + 10*x) - 2*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/125

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IntegrateAlgebraic [A]  time = 0.05, size = 52, normalized size = 0.93 \begin {gather*} -\frac {1}{25} \sqrt {1-2 x} (5 (1-2 x)-2)-\frac {2}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[1 - 2*x]*(2 + 3*x))/(3 + 5*x),x]

[Out]

-1/25*((-2 + 5*(1 - 2*x))*Sqrt[1 - 2*x]) - (2*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/25

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fricas [A]  time = 1.18, size = 51, normalized size = 0.91 \begin {gather*} \frac {1}{125} \, \sqrt {11} \sqrt {5} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + \frac {1}{25} \, {\left (10 \, x - 3\right )} \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)*(1-2*x)^(1/2)/(3+5*x),x, algorithm="fricas")

[Out]

1/125*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 1/25*(10*x - 3)*sqrt(-2*x
+ 1)

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giac [A]  time = 1.20, size = 58, normalized size = 1.04 \begin {gather*} -\frac {1}{5} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {2}{25} \, \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)*(1-2*x)^(1/2)/(3+5*x),x, algorithm="giac")

[Out]

-1/5*(-2*x + 1)^(3/2) + 1/125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x +
1))) + 2/25*sqrt(-2*x + 1)

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maple [A]  time = 0.00, size = 38, normalized size = 0.68 \begin {gather*} -\frac {2 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{125}-\frac {\left (-2 x +1\right )^{\frac {3}{2}}}{5}+\frac {2 \sqrt {-2 x +1}}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x+2)*(-2*x+1)^(1/2)/(5*x+3),x)

[Out]

-1/5*(-2*x+1)^(3/2)-2/125*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)+2/25*(-2*x+1)^(1/2)

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maxima [A]  time = 1.20, size = 55, normalized size = 0.98 \begin {gather*} -\frac {1}{5} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {1}{125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {2}{25} \, \sqrt {-2 \, x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)*(1-2*x)^(1/2)/(3+5*x),x, algorithm="maxima")

[Out]

-1/5*(-2*x + 1)^(3/2) + 1/125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 2/2
5*sqrt(-2*x + 1)

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mupad [B]  time = 0.07, size = 37, normalized size = 0.66 \begin {gather*} \frac {2\,\sqrt {1-2\,x}}{25}-\frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{125}-\frac {{\left (1-2\,x\right )}^{3/2}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1 - 2*x)^(1/2)*(3*x + 2))/(5*x + 3),x)

[Out]

(2*(1 - 2*x)^(1/2))/25 - (2*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/125 - (1 - 2*x)^(3/2)/5

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sympy [A]  time = 5.36, size = 88, normalized size = 1.57 \begin {gather*} - \frac {\left (1 - 2 x\right )^{\frac {3}{2}}}{5} + \frac {2 \sqrt {1 - 2 x}}{25} + \frac {22 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 < - \frac {11}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 > - \frac {11}{5} \end {cases}\right )}{25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)*(1-2*x)**(1/2)/(3+5*x),x)

[Out]

-(1 - 2*x)**(3/2)/5 + 2*sqrt(1 - 2*x)/25 + 22*Piecewise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x -
1 < -11/5), (-sqrt(55)*atanh(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 > -11/5))/25

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