3.19.41 \(\int \frac {\sqrt {1-2 x} (2+3 x)^2}{(3+5 x)^2} \, dx\) [1841]

Optimal. Leaf size=74 \[ \frac {26}{275} \sqrt {1-2 x}-\frac {3}{25} (1-2 x)^{3/2}-\frac {(1-2 x)^{3/2}}{275 (3+5 x)}-\frac {26 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \]

[Out]

-3/25*(1-2*x)^(3/2)-1/275*(1-2*x)^(3/2)/(3+5*x)-26/1375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+26/275*(
1-2*x)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {91, 81, 52, 65, 212} \begin {gather*} -\frac {(1-2 x)^{3/2}}{275 (5 x+3)}-\frac {3}{25} (1-2 x)^{3/2}+\frac {26}{275} \sqrt {1-2 x}-\frac {26 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(26*Sqrt[1 - 2*x])/275 - (3*(1 - 2*x)^(3/2))/25 - (1 - 2*x)^(3/2)/(275*(3 + 5*x)) - (26*ArcTanh[Sqrt[5/11]*Sqr
t[1 - 2*x]])/(25*Sqrt[55])

Rule 52

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 65

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 81

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 91

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*c - a*d
)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d*e - c*f)*(n + 1))), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 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)^2}{(3+5 x)^2} \, dx &=-\frac {(1-2 x)^{3/2}}{275 (3+5 x)}+\frac {1}{275} \int \frac {\sqrt {1-2 x} (362+495 x)}{3+5 x} \, dx\\ &=-\frac {3}{25} (1-2 x)^{3/2}-\frac {(1-2 x)^{3/2}}{275 (3+5 x)}+\frac {13}{55} \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx\\ &=\frac {26}{275} \sqrt {1-2 x}-\frac {3}{25} (1-2 x)^{3/2}-\frac {(1-2 x)^{3/2}}{275 (3+5 x)}+\frac {13}{25} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {26}{275} \sqrt {1-2 x}-\frac {3}{25} (1-2 x)^{3/2}-\frac {(1-2 x)^{3/2}}{275 (3+5 x)}-\frac {13}{25} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {26}{275} \sqrt {1-2 x}-\frac {3}{25} (1-2 x)^{3/2}-\frac {(1-2 x)^{3/2}}{275 (3+5 x)}-\frac {26 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 58, normalized size = 0.78 \begin {gather*} \frac {\sqrt {1-2 x} \left (-2+15 x+30 x^2\right )}{25 (3+5 x)}-\frac {26 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{25 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(-2 + 15*x + 30*x^2))/(25*(3 + 5*x)) - (26*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(25*Sqrt[55])

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Maple [A]
time = 0.10, size = 54, normalized size = 0.73

method	result	size
risch	\(-\frac {60 x^{3}-19 x +2}{25 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {26 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}\)	\(46\)
derivativedivides	\(-\frac {3 \left (1-2 x \right )^{\frac {3}{2}}}{25}+\frac {12 \sqrt {1-2 x}}{125}+\frac {2 \sqrt {1-2 x}}{625 \left (-\frac {6}{5}-2 x \right )}-\frac {26 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}\)	\(54\)
default	\(-\frac {3 \left (1-2 x \right )^{\frac {3}{2}}}{25}+\frac {12 \sqrt {1-2 x}}{125}+\frac {2 \sqrt {1-2 x}}{625 \left (-\frac {6}{5}-2 x \right )}-\frac {26 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1375}\)	\(54\)
trager	\(\frac {\left (30 x^{2}+15 x -2\right ) \sqrt {1-2 x}}{75+125 x}-\frac {13 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x -8 \RootOf \left (\textit {\_Z}^{2}-55\right )-55 \sqrt {1-2 x}}{3+5 x}\right )}{1375}\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/25*(1-2*x)^(3/2)+12/125*(1-2*x)^(1/2)+2/625*(1-2*x)^(1/2)/(-6/5-2*x)-26/1375*arctanh(1/11*55^(1/2)*(1-2*x)^
(1/2))*55^(1/2)

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Maxima [A]
time = 0.50, size = 71, normalized size = 0.96 \begin {gather*} -\frac {3}{25} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {13}{1375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {12}{125} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{125 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/25*(-2*x + 1)^(3/2) + 13/1375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) +
12/125*sqrt(-2*x + 1) - 1/125*sqrt(-2*x + 1)/(5*x + 3)

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Fricas [A]
time = 0.79, size = 64, normalized size = 0.86 \begin {gather*} \frac {13 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (30 \, x^{2} + 15 \, x - 2\right )} \sqrt {-2 \, x + 1}}{1375 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1375*(13*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(30*x^2 + 15*x - 2)*sqrt
(-2*x + 1))/(5*x + 3)

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Sympy [A]
time = 53.47, size = 204, normalized size = 2.76 \begin {gather*} - \frac {3 \left (1 - 2 x\right )^{\frac {3}{2}}}{25} + \frac {12 \sqrt {1 - 2 x}}{125} - \frac {44 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{125} + \frac {128 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: x < - \frac {3}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: x > - \frac {3}{5} \end {cases}\right )}{125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3*(1 - 2*x)**(3/2)/25 + 12*sqrt(1 - 2*x)/125 - 44*Piecewise((sqrt(55)*(-log(sqrt(55)*sqrt(1 - 2*x)/11 - 1)/4
+ log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/4 - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/
11 - 1)))/605, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/125 + 128*Piecewise((-sqrt(55)*a
coth(sqrt(55)*sqrt(1 - 2*x)/11)/55, x < -3/5), (-sqrt(55)*atanh(sqrt(55)*sqrt(1 - 2*x)/11)/55, x > -3/5))/125

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Giac [A]
time = 1.81, size = 74, normalized size = 1.00 \begin {gather*} -\frac {3}{25} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {13}{1375} \, \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 {12}{125} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{125 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/25*(-2*x + 1)^(3/2) + 13/1375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x
+ 1))) + 12/125*sqrt(-2*x + 1) - 1/125*sqrt(-2*x + 1)/(5*x + 3)

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Mupad [B]
time = 0.07, size = 55, normalized size = 0.74 \begin {gather*} \frac {12\,\sqrt {1-2\,x}}{125}-\frac {2\,\sqrt {1-2\,x}}{625\,\left (2\,x+\frac {6}{5}\right )}-\frac {3\,{\left (1-2\,x\right )}^{3/2}}{25}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,26{}\mathrm {i}}{1375} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*26i)/1375 - (2*(1 - 2*x)^(1/2))/(625*(2*x + 6/5)) + (12*(1 -
2*x)^(1/2))/125 - (3*(1 - 2*x)^(3/2))/25

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