3.21.52 \(\int \frac {2+3 x}{\sqrt {1-2 x} (3+5 x)^2} \, dx\) [2052]

Optimal. Leaf size=48 \[ -\frac {\sqrt {1-2 x}}{55 (3+5 x)}-\frac {68 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}} \]

[Out]

-68/3025*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1/55*(1-2*x)^(1/2)/(3+5*x)

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Rubi [A]
time = 0.01, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {79, 65, 212} \begin {gather*} -\frac {\sqrt {1-2 x}}{55 (5 x+3)}-\frac {68 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/55*Sqrt[1 - 2*x]/(3 + 5*x) - (68*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(55*Sqrt[55])

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 79

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] || L
tQ[p, n]))))

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 {2+3 x}{\sqrt {1-2 x} (3+5 x)^2} \, dx &=-\frac {\sqrt {1-2 x}}{55 (3+5 x)}+\frac {34}{55} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {\sqrt {1-2 x}}{55 (3+5 x)}-\frac {34}{55} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {\sqrt {1-2 x}}{55 (3+5 x)}-\frac {68 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}}\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 48, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {1-2 x}}{55 (3+5 x)}-\frac {68 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{55 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/55*Sqrt[1 - 2*x]/(3 + 5*x) - (68*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(55*Sqrt[55])

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Maple [A]
time = 0.11, size = 36, normalized size = 0.75

method	result	size
derivativedivides	\(\frac {2 \sqrt {1-2 x}}{275 \left (-\frac {6}{5}-2 x \right )}-\frac {68 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3025}\)	\(36\)
default	\(\frac {2 \sqrt {1-2 x}}{275 \left (-\frac {6}{5}-2 x \right )}-\frac {68 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3025}\)	\(36\)
risch	\(\frac {-1+2 x}{55 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {68 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{3025}\)	\(41\)
trager	\(-\frac {\sqrt {1-2 x}}{55 \left (3+5 x \right )}-\frac {34 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x -8 \RootOf \left (\textit {\_Z}^{2}-55\right )-55 \sqrt {1-2 x}}{3+5 x}\right )}{3025}\)	\(63\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/275*(1-2*x)^(1/2)/(-6/5-2*x)-68/3025*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]
time = 0.49, size = 53, normalized size = 1.10 \begin {gather*} \frac {34}{3025} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {\sqrt {-2 \, x + 1}}{55 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

34/3025*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/55*sqrt(-2*x + 1)/(5*x
+ 3)

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Fricas [A]
time = 0.81, size = 54, normalized size = 1.12 \begin {gather*} \frac {34 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, \sqrt {-2 \, x + 1}}{3025 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3025*(34*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*sqrt(-2*x + 1))/(5*x + 3
)

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 122 vs. \(2 (39) = 78\).
time = 223.96, size = 190, normalized size = 3.96 \begin {gather*} \frac {4 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (-1 + \frac {\sqrt {55}}{5 \sqrt {1 - 2 x}} \right )}}{4} + \frac {\log {\left (1 + \frac {\sqrt {55}}{5 \sqrt {1 - 2 x}} \right )}}{4} - \frac {1}{4 \cdot \left (1 + \frac {\sqrt {55}}{5 \sqrt {1 - 2 x}}\right )} - \frac {1}{4 \left (-1 + \frac {\sqrt {55}}{5 \sqrt {1 - 2 x}}\right )}\right )}{275} & \text {for}\: \frac {1}{\sqrt {1 - 2 x}} > - \frac {\sqrt {55}}{11} \wedge \frac {1}{\sqrt {1 - 2 x}} < \frac {\sqrt {55}}{11} \end {cases}\right )}{11} + \frac {14 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55}}{5 \sqrt {1 - 2 x}} \right )}}{55} & \text {for}\: \frac {1}{1 - 2 x} > \frac {5}{11} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55}}{5 \sqrt {1 - 2 x}} \right )}}{55} & \text {for}\: \frac {1}{1 - 2 x} < \frac {5}{11} \end {cases}\right )}{11} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*Piecewise((sqrt(55)*(-log(-1 + sqrt(55)/(5*sqrt(1 - 2*x)))/4 + log(1 + sqrt(55)/(5*sqrt(1 - 2*x)))/4 - 1/(4*
(1 + sqrt(55)/(5*sqrt(1 - 2*x)))) - 1/(4*(-1 + sqrt(55)/(5*sqrt(1 - 2*x)))))/275, (1/sqrt(1 - 2*x) > -sqrt(55)
/11) & (1/sqrt(1 - 2*x) < sqrt(55)/11)))/11 + 14*Piecewise((-sqrt(55)*acoth(sqrt(55)/(5*sqrt(1 - 2*x)))/55, 1/
(1 - 2*x) > 5/11), (-sqrt(55)*atanh(sqrt(55)/(5*sqrt(1 - 2*x)))/55, 1/(1 - 2*x) < 5/11))/11

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Giac [A]
time = 1.71, size = 56, normalized size = 1.17 \begin {gather*} \frac {34}{3025} \, \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 {\sqrt {-2 \, x + 1}}{55 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

34/3025*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/55*sqrt(-2*x
+ 1)/(5*x + 3)

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Mupad [B]
time = 0.06, size = 35, normalized size = 0.73 \begin {gather*} -\frac {68\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{3025}-\frac {2\,\sqrt {1-2\,x}}{275\,\left (2\,x+\frac {6}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (68*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/3025 - (2*(1 - 2*x)^(1/2))/(275*(2*x + 6/5))

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