3.22.74 \(\int \frac {2+3 x}{(1-2 x)^{5/2} (3+5 x)} \, dx\) [2174]

Optimal. Leaf size=56 \[ \frac {7}{33 (1-2 x)^{3/2}}+\frac {2}{121 \sqrt {1-2 x}}-\frac {2}{121} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

7/33/(1-2*x)^(3/2)-2/1331*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+2/121/(1-2*x)^(1/2)

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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 = {79, 53, 65, 212} \begin {gather*} \frac {2}{121 \sqrt {1-2 x}}+\frac {7}{33 (1-2 x)^{3/2}}-\frac {2}{121} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

7/(33*(1 - 2*x)^(3/2)) + 2/(121*Sqrt[1 - 2*x]) - (2*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/121

Rule 53

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*((m + n + 2)/((b*c - a*d)*(m + 1))), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 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}{(1-2 x)^{5/2} (3+5 x)} \, dx &=\frac {7}{33 (1-2 x)^{3/2}}+\frac {1}{11} \int \frac {1}{(1-2 x)^{3/2} (3+5 x)} \, dx\\ &=\frac {7}{33 (1-2 x)^{3/2}}+\frac {2}{121 \sqrt {1-2 x}}+\frac {5}{121} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {7}{33 (1-2 x)^{3/2}}+\frac {2}{121 \sqrt {1-2 x}}-\frac {5}{121} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {7}{33 (1-2 x)^{3/2}}+\frac {2}{121 \sqrt {1-2 x}}-\frac {2}{121} \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 45, normalized size = 0.80 \begin {gather*} \frac {\frac {913-132 x}{(1-2 x)^{3/2}}-6 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3993} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((913 - 132*x)/(1 - 2*x)^(3/2) - 6*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/3993

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Maple [A]
time = 0.11, size = 38, normalized size = 0.68

method	result	size
derivativedivides	\(\frac {7}{33 \left (1-2 x \right )^{\frac {3}{2}}}-\frac {2 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1331}+\frac {2}{121 \sqrt {1-2 x}}\)	\(38\)
default	\(\frac {7}{33 \left (1-2 x \right )^{\frac {3}{2}}}-\frac {2 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{1331}+\frac {2}{121 \sqrt {1-2 x}}\)	\(38\)
trager	\(-\frac {\left (12 x -83\right ) \sqrt {1-2 x}}{363 \left (-1+2 x \right )^{2}}+\frac {\RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \RootOf \left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{1331}\)	\(67\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

7/33/(1-2*x)^(3/2)-2/1331*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+2/121/(1-2*x)^(1/2)

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Maxima [A]
time = 0.51, size = 51, normalized size = 0.91 \begin {gather*} \frac {1}{1331} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {12 \, x - 83}{363 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1331*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 1/363*(12*x - 83)/(-2*x +
1)^(3/2)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 75 vs. \(2 (37) = 74\).
time = 1.04, size = 75, normalized size = 1.34 \begin {gather*} \frac {3 \, \sqrt {11} \sqrt {5} {\left (4 \, x^{2} - 4 \, x + 1\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - 11 \, {\left (12 \, x - 83\right )} \sqrt {-2 \, x + 1}}{3993 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3993*(3*sqrt(11)*sqrt(5)*(4*x^2 - 4*x + 1)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) - 11*(
12*x - 83)*sqrt(-2*x + 1))/(4*x^2 - 4*x + 1)

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Sympy [A]
time = 12.92, size = 83, normalized size = 1.48 \begin {gather*} \frac {10 \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 )}{121} + \frac {2}{121 \sqrt {1 - 2 x}} + \frac {7}{33 \left (1 - 2 x\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

10*Piecewise((-sqrt(55)*acoth(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))/121 + 2/(121*sqrt(1 - 2*x)) + 7/(33*(1 - 2*x)**(3/2))

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Giac [A]
time = 1.79, size = 61, normalized size = 1.09 \begin {gather*} \frac {1}{1331} \, \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 \, x - 83}{363 \, {\left (2 \, x - 1\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)^(5/2)/(3+5*x),x, algorithm="giac")

[Out]

1/1331*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 1/363*(12*x - 83
)/((2*x - 1)*sqrt(-2*x + 1))

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Mupad [B]
time = 0.07, size = 33, normalized size = 0.59 \begin {gather*} -\frac {\frac {4\,x}{121}-\frac {83}{363}}{{\left (1-2\,x\right )}^{3/2}}-\frac {2\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{1331} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- ((4*x)/121 - 83/363)/(1 - 2*x)^(3/2) - (2*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/1331

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