3.2174 \(\int \frac{2+3 x}{(1-2 x)^{5/2} (3+5 x)} \, dx\)

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

[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

________________________________________________________________________________________

Rubi [A]  time = 0.0137696, antiderivative size = 56, normalized size of antiderivative = 1., 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 = {78, 51, 63, 206} \[ \frac{2}{121 \sqrt{1-2 x}}+\frac{7}{33 (1-2 x)^{3/2}}-\frac{2}{121} \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

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 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]))))

Rule 51

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 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 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{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} \operatorname{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*}

Mathematica [C]  time = 0.0130728, size = 38, normalized size = 0.68 \[ \frac{(6-12 x) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};-\frac{5}{11} (2 x-1)\right )+77}{363 (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(77 + (6 - 12*x)*Hypergeometric2F1[-1/2, 1, 1/2, (-5*(-1 + 2*x))/11])/(363*(1 - 2*x)^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.008, size = 38, normalized size = 0.7 \begin{align*}{\frac{7}{33} \left ( 1-2\,x \right ) ^{-{\frac{3}{2}}}}-{\frac{2\,\sqrt{55}}{1331}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) }+{\frac{2}{121}{\frac{1}{\sqrt{1-2\,x}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

________________________________________________________________________________________

Maxima [A]  time = 1.78161, size = 69, normalized size = 1.23 \begin{align*} \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{align*}

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)

________________________________________________________________________________________

Fricas [B]  time = 1.39125, size = 211, normalized size = 3.77 \begin{align*} \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{align*}

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)

________________________________________________________________________________________

Sympy [A]  time = 15.3294, size = 90, normalized size = 1.61 \begin{align*} \frac{10 \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 )}{121} + \frac{2}{121 \sqrt{1 - 2 x}} + \frac{7}{33 \left (1 - 2 x\right )^{\frac{3}{2}}} \end{align*}

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, 2*x - 1 < -11/5), (-sqrt(55)*atanh(sqrt(55)*sqrt(
1 - 2*x)/11)/55, 2*x - 1 > -11/5))/121 + 2/(121*sqrt(1 - 2*x)) + 7/(33*(1 - 2*x)**(3/2))

________________________________________________________________________________________

Giac [A]  time = 2.51851, size = 82, normalized size = 1.46 \begin{align*} \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{align*}

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))