3.1903 \(\int \frac{(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)} \, dx\)

Optimal. Leaf size=77 \[ \frac{7 \sqrt{1-2 x}}{3 (3 x+2)}+\frac{64}{3} \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-22 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

[Out]

(7*Sqrt[1 - 2*x])/(3*(2 + 3*x)) + (64*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/3 - 22*Sqrt[11/5]*ArcTanh[Sq
rt[5/11]*Sqrt[1 - 2*x]]

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Rubi [A]  time = 0.0246948, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {98, 156, 63, 206} \[ \frac{7 \sqrt{1-2 x}}{3 (3 x+2)}+\frac{64}{3} \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-22 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Sqrt[1 - 2*x])/(3*(2 + 3*x)) + (64*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/3 - 22*Sqrt[11/5]*ArcTanh[Sq
rt[5/11]*Sqrt[1 - 2*x]]

Rule 98

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

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, 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{(1-2 x)^{3/2}}{(2+3 x)^2 (3+5 x)} \, dx &=\frac{7 \sqrt{1-2 x}}{3 (2+3 x)}+\frac{1}{3} \int \frac{54-31 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=\frac{7 \sqrt{1-2 x}}{3 (2+3 x)}-\frac{224}{3} \int \frac{1}{\sqrt{1-2 x} (2+3 x)} \, dx+121 \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=\frac{7 \sqrt{1-2 x}}{3 (2+3 x)}+\frac{224}{3} \operatorname{Subst}\left (\int \frac{1}{\frac{7}{2}-\frac{3 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )-121 \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=\frac{7 \sqrt{1-2 x}}{3 (2+3 x)}+\frac{64}{3} \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-22 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0563793, size = 75, normalized size = 0.97 \[ \frac{7 \sqrt{1-2 x}}{9 x+6}+\frac{64}{3} \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-22 \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Sqrt[1 - 2*x])/(6 + 9*x) + (64*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/3 - 22*Sqrt[11/5]*ArcTanh[Sqrt[5
/11]*Sqrt[1 - 2*x]]

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Maple [A]  time = 0.01, size = 54, normalized size = 0.7 \begin{align*} -{\frac{14}{9}\sqrt{1-2\,x} \left ( -2\,x-{\frac{4}{3}} \right ) ^{-1}}+{\frac{64\,\sqrt{21}}{9}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) }-{\frac{22\,\sqrt{55}}{5}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-14/9*(1-2*x)^(1/2)/(-2*x-4/3)+64/9*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-22/5*arctanh(1/11*55^(1/2)*(1
-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.81709, size = 120, normalized size = 1.56 \begin{align*} \frac{11}{5} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{32}{9} \, \sqrt{21} \log \left (-\frac{\sqrt{21} - 3 \, \sqrt{-2 \, x + 1}}{\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}}\right ) + \frac{7 \, \sqrt{-2 \, x + 1}}{3 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/5*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 32/9*sqrt(21)*log(-(sqrt(21)
- 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 7/3*sqrt(-2*x + 1)/(3*x + 2)

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Fricas [A]  time = 1.48089, size = 296, normalized size = 3.84 \begin{align*} \frac{99 \, \sqrt{11} \sqrt{5}{\left (3 \, x + 2\right )} \log \left (\frac{\sqrt{11} \sqrt{5} \sqrt{-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 160 \, \sqrt{7} \sqrt{3}{\left (3 \, x + 2\right )} \log \left (-\frac{\sqrt{7} \sqrt{3} \sqrt{-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 105 \, \sqrt{-2 \, x + 1}}{45 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/45*(99*sqrt(11)*sqrt(5)*(3*x + 2)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 160*sqrt(7)*s
qrt(3)*(3*x + 2)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) + 105*sqrt(-2*x + 1))/(3*x + 2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 2.03985, size = 128, normalized size = 1.66 \begin{align*} \frac{11}{5} \, \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{32}{9} \, \sqrt{21} \log \left (\frac{{\left | -2 \, \sqrt{21} + 6 \, \sqrt{-2 \, x + 1} \right |}}{2 \,{\left (\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}\right )}}\right ) + \frac{7 \, \sqrt{-2 \, x + 1}}{3 \,{\left (3 \, x + 2\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/5*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 32/9*sqrt(21)*log(
1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 7/3*sqrt(-2*x + 1)/(3*x + 2)