3.1902 \(\int \frac{(1-2 x)^{3/2}}{(2+3 x) (3+5 x)} \, dx\)

Optimal. Leaf size=72 \[ -\frac{4}{15} \sqrt{1-2 x}+\frac{14}{3} \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{22}{5} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

[Out]

(-4*Sqrt[1 - 2*x])/15 + (14*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/3 - (22*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*
Sqrt[1 - 2*x]])/5

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Rubi [A]  time = 0.0233582, antiderivative size = 72, 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 = {84, 156, 63, 206} \[ -\frac{4}{15} \sqrt{1-2 x}+\frac{14}{3} \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{22}{5} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*Sqrt[1 - 2*x])/15 + (14*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/3 - (22*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*
Sqrt[1 - 2*x]])/5

Rule 84

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

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) (3+5 x)} \, dx &=-\frac{4}{15} \sqrt{1-2 x}+\frac{1}{15} \int \frac{-9-136 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=-\frac{4}{15} \sqrt{1-2 x}-\frac{49}{3} \int \frac{1}{\sqrt{1-2 x} (2+3 x)} \, dx+\frac{121}{5} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=-\frac{4}{15} \sqrt{1-2 x}+\frac{49}{3} \operatorname{Subst}\left (\int \frac{1}{\frac{7}{2}-\frac{3 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )-\frac{121}{5} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{4}{15} \sqrt{1-2 x}+\frac{14}{3} \sqrt{\frac{7}{3}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{22}{5} \sqrt{\frac{11}{5}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0190875, size = 66, normalized size = 0.92 \[ -\frac{2}{225} \left (30 \sqrt{1-2 x}-175 \sqrt{21} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )+99 \sqrt{55} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(30*Sqrt[1 - 2*x] - 175*Sqrt[21]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + 99*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1
- 2*x]]))/225

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Maple [A]  time = 0.008, size = 47, normalized size = 0.7 \begin{align*} -{\frac{22\,\sqrt{55}}{25}{\it Artanh} \left ({\frac{\sqrt{55}}{11}\sqrt{1-2\,x}} \right ) }+{\frac{14\,\sqrt{21}}{9}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) }-{\frac{4}{15}\sqrt{1-2\,x}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-22/25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+14/9*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-4/15*(1
-2*x)^(1/2)

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Maxima [A]  time = 1.90942, size = 111, normalized size = 1.54 \begin{align*} \frac{11}{25} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{7}{9} \, \sqrt{21} \log \left (-\frac{\sqrt{21} - 3 \, \sqrt{-2 \, x + 1}}{\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}}\right ) - \frac{4}{15} \, \sqrt{-2 \, x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/25*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 7/9*sqrt(21)*log(-(sqrt(21)
- 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 4/15*sqrt(-2*x + 1)

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Fricas [A]  time = 1.38648, size = 251, normalized size = 3.49 \begin{align*} \frac{11}{25} \, \sqrt{11} \sqrt{5} \log \left (\frac{\sqrt{11} \sqrt{5} \sqrt{-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + \frac{7}{9} \, \sqrt{7} \sqrt{3} \log \left (-\frac{\sqrt{7} \sqrt{3} \sqrt{-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) - \frac{4}{15} \, \sqrt{-2 \, x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/25*sqrt(11)*sqrt(5)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 7/9*sqrt(7)*sqrt(3)*log(-(
sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) - 4/15*sqrt(-2*x + 1)

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Sympy [A]  time = 11.9247, size = 146, normalized size = 2.03 \begin{align*} - \frac{4 \sqrt{1 - 2 x}}{15} - \frac{98 \left (\begin{cases} - \frac{\sqrt{21} \operatorname{acoth}{\left (\frac{\sqrt{21} \sqrt{1 - 2 x}}{7} \right )}}{21} & \text{for}\: 2 x - 1 < - \frac{7}{3} \\- \frac{\sqrt{21} \operatorname{atanh}{\left (\frac{\sqrt{21} \sqrt{1 - 2 x}}{7} \right )}}{21} & \text{for}\: 2 x - 1 > - \frac{7}{3} \end{cases}\right )}{3} + \frac{242 \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 )}{5} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*sqrt(1 - 2*x)/15 - 98*Piecewise((-sqrt(21)*acoth(sqrt(21)*sqrt(1 - 2*x)/7)/21, 2*x - 1 < -7/3), (-sqrt(21)*
atanh(sqrt(21)*sqrt(1 - 2*x)/7)/21, 2*x - 1 > -7/3))/3 + 242*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))/5

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Giac [A]  time = 2.50086, size = 119, normalized size = 1.65 \begin{align*} \frac{11}{25} \, \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{7}{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{4}{15} \, \sqrt{-2 \, x + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

11/25*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 7/9*sqrt(21)*log(
1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 4/15*sqrt(-2*x + 1)