3.1914 \(\int \frac {(1-2 x)^{3/2}}{(2+3 x) (3+5 x)^2} \, dx\)

Optimal. Leaf size=77 \[ -\frac {11 \sqrt {1-2 x}}{5 (5 x+3)}-14 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {72}{5} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

72/25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-14/3*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)-11/5*(1-
2*x)^(1/2)/(3+5*x)

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Rubi [A]  time = 0.02, antiderivative size = 77, normalized size of antiderivative = 1.00, 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} \[ -\frac {11 \sqrt {1-2 x}}{5 (5 x+3)}-14 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {72}{5} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*Sqrt[1 - 2*x])/(5*(3 + 5*x)) - 14*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + (72*Sqrt[11/5]*ArcTanh[Sqr
t[5/11]*Sqrt[1 - 2*x]])/5

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 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 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)^2} \, dx &=-\frac {11 \sqrt {1-2 x}}{5 (3+5 x)}-\frac {1}{5} \int \frac {57-37 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=-\frac {11 \sqrt {1-2 x}}{5 (3+5 x)}+49 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx-\frac {396}{5} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {11 \sqrt {1-2 x}}{5 (3+5 x)}-49 \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )+\frac {396}{5} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {11 \sqrt {1-2 x}}{5 (3+5 x)}-14 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {72}{5} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 76, normalized size = 0.99 \[ \frac {1}{25} \left (72 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )-\frac {55 \sqrt {1-2 x}}{5 x+3}\right )-14 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-14*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + ((-55*Sqrt[1 - 2*x])/(3 + 5*x) + 72*Sqrt[55]*ArcTanh[Sqrt[5/1
1]*Sqrt[1 - 2*x]])/25

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fricas [A]  time = 1.24, size = 102, normalized size = 1.32 \[ \frac {108 \, \sqrt {11} \sqrt {5} {\left (5 \, x + 3\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 175 \, \sqrt {7} \sqrt {3} {\left (5 \, x + 3\right )} \log \left (\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} + 3 \, x - 5}{3 \, x + 2}\right ) - 165 \, \sqrt {-2 \, x + 1}}{75 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/75*(108*sqrt(11)*sqrt(5)*(5*x + 3)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) + 175*sqrt(7)
*sqrt(3)*(5*x + 3)*log((sqrt(7)*sqrt(3)*sqrt(-2*x + 1) + 3*x - 5)/(3*x + 2)) - 165*sqrt(-2*x + 1))/(5*x + 3)

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giac [A]  time = 0.91, size = 95, normalized size = 1.23 \[ -\frac {36}{25} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) + \frac {7}{3} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {11 \, \sqrt {-2 \, x + 1}}{5 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-36/25*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 7/3*sqrt(21)*log
(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 11/5*sqrt(-2*x + 1)/(5*x + 3)

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maple [A]  time = 0.01, size = 54, normalized size = 0.70 \[ -\frac {14 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{3}+\frac {72 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{25}+\frac {22 \sqrt {-2 x +1}}{25 \left (-2 x -\frac {6}{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

22/25*(-2*x+1)^(1/2)/(-2*x-6/5)+72/25*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)-14/3*arctanh(1/7*21^(1/2)
*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.28, size = 89, normalized size = 1.16 \[ -\frac {36}{25} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {7}{3} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {11 \, \sqrt {-2 \, x + 1}}{5 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-36/25*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 7/3*sqrt(21)*log(-(sqrt(21
) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 11/5*sqrt(-2*x + 1)/(5*x + 3)

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mupad [B]  time = 0.10, size = 53, normalized size = 0.69 \[ \frac {72\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{25}-\frac {14\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{3}-\frac {22\,\sqrt {1-2\,x}}{25\,\left (2\,x+\frac {6}{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(72*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/25 - (14*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/3 -
(22*(1 - 2*x)^(1/2))/(25*(2*x + 6/5))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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