3.1975 \(\int \frac {(1-2 x)^{5/2}}{(2+3 x)^3 (3+5 x)} \, dx\)

Optimal. Leaf size=95 \[ \frac {7 (1-2 x)^{3/2}}{6 (3 x+2)^2}+\frac {49 \sqrt {1-2 x}}{2 (3 x+2)}+235 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-242 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

7/6*(1-2*x)^(3/2)/(2+3*x)^2-242/5*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+235/3*arctanh(1/7*21^(1/2)*(1-
2*x)^(1/2))*21^(1/2)+49/2*(1-2*x)^(1/2)/(2+3*x)

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Rubi [A]  time = 0.04, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {98, 149, 156, 63, 206} \[ \frac {7 (1-2 x)^{3/2}}{6 (3 x+2)^2}+\frac {49 \sqrt {1-2 x}}{2 (3 x+2)}+235 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-242 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*(1 - 2*x)^(3/2))/(6*(2 + 3*x)^2) + (49*Sqrt[1 - 2*x])/(2*(2 + 3*x)) + 235*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[
1 - 2*x]] - 242*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*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 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 149

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(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 - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegerQ[m]

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)^{5/2}}{(2+3 x)^3 (3+5 x)} \, dx &=\frac {7 (1-2 x)^{3/2}}{6 (2+3 x)^2}+\frac {1}{6} \int \frac {(129-27 x) \sqrt {1-2 x}}{(2+3 x)^2 (3+5 x)} \, dx\\ &=\frac {7 (1-2 x)^{3/2}}{6 (2+3 x)^2}+\frac {49 \sqrt {1-2 x}}{2 (2+3 x)}-\frac {1}{18} \int \frac {-3501+2151 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=\frac {7 (1-2 x)^{3/2}}{6 (2+3 x)^2}+\frac {49 \sqrt {1-2 x}}{2 (2+3 x)}-\frac {1645}{2} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+1331 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {7 (1-2 x)^{3/2}}{6 (2+3 x)^2}+\frac {49 \sqrt {1-2 x}}{2 (2+3 x)}+\frac {1645}{2} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-1331 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {7 (1-2 x)^{3/2}}{6 (2+3 x)^2}+\frac {49 \sqrt {1-2 x}}{2 (2+3 x)}+235 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-242 \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 = 0.09, size = 80, normalized size = 0.84 \[ \frac {7 \sqrt {1-2 x} (61 x+43)}{6 (3 x+2)^2}+235 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-242 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Sqrt[1 - 2*x]*(43 + 61*x))/(6*(2 + 3*x)^2) + 235*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] - 242*Sqrt[11/5
]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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fricas [A]  time = 1.02, size = 122, normalized size = 1.28 \[ \frac {726 \, \sqrt {11} \sqrt {5} {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 1175 \, \sqrt {7} \sqrt {3} {\left (9 \, x^{2} + 12 \, x + 4\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 35 \, {\left (61 \, x + 43\right )} \sqrt {-2 \, x + 1}}{30 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/30*(726*sqrt(11)*sqrt(5)*(9*x^2 + 12*x + 4)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 117
5*sqrt(7)*sqrt(3)*(9*x^2 + 12*x + 4)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) + 35*(61*x + 4
3)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)

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giac [A]  time = 1.12, size = 107, normalized size = 1.13 \[ \frac {121}{5} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {235}{6} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {7 \, {\left (61 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 147 \, \sqrt {-2 \, x + 1}\right )}}{12 \, {\left (3 \, x + 2\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

121/5*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 235/6*sqrt(21)*lo
g(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 7/12*(61*(-2*x + 1)^(3/2) - 147*sqr
t(-2*x + 1))/(3*x + 2)^2

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maple [A]  time = 0.01, size = 66, normalized size = 0.69 \[ \frac {235 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{3}-\frac {242 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{5}-\frac {126 \left (\frac {61 \left (-2 x +1\right )^{\frac {3}{2}}}{54}-\frac {49 \sqrt {-2 x +1}}{18}\right )}{\left (-6 x -4\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-242/5*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)-126*(61/54*(-2*x+1)^(3/2)-49/18*(-2*x+1)^(1/2))/(-6*x-4)
^2+235/3*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.36, size = 110, normalized size = 1.16 \[ \frac {121}{5} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {235}{6} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {7 \, {\left (61 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 147 \, \sqrt {-2 \, x + 1}\right )}}{3 \, {\left (9 \, {\left (2 \, x - 1\right )}^{2} + 84 \, x + 7\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

121/5*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 235/6*sqrt(21)*log(-(sqrt(2
1) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 7/3*(61*(-2*x + 1)^(3/2) - 147*sqrt(-2*x + 1))/(9*(2*x
- 1)^2 + 84*x + 7)

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mupad [B]  time = 1.23, size = 71, normalized size = 0.75 \[ \frac {235\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{3}-\frac {242\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{5}+\frac {\frac {343\,\sqrt {1-2\,x}}{9}-\frac {427\,{\left (1-2\,x\right )}^{3/2}}{27}}{\frac {28\,x}{3}+{\left (2\,x-1\right )}^2+\frac {7}{9}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(235*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/3 - (242*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/5 +
((343*(1 - 2*x)^(1/2))/9 - (427*(1 - 2*x)^(3/2))/27)/((28*x)/3 + (2*x - 1)^2 + 7/9)

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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)**(5/2)/(2+3*x)**3/(3+5*x),x)

[Out]

Timed out

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