3.1996 \(\int \frac {(1-2 x)^{5/2}}{(2+3 x) (3+5 x)^3} \, dx\)

Optimal. Leaf size=97 \[ -\frac {11 (1-2 x)^{3/2}}{10 (5 x+3)^2}+\frac {803 \sqrt {1-2 x}}{50 (5 x+3)}+98 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {2523}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

-11/10*(1-2*x)^(3/2)/(3+5*x)^2-2523/125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+98/3*arctanh(1/7*21^(1/2
)*(1-2*x)^(1/2))*21^(1/2)+803/50*(1-2*x)^(1/2)/(3+5*x)

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Rubi [A]  time = 0.03, antiderivative size = 97, 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} \[ -\frac {11 (1-2 x)^{3/2}}{10 (5 x+3)^2}+\frac {803 \sqrt {1-2 x}}{50 (5 x+3)}+98 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {2523}{25} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-11*(1 - 2*x)^(3/2))/(10*(3 + 5*x)^2) + (803*Sqrt[1 - 2*x])/(50*(3 + 5*x)) + 98*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*S
qrt[1 - 2*x]] - (2523*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/25

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+5 x)^3} \, dx &=-\frac {11 (1-2 x)^{3/2}}{10 (3+5 x)^2}-\frac {1}{10} \int \frac {(136-41 x) \sqrt {1-2 x}}{(2+3 x) (3+5 x)^2} \, dx\\ &=-\frac {11 (1-2 x)^{3/2}}{10 (3+5 x)^2}+\frac {803 \sqrt {1-2 x}}{50 (3+5 x)}-\frac {1}{50} \int \frac {-4056+2491 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=-\frac {11 (1-2 x)^{3/2}}{10 (3+5 x)^2}+\frac {803 \sqrt {1-2 x}}{50 (3+5 x)}-343 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+\frac {27753}{50} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {11 (1-2 x)^{3/2}}{10 (3+5 x)^2}+\frac {803 \sqrt {1-2 x}}{50 (3+5 x)}+343 \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-\frac {27753}{50} \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {11 (1-2 x)^{3/2}}{10 (3+5 x)^2}+\frac {803 \sqrt {1-2 x}}{50 (3+5 x)}+98 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-\frac {2523}{25} \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 = 0.15, size = 81, normalized size = 0.84 \[ 98 \sqrt {\frac {7}{3}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )+\frac {1}{250} \left (\frac {55 \sqrt {1-2 x} (375 x+214)}{(5 x+3)^2}-5046 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

98*Sqrt[7/3]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]] + ((55*Sqrt[1 - 2*x]*(214 + 375*x))/(3 + 5*x)^2 - 5046*Sqrt[55]*
ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/250

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fricas [A]  time = 0.93, size = 122, normalized size = 1.26 \[ \frac {7569 \, \sqrt {11} \sqrt {5} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 12250 \, \sqrt {7} \sqrt {3} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 165 \, {\left (375 \, x + 214\right )} \sqrt {-2 \, x + 1}}{750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/750*(7569*sqrt(11)*sqrt(5)*(25*x^2 + 30*x + 9)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) +
12250*sqrt(7)*sqrt(3)*(25*x^2 + 30*x + 9)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) + 165*(37
5*x + 214)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 0.95, size = 107, normalized size = 1.10 \[ \frac {2523}{250} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {49}{3} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {11 \, {\left (375 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 803 \, \sqrt {-2 \, x + 1}\right )}}{100 \, {\left (5 \, x + 3\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2523/250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 49/3*sqrt(21)*
log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 11/100*(375*(-2*x + 1)^(3/2) - 80
3*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 66, normalized size = 0.68 \[ \frac {98 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{3}-\frac {2523 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{125}+\frac {-165 \left (-2 x +1\right )^{\frac {3}{2}}+\frac {8833 \sqrt {-2 x +1}}{25}}{\left (-10 x -6\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

550*(-3/10*(-2*x+1)^(3/2)+803/1250*(-2*x+1)^(1/2))/(-10*x-6)^2-2523/125*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*
55^(1/2)+98/3*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.28, size = 110, normalized size = 1.13 \[ \frac {2523}{250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {49}{3} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {11 \, {\left (375 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 803 \, \sqrt {-2 \, x + 1}\right )}}{25 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2523/250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 49/3*sqrt(21)*log(-(sqrt
(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 11/25*(375*(-2*x + 1)^(3/2) - 803*sqrt(-2*x + 1))/(2
5*(2*x - 1)^2 + 220*x + 11)

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mupad [B]  time = 1.22, size = 71, normalized size = 0.73 \[ \frac {98\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{3}-\frac {2523\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{125}+\frac {\frac {8833\,\sqrt {1-2\,x}}{625}-\frac {33\,{\left (1-2\,x\right )}^{3/2}}{5}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(98*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/3 - (2523*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/125
+ ((8833*(1 - 2*x)^(1/2))/625 - (33*(1 - 2*x)^(3/2))/5)/((44*x)/5 + (2*x - 1)^2 + 11/25)

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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+5*x)**3,x)

[Out]

Timed out

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