3.2198 \(\int \frac{1}{(1-2 x)^{5/2} (2+3 x)^2 (3+5 x)^3} \, dx\)

Optimal. Leaf size=152 \[ -\frac{7554245}{5021863 \sqrt{1-2 x}}+\frac{32765}{1694 (1-2 x)^{3/2} (5 x+3)}-\frac{667615}{195657 (1-2 x)^{3/2}}+\frac{3}{7 (1-2 x)^{3/2} (3 x+2) (5 x+3)^2}-\frac{505}{154 (1-2 x)^{3/2} (5 x+3)^2}+\frac{17820}{343} \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{738625 \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{14641} \]

[Out]

-667615/(195657*(1 - 2*x)^(3/2)) - 7554245/(5021863*Sqrt[1 - 2*x]) - 505/(154*(1 - 2*x)^(3/2)*(3 + 5*x)^2) + 3
/(7*(1 - 2*x)^(3/2)*(2 + 3*x)*(3 + 5*x)^2) + 32765/(1694*(1 - 2*x)^(3/2)*(3 + 5*x)) + (17820*Sqrt[3/7]*ArcTanh
[Sqrt[3/7]*Sqrt[1 - 2*x]])/343 - (738625*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/14641

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Rubi [A]  time = 0.068985, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 6, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {103, 151, 152, 156, 63, 206} \[ -\frac{7554245}{5021863 \sqrt{1-2 x}}+\frac{32765}{1694 (1-2 x)^{3/2} (5 x+3)}-\frac{667615}{195657 (1-2 x)^{3/2}}+\frac{3}{7 (1-2 x)^{3/2} (3 x+2) (5 x+3)^2}-\frac{505}{154 (1-2 x)^{3/2} (5 x+3)^2}+\frac{17820}{343} \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{738625 \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{14641} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-667615/(195657*(1 - 2*x)^(3/2)) - 7554245/(5021863*Sqrt[1 - 2*x]) - 505/(154*(1 - 2*x)^(3/2)*(3 + 5*x)^2) + 3
/(7*(1 - 2*x)^(3/2)*(2 + 3*x)*(3 + 5*x)^2) + 32765/(1694*(1 - 2*x)^(3/2)*(3 + 5*x)) + (17820*Sqrt[3/7]*ArcTanh
[Sqrt[3/7]*Sqrt[1 - 2*x]])/343 - (738625*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/14641

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 151

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 + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 152

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 + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

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}{(1-2 x)^{5/2} (2+3 x)^2 (3+5 x)^3} \, dx &=\frac{3}{7 (1-2 x)^{3/2} (2+3 x) (3+5 x)^2}+\frac{1}{7} \int \frac{20-135 x}{(1-2 x)^{5/2} (2+3 x) (3+5 x)^3} \, dx\\ &=-\frac{505}{154 (1-2 x)^{3/2} (3+5 x)^2}+\frac{3}{7 (1-2 x)^{3/2} (2+3 x) (3+5 x)^2}-\frac{1}{154} \int \frac{190-10605 x}{(1-2 x)^{5/2} (2+3 x) (3+5 x)^2} \, dx\\ &=-\frac{505}{154 (1-2 x)^{3/2} (3+5 x)^2}+\frac{3}{7 (1-2 x)^{3/2} (2+3 x) (3+5 x)^2}+\frac{32765}{1694 (1-2 x)^{3/2} (3+5 x)}+\frac{\int \frac{-88070-491475 x}{(1-2 x)^{5/2} (2+3 x) (3+5 x)} \, dx}{1694}\\ &=-\frac{667615}{195657 (1-2 x)^{3/2}}-\frac{505}{154 (1-2 x)^{3/2} (3+5 x)^2}+\frac{3}{7 (1-2 x)^{3/2} (2+3 x) (3+5 x)^2}+\frac{32765}{1694 (1-2 x)^{3/2} (3+5 x)}-\frac{\int \frac{-1844985+\frac{30042675 x}{2}}{(1-2 x)^{3/2} (2+3 x) (3+5 x)} \, dx}{195657}\\ &=-\frac{667615}{195657 (1-2 x)^{3/2}}-\frac{7554245}{5021863 \sqrt{1-2 x}}-\frac{505}{154 (1-2 x)^{3/2} (3+5 x)^2}+\frac{3}{7 (1-2 x)^{3/2} (2+3 x) (3+5 x)^2}+\frac{32765}{1694 (1-2 x)^{3/2} (3+5 x)}+\frac{2 \int \frac{\frac{278040255}{2}-\frac{339941025 x}{4}}{\sqrt{1-2 x} (2+3 x) (3+5 x)} \, dx}{15065589}\\ &=-\frac{667615}{195657 (1-2 x)^{3/2}}-\frac{7554245}{5021863 \sqrt{1-2 x}}-\frac{505}{154 (1-2 x)^{3/2} (3+5 x)^2}+\frac{3}{7 (1-2 x)^{3/2} (2+3 x) (3+5 x)^2}+\frac{32765}{1694 (1-2 x)^{3/2} (3+5 x)}-\frac{26730}{343} \int \frac{1}{\sqrt{1-2 x} (2+3 x)} \, dx+\frac{3693125 \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx}{29282}\\ &=-\frac{667615}{195657 (1-2 x)^{3/2}}-\frac{7554245}{5021863 \sqrt{1-2 x}}-\frac{505}{154 (1-2 x)^{3/2} (3+5 x)^2}+\frac{3}{7 (1-2 x)^{3/2} (2+3 x) (3+5 x)^2}+\frac{32765}{1694 (1-2 x)^{3/2} (3+5 x)}+\frac{26730}{343} \operatorname{Subst}\left (\int \frac{1}{\frac{7}{2}-\frac{3 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )-\frac{3693125 \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )}{29282}\\ &=-\frac{667615}{195657 (1-2 x)^{3/2}}-\frac{7554245}{5021863 \sqrt{1-2 x}}-\frac{505}{154 (1-2 x)^{3/2} (3+5 x)^2}+\frac{3}{7 (1-2 x)^{3/2} (2+3 x) (3+5 x)^2}+\frac{32765}{1694 (1-2 x)^{3/2} (3+5 x)}+\frac{17820}{343} \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{738625 \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{14641}\\ \end{align*}

Mathematica [C]  time = 0.0482841, size = 78, normalized size = 0.51 \[ \frac{-15812280 \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{3}{7}-\frac{6 x}{7}\right )[phone] \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};-\frac{5}{11} (2 x-1)\right )+\frac{231 \left (491475 x^2+605870 x+186206\right )}{(3 x+2) (5 x+3)^2}}{391314 (1-2 x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((231*(186206 + 605870*x + 491475*x^2))/((2 + 3*x)*(3 + 5*x)^2) - 15812280*Hypergeometric2F1[-3/2, 1, -1/2, 3/
7 - (6*x)/7] + 14477050*Hypergeometric2F1[-3/2, 1, -1/2, (-5*(-1 + 2*x))/11])/(391314*(1 - 2*x)^(3/2))

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Maple [A]  time = 0.018, size = 100, normalized size = 0.7 \begin{align*} -{\frac{162}{343}\sqrt{1-2\,x} \left ( -2\,x-{\frac{4}{3}} \right ) ^{-1}}+{\frac{17820\,\sqrt{21}}{2401}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) }+{\frac{32}{195657} \left ( 1-2\,x \right ) ^{-{\frac{3}{2}}}}+{\frac{5472}{5021863}{\frac{1}{\sqrt{1-2\,x}}}}+{\frac{156250}{14641\, \left ( -10\,x-6 \right ) ^{2}} \left ( -{\frac{121}{50} \left ( 1-2\,x \right ) ^{{\frac{3}{2}}}}+{\frac{1309}{250}\sqrt{1-2\,x}} \right ) }-{\frac{738625\,\sqrt{55}}{161051}{\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/(1-2*x)^(5/2)/(2+3*x)^2/(3+5*x)^3,x)

[Out]

-162/343*(1-2*x)^(1/2)/(-2*x-4/3)+17820/2401*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21^(1/2)+32/195657/(1-2*x)^(3
/2)+5472/5021863/(1-2*x)^(1/2)+156250/14641*(-121/50*(1-2*x)^(3/2)+1309/250*(1-2*x)^(1/2))/(-10*x-6)^2-738625/
161051*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 3.06941, size = 197, normalized size = 1.3 \begin{align*} \frac{738625}{322102} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{8910}{2401} \, \sqrt{21} \log \left (-\frac{\sqrt{21} - 3 \, \sqrt{-2 \, x + 1}}{\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}}\right ) - \frac{1699705125 \,{\left (2 \, x - 1\right )}^{4} + 7589204550 \,{\left (2 \, x - 1\right )}^{3} + 8458535305 \,{\left (2 \, x - 1\right )}^{2} - 22225280 \, x + 13199648}{15065589 \,{\left (75 \,{\left (-2 \, x + 1\right )}^{\frac{9}{2}} - 505 \,{\left (-2 \, x + 1\right )}^{\frac{7}{2}} + 1133 \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} - 847 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

738625/322102*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 8910/2401*sqrt(21)*
log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 1/15065589*(1699705125*(2*x - 1)^4 + 75892
04550*(2*x - 1)^3 + 8458535305*(2*x - 1)^2 - 22225280*x + 13199648)/(75*(-2*x + 1)^(9/2) - 505*(-2*x + 1)^(7/2
) + 1133*(-2*x + 1)^(5/2) - 847*(-2*x + 1)^(3/2))

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Fricas [A]  time = 1.0916, size = 599, normalized size = 3.94 \begin{align*} \frac{5320315875 \, \sqrt{11} \sqrt{5}{\left (300 \, x^{5} + 260 \, x^{4} - 137 \, x^{3} - 136 \, x^{2} + 15 \, x + 18\right )} \log \left (\frac{\sqrt{11} \sqrt{5} \sqrt{-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 8609786460 \, \sqrt{7} \sqrt{3}{\left (300 \, x^{5} + 260 \, x^{4} - 137 \, x^{3} - 136 \, x^{2} + 15 \, x + 18\right )} \log \left (-\frac{\sqrt{7} \sqrt{3} \sqrt{-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 77 \,{\left (6798820500 \, x^{4} + 1580768100 \, x^{3} - 4110847595 \, x^{2} - 479695050 \, x + 645558882\right )} \sqrt{-2 \, x + 1}}{2320100706 \,{\left (300 \, x^{5} + 260 \, x^{4} - 137 \, x^{3} - 136 \, x^{2} + 15 \, x + 18\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2320100706*(5320315875*sqrt(11)*sqrt(5)*(300*x^5 + 260*x^4 - 137*x^3 - 136*x^2 + 15*x + 18)*log((sqrt(11)*sq
rt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) + 8609786460*sqrt(7)*sqrt(3)*(300*x^5 + 260*x^4 - 137*x^3 - 136*x^2
+ 15*x + 18)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) + 77*(6798820500*x^4 + 1580768100*x^3
- 4110847595*x^2 - 479695050*x + 645558882)*sqrt(-2*x + 1))/(300*x^5 + 260*x^4 - 137*x^3 - 136*x^2 + 15*x + 1
8)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 1.81635, size = 194, normalized size = 1.28 \begin{align*} \frac{738625}{322102} \, \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{8910}{2401} \, \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{64 \,{\left (513 \, x - 295\right )}}{15065589 \,{\left (2 \, x - 1\right )} \sqrt{-2 \, x + 1}} + \frac{243 \, \sqrt{-2 \, x + 1}}{343 \,{\left (3 \, x + 2\right )}} - \frac{625 \,{\left (55 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 119 \, \sqrt{-2 \, x + 1}\right )}}{5324 \,{\left (5 \, x + 3\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

738625/322102*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 8910/2401
*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 64/15065589*(513*x - 29
5)/((2*x - 1)*sqrt(-2*x + 1)) + 243/343*sqrt(-2*x + 1)/(3*x + 2) - 625/5324*(55*(-2*x + 1)^(3/2) - 119*sqrt(-2
*x + 1))/(5*x + 3)^2