3.2067 \(\int \frac{1}{\sqrt{1-2 x} (2+3 x)^3 (3+5 x)^3} \, dx\)

Optimal. Leaf size=153 \[ \frac{1177080 \sqrt{1-2 x}}{5929 (5 x+3)}-\frac{35495 \sqrt{1-2 x}}{1078 (5 x+3)^2}+\frac{429 \sqrt{1-2 x}}{98 (3 x+2) (5 x+3)^2}+\frac{3 \sqrt{1-2 x}}{14 (3 x+2)^2 (5 x+3)^2}+\frac{134217}{49} \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{321825}{121} \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

[Out]

(-35495*Sqrt[1 - 2*x])/(1078*(3 + 5*x)^2) + (3*Sqrt[1 - 2*x])/(14*(2 + 3*x)^2*(3 + 5*x)^2) + (429*Sqrt[1 - 2*x
])/(98*(2 + 3*x)*(3 + 5*x)^2) + (1177080*Sqrt[1 - 2*x])/(5929*(3 + 5*x)) + (134217*Sqrt[3/7]*ArcTanh[Sqrt[3/7]
*Sqrt[1 - 2*x]])/49 - (321825*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/121

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Rubi [A]  time = 0.0592094, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {103, 151, 156, 63, 206} \[ \frac{1177080 \sqrt{1-2 x}}{5929 (5 x+3)}-\frac{35495 \sqrt{1-2 x}}{1078 (5 x+3)^2}+\frac{429 \sqrt{1-2 x}}{98 (3 x+2) (5 x+3)^2}+\frac{3 \sqrt{1-2 x}}{14 (3 x+2)^2 (5 x+3)^2}+\frac{134217}{49} \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{321825}{121} \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-35495*Sqrt[1 - 2*x])/(1078*(3 + 5*x)^2) + (3*Sqrt[1 - 2*x])/(14*(2 + 3*x)^2*(3 + 5*x)^2) + (429*Sqrt[1 - 2*x
])/(98*(2 + 3*x)*(3 + 5*x)^2) + (1177080*Sqrt[1 - 2*x])/(5929*(3 + 5*x)) + (134217*Sqrt[3/7]*ArcTanh[Sqrt[3/7]
*Sqrt[1 - 2*x]])/49 - (321825*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/121

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 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}{\sqrt{1-2 x} (2+3 x)^3 (3+5 x)^3} \, dx &=\frac{3 \sqrt{1-2 x}}{14 (2+3 x)^2 (3+5 x)^2}+\frac{1}{14} \int \frac{73-105 x}{\sqrt{1-2 x} (2+3 x)^2 (3+5 x)^3} \, dx\\ &=\frac{3 \sqrt{1-2 x}}{14 (2+3 x)^2 (3+5 x)^2}+\frac{429 \sqrt{1-2 x}}{98 (2+3 x) (3+5 x)^2}+\frac{1}{98} \int \frac{7763-10725 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)^3} \, dx\\ &=-\frac{35495 \sqrt{1-2 x}}{1078 (3+5 x)^2}+\frac{3 \sqrt{1-2 x}}{14 (2+3 x)^2 (3+5 x)^2}+\frac{429 \sqrt{1-2 x}}{98 (2+3 x) (3+5 x)^2}-\frac{\int \frac{558318-638910 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)^2} \, dx}{2156}\\ &=-\frac{35495 \sqrt{1-2 x}}{1078 (3+5 x)^2}+\frac{3 \sqrt{1-2 x}}{14 (2+3 x)^2 (3+5 x)^2}+\frac{429 \sqrt{1-2 x}}{98 (2+3 x) (3+5 x)^2}+\frac{1177080 \sqrt{1-2 x}}{5929 (3+5 x)}+\frac{\int \frac{23063874-14124960 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)} \, dx}{23716}\\ &=-\frac{35495 \sqrt{1-2 x}}{1078 (3+5 x)^2}+\frac{3 \sqrt{1-2 x}}{14 (2+3 x)^2 (3+5 x)^2}+\frac{429 \sqrt{1-2 x}}{98 (2+3 x) (3+5 x)^2}+\frac{1177080 \sqrt{1-2 x}}{5929 (3+5 x)}-\frac{402651}{98} \int \frac{1}{\sqrt{1-2 x} (2+3 x)} \, dx+\frac{1609125}{242} \int \frac{1}{\sqrt{1-2 x} (3+5 x)} \, dx\\ &=-\frac{35495 \sqrt{1-2 x}}{1078 (3+5 x)^2}+\frac{3 \sqrt{1-2 x}}{14 (2+3 x)^2 (3+5 x)^2}+\frac{429 \sqrt{1-2 x}}{98 (2+3 x) (3+5 x)^2}+\frac{1177080 \sqrt{1-2 x}}{5929 (3+5 x)}+\frac{402651}{98} \operatorname{Subst}\left (\int \frac{1}{\frac{7}{2}-\frac{3 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )-\frac{1609125}{242} \operatorname{Subst}\left (\int \frac{1}{\frac{11}{2}-\frac{5 x^2}{2}} \, dx,x,\sqrt{1-2 x}\right )\\ &=-\frac{35495 \sqrt{1-2 x}}{1078 (3+5 x)^2}+\frac{3 \sqrt{1-2 x}}{14 (2+3 x)^2 (3+5 x)^2}+\frac{429 \sqrt{1-2 x}}{98 (2+3 x) (3+5 x)^2}+\frac{1177080 \sqrt{1-2 x}}{5929 (3+5 x)}+\frac{134217}{49} \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right )-\frac{321825}{121} \sqrt{\frac{5}{11}} \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )\\ \end{align*}

Mathematica [A]  time = 0.0809617, size = 112, normalized size = 0.73 \[ \frac{11 \sqrt{1-2 x} \left (105937200 x^3+201297915 x^2+127303347 x+26794499\right )-31538850 \sqrt{55} \left (15 x^2+19 x+6\right )^2 \tanh ^{-1}\left (\sqrt{\frac{5}{11}} \sqrt{1-2 x}\right )}{130438 (3 x+2)^2 (5 x+3)^2}+\frac{134217}{49} \sqrt{\frac{3}{7}} \tanh ^{-1}\left (\sqrt{\frac{3}{7}} \sqrt{1-2 x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(134217*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/49 + (11*Sqrt[1 - 2*x]*(26794499 + 127303347*x + 201297915
*x^2 + 105937200*x^3) - 31538850*Sqrt[55]*(6 + 19*x + 15*x^2)^2*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(130438*(2
+ 3*x)^2*(3 + 5*x)^2)

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Maple [A]  time = 0.012, size = 94, normalized size = 0.6 \begin{align*} -972\,{\frac{1}{ \left ( -6\,x-4 \right ) ^{2}} \left ({\frac{71\, \left ( 1-2\,x \right ) ^{3/2}}{196}}-{\frac{215\,\sqrt{1-2\,x}}{252}} \right ) }+{\frac{134217\,\sqrt{21}}{343}{\it Artanh} \left ({\frac{\sqrt{21}}{7}\sqrt{1-2\,x}} \right ) }+62500\,{\frac{1}{ \left ( -10\,x-6 \right ) ^{2}} \left ( -{\frac{39\, \left ( 1-2\,x \right ) ^{3/2}}{2420}}+{\frac{193\,\sqrt{1-2\,x}}{5500}} \right ) }-{\frac{321825\,\sqrt{55}}{1331}{\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/(2+3*x)^3/(3+5*x)^3/(1-2*x)^(1/2),x)

[Out]

-972*(71/196*(1-2*x)^(3/2)-215/252*(1-2*x)^(1/2))/(-6*x-4)^2+134217/343*arctanh(1/7*21^(1/2)*(1-2*x)^(1/2))*21
^(1/2)+62500*(-39/2420*(1-2*x)^(3/2)+193/5500*(1-2*x)^(1/2))/(-10*x-6)^2-321825/1331*arctanh(1/11*55^(1/2)*(1-
2*x)^(1/2))*55^(1/2)

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Maxima [A]  time = 1.78922, size = 197, normalized size = 1.29 \begin{align*} \frac{321825}{2662} \, \sqrt{55} \log \left (-\frac{\sqrt{55} - 5 \, \sqrt{-2 \, x + 1}}{\sqrt{55} + 5 \, \sqrt{-2 \, x + 1}}\right ) - \frac{134217}{686} \, \sqrt{21} \log \left (-\frac{\sqrt{21} - 3 \, \sqrt{-2 \, x + 1}}{\sqrt{21} + 3 \, \sqrt{-2 \, x + 1}}\right ) - \frac{2 \,{\left (52968600 \,{\left (-2 \, x + 1\right )}^{\frac{7}{2}} - 360203715 \,{\left (-2 \, x + 1\right )}^{\frac{5}{2}} + 816108324 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} - 616051205 \, \sqrt{-2 \, x + 1}\right )}}{5929 \,{\left (225 \,{\left (2 \, x - 1\right )}^{4} + 2040 \,{\left (2 \, x - 1\right )}^{3} + 6934 \,{\left (2 \, x - 1\right )}^{2} + 20944 \, x - 4543\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

321825/2662*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 134217/686*sqrt(21)*l
og(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 2/5929*(52968600*(-2*x + 1)^(7/2) - 3602037
15*(-2*x + 1)^(5/2) + 816108324*(-2*x + 1)^(3/2) - 616051205*sqrt(-2*x + 1))/(225*(2*x - 1)^4 + 2040*(2*x - 1)
^3 + 6934*(2*x - 1)^2 + 20944*x - 4543)

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Fricas [A]  time = 1.98197, size = 528, normalized size = 3.45 \begin{align*} \frac{110385975 \, \sqrt{11} \sqrt{5}{\left (225 \, x^{4} + 570 \, x^{3} + 541 \, x^{2} + 228 \, x + 36\right )} \log \left (\frac{\sqrt{11} \sqrt{5} \sqrt{-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 178642827 \, \sqrt{7} \sqrt{3}{\left (225 \, x^{4} + 570 \, x^{3} + 541 \, x^{2} + 228 \, x + 36\right )} \log \left (-\frac{\sqrt{7} \sqrt{3} \sqrt{-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 77 \,{\left (105937200 \, x^{3} + 201297915 \, x^{2} + 127303347 \, x + 26794499\right )} \sqrt{-2 \, x + 1}}{913066 \,{\left (225 \, x^{4} + 570 \, x^{3} + 541 \, x^{2} + 228 \, x + 36\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/913066*(110385975*sqrt(11)*sqrt(5)*(225*x^4 + 570*x^3 + 541*x^2 + 228*x + 36)*log((sqrt(11)*sqrt(5)*sqrt(-2*
x + 1) + 5*x - 8)/(5*x + 3)) + 178642827*sqrt(7)*sqrt(3)*(225*x^4 + 570*x^3 + 541*x^2 + 228*x + 36)*log(-(sqrt
(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x + 5)/(3*x + 2)) + 77*(105937200*x^3 + 201297915*x^2 + 127303347*x + 26794499)
*sqrt(-2*x + 1))/(225*x^4 + 570*x^3 + 541*x^2 + 228*x + 36)

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

[Out]

Exception raised: ValueError

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Giac [A]  time = 2.28317, size = 200, normalized size = 1.31 \begin{align*} \frac{321825}{2662} \, \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{134217}{686} \, \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{2 \,{\left (52968600 \,{\left (2 \, x - 1\right )}^{3} \sqrt{-2 \, x + 1} + 360203715 \,{\left (2 \, x - 1\right )}^{2} \sqrt{-2 \, x + 1} - 816108324 \,{\left (-2 \, x + 1\right )}^{\frac{3}{2}} + 616051205 \, \sqrt{-2 \, x + 1}\right )}}{5929 \,{\left (15 \,{\left (2 \, x - 1\right )}^{2} + 136 \, x + 9\right )}^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

321825/2662*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 134217/686*
sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 2/5929*(52968600*(2*x -
1)^3*sqrt(-2*x + 1) + 360203715*(2*x - 1)^2*sqrt(-2*x + 1) - 816108324*(-2*x + 1)^(3/2) + 616051205*sqrt(-2*x
+ 1))/(15*(2*x - 1)^2 + 136*x + 9)^2