3.20.6 \(\int \frac {1}{\sqrt {1-2 x} (2+3 x)^4 (3+5 x)} \, dx\)

Optimal. Leaf size=117 \[ \frac {3840 \sqrt {1-2 x}}{343 (3 x+2)}+\frac {55 \sqrt {1-2 x}}{49 (3 x+2)^2}+\frac {\sqrt {1-2 x}}{7 (3 x+2)^3}+\frac {88310}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-250 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

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Rubi [A]  time = 0.05, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 8, 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} \begin {gather*} \frac {3840 \sqrt {1-2 x}}{343 (3 x+2)}+\frac {55 \sqrt {1-2 x}}{49 (3 x+2)^2}+\frac {\sqrt {1-2 x}}{7 (3 x+2)^3}+\frac {88310}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-250 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[1 - 2*x]*(2 + 3*x)^4*(3 + 5*x)),x]

[Out]

Sqrt[1 - 2*x]/(7*(2 + 3*x)^3) + (55*Sqrt[1 - 2*x])/(49*(2 + 3*x)^2) + (3840*Sqrt[1 - 2*x])/(343*(2 + 3*x)) + (
88310*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/343 - 250*Sqrt[5/11]*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 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 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)^4 (3+5 x)} \, dx &=\frac {\sqrt {1-2 x}}{7 (2+3 x)^3}+\frac {1}{21} \int \frac {60-75 x}{\sqrt {1-2 x} (2+3 x)^3 (3+5 x)} \, dx\\ &=\frac {\sqrt {1-2 x}}{7 (2+3 x)^3}+\frac {55 \sqrt {1-2 x}}{49 (2+3 x)^2}+\frac {1}{294} \int \frac {4380-4950 x}{\sqrt {1-2 x} (2+3 x)^2 (3+5 x)} \, dx\\ &=\frac {\sqrt {1-2 x}}{7 (2+3 x)^3}+\frac {55 \sqrt {1-2 x}}{49 (2+3 x)^2}+\frac {3840 \sqrt {1-2 x}}{343 (2+3 x)}+\frac {\int \frac {188130-115200 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx}{2058}\\ &=\frac {\sqrt {1-2 x}}{7 (2+3 x)^3}+\frac {55 \sqrt {1-2 x}}{49 (2+3 x)^2}+\frac {3840 \sqrt {1-2 x}}{343 (2+3 x)}-\frac {132465}{343} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+625 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {\sqrt {1-2 x}}{7 (2+3 x)^3}+\frac {55 \sqrt {1-2 x}}{49 (2+3 x)^2}+\frac {3840 \sqrt {1-2 x}}{343 (2+3 x)}+\frac {132465}{343} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-625 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {\sqrt {1-2 x}}{7 (2+3 x)^3}+\frac {55 \sqrt {1-2 x}}{49 (2+3 x)^2}+\frac {3840 \sqrt {1-2 x}}{343 (2+3 x)}+\frac {88310}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-250 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 87, normalized size = 0.74 \begin {gather*} \frac {3 \sqrt {1-2 x} \left (11520 x^2+15745 x+5393\right )}{343 (3 x+2)^3}+\frac {88310}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-250 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[1 - 2*x]*(2 + 3*x)^4*(3 + 5*x)),x]

[Out]

(3*Sqrt[1 - 2*x]*(5393 + 15745*x + 11520*x^2))/(343*(2 + 3*x)^3) + (88310*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 -
2*x]])/343 - 250*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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IntegrateAlgebraic [A]  time = 0.51, size = 99, normalized size = 0.85 \begin {gather*} -\frac {12 \sqrt {1-2 x} \left (5760 (1-2 x)^2-27265 (1-2 x)+32291\right )}{343 (3 (1-2 x)-7)^3}+\frac {88310}{343} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-250 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/(Sqrt[1 - 2*x]*(2 + 3*x)^4*(3 + 5*x)),x]

[Out]

(-12*(32291 - 27265*(1 - 2*x) + 5760*(1 - 2*x)^2)*Sqrt[1 - 2*x])/(343*(-7 + 3*(1 - 2*x))^3) + (88310*Sqrt[3/7]
*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/343 - 250*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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fricas [A]  time = 1.72, size = 142, normalized size = 1.21 \begin {gather*} \frac {300125 \, \sqrt {11} \sqrt {5} {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) + 485705 \, \sqrt {7} \sqrt {3} {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )} \log \left (-\frac {\sqrt {7} \sqrt {3} \sqrt {-2 \, x + 1} - 3 \, x + 5}{3 \, x + 2}\right ) + 231 \, {\left (11520 \, x^{2} + 15745 \, x + 5393\right )} \sqrt {-2 \, x + 1}}{26411 \, {\left (27 \, x^{3} + 54 \, x^{2} + 36 \, x + 8\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2+3*x)^4/(3+5*x)/(1-2*x)^(1/2),x, algorithm="fricas")

[Out]

1/26411*(300125*sqrt(11)*sqrt(5)*(27*x^3 + 54*x^2 + 36*x + 8)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/
(5*x + 3)) + 485705*sqrt(7)*sqrt(3)*(27*x^3 + 54*x^2 + 36*x + 8)*log(-(sqrt(7)*sqrt(3)*sqrt(-2*x + 1) - 3*x +
5)/(3*x + 2)) + 231*(11520*x^2 + 15745*x + 5393)*sqrt(-2*x + 1))/(27*x^3 + 54*x^2 + 36*x + 8)

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giac [A]  time = 1.23, size = 123, normalized size = 1.05 \begin {gather*} \frac {125}{11} \, \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 {44155}{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 {3 \, {\left (5760 \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - 27265 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 32291 \, \sqrt {-2 \, x + 1}\right )}}{686 \, {\left (3 \, x + 2\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2+3*x)^4/(3+5*x)/(1-2*x)^(1/2),x, algorithm="giac")

[Out]

125/11*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 44155/2401*sqrt(
21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 3/686*(5760*(2*x - 1)^2*sqrt(
-2*x + 1) - 27265*(-2*x + 1)^(3/2) + 32291*sqrt(-2*x + 1))/(3*x + 2)^3

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maple [A]  time = 0.01, size = 75, normalized size = 0.64 \begin {gather*} \frac {88310 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{2401}-\frac {250 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{11}-\frac {162 \left (\frac {1280 \left (-2 x +1\right )^{\frac {5}{2}}}{1029}-\frac {7790 \left (-2 x +1\right )^{\frac {3}{2}}}{1323}+\frac {1318 \sqrt {-2 x +1}}{189}\right )}{\left (-6 x -4\right )^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(3*x+2)^4/(5*x+3)/(-2*x+1)^(1/2),x)

[Out]

-250/11*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)-162*(1280/1029*(-2*x+1)^(5/2)-7790/1323*(-2*x+1)^(3/2)+
1318/189*(-2*x+1)^(1/2))/(-6*x-4)^3+88310/2401*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.23, size = 128, normalized size = 1.09 \begin {gather*} \frac {125}{11} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {44155}{2401} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) + \frac {12 \, {\left (5760 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 27265 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 32291 \, \sqrt {-2 \, x + 1}\right )}}{343 \, {\left (27 \, {\left (2 \, x - 1\right )}^{3} + 189 \, {\left (2 \, x - 1\right )}^{2} + 882 \, x - 98\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2+3*x)^4/(3+5*x)/(1-2*x)^(1/2),x, algorithm="maxima")

[Out]

125/11*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 44155/2401*sqrt(21)*log(-(
sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) + 12/343*(5760*(-2*x + 1)^(5/2) - 27265*(-2*x + 1)
^(3/2) + 32291*sqrt(-2*x + 1))/(27*(2*x - 1)^3 + 189*(2*x - 1)^2 + 882*x - 98)

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mupad [B]  time = 0.10, size = 89, normalized size = 0.76 \begin {gather*} \frac {88310\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{2401}-\frac {250\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{11}+\frac {\frac {2636\,\sqrt {1-2\,x}}{63}-\frac {15580\,{\left (1-2\,x\right )}^{3/2}}{441}+\frac {2560\,{\left (1-2\,x\right )}^{5/2}}{343}}{\frac {98\,x}{3}+7\,{\left (2\,x-1\right )}^2+{\left (2\,x-1\right )}^3-\frac {98}{27}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((1 - 2*x)^(1/2)*(3*x + 2)^4*(5*x + 3)),x)

[Out]

(88310*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/2401 - (250*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11)
)/11 + ((2636*(1 - 2*x)^(1/2))/63 - (15580*(1 - 2*x)^(3/2))/441 + (2560*(1 - 2*x)^(5/2))/343)/((98*x)/3 + 7*(2
*x - 1)^2 + (2*x - 1)^3 - 98/27)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(2+3*x)**4/(3+5*x)/(1-2*x)**(1/2),x)

[Out]

Timed out

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