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

Optimal. Leaf size=97 \[ \frac {219 \sqrt {1-2 x}}{98 (3 x+2)}+\frac {3 \sqrt {1-2 x}}{14 (3 x+2)^2}+\frac {2523}{49} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-50 \sqrt {\frac {5}{11}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

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Rubi [A]  time = 0.04, 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 = {103, 151, 156, 63, 206} \begin {gather*} \frac {219 \sqrt {1-2 x}}{98 (3 x+2)}+\frac {3 \sqrt {1-2 x}}{14 (3 x+2)^2}+\frac {2523}{49} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-50 \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)^3*(3 + 5*x)),x]

[Out]

(3*Sqrt[1 - 2*x])/(14*(2 + 3*x)^2) + (219*Sqrt[1 - 2*x])/(98*(2 + 3*x)) + (2523*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sq
rt[1 - 2*x]])/49 - 50*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)^3 (3+5 x)} \, dx &=\frac {3 \sqrt {1-2 x}}{14 (2+3 x)^2}+\frac {1}{14} \int \frac {43-45 x}{\sqrt {1-2 x} (2+3 x)^2 (3+5 x)} \, dx\\ &=\frac {3 \sqrt {1-2 x}}{14 (2+3 x)^2}+\frac {219 \sqrt {1-2 x}}{98 (2+3 x)}+\frac {1}{98} \int \frac {1793-1095 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)} \, dx\\ &=\frac {3 \sqrt {1-2 x}}{14 (2+3 x)^2}+\frac {219 \sqrt {1-2 x}}{98 (2+3 x)}-\frac {7569}{98} \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx+125 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {3 \sqrt {1-2 x}}{14 (2+3 x)^2}+\frac {219 \sqrt {1-2 x}}{98 (2+3 x)}+\frac {7569}{98} \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )-125 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=\frac {3 \sqrt {1-2 x}}{14 (2+3 x)^2}+\frac {219 \sqrt {1-2 x}}{98 (2+3 x)}+\frac {2523}{49} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-50 \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 = 0.07, size = 82, normalized size = 0.85 \begin {gather*} \frac {9 \sqrt {1-2 x} (73 x+51)}{98 (3 x+2)^2}+\frac {2523}{49} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-50 \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)^3*(3 + 5*x)),x]

[Out]

(9*Sqrt[1 - 2*x]*(51 + 73*x))/(98*(2 + 3*x)^2) + (2523*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/49 - 50*Sqr
t[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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IntegrateAlgebraic [A]  time = 0.27, size = 90, normalized size = 0.93 \begin {gather*} -\frac {9 \sqrt {1-2 x} (73 (1-2 x)-175)}{49 (3 (1-2 x)-7)^2}+\frac {2523}{49} \sqrt {\frac {3}{7}} \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )-50 \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)^3*(3 + 5*x)),x]

[Out]

(-9*(-175 + 73*(1 - 2*x))*Sqrt[1 - 2*x])/(49*(-7 + 3*(1 - 2*x))^2) + (2523*Sqrt[3/7]*ArcTanh[Sqrt[3/7]*Sqrt[1
- 2*x]])/49 - 50*Sqrt[5/11]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]]

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fricas [A]  time = 1.40, size = 122, normalized size = 1.26 \begin {gather*} \frac {17150 \, \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 ) + 27753 \, \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 ) + 693 \, {\left (73 \, x + 51\right )} \sqrt {-2 \, x + 1}}{7546 \, {\left (9 \, x^{2} + 12 \, x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7546*(17150*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)) +
27753*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)) + 693*(73
*x + 51)*sqrt(-2*x + 1))/(9*x^2 + 12*x + 4)

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giac [A]  time = 1.21, size = 107, normalized size = 1.10 \begin {gather*} \frac {25}{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 {2523}{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 {9 \, {\left (73 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 175 \, \sqrt {-2 \, x + 1}\right )}}{196 \, {\left (3 \, x + 2\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/11*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 2523/686*sqrt(21)
*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 9/196*(73*(-2*x + 1)^(3/2) - 175
*sqrt(-2*x + 1))/(3*x + 2)^2

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maple [A]  time = 0.01, size = 66, normalized size = 0.68 \begin {gather*} \frac {2523 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{343}-\frac {50 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{11}-\frac {162 \left (\frac {73 \left (-2 x +1\right )^{\frac {3}{2}}}{882}-\frac {25 \sqrt {-2 x +1}}{126}\right )}{\left (-6 x -4\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-50/11*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)-162*(73/882*(-2*x+1)^(3/2)-25/126*(-2*x+1)^(1/2))/(-6*x-
4)^2+2523/343*arctanh(1/7*21^(1/2)*(-2*x+1)^(1/2))*21^(1/2)

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maxima [A]  time = 1.37, size = 110, normalized size = 1.13 \begin {gather*} \frac {25}{11} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {2523}{686} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {9 \, {\left (73 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 175 \, \sqrt {-2 \, x + 1}\right )}}{49 \, {\left (9 \, {\left (2 \, x - 1\right )}^{2} + 84 \, x + 7\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

25/11*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 2523/686*sqrt(21)*log(-(sqr
t(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x + 1))) - 9/49*(73*(-2*x + 1)^(3/2) - 175*sqrt(-2*x + 1))/(9*
(2*x - 1)^2 + 84*x + 7)

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mupad [B]  time = 1.27, size = 71, normalized size = 0.73 \begin {gather*} \frac {2523\,\sqrt {21}\,\mathrm {atanh}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}}{7}\right )}{343}-\frac {50\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{11}+\frac {\frac {25\,\sqrt {1-2\,x}}{7}-\frac {73\,{\left (1-2\,x\right )}^{3/2}}{49}}{\frac {28\,x}{3}+{\left (2\,x-1\right )}^2+\frac {7}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2523*21^(1/2)*atanh((21^(1/2)*(1 - 2*x)^(1/2))/7))/343 - (50*55^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/1
1 + ((25*(1 - 2*x)^(1/2))/7 - (73*(1 - 2*x)^(3/2))/49)/((28*x)/3 + (2*x - 1)^2 + 7/9)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: MellinTransformStripError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: MellinTransformStripError

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