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

Optimal. Leaf size=100 \[ \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)}-\frac {36 \sqrt {1-2 x} (3 x+2)^2}{605 (5 x+3)}+\frac {27 \sqrt {1-2 x} (265 x+792)}{3025}-\frac {54 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \]

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Rubi [A]  time = 0.03, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {98, 149, 147, 63, 206} \begin {gather*} \frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} (5 x+3)}-\frac {36 \sqrt {1-2 x} (3 x+2)^2}{605 (5 x+3)}+\frac {27 \sqrt {1-2 x} (265 x+792)}{3025}-\frac {54 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-36*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(605*(3 + 5*x)) + (7*(2 + 3*x)^3)/(11*Sqrt[1 - 2*x]*(3 + 5*x)) + (27*Sqrt[1 -
2*x]*(792 + 265*x))/3025 - (54*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3025*Sqrt[55])

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 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 0]

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 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 {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^2} \, dx &=\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)}-\frac {1}{11} \int \frac {(2+3 x)^2 (117+207 x)}{\sqrt {1-2 x} (3+5 x)^2} \, dx\\ &=-\frac {36 \sqrt {1-2 x} (2+3 x)^2}{605 (3+5 x)}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)}-\frac {1}{605} \int \frac {(2+3 x) (4266+7155 x)}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {36 \sqrt {1-2 x} (2+3 x)^2}{605 (3+5 x)}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)}+\frac {27 \sqrt {1-2 x} (792+265 x)}{3025}+\frac {27 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{3025}\\ &=-\frac {36 \sqrt {1-2 x} (2+3 x)^2}{605 (3+5 x)}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)}+\frac {27 \sqrt {1-2 x} (792+265 x)}{3025}-\frac {27 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{3025}\\ &=-\frac {36 \sqrt {1-2 x} (2+3 x)^2}{605 (3+5 x)}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} (3+5 x)}+\frac {27 \sqrt {1-2 x} (792+265 x)}{3025}-\frac {54 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.13, size = 91, normalized size = 0.91 \begin {gather*} \frac {\frac {252 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {5}{11} (1-2 x)\right )}{\sqrt {1-2 x}}+\frac {11 \left (-7425 x^3-51975 x^2+31095 x+35764\right )}{\sqrt {1-2 x} (5 x+3)}+18 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15125} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((11*(35764 + 31095*x - 51975*x^2 - 7425*x^3))/(Sqrt[1 - 2*x]*(3 + 5*x)) + 18*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt
[1 - 2*x]] + (252*Hypergeometric2F1[-1/2, 1, 1/2, (5*(1 - 2*x))/11])/Sqrt[1 - 2*x])/15125

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IntegrateAlgebraic [A]  time = 0.12, size = 79, normalized size = 0.79 \begin {gather*} \frac {-16335 (1-2 x)^3+277695 (1-2 x)^2-231741 (1-2 x)-660275}{12100 (5 (1-2 x)-11) \sqrt {1-2 x}}-\frac {54 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3025 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-660275 - 231741*(1 - 2*x) + 277695*(1 - 2*x)^2 - 16335*(1 - 2*x)^3)/(12100*(-11 + 5*(1 - 2*x))*Sqrt[1 - 2*x]
) - (54*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3025*Sqrt[55])

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fricas [A]  time = 1.21, size = 75, normalized size = 0.75 \begin {gather*} \frac {27 \, \sqrt {55} {\left (10 \, x^{2} + x - 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (16335 \, x^{3} + 114345 \, x^{2} - 68661 \, x - 78832\right )} \sqrt {-2 \, x + 1}}{166375 \, {\left (10 \, x^{2} + x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/166375*(27*sqrt(55)*(10*x^2 + x - 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(16335*x^3 + 11
4345*x^2 - 68661*x - 78832)*sqrt(-2*x + 1))/(10*x^2 + x - 3)

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giac [A]  time = 1.26, size = 86, normalized size = 0.86 \begin {gather*} -\frac {27}{100} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {27}{166375} \, \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 {999}{250} \, \sqrt {-2 \, x + 1} - \frac {1500633 \, x + 900371}{30250 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/100*(-2*x + 1)^(3/2) + 27/166375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(
-2*x + 1))) + 999/250*sqrt(-2*x + 1) - 1/30250*(1500633*x + 900371)/(5*(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))

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maple [A]  time = 0.01, size = 63, normalized size = 0.63 \begin {gather*} -\frac {54 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{166375}-\frac {27 \left (-2 x +1\right )^{\frac {3}{2}}}{100}+\frac {999 \sqrt {-2 x +1}}{250}+\frac {2401}{484 \sqrt {-2 x +1}}+\frac {2 \sqrt {-2 x +1}}{75625 \left (-2 x -\frac {6}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/100*(-2*x+1)^(3/2)+999/250*(-2*x+1)^(1/2)+2401/484/(-2*x+1)^(1/2)+2/75625*(-2*x+1)^(1/2)/(-2*x-6/5)-54/166
375*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.24, size = 83, normalized size = 0.83 \begin {gather*} -\frac {27}{100} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {27}{166375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {999}{250} \, \sqrt {-2 \, x + 1} - \frac {1500633 \, x + 900371}{30250 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 11 \, \sqrt {-2 \, x + 1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/100*(-2*x + 1)^(3/2) + 27/166375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))
) + 999/250*sqrt(-2*x + 1) - 1/30250*(1500633*x + 900371)/(5*(-2*x + 1)^(3/2) - 11*sqrt(-2*x + 1))

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mupad [B]  time = 0.07, size = 66, normalized size = 0.66 \begin {gather*} \frac {\frac {1500633\,x}{151250}+\frac {900371}{151250}}{\frac {11\,\sqrt {1-2\,x}}{5}-{\left (1-2\,x\right )}^{3/2}}+\frac {999\,\sqrt {1-2\,x}}{250}-\frac {27\,{\left (1-2\,x\right )}^{3/2}}{100}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,54{}\mathrm {i}}{166375} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((1500633*x)/151250 + 900371/151250)/((11*(1 - 2*x)^(1/2))/5 - (1 - 2*x)^(3/2)) + (55^(1/2)*atan((55^(1/2)*(1
- 2*x)^(1/2)*1i)/11)*54i)/166375 + (999*(1 - 2*x)^(1/2))/250 - (27*(1 - 2*x)^(3/2))/100

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

[Out]

Timed out

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