3.2182 \(\int \frac {(2+3 x)^4}{(1-2 x)^{5/2} (3+5 x)^2} \, dx\)

Optimal. Leaf size=100 \[ \frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)}-\frac {38 (3 x+2)^2}{1815 \sqrt {1-2 x} (5 x+3)}-\frac {3 (40912-24739 x)}{33275 \sqrt {1-2 x}}-\frac {274 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{33275 \sqrt {55}} \]

[Out]

7/33*(2+3*x)^3/(1-2*x)^(3/2)/(3+5*x)-274/1830125*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-3/33275*(40912-
24739*x)/(1-2*x)^(1/2)-38/1815*(2+3*x)^2/(3+5*x)/(1-2*x)^(1/2)

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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, 146, 63, 206} \[ \frac {7 (3 x+2)^3}{33 (1-2 x)^{3/2} (5 x+3)}-\frac {38 (3 x+2)^2}{1815 \sqrt {1-2 x} (5 x+3)}-\frac {3 (40912-24739 x)}{33275 \sqrt {1-2 x}}-\frac {274 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{33275 \sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(40912 - 24739*x))/(33275*Sqrt[1 - 2*x]) - (38*(2 + 3*x)^2)/(1815*Sqrt[1 - 2*x]*(3 + 5*x)) + (7*(2 + 3*x)^
3)/(33*(1 - 2*x)^(3/2)*(3 + 5*x)) - (274*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(33275*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 146

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol] :
> Simp[((a^2*d*f*h*(n + 2) + b^2*d*e*g*(m + n + 3) + a*b*(c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b*f*h*(
b*c - a*d)*(m + 1)*x)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/(b^2*d*(b*c - a*d)*(m + 1)*(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*(b*c - a*d)*(m +
1)*(m + n + 3)), Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && ((Ge
Q[m, -2] && LtQ[m, -1]) || SumSimplerQ[m, 1]) && NeQ[m, -1] && 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)^{5/2} (3+5 x)^2} \, dx &=\frac {7 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {1}{33} \int \frac {(2+3 x)^2 (113+201 x)}{(1-2 x)^{3/2} (3+5 x)^2} \, dx\\ &=-\frac {38 (2+3 x)^2}{1815 \sqrt {1-2 x} (3+5 x)}+\frac {7 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {\int \frac {(2+3 x) (3966+6747 x)}{(1-2 x)^{3/2} (3+5 x)} \, dx}{1815}\\ &=-\frac {3 (40912-24739 x)}{33275 \sqrt {1-2 x}}-\frac {38 (2+3 x)^2}{1815 \sqrt {1-2 x} (3+5 x)}+\frac {7 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)}+\frac {137 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{33275}\\ &=-\frac {3 (40912-24739 x)}{33275 \sqrt {1-2 x}}-\frac {38 (2+3 x)^2}{1815 \sqrt {1-2 x} (3+5 x)}+\frac {7 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {137 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{33275}\\ &=-\frac {3 (40912-24739 x)}{33275 \sqrt {1-2 x}}-\frac {38 (2+3 x)^2}{1815 \sqrt {1-2 x} (3+5 x)}+\frac {7 (2+3 x)^3}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {274 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{33275 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.08, size = 89, normalized size = 0.89 \[ -\frac {-288 \left (10 x^2+x-3\right ) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {5}{11} (1-2 x)\right )-266 (5 x+3) \, _2F_1\left (-\frac {3}{2},1;-\frac {1}{2};\frac {5}{11} (1-2 x)\right )+33 \left (22275 x^3-63855 x^2-24619 x+13028\right )}{45375 (1-2 x)^{3/2} (5 x+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/45375*(33*(13028 - 24619*x - 63855*x^2 + 22275*x^3) - 266*(3 + 5*x)*Hypergeometric2F1[-3/2, 1, -1/2, (5*(1
- 2*x))/11] - 288*(-3 + x + 10*x^2)*Hypergeometric2F1[-1/2, 1, 1/2, (5*(1 - 2*x))/11])/((1 - 2*x)^(3/2)*(3 + 5
*x))

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fricas [A]  time = 0.62, size = 89, normalized size = 0.89 \[ \frac {411 \, \sqrt {55} {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (1617165 \, x^{3} - 4634229 \, x^{2} - 1790101 \, x + 943584\right )} \sqrt {-2 \, x + 1}}{5490375 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5490375*(411*sqrt(55)*(20*x^3 - 8*x^2 - 7*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(16
17165*x^3 - 4634229*x^2 - 1790101*x + 943584)*sqrt(-2*x + 1))/(20*x^3 - 8*x^2 - 7*x + 3)

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giac [A]  time = 1.26, size = 86, normalized size = 0.86 \[ \frac {137}{1830125} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {81}{100} \, \sqrt {-2 \, x + 1} - \frac {343 \, {\left (372 \, x - 109\right )}}{15972 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} - \frac {\sqrt {-2 \, x + 1}}{33275 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

137/1830125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 81/100*sqrt
(-2*x + 1) - 343/15972*(372*x - 109)/((2*x - 1)*sqrt(-2*x + 1)) - 1/33275*sqrt(-2*x + 1)/(5*x + 3)

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maple [A]  time = 0.02, size = 63, normalized size = 0.63 \[ -\frac {274 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{1830125}-\frac {81 \sqrt {-2 x +1}}{100}+\frac {2401}{1452 \left (-2 x +1\right )^{\frac {3}{2}}}-\frac {10633}{2662 \sqrt {-2 x +1}}+\frac {2 \sqrt {-2 x +1}}{166375 \left (-2 x -\frac {6}{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-81/100*(-2*x+1)^(1/2)+2401/1452/(-2*x+1)^(3/2)-10633/2662/(-2*x+1)^(1/2)+2/166375*(-2*x+1)^(1/2)/(-2*x-6/5)-2
74/1830125*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.08, size = 83, normalized size = 0.83 \[ \frac {137}{1830125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {81}{100} \, \sqrt {-2 \, x + 1} - \frac {3987363 \, {\left (2 \, x - 1\right )}^{2} + 20845825 \, x - 6791400}{199650 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 11 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

137/1830125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 81/100*sqrt(-2*x + 1)
- 1/199650*(3987363*(2*x - 1)^2 + 20845825*x - 6791400)/(5*(-2*x + 1)^(5/2) - 11*(-2*x + 1)^(3/2))

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mupad [B]  time = 0.06, size = 64, normalized size = 0.64 \[ \frac {\frac {75803\,x}{3630}+\frac {1329121\,{\left (2\,x-1\right )}^2}{332750}-\frac {4116}{605}}{\frac {11\,{\left (1-2\,x\right )}^{3/2}}{5}-{\left (1-2\,x\right )}^{5/2}}-\frac {274\,\sqrt {55}\,\mathrm {atanh}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}}{11}\right )}{1830125}-\frac {81\,\sqrt {1-2\,x}}{100} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((75803*x)/3630 + (1329121*(2*x - 1)^2)/332750 - 4116/605)/((11*(1 - 2*x)^(3/2))/5 - (1 - 2*x)^(5/2)) - (274*5
5^(1/2)*atanh((55^(1/2)*(1 - 2*x)^(1/2))/11))/1830125 - (81*(1 - 2*x)^(1/2))/100

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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