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

Optimal. Leaf size=120 \[ \frac {7 (3 x+2)^4}{33 (1-2 x)^{3/2} (5 x+3)}-\frac {38 (3 x+2)^3}{1815 \sqrt {1-2 x} (5 x+3)}-\frac {7588 (3 x+2)^2}{6655 \sqrt {1-2 x}}-\frac {6 \sqrt {1-2 x} (38025 x+114092)}{33275}-\frac {68 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{33275 \sqrt {55}} \]

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Rubi [A]  time = 0.04, antiderivative size = 120, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {98, 149, 150, 147, 63, 206} \begin {gather*} \frac {7 (3 x+2)^4}{33 (1-2 x)^{3/2} (5 x+3)}-\frac {38 (3 x+2)^3}{1815 \sqrt {1-2 x} (5 x+3)}-\frac {7588 (3 x+2)^2}{6655 \sqrt {1-2 x}}-\frac {6 \sqrt {1-2 x} (38025 x+114092)}{33275}-\frac {68 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{33275 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-7588*(2 + 3*x)^2)/(6655*Sqrt[1 - 2*x]) - (38*(2 + 3*x)^3)/(1815*Sqrt[1 - 2*x]*(3 + 5*x)) + (7*(2 + 3*x)^4)/(
33*(1 - 2*x)^(3/2)*(3 + 5*x)) - (6*Sqrt[1 - 2*x]*(114092 + 38025*x))/33275 - (68*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 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 150

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] && IntegersQ[2*m, 2*n, 2*p]

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)^5}{(1-2 x)^{5/2} (3+5 x)^2} \, dx &=\frac {7 (2+3 x)^4}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {1}{33} \int \frac {(2+3 x)^3 (176+306 x)}{(1-2 x)^{3/2} (3+5 x)^2} \, dx\\ &=-\frac {38 (2+3 x)^3}{1815 \sqrt {1-2 x} (3+5 x)}+\frac {7 (2+3 x)^4}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {\int \frac {(2+3 x)^2 (6162+10440 x)}{(1-2 x)^{3/2} (3+5 x)} \, dx}{1815}\\ &=-\frac {7588 (2+3 x)^2}{6655 \sqrt {1-2 x}}-\frac {38 (2+3 x)^3}{1815 \sqrt {1-2 x} (3+5 x)}+\frac {7 (2+3 x)^4}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {\int \frac {(-410772-684450 x) (2+3 x)}{\sqrt {1-2 x} (3+5 x)} \, dx}{19965}\\ &=-\frac {7588 (2+3 x)^2}{6655 \sqrt {1-2 x}}-\frac {38 (2+3 x)^3}{1815 \sqrt {1-2 x} (3+5 x)}+\frac {7 (2+3 x)^4}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {6 \sqrt {1-2 x} (114092+38025 x)}{33275}+\frac {34 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{33275}\\ &=-\frac {7588 (2+3 x)^2}{6655 \sqrt {1-2 x}}-\frac {38 (2+3 x)^3}{1815 \sqrt {1-2 x} (3+5 x)}+\frac {7 (2+3 x)^4}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {6 \sqrt {1-2 x} (114092+38025 x)}{33275}-\frac {34 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{33275}\\ &=-\frac {7588 (2+3 x)^2}{6655 \sqrt {1-2 x}}-\frac {38 (2+3 x)^3}{1815 \sqrt {1-2 x} (3+5 x)}+\frac {7 (2+3 x)^4}{33 (1-2 x)^{3/2} (3+5 x)}-\frac {6 \sqrt {1-2 x} (114092+38025 x)}{33275}-\frac {68 \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 = 94, normalized size = 0.78 \begin {gather*} -\frac {-270 \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 (111375 x^4+1113750 x^3-1975050 x^2-734880 x+496226\right )}{226875 (1-2 x)^{3/2} (5 x+3)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/226875*(33*(496226 - 734880*x - 1975050*x^2 + 1113750*x^3 + 111375*x^4) - 266*(3 + 5*x)*Hypergeometric2F1[-
3/2, 1, -1/2, (5*(1 - 2*x))/11] - 270*(-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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IntegrateAlgebraic [A]  time = 0.13, size = 88, normalized size = 0.73 \begin {gather*} \frac {1617165 (1-2 x)^4-38811960 (1-2 x)^3-7976382 (1-2 x)^2+211288000 (1-2 x)-50841175}{798600 (5 (1-2 x)-11) (1-2 x)^{3/2}}-\frac {68 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{33275 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-50841175 + 211288000*(1 - 2*x) - 7976382*(1 - 2*x)^2 - 38811960*(1 - 2*x)^3 + 1617165*(1 - 2*x)^4)/(798600*(
-11 + 5*(1 - 2*x))*(1 - 2*x)^(3/2)) - (68*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(33275*Sqrt[55])

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fricas [A]  time = 1.19, size = 94, normalized size = 0.78 \begin {gather*} \frac {102 \, \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^{4} + 16171650 \, x^{3} - 28677318 \, x^{2} - 10671002 \, x + 7204728\right )} \sqrt {-2 \, x + 1}}{5490375 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5490375*(102*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^4 + 16171650*x^3 - 28677318*x^2 - 10671002*x + 7204728)*sqrt(-2*x + 1))/(20*x^3 - 8*x^2 - 7*x + 3)

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giac [A]  time = 1.29, size = 95, normalized size = 0.79 \begin {gather*} \frac {81}{200} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {34}{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 {8829}{1000} \, \sqrt {-2 \, x + 1} - \frac {2401 \, {\left (285 \, x - 104\right )}}{15972 \, {\left (2 \, x - 1\right )} \sqrt {-2 \, x + 1}} - \frac {\sqrt {-2 \, x + 1}}{166375 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

81/200*(-2*x + 1)^(3/2) + 34/1830125*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(
-2*x + 1))) - 8829/1000*sqrt(-2*x + 1) - 2401/15972*(285*x - 104)/((2*x - 1)*sqrt(-2*x + 1)) - 1/166375*sqrt(-
2*x + 1)/(5*x + 3)

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maple [A]  time = 0.02, size = 72, normalized size = 0.60 \begin {gather*} -\frac {68 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{1830125}+\frac {81 \left (-2 x +1\right )^{\frac {3}{2}}}{200}-\frac {8829 \sqrt {-2 x +1}}{1000}+\frac {16807}{2904 \left (-2 x +1\right )^{\frac {3}{2}}}-\frac {228095}{10648 \sqrt {-2 x +1}}+\frac {2 \sqrt {-2 x +1}}{831875 \left (-2 x -\frac {6}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

81/200*(-2*x+1)^(3/2)-8829/1000*(-2*x+1)^(1/2)+16807/2904/(-2*x+1)^(3/2)-228095/10648/(-2*x+1)^(1/2)+2/831875*
(-2*x+1)^(1/2)/(-2*x-6/5)-68/1830125*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.32, size = 92, normalized size = 0.77 \begin {gather*} \frac {81}{200} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {34}{1830125} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {8829}{1000} \, \sqrt {-2 \, x + 1} - \frac {427678077 \, {\left (2 \, x - 1\right )}^{2} + 2112880000 \, x - 802234125}{3993000 \, {\left (5 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 11 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

81/200*(-2*x + 1)^(3/2) + 34/1830125*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))
) - 8829/1000*sqrt(-2*x + 1) - 1/3993000*(427678077*(2*x - 1)^2 + 2112880000*x - 802234125)/(5*(-2*x + 1)^(5/2
) - 11*(-2*x + 1)^(3/2))

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mupad [B]  time = 1.19, size = 75, normalized size = 0.62 \begin {gather*} \frac {\frac {38416\,x}{363}+\frac {142559359\,{\left (2\,x-1\right )}^2}{6655000}-\frac {194481}{4840}}{\frac {11\,{\left (1-2\,x\right )}^{3/2}}{5}-{\left (1-2\,x\right )}^{5/2}}-\frac {8829\,\sqrt {1-2\,x}}{1000}+\frac {81\,{\left (1-2\,x\right )}^{3/2}}{200}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,68{}\mathrm {i}}{1830125} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((38416*x)/363 + (142559359*(2*x - 1)^2)/6655000 - 194481/4840)/((11*(1 - 2*x)^(3/2))/5 - (1 - 2*x)^(5/2)) + (
55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*68i)/1830125 - (8829*(1 - 2*x)^(1/2))/1000 + (81*(1 - 2*x)^(3/
2))/200

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

[Out]

Timed out

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