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

Optimal. Leaf size=127 \[ \frac {7 (3 x+2)^4}{11 \sqrt {1-2 x} (5 x+3)^2}-\frac {71 \sqrt {1-2 x} (3 x+2)^3}{1210 (5 x+3)^2}-\frac {1344 \sqrt {1-2 x} (3 x+2)^2}{33275 (5 x+3)}+\frac {441 \sqrt {1-2 x} (1125 x+3344)}{332750}-\frac {4557 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{166375 \sqrt {55}} \]

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Rubi [A]  time = 0.04, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, 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)^4}{11 \sqrt {1-2 x} (5 x+3)^2}-\frac {71 \sqrt {1-2 x} (3 x+2)^3}{1210 (5 x+3)^2}-\frac {1344 \sqrt {1-2 x} (3 x+2)^2}{33275 (5 x+3)}+\frac {441 \sqrt {1-2 x} (1125 x+3344)}{332750}-\frac {4557 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{166375 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-71*Sqrt[1 - 2*x]*(2 + 3*x)^3)/(1210*(3 + 5*x)^2) + (7*(2 + 3*x)^4)/(11*Sqrt[1 - 2*x]*(3 + 5*x)^2) - (1344*Sq
rt[1 - 2*x]*(2 + 3*x)^2)/(33275*(3 + 5*x)) + (441*Sqrt[1 - 2*x]*(3344 + 1125*x))/332750 - (4557*ArcTanh[Sqrt[5
/11]*Sqrt[1 - 2*x]])/(166375*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)^5}{(1-2 x)^{3/2} (3+5 x)^3} \, dx &=\frac {7 (2+3 x)^4}{11 \sqrt {1-2 x} (3+5 x)^2}-\frac {1}{11} \int \frac {(2+3 x)^3 (110+207 x)}{\sqrt {1-2 x} (3+5 x)^3} \, dx\\ &=-\frac {71 \sqrt {1-2 x} (2+3 x)^3}{1210 (3+5 x)^2}+\frac {7 (2+3 x)^4}{11 \sqrt {1-2 x} (3+5 x)^2}-\frac {\int \frac {(2+3 x)^2 (8043+14301 x)}{\sqrt {1-2 x} (3+5 x)^2} \, dx}{1210}\\ &=-\frac {71 \sqrt {1-2 x} (2+3 x)^3}{1210 (3+5 x)^2}+\frac {7 (2+3 x)^4}{11 \sqrt {1-2 x} (3+5 x)^2}-\frac {1344 \sqrt {1-2 x} (2+3 x)^2}{33275 (3+5 x)}-\frac {\int \frac {(2+3 x) (293118+496125 x)}{\sqrt {1-2 x} (3+5 x)} \, dx}{66550}\\ &=-\frac {71 \sqrt {1-2 x} (2+3 x)^3}{1210 (3+5 x)^2}+\frac {7 (2+3 x)^4}{11 \sqrt {1-2 x} (3+5 x)^2}-\frac {1344 \sqrt {1-2 x} (2+3 x)^2}{33275 (3+5 x)}+\frac {441 \sqrt {1-2 x} (3344+1125 x)}{332750}+\frac {4557 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{332750}\\ &=-\frac {71 \sqrt {1-2 x} (2+3 x)^3}{1210 (3+5 x)^2}+\frac {7 (2+3 x)^4}{11 \sqrt {1-2 x} (3+5 x)^2}-\frac {1344 \sqrt {1-2 x} (2+3 x)^2}{33275 (3+5 x)}+\frac {441 \sqrt {1-2 x} (3344+1125 x)}{332750}-\frac {4557 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{332750}\\ &=-\frac {71 \sqrt {1-2 x} (2+3 x)^3}{1210 (3+5 x)^2}+\frac {7 (2+3 x)^4}{11 \sqrt {1-2 x} (3+5 x)^2}-\frac {1344 \sqrt {1-2 x} (2+3 x)^2}{33275 (3+5 x)}+\frac {441 \sqrt {1-2 x} (3344+1125 x)}{332750}-\frac {4557 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{166375 \sqrt {55}}\\ \end {align*}

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Mathematica [C]  time = 0.14, size = 96, normalized size = 0.76 \begin {gather*} \frac {\frac {2415 \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {5}{11} (1-2 x)\right )}{\sqrt {1-2 x}}-\frac {55 \left (490050 x^4+3822390 x^3-68385 x^2-3786773 x-1485319\right )}{\sqrt {1-2 x} (5 x+3)^2}-609 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1663750} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-55*(-1485319 - 3786773*x - 68385*x^2 + 3822390*x^3 + 490050*x^4))/(Sqrt[1 - 2*x]*(3 + 5*x)^2) - 609*Sqrt[55
]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]] + (2415*Hypergeometric2F1[-1/2, 1, 1/2, (5*(1 - 2*x))/11])/Sqrt[1 - 2*x])/
1663750

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IntegrateAlgebraic [A]  time = 0.19, size = 88, normalized size = 0.69 \begin {gather*} \frac {-2695275 (1-2 x)^4+52827390 (1-2 x)^3-140781900 (1-2 x)^2-32813242 (1-2 x)[phone]}{665500 (5 (1-2 x)-11)^2 \sqrt {1-2 x}}-\frac {4557 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{166375 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(254205875 - 32813242*(1 - 2*x) - 140781900*(1 - 2*x)^2 + 52827390*(1 - 2*x)^3 - 2695275*(1 - 2*x)^4)/(665500*
(-11 + 5*(1 - 2*x))^2*Sqrt[1 - 2*x]) - (4557*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(166375*Sqrt[55])

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fricas [A]  time = 1.15, size = 94, normalized size = 0.74 \begin {gather*} \frac {4557 \, \sqrt {55} {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (5390550 \, x^{4} + 42046290 \, x^{3} - 764310 \, x^{2} - 41668993 \, x - 16342856\right )} \sqrt {-2 \, x + 1}}{18301250 \, {\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/18301250*(4557*sqrt(55)*(50*x^3 + 35*x^2 - 12*x - 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55
*(5390550*x^4 + 42046290*x^3 - 764310*x^2 - 41668993*x - 16342856)*sqrt(-2*x + 1))/(50*x^3 + 35*x^2 - 12*x - 9
)

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giac [A]  time = 1.30, size = 95, normalized size = 0.75 \begin {gather*} -\frac {81}{500} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {4557}{18301250} \, \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 {1539}{625} \, \sqrt {-2 \, x + 1} + \frac {16807}{5324 \, \sqrt {-2 \, x + 1}} + \frac {1685 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 3729 \, \sqrt {-2 \, x + 1}}{3327500 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-81/500*(-2*x + 1)^(3/2) + 4557/18301250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*s
qrt(-2*x + 1))) + 1539/625*sqrt(-2*x + 1) + 16807/5324/sqrt(-2*x + 1) + 1/3327500*(1685*(-2*x + 1)^(3/2) - 372
9*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.02, size = 75, normalized size = 0.59 \begin {gather*} -\frac {4557 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{9150625}-\frac {81 \left (-2 x +1\right )^{\frac {3}{2}}}{500}+\frac {1539 \sqrt {-2 x +1}}{625}+\frac {16807}{5324 \sqrt {-2 x +1}}+\frac {\frac {337 \left (-2 x +1\right )^{\frac {3}{2}}}{166375}-\frac {339 \sqrt {-2 x +1}}{75625}}{\left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-81/500*(-2*x+1)^(3/2)+1539/625*(-2*x+1)^(1/2)+16807/5324/(-2*x+1)^(1/2)+4/33275*(337/20*(-2*x+1)^(3/2)-3729/1
00*(-2*x+1)^(1/2))/(-10*x-6)^2-4557/9150625*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.19, size = 101, normalized size = 0.80 \begin {gather*} -\frac {81}{500} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {4557}{18301250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {1539}{625} \, \sqrt {-2 \, x + 1} + \frac {262616115 \, {\left (2 \, x - 1\right )}^{2} + 2310992332 \, x + 115533209}{3327500 \, {\left (25 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - 110 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 121 \, \sqrt {-2 \, x + 1}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-81/500*(-2*x + 1)^(3/2) + 4557/18301250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x +
1))) + 1539/625*sqrt(-2*x + 1) + 1/3327500*(262616115*(2*x - 1)^2 + 2310992332*x + 115533209)/(25*(-2*x + 1)^
(5/2) - 110*(-2*x + 1)^(3/2) + 121*sqrt(-2*x + 1))

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mupad [B]  time = 0.07, size = 82, normalized size = 0.65 \begin {gather*} \frac {1539\,\sqrt {1-2\,x}}{625}-\frac {81\,{\left (1-2\,x\right )}^{3/2}}{500}+\frac {\frac {52522553\,x}{1890625}+\frac {52523223\,{\left (2\,x-1\right )}^2}{16637500}+\frac {10503019}{7562500}}{\frac {121\,\sqrt {1-2\,x}}{25}-\frac {22\,{\left (1-2\,x\right )}^{3/2}}{5}+{\left (1-2\,x\right )}^{5/2}}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,4557{}\mathrm {i}}{9150625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*4557i)/9150625 + (1539*(1 - 2*x)^(1/2))/625 - (81*(1 - 2*x)^(
3/2))/500 + ((52522553*x)/1890625 + (52523223*(2*x - 1)^2)/16637500 + 10503019/7562500)/((121*(1 - 2*x)^(1/2))
/25 - (22*(1 - 2*x)^(3/2))/5 + (1 - 2*x)^(5/2))

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

[Out]

Timed out

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