3.1964 \(\int \frac {(1-2 x)^{5/2} (3+5 x)^3}{(2+3 x)^5} \, dx\)

Optimal. Leaf size=154 \[ -\frac {55 \sqrt {1-2 x} (5 x+3)^3}{24 (3 x+2)^2}+\frac {55 (1-2 x)^{3/2} (5 x+3)^3}{54 (3 x+2)^3}-\frac {(1-2 x)^{5/2} (5 x+3)^3}{12 (3 x+2)^4}-\frac {2255 \sqrt {1-2 x} (5 x+3)^2}{378 (3 x+2)}+\frac {275 \sqrt {1-2 x} (4595 x+1123)}{13608}+\frac {645865 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{6804 \sqrt {21}} \]

[Out]

-1/12*(1-2*x)^(5/2)*(3+5*x)^3/(2+3*x)^4+55/54*(1-2*x)^(3/2)*(3+5*x)^3/(2+3*x)^3+645865/142884*arctanh(1/7*21^(
1/2)*(1-2*x)^(1/2))*21^(1/2)-2255/378*(3+5*x)^2*(1-2*x)^(1/2)/(2+3*x)-55/24*(3+5*x)^3*(1-2*x)^(1/2)/(2+3*x)^2+
275/13608*(1123+4595*x)*(1-2*x)^(1/2)

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Rubi [A]  time = 0.06, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {97, 12, 149, 147, 63, 206} \[ -\frac {55 \sqrt {1-2 x} (5 x+3)^3}{24 (3 x+2)^2}+\frac {55 (1-2 x)^{3/2} (5 x+3)^3}{54 (3 x+2)^3}-\frac {(1-2 x)^{5/2} (5 x+3)^3}{12 (3 x+2)^4}-\frac {2255 \sqrt {1-2 x} (5 x+3)^2}{378 (3 x+2)}+\frac {275 \sqrt {1-2 x} (4595 x+1123)}{13608}+\frac {645865 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{6804 \sqrt {21}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2255*Sqrt[1 - 2*x]*(3 + 5*x)^2)/(378*(2 + 3*x)) - ((1 - 2*x)^(5/2)*(3 + 5*x)^3)/(12*(2 + 3*x)^4) + (55*(1 -
2*x)^(3/2)*(3 + 5*x)^3)/(54*(2 + 3*x)^3) - (55*Sqrt[1 - 2*x]*(3 + 5*x)^3)/(24*(2 + 3*x)^2) + (275*Sqrt[1 - 2*x
]*(1123 + 4595*x))/13608 + (645865*ArcTanh[Sqrt[3/7]*Sqrt[1 - 2*x]])/(6804*Sqrt[21])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 97

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a + b
*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (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 {(1-2 x)^{5/2} (3+5 x)^3}{(2+3 x)^5} \, dx &=-\frac {(1-2 x)^{5/2} (3+5 x)^3}{12 (2+3 x)^4}+\frac {1}{12} \int -\frac {55 (1-2 x)^{3/2} x (3+5 x)^2}{(2+3 x)^4} \, dx\\ &=-\frac {(1-2 x)^{5/2} (3+5 x)^3}{12 (2+3 x)^4}-\frac {55}{12} \int \frac {(1-2 x)^{3/2} x (3+5 x)^2}{(2+3 x)^4} \, dx\\ &=-\frac {(1-2 x)^{5/2} (3+5 x)^3}{12 (2+3 x)^4}+\frac {55 (1-2 x)^{3/2} (3+5 x)^3}{54 (2+3 x)^3}+\frac {55}{108} \int \frac {\sqrt {1-2 x} (3+5 x)^2 (15+36 x)}{(2+3 x)^3} \, dx\\ &=-\frac {(1-2 x)^{5/2} (3+5 x)^3}{12 (2+3 x)^4}+\frac {55 (1-2 x)^{3/2} (3+5 x)^3}{54 (2+3 x)^3}-\frac {55 \sqrt {1-2 x} (3+5 x)^3}{24 (2+3 x)^2}-\frac {55}{648} \int \frac {(3+5 x)^2 (-126+549 x)}{\sqrt {1-2 x} (2+3 x)^2} \, dx\\ &=-\frac {2255 \sqrt {1-2 x} (3+5 x)^2}{378 (2+3 x)}-\frac {(1-2 x)^{5/2} (3+5 x)^3}{12 (2+3 x)^4}+\frac {55 (1-2 x)^{3/2} (3+5 x)^3}{54 (2+3 x)^3}-\frac {55 \sqrt {1-2 x} (3+5 x)^3}{24 (2+3 x)^2}-\frac {55 \int \frac {(3+5 x) (-7659+41355 x)}{\sqrt {1-2 x} (2+3 x)} \, dx}{13608}\\ &=-\frac {2255 \sqrt {1-2 x} (3+5 x)^2}{378 (2+3 x)}-\frac {(1-2 x)^{5/2} (3+5 x)^3}{12 (2+3 x)^4}+\frac {55 (1-2 x)^{3/2} (3+5 x)^3}{54 (2+3 x)^3}-\frac {55 \sqrt {1-2 x} (3+5 x)^3}{24 (2+3 x)^2}+\frac {275 \sqrt {1-2 x} (1123+4595 x)}{13608}-\frac {645865 \int \frac {1}{\sqrt {1-2 x} (2+3 x)} \, dx}{13608}\\ &=-\frac {2255 \sqrt {1-2 x} (3+5 x)^2}{378 (2+3 x)}-\frac {(1-2 x)^{5/2} (3+5 x)^3}{12 (2+3 x)^4}+\frac {55 (1-2 x)^{3/2} (3+5 x)^3}{54 (2+3 x)^3}-\frac {55 \sqrt {1-2 x} (3+5 x)^3}{24 (2+3 x)^2}+\frac {275 \sqrt {1-2 x} (1123+4595 x)}{13608}+\frac {645865 \operatorname {Subst}\left (\int \frac {1}{\frac {7}{2}-\frac {3 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{13608}\\ &=-\frac {2255 \sqrt {1-2 x} (3+5 x)^2}{378 (2+3 x)}-\frac {(1-2 x)^{5/2} (3+5 x)^3}{12 (2+3 x)^4}+\frac {55 (1-2 x)^{3/2} (3+5 x)^3}{54 (2+3 x)^3}-\frac {55 \sqrt {1-2 x} (3+5 x)^3}{24 (2+3 x)^2}+\frac {275 \sqrt {1-2 x} (1123+4595 x)}{13608}+\frac {645865 \tanh ^{-1}\left (\sqrt {\frac {3}{7}} \sqrt {1-2 x}\right )}{6804 \sqrt {21}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 59, normalized size = 0.38 \[ \frac {(1-2 x)^{7/2} \left (1033384 (3 x+2)^4 \, _2F_1\left (3,\frac {7}{2};\frac {9}{2};\frac {3}{7}-\frac {6 x}{7}\right )-2401 \left (73500 x^2+98419 x+32939\right )\right )}{12706092 (3 x+2)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1 - 2*x)^(7/2)*(-2401*(32939 + 98419*x + 73500*x^2) + 1033384*(2 + 3*x)^4*Hypergeometric2F1[3, 7/2, 9/2, 3/7
- (6*x)/7]))/(12706092*(2 + 3*x)^4)

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fricas [A]  time = 0.92, size = 110, normalized size = 0.71 \[ \frac {645865 \, \sqrt {21} {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )} \log \left (\frac {3 \, x - \sqrt {21} \sqrt {-2 \, x + 1} - 5}{3 \, x + 2}\right ) + 21 \, {\left (1512000 \, x^{5} - 8215200 \, x^{4} - 32946525 \, x^{3} - 39158517 \, x^{2} - 19526798 \, x - 3553918\right )} \sqrt {-2 \, x + 1}}{285768 \, {\left (81 \, x^{4} + 216 \, x^{3} + 216 \, x^{2} + 96 \, x + 16\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/285768*(645865*sqrt(21)*(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)*log((3*x - sqrt(21)*sqrt(-2*x + 1) - 5)/(3*
x + 2)) + 21*(1512000*x^5 - 8215200*x^4 - 32946525*x^3 - 39158517*x^2 - 19526798*x - 3553918)*sqrt(-2*x + 1))/
(81*x^4 + 216*x^3 + 216*x^2 + 96*x + 16)

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giac [A]  time = 0.93, size = 118, normalized size = 0.77 \[ -\frac {500}{729} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {645865}{285768} \, \sqrt {21} \log \left (\frac {{\left | -2 \, \sqrt {21} + 6 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {7600}{729} \, \sqrt {-2 \, x + 1} - \frac {12957975 \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + 88621827 \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - 202092905 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + 153662285 \, \sqrt {-2 \, x + 1}}{326592 \, {\left (3 \, x + 2\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-500/729*(-2*x + 1)^(3/2) - 645865/285768*sqrt(21)*log(1/2*abs(-2*sqrt(21) + 6*sqrt(-2*x + 1))/(sqrt(21) + 3*s
qrt(-2*x + 1))) - 7600/729*sqrt(-2*x + 1) - 1/326592*(12957975*(2*x - 1)^3*sqrt(-2*x + 1) + 88621827*(2*x - 1)
^2*sqrt(-2*x + 1) - 202092905*(-2*x + 1)^(3/2) + 153662285*sqrt(-2*x + 1))/(3*x + 2)^4

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maple [A]  time = 0.01, size = 84, normalized size = 0.55 \[ \frac {645865 \sqrt {21}\, \arctanh \left (\frac {\sqrt {21}\, \sqrt {-2 x +1}}{7}\right )}{142884}-\frac {500 \left (-2 x +1\right )^{\frac {3}{2}}}{729}-\frac {7600 \sqrt {-2 x +1}}{729}-\frac {4 \left (-\frac {159975 \left (-2 x +1\right )^{\frac {7}{2}}}{112}+\frac {4220087 \left (-2 x +1\right )^{\frac {5}{2}}}{432}-\frac {28870415 \left (-2 x +1\right )^{\frac {3}{2}}}{1296}+\frac {21951755 \sqrt {-2 x +1}}{1296}\right )}{9 \left (-6 x -4\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-500/729*(-2*x+1)^(3/2)-7600/729*(-2*x+1)^(1/2)-4/9*(-159975/112*(-2*x+1)^(7/2)+4220087/432*(-2*x+1)^(5/2)-288
70415/1296*(-2*x+1)^(3/2)[phone]/1296*(-2*x+1)^(1/2))/(-6*x-4)^4+645865/142884*arctanh(1/7*21^(1/2)*(-2*x+1)
^(1/2))*21^(1/2)

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maxima [A]  time = 1.09, size = 128, normalized size = 0.83 \[ -\frac {500}{729} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {645865}{285768} \, \sqrt {21} \log \left (-\frac {\sqrt {21} - 3 \, \sqrt {-2 \, x + 1}}{\sqrt {21} + 3 \, \sqrt {-2 \, x + 1}}\right ) - \frac {7600}{729} \, \sqrt {-2 \, x + 1} + \frac {12957975 \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - 88621827 \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + 202092905 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 153662285 \, \sqrt {-2 \, x + 1}}{20412 \, {\left (81 \, {\left (2 \, x - 1\right )}^{4} + 756 \, {\left (2 \, x - 1\right )}^{3} + 2646 \, {\left (2 \, x - 1\right )}^{2} + 8232 \, x - 1715\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-500/729*(-2*x + 1)^(3/2) - 645865/285768*sqrt(21)*log(-(sqrt(21) - 3*sqrt(-2*x + 1))/(sqrt(21) + 3*sqrt(-2*x
+ 1))) - 7600/729*sqrt(-2*x + 1) + 1/20412*(12957975*(-2*x + 1)^(7/2) - 88621827*(-2*x + 1)^(5/2) + 202092905*
(-2*x + 1)^(3/2) - 153662285*sqrt(-2*x + 1))/(81*(2*x - 1)^4 + 756*(2*x - 1)^3 + 2646*(2*x - 1)^2 + 8232*x - 1
715)

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mupad [B]  time = 0.07, size = 110, normalized size = 0.71 \[ -\frac {7600\,\sqrt {1-2\,x}}{729}-\frac {500\,{\left (1-2\,x\right )}^{3/2}}{729}-\frac {\frac {21951755\,\sqrt {1-2\,x}}{236196}-\frac {28870415\,{\left (1-2\,x\right )}^{3/2}}{236196}+\frac {4220087\,{\left (1-2\,x\right )}^{5/2}}{78732}-\frac {1975\,{\left (1-2\,x\right )}^{7/2}}{252}}{\frac {2744\,x}{27}+\frac {98\,{\left (2\,x-1\right )}^2}{3}+\frac {28\,{\left (2\,x-1\right )}^3}{3}+{\left (2\,x-1\right )}^4-\frac {1715}{81}}-\frac {\sqrt {21}\,\mathrm {atan}\left (\frac {\sqrt {21}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{7}\right )\,645865{}\mathrm {i}}{142884} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (21^(1/2)*atan((21^(1/2)*(1 - 2*x)^(1/2)*1i)/7)*645865i)/142884 - (7600*(1 - 2*x)^(1/2))/729 - (500*(1 - 2*x
)^(3/2))/729 - ((21951755*(1 - 2*x)^(1/2))/236196 - (28870415*(1 - 2*x)^(3/2))/236196 + (4220087*(1 - 2*x)^(5/
2))/78732 - (1975*(1 - 2*x)^(7/2))/252)/((2744*x)/27 + (98*(2*x - 1)^2)/3 + (28*(2*x - 1)^3)/3 + (2*x - 1)^4 -
1715/81)

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

[Out]

Timed out

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