3.20.10 \(\int \frac {(1-2 x)^{3/2} (2+3 x)^3}{(3+5 x)^2} \, dx\) [1910]

Optimal. Leaf size=108 \[ \frac {192 \sqrt {1-2 x}}{3125}+\frac {27}{175} (1-2 x)^{3/2} (2+3 x)^2-\frac {(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}-\frac {6}{625} (1-2 x)^{3/2} (29+9 x)-\frac {192 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125} \]

[Out]

27/175*(1-2*x)^(3/2)*(2+3*x)^2-1/5*(1-2*x)^(3/2)*(2+3*x)^3/(3+5*x)-6/625*(1-2*x)^(3/2)*(29+9*x)-192/15625*arct
anh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)+192/3125*(1-2*x)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 108, 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 = {99, 158, 152, 52, 65, 212} \begin {gather*} -\frac {(1-2 x)^{3/2} (3 x+2)^3}{5 (5 x+3)}+\frac {27}{175} (1-2 x)^{3/2} (3 x+2)^2-\frac {6}{625} (1-2 x)^{3/2} (9 x+29)+\frac {192 \sqrt {1-2 x}}{3125}-\frac {192 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(192*Sqrt[1 - 2*x])/3125 + (27*(1 - 2*x)^(3/2)*(2 + 3*x)^2)/175 - ((1 - 2*x)^(3/2)*(2 + 3*x)^3)/(5*(3 + 5*x))
- (6*(1 - 2*x)^(3/2)*(29 + 9*x))/625 - (192*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/3125

Rule 52

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^n/(b*(
m + n + 1))), x] + Dist[n*((b*c - a*d)/(b*(m + n + 1))), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 65

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 99

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 152

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 158

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
 /; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin {align*} \int \frac {(1-2 x)^{3/2} (2+3 x)^3}{(3+5 x)^2} \, dx &=-\frac {(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}+\frac {1}{5} \int \frac {(3-27 x) \sqrt {1-2 x} (2+3 x)^2}{3+5 x} \, dx\\ &=\frac {27}{175} (1-2 x)^{3/2} (2+3 x)^2-\frac {(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}-\frac {1}{175} \int \frac {(-210-126 x) \sqrt {1-2 x} (2+3 x)}{3+5 x} \, dx\\ &=\frac {27}{175} (1-2 x)^{3/2} (2+3 x)^2-\frac {(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}-\frac {6}{625} (1-2 x)^{3/2} (29+9 x)+\frac {96}{625} \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx\\ &=\frac {192 \sqrt {1-2 x}}{3125}+\frac {27}{175} (1-2 x)^{3/2} (2+3 x)^2-\frac {(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}-\frac {6}{625} (1-2 x)^{3/2} (29+9 x)+\frac {1056 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{3125}\\ &=\frac {192 \sqrt {1-2 x}}{3125}+\frac {27}{175} (1-2 x)^{3/2} (2+3 x)^2-\frac {(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}-\frac {6}{625} (1-2 x)^{3/2} (29+9 x)-\frac {1056 \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{3125}\\ &=\frac {192 \sqrt {1-2 x}}{3125}+\frac {27}{175} (1-2 x)^{3/2} (2+3 x)^2-\frac {(1-2 x)^{3/2} (2+3 x)^3}{5 (3+5 x)}-\frac {6}{625} (1-2 x)^{3/2} (29+9 x)-\frac {192 \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 68, normalized size = 0.63 \begin {gather*} \frac {-\frac {5 \sqrt {1-2 x} \left (8738-27640 x-57165 x^2+62100 x^3+67500 x^4\right )}{3+5 x}-1344 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{109375} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-5*Sqrt[1 - 2*x]*(8738 - 27640*x - 57165*x^2 + 62100*x^3 + 67500*x^4))/(3 + 5*x) - 1344*Sqrt[55]*ArcTanh[Sqr
t[5/11]*Sqrt[1 - 2*x]])/109375

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Maple [A]
time = 0.10, size = 72, normalized size = 0.67

method result size
risch \(\frac {135000 x^{5}+56700 x^{4}-176430 x^{3}+1885 x^{2}+45116 x -8738}{21875 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {192 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15625}\) \(61\)
derivativedivides \(\frac {27 \left (1-2 x \right )^{\frac {7}{2}}}{350}-\frac {351 \left (1-2 x \right )^{\frac {5}{2}}}{1250}+\frac {6 \left (1-2 x \right )^{\frac {3}{2}}}{625}+\frac {194 \sqrt {1-2 x}}{3125}+\frac {22 \sqrt {1-2 x}}{15625 \left (-\frac {6}{5}-2 x \right )}-\frac {192 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15625}\) \(72\)
default \(\frac {27 \left (1-2 x \right )^{\frac {7}{2}}}{350}-\frac {351 \left (1-2 x \right )^{\frac {5}{2}}}{1250}+\frac {6 \left (1-2 x \right )^{\frac {3}{2}}}{625}+\frac {194 \sqrt {1-2 x}}{3125}+\frac {22 \sqrt {1-2 x}}{15625 \left (-\frac {6}{5}-2 x \right )}-\frac {192 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{15625}\) \(72\)
trager \(-\frac {\left (67500 x^{4}+62100 x^{3}-57165 x^{2}-27640 x +8738\right ) \sqrt {1-2 x}}{21875 \left (3+5 x \right )}-\frac {96 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x -8 \RootOf \left (\textit {\_Z}^{2}-55\right )-55 \sqrt {1-2 x}}{3+5 x}\right )}{15625}\) \(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/350*(1-2*x)^(7/2)-351/1250*(1-2*x)^(5/2)+6/625*(1-2*x)^(3/2)+194/3125*(1-2*x)^(1/2)+22/15625*(1-2*x)^(1/2)/
(-6/5-2*x)-192/15625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]
time = 0.52, size = 89, normalized size = 0.82 \begin {gather*} \frac {27}{350} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {351}{1250} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {6}{625} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {96}{15625} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {194}{3125} \, \sqrt {-2 \, x + 1} - \frac {11 \, \sqrt {-2 \, x + 1}}{3125 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/350*(-2*x + 1)^(7/2) - 351/1250*(-2*x + 1)^(5/2) + 6/625*(-2*x + 1)^(3/2) + 96/15625*sqrt(55)*log(-(sqrt(55
) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 194/3125*sqrt(-2*x + 1) - 11/3125*sqrt(-2*x + 1)/(5*x +
 3)

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Fricas [A]
time = 0.90, size = 80, normalized size = 0.74 \begin {gather*} \frac {672 \, \sqrt {11} \sqrt {5} {\left (5 \, x + 3\right )} \log \left (\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} + 5 \, x - 8}{5 \, x + 3}\right ) - 5 \, {\left (67500 \, x^{4} + 62100 \, x^{3} - 57165 \, x^{2} - 27640 \, x + 8738\right )} \sqrt {-2 \, x + 1}}{109375 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/109375*(672*sqrt(11)*sqrt(5)*(5*x + 3)*log((sqrt(11)*sqrt(5)*sqrt(-2*x + 1) + 5*x - 8)/(5*x + 3)) - 5*(67500
*x^4 + 62100*x^3 - 57165*x^2 - 27640*x + 8738)*sqrt(-2*x + 1))/(5*x + 3)

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Sympy [F(-1)] Timed out
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((1-2*x)**(3/2)*(2+3*x)**3/(3+5*x)**2,x)

[Out]

Timed out

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Giac [A]
time = 0.54, size = 106, normalized size = 0.98 \begin {gather*} -\frac {27}{350} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {351}{1250} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {6}{625} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {96}{15625} \, \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 {194}{3125} \, \sqrt {-2 \, x + 1} - \frac {11 \, \sqrt {-2 \, x + 1}}{3125 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/350*(2*x - 1)^3*sqrt(-2*x + 1) - 351/1250*(2*x - 1)^2*sqrt(-2*x + 1) + 6/625*(-2*x + 1)^(3/2) + 96/15625*s
qrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 194/3125*sqrt(-2*x + 1)
- 11/3125*sqrt(-2*x + 1)/(5*x + 3)

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Mupad [B]
time = 0.06, size = 73, normalized size = 0.68 \begin {gather*} \frac {194\,\sqrt {1-2\,x}}{3125}-\frac {22\,\sqrt {1-2\,x}}{15625\,\left (2\,x+\frac {6}{5}\right )}+\frac {6\,{\left (1-2\,x\right )}^{3/2}}{625}-\frac {351\,{\left (1-2\,x\right )}^{5/2}}{1250}+\frac {27\,{\left (1-2\,x\right )}^{7/2}}{350}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,192{}\mathrm {i}}{15625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*192i)/15625 - (22*(1 - 2*x)^(1/2))/(15625*(2*x + 6/5)) + (194
*(1 - 2*x)^(1/2))/3125 + (6*(1 - 2*x)^(3/2))/625 - (351*(1 - 2*x)^(5/2))/1250 + (27*(1 - 2*x)^(7/2))/350

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