3.20.94 \(\int \frac {(1-2 x)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx\) [1994]

Optimal. Leaf size=96 \[ -\frac {63}{125} \sqrt {1-2 x}-\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}+\frac {63}{125} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \]

[Out]

-21/275*(1-2*x)^(3/2)-1/110*(1-2*x)^(7/2)/(3+5*x)^2-63/550*(1-2*x)^(5/2)/(3+5*x)+63/625*arctanh(1/11*55^(1/2)*
(1-2*x)^(1/2))*55^(1/2)-63/125*(1-2*x)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {79, 43, 52, 65, 212} \begin {gather*} -\frac {(1-2 x)^{7/2}}{110 (5 x+3)^2}-\frac {63 (1-2 x)^{5/2}}{550 (5 x+3)}-\frac {21}{275} (1-2 x)^{3/2}-\frac {63}{125} \sqrt {1-2 x}+\frac {63}{125} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-63*Sqrt[1 - 2*x])/125 - (21*(1 - 2*x)^(3/2))/275 - (1 - 2*x)^(7/2)/(110*(3 + 5*x)^2) - (63*(1 - 2*x)^(5/2))/
(550*(3 + 5*x)) + (63*Sqrt[11/5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/125

Rule 43

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

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 79

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(-(b*e - a*f
))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1
) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e,
f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || L
tQ[p, n]))))

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)^{5/2} (2+3 x)}{(3+5 x)^3} \, dx &=-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}+\frac {63}{110} \int \frac {(1-2 x)^{5/2}}{(3+5 x)^2} \, dx\\ &=-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}-\frac {63}{110} \int \frac {(1-2 x)^{3/2}}{3+5 x} \, dx\\ &=-\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}-\frac {63}{50} \int \frac {\sqrt {1-2 x}}{3+5 x} \, dx\\ &=-\frac {63}{125} \sqrt {1-2 x}-\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}-\frac {693}{250} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {63}{125} \sqrt {1-2 x}-\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}+\frac {693}{250} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {63}{125} \sqrt {1-2 x}-\frac {21}{275} (1-2 x)^{3/2}-\frac {(1-2 x)^{7/2}}{110 (3+5 x)^2}-\frac {63 (1-2 x)^{5/2}}{550 (3+5 x)}+\frac {63}{125} \sqrt {\frac {11}{5}} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )\\ \end {align*}

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Mathematica [A]
time = 0.17, size = 63, normalized size = 0.66 \begin {gather*} \frac {\frac {5 \sqrt {1-2 x} \left (-1394-3795 x-2280 x^2+400 x^3\right )}{(3+5 x)^2}+126 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1250} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((5*Sqrt[1 - 2*x]*(-1394 - 3795*x - 2280*x^2 + 400*x^3))/(3 + 5*x)^2 + 126*Sqrt[55]*ArcTanh[Sqrt[5/11]*Sqrt[1
- 2*x]])/1250

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Maple [A]
time = 0.10, size = 66, normalized size = 0.69

method	result	size
risch	\(-\frac {800 x^{4}-4960 x^{3}-5310 x^{2}+1007 x +1394}{250 \left (3+5 x \right )^{2} \sqrt {1-2 x}}+\frac {63 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{625}\)	\(56\)
derivativedivides	\(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{125}-\frac {256 \sqrt {1-2 x}}{625}-\frac {44 \left (-\frac {57 \left (1-2 x \right )^{\frac {3}{2}}}{20}+\frac {649 \sqrt {1-2 x}}{100}\right )}{25 \left (-6-10 x \right )^{2}}+\frac {63 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{625}\)	\(66\)
default	\(-\frac {4 \left (1-2 x \right )^{\frac {3}{2}}}{125}-\frac {256 \sqrt {1-2 x}}{625}-\frac {44 \left (-\frac {57 \left (1-2 x \right )^{\frac {3}{2}}}{20}+\frac {649 \sqrt {1-2 x}}{100}\right )}{25 \left (-6-10 x \right )^{2}}+\frac {63 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{625}\)	\(66\)
trager	\(\frac {\left (400 x^{3}-2280 x^{2}-3795 x -1394\right ) \sqrt {1-2 x}}{250 \left (3+5 x \right )^{2}}-\frac {63 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x +55 \sqrt {1-2 x}-8 \RootOf \left (\textit {\_Z}^{2}-55\right )}{3+5 x}\right )}{1250}\)	\(77\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/125*(1-2*x)^(3/2)-256/625*(1-2*x)^(1/2)-44/25*(-57/20*(1-2*x)^(3/2)+649/100*(1-2*x)^(1/2))/(-6-10*x)^2+63/6
25*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]
time = 0.48, size = 92, normalized size = 0.96 \begin {gather*} -\frac {4}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {63}{1250} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {256}{625} \, \sqrt {-2 \, x + 1} + \frac {11 \, {\left (285 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 649 \, \sqrt {-2 \, x + 1}\right )}}{625 \, {\left (25 \, {\left (2 \, x - 1\right )}^{2} + 220 \, x + 11\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/125*(-2*x + 1)^(3/2) - 63/1250*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) -
256/625*sqrt(-2*x + 1) + 11/625*(285*(-2*x + 1)^(3/2) - 649*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x + 11)

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Fricas [A]
time = 0.82, size = 86, normalized size = 0.90 \begin {gather*} \frac {63 \, \sqrt {11} \sqrt {5} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (-\frac {\sqrt {11} \sqrt {5} \sqrt {-2 \, x + 1} - 5 \, x + 8}{5 \, x + 3}\right ) + 5 \, {\left (400 \, x^{3} - 2280 \, x^{2} - 3795 \, x - 1394\right )} \sqrt {-2 \, x + 1}}{1250 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1250*(63*sqrt(11)*sqrt(5)*(25*x^2 + 30*x + 9)*log(-(sqrt(11)*sqrt(5)*sqrt(-2*x + 1) - 5*x + 8)/(5*x + 3)) +
5*(400*x^3 - 2280*x^2 - 3795*x - 1394)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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

[Out]

Timed out

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Giac [A]
time = 0.57, size = 86, normalized size = 0.90 \begin {gather*} -\frac {4}{125} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - \frac {63}{1250} \, \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 {256}{625} \, \sqrt {-2 \, x + 1} + \frac {11 \, {\left (285 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 649 \, \sqrt {-2 \, x + 1}\right )}}{2500 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/125*(-2*x + 1)^(3/2) - 63/1250*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*
x + 1))) - 256/625*sqrt(-2*x + 1) + 11/2500*(285*(-2*x + 1)^(3/2) - 649*sqrt(-2*x + 1))/(5*x + 3)^2

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Mupad [B]
time = 0.08, size = 74, normalized size = 0.77 \begin {gather*} -\frac {256\,\sqrt {1-2\,x}}{625}-\frac {4\,{\left (1-2\,x\right )}^{3/2}}{125}-\frac {\frac {7139\,\sqrt {1-2\,x}}{15625}-\frac {627\,{\left (1-2\,x\right )}^{3/2}}{3125}}{\frac {44\,x}{5}+{\left (2\,x-1\right )}^2+\frac {11}{25}}-\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,63{}\mathrm {i}}{625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*63i)/625 - (256*(1 - 2*x)^(1/2))/625 - (4*(1 - 2*x)^(3/2))/
125 - ((7139*(1 - 2*x)^(1/2))/15625 - (627*(1 - 2*x)^(3/2))/3125)/((44*x)/5 + (2*x - 1)^2 + 11/25)

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