3.19.40 \(\int \frac {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx\) [1840]

Optimal. Leaf size=88 \[ -\frac {294}{625} \sqrt {1-2 x}+\frac {21}{125} \sqrt {1-2 x} (2+3 x)^2-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}-\frac {196 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}} \]

[Out]

-196/34375*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-294/625*(1-2*x)^(1/2)+21/125*(2+3*x)^2*(1-2*x)^(1/2)-
1/5*(2+3*x)^3*(1-2*x)^(1/2)/(3+5*x)

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Rubi [A]
time = 0.02, antiderivative size = 88, 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, 12, 81, 65, 212} \begin {gather*} -\frac {\sqrt {1-2 x} (3 x+2)^3}{5 (5 x+3)}+\frac {21}{125} \sqrt {1-2 x} (3 x+2)^2-\frac {294}{625} \sqrt {1-2 x}-\frac {196 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-294*Sqrt[1 - 2*x])/625 + (21*Sqrt[1 - 2*x]*(2 + 3*x)^2)/125 - (Sqrt[1 - 2*x]*(2 + 3*x)^3)/(5*(3 + 5*x)) - (1
96*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(625*Sqrt[55])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 81

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

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 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 {\sqrt {1-2 x} (2+3 x)^3}{(3+5 x)^2} \, dx &=-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}+\frac {1}{5} \int \frac {(7-21 x) (2+3 x)^2}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {21}{125} \sqrt {1-2 x} (2+3 x)^2-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}-\frac {1}{125} \int -\frac {98 (2+3 x)}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {21}{125} \sqrt {1-2 x} (2+3 x)^2-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}+\frac {98}{125} \int \frac {2+3 x}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {294}{625} \sqrt {1-2 x}+\frac {21}{125} \sqrt {1-2 x} (2+3 x)^2-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}+\frac {98}{625} \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {294}{625} \sqrt {1-2 x}+\frac {21}{125} \sqrt {1-2 x} (2+3 x)^2-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}-\frac {98}{625} \text {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )\\ &=-\frac {294}{625} \sqrt {1-2 x}+\frac {21}{125} \sqrt {1-2 x} (2+3 x)^2-\frac {\sqrt {1-2 x} (2+3 x)^3}{5 (3+5 x)}-\frac {196 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 63, normalized size = 0.72 \begin {gather*} \frac {\sqrt {1-2 x} \left (-622-90 x+2385 x^2+1350 x^3\right )}{625 (3+5 x)}-\frac {196 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{625 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(-622 - 90*x + 2385*x^2 + 1350*x^3))/(625*(3 + 5*x)) - (196*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/
(625*Sqrt[55])

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Maple [A]
time = 0.12, size = 63, normalized size = 0.72

method	result	size
risch	\(-\frac {2700 x^{4}+3420 x^{3}-2565 x^{2}-1154 x +622}{625 \left (3+5 x \right ) \sqrt {1-2 x}}-\frac {196 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{34375}\)	\(56\)
derivativedivides	\(\frac {27 \left (1-2 x \right )^{\frac {5}{2}}}{250}-\frac {117 \left (1-2 x \right )^{\frac {3}{2}}}{250}+\frac {18 \sqrt {1-2 x}}{625}+\frac {2 \sqrt {1-2 x}}{3125 \left (-\frac {6}{5}-2 x \right )}-\frac {196 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{34375}\)	\(63\)
default	\(\frac {27 \left (1-2 x \right )^{\frac {5}{2}}}{250}-\frac {117 \left (1-2 x \right )^{\frac {3}{2}}}{250}+\frac {18 \sqrt {1-2 x}}{625}+\frac {2 \sqrt {1-2 x}}{3125 \left (-\frac {6}{5}-2 x \right )}-\frac {196 \arctanh \left (\frac {\sqrt {55}\, \sqrt {1-2 x}}{11}\right ) \sqrt {55}}{34375}\)	\(63\)
trager	\(\frac {\left (1350 x^{3}+2385 x^{2}-90 x -622\right ) \sqrt {1-2 x}}{1875+3125 x}-\frac {98 \RootOf \left (\textit {\_Z}^{2}-55\right ) \ln \left (-\frac {5 \RootOf \left (\textit {\_Z}^{2}-55\right ) x -8 \RootOf \left (\textit {\_Z}^{2}-55\right )-55 \sqrt {1-2 x}}{3+5 x}\right )}{34375}\)	\(78\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/250*(1-2*x)^(5/2)-117/250*(1-2*x)^(3/2)+18/625*(1-2*x)^(1/2)+2/3125*(1-2*x)^(1/2)/(-6/5-2*x)-196/34375*arct
anh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)

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Maxima [A]
time = 0.52, size = 80, normalized size = 0.91 \begin {gather*} \frac {27}{250} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {117}{250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {98}{34375} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {18}{625} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{625 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/250*(-2*x + 1)^(5/2) - 117/250*(-2*x + 1)^(3/2) + 98/34375*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqr
t(55) + 5*sqrt(-2*x + 1))) + 18/625*sqrt(-2*x + 1) - 1/625*sqrt(-2*x + 1)/(5*x + 3)

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Fricas [A]
time = 0.94, size = 69, normalized size = 0.78 \begin {gather*} \frac {98 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) + 55 \, {\left (1350 \, x^{3} + 2385 \, x^{2} - 90 \, x - 622\right )} \sqrt {-2 \, x + 1}}{34375 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/34375*(98*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(1350*x^3 + 2385*x^2 -
90*x - 622)*sqrt(-2*x + 1))/(5*x + 3)

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Sympy [A]
time = 59.48, size = 216, normalized size = 2.45 \begin {gather*} \frac {27 \left (1 - 2 x\right )^{\frac {5}{2}}}{250} - \frac {117 \left (1 - 2 x\right )^{\frac {3}{2}}}{250} + \frac {18 \sqrt {1 - 2 x}}{625} - \frac {44 \left (\begin {cases} \frac {\sqrt {55} \left (- \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1 \right )}}{4} + \frac {\log {\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1 \right )}}{4} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} + 1\right )} - \frac {1}{4 \left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} - 1\right )}\right )}{605} & \text {for}\: \sqrt {1 - 2 x} > - \frac {\sqrt {55}}{5} \wedge \sqrt {1 - 2 x} < \frac {\sqrt {55}}{5} \end {cases}\right )}{625} + \frac {194 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: x < - \frac {3}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: x > - \frac {3}{5} \end {cases}\right )}{625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27*(1 - 2*x)**(5/2)/250 - 117*(1 - 2*x)**(3/2)/250 + 18*sqrt(1 - 2*x)/625 - 44*Piecewise((sqrt(55)*(-log(sqrt(
55)*sqrt(1 - 2*x)/11 - 1)/4 + log(sqrt(55)*sqrt(1 - 2*x)/11 + 1)/4 - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 + 1)) - 1
/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (sqrt(1 - 2*x) > -sqrt(55)/5) & (sqrt(1 - 2*x) < sqrt(55)/5)))/625
+ 194*Piecewise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)/11)/55, x < -3/5), (-sqrt(55)*atanh(sqrt(55)*sqrt(1 -
2*x)/11)/55, x > -3/5))/625

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Giac [A]
time = 1.29, size = 90, normalized size = 1.02 \begin {gather*} \frac {27}{250} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {117}{250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {98}{34375} \, \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 {18}{625} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{625 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/250*(2*x - 1)^2*sqrt(-2*x + 1) - 117/250*(-2*x + 1)^(3/2) + 98/34375*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*
sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 18/625*sqrt(-2*x + 1) - 1/625*sqrt(-2*x + 1)/(5*x + 3)

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Mupad [B]
time = 0.05, size = 64, normalized size = 0.73 \begin {gather*} \frac {18\,\sqrt {1-2\,x}}{625}-\frac {2\,\sqrt {1-2\,x}}{3125\,\left (2\,x+\frac {6}{5}\right )}-\frac {117\,{\left (1-2\,x\right )}^{3/2}}{250}+\frac {27\,{\left (1-2\,x\right )}^{5/2}}{250}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,196{}\mathrm {i}}{34375} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*196i)/34375 - (2*(1 - 2*x)^(1/2))/(3125*(2*x + 6/5)) + (18*(1
- 2*x)^(1/2))/625 - (117*(1 - 2*x)^(3/2))/250 + (27*(1 - 2*x)^(5/2))/250

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