3.17.100 \(\int \frac {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^2} \, dx\)

Optimal. Leaf size=113 \[ -\frac {\sqrt {1-2 x} (3 x+2)^4}{5 (5 x+3)}+\frac {27}{175} \sqrt {1-2 x} (3 x+2)^3+\frac {12}{625} \sqrt {1-2 x} (3 x+2)^2-\frac {3 \sqrt {1-2 x} (375 x+1256)}{3125}-\frac {262 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}} \]

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Rubi [A]  time = 0.04, antiderivative size = 113, 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 = {97, 153, 147, 63, 206} \begin {gather*} -\frac {\sqrt {1-2 x} (3 x+2)^4}{5 (5 x+3)}+\frac {27}{175} \sqrt {1-2 x} (3 x+2)^3+\frac {12}{625} \sqrt {1-2 x} (3 x+2)^2-\frac {3 \sqrt {1-2 x} (375 x+1256)}{3125}-\frac {262 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(Sqrt[1 - 2*x]*(2 + 3*x)^4)/(3 + 5*x)^2,x]

[Out]

(12*Sqrt[1 - 2*x]*(2 + 3*x)^2)/625 + (27*Sqrt[1 - 2*x]*(2 + 3*x)^3)/175 - (Sqrt[1 - 2*x]*(2 + 3*x)^4)/(5*(3 +
5*x)) - (3*Sqrt[1 - 2*x]*(1256 + 375*x))/3125 - (262*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3125*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 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 153

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 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 {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^2} \, dx &=-\frac {\sqrt {1-2 x} (2+3 x)^4}{5 (3+5 x)}+\frac {1}{5} \int \frac {(10-27 x) (2+3 x)^3}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {27}{175} \sqrt {1-2 x} (2+3 x)^3-\frac {\sqrt {1-2 x} (2+3 x)^4}{5 (3+5 x)}-\frac {1}{175} \int \frac {(2+3 x)^2 (-133+84 x)}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=\frac {12}{625} \sqrt {1-2 x} (2+3 x)^2+\frac {27}{175} \sqrt {1-2 x} (2+3 x)^3-\frac {\sqrt {1-2 x} (2+3 x)^4}{5 (3+5 x)}+\frac {\int \frac {(2+3 x) (5642+7875 x)}{\sqrt {1-2 x} (3+5 x)} \, dx}{4375}\\ &=\frac {12}{625} \sqrt {1-2 x} (2+3 x)^2+\frac {27}{175} \sqrt {1-2 x} (2+3 x)^3-\frac {\sqrt {1-2 x} (2+3 x)^4}{5 (3+5 x)}-\frac {3 \sqrt {1-2 x} (1256+375 x)}{3125}+\frac {131 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{3125}\\ &=\frac {12}{625} \sqrt {1-2 x} (2+3 x)^2+\frac {27}{175} \sqrt {1-2 x} (2+3 x)^3-\frac {\sqrt {1-2 x} (2+3 x)^4}{5 (3+5 x)}-\frac {3 \sqrt {1-2 x} (1256+375 x)}{3125}-\frac {131 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{3125}\\ &=\frac {12}{625} \sqrt {1-2 x} (2+3 x)^2+\frac {27}{175} \sqrt {1-2 x} (2+3 x)^3-\frac {\sqrt {1-2 x} (2+3 x)^4}{5 (3+5 x)}-\frac {3 \sqrt {1-2 x} (1256+375 x)}{3125}-\frac {262 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 68, normalized size = 0.60 \begin {gather*} \frac {\sqrt {1-2 x} \left (101250 x^4+258525 x^3+206415 x^2-52485 x-63088\right )}{21875 (5 x+3)}-\frac {262 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[1 - 2*x]*(2 + 3*x)^4)/(3 + 5*x)^2,x]

[Out]

(Sqrt[1 - 2*x]*(-63088 - 52485*x + 206415*x^2 + 258525*x^3 + 101250*x^4))/(21875*(3 + 5*x)) - (262*ArcTanh[Sqr
t[5/11]*Sqrt[1 - 2*x]])/(3125*Sqrt[55])

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IntegrateAlgebraic [A]  time = 0.12, size = 88, normalized size = 0.78 \begin {gather*} -\frac {\sqrt {1-2 x} \left (50625 (1-2 x)^4-461025 (1-2 x)^3+1492155 (1-2 x)^2-1593795 (1-2 x)+7336\right )}{87500 (5 (1-2 x)-11)}-\frac {262 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{3125 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(Sqrt[1 - 2*x]*(2 + 3*x)^4)/(3 + 5*x)^2,x]

[Out]

-1/87500*((7336 - 1593795*(1 - 2*x) + 1492155*(1 - 2*x)^2 - 461025*(1 - 2*x)^3 + 50625*(1 - 2*x)^4)*Sqrt[1 - 2
*x])/(-11 + 5*(1 - 2*x)) - (262*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(3125*Sqrt[55])

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fricas [A]  time = 1.57, size = 74, normalized size = 0.65 \begin {gather*} \frac {917 \, \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 (101250 \, x^{4} + 258525 \, x^{3} + 206415 \, x^{2} - 52485 \, x - 63088\right )} \sqrt {-2 \, x + 1}}{1203125 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4*(1-2*x)^(1/2)/(3+5*x)^2,x, algorithm="fricas")

[Out]

1/1203125*(917*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) + 55*(101250*x^4 + 258525
*x^3 + 206415*x^2 - 52485*x - 63088)*sqrt(-2*x + 1))/(5*x + 3)

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giac [A]  time = 1.31, size = 106, normalized size = 0.94 \begin {gather*} \frac {81}{700} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} + \frac {999}{1250} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} - \frac {4131}{2500} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {131}{171875} \, \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 {24}{3125} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{3125 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4*(1-2*x)^(1/2)/(3+5*x)^2,x, algorithm="giac")

[Out]

81/700*(2*x - 1)^3*sqrt(-2*x + 1) + 999/1250*(2*x - 1)^2*sqrt(-2*x + 1) - 4131/2500*(-2*x + 1)^(3/2) + 131/171
875*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 24/3125*sqrt(-2*x +
1) - 1/3125*sqrt(-2*x + 1)/(5*x + 3)

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maple [A]  time = 0.01, size = 72, normalized size = 0.64 \begin {gather*} -\frac {262 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{171875}-\frac {81 \left (-2 x +1\right )^{\frac {7}{2}}}{700}+\frac {999 \left (-2 x +1\right )^{\frac {5}{2}}}{1250}-\frac {4131 \left (-2 x +1\right )^{\frac {3}{2}}}{2500}+\frac {24 \sqrt {-2 x +1}}{3125}+\frac {2 \sqrt {-2 x +1}}{15625 \left (-2 x -\frac {6}{5}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((3*x+2)^4*(-2*x+1)^(1/2)/(5*x+3)^2,x)

[Out]

-81/700*(-2*x+1)^(7/2)+999/1250*(-2*x+1)^(5/2)-4131/2500*(-2*x+1)^(3/2)+24/3125*(-2*x+1)^(1/2)+2/15625*(-2*x+1
)^(1/2)/(-2*x-6/5)-262/171875*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.49, size = 89, normalized size = 0.79 \begin {gather*} -\frac {81}{700} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} + \frac {999}{1250} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} - \frac {4131}{2500} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {131}{171875} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) + \frac {24}{3125} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{3125 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)^4*(1-2*x)^(1/2)/(3+5*x)^2,x, algorithm="maxima")

[Out]

-81/700*(-2*x + 1)^(7/2) + 999/1250*(-2*x + 1)^(5/2) - 4131/2500*(-2*x + 1)^(3/2) + 131/171875*sqrt(55)*log(-(
sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) + 24/3125*sqrt(-2*x + 1) - 1/3125*sqrt(-2*x + 1)/(
5*x + 3)

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mupad [B]  time = 1.17, size = 73, normalized size = 0.65 \begin {gather*} \frac {24\,\sqrt {1-2\,x}}{3125}-\frac {2\,\sqrt {1-2\,x}}{15625\,\left (2\,x+\frac {6}{5}\right )}-\frac {4131\,{\left (1-2\,x\right )}^{3/2}}{2500}+\frac {999\,{\left (1-2\,x\right )}^{5/2}}{1250}-\frac {81\,{\left (1-2\,x\right )}^{7/2}}{700}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,262{}\mathrm {i}}{171875} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1 - 2*x)^(1/2)*(3*x + 2)^4)/(5*x + 3)^2,x)

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*262i)/171875 - (2*(1 - 2*x)^(1/2))/(15625*(2*x + 6/5)) + (24*
(1 - 2*x)^(1/2))/3125 - (4131*(1 - 2*x)^(3/2))/2500 + (999*(1 - 2*x)^(5/2))/1250 - (81*(1 - 2*x)^(7/2))/700

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sympy [A]  time = 162.62, size = 214, normalized size = 1.89 \begin {gather*} - \frac {81 \left (1 - 2 x\right )^{\frac {7}{2}}}{700} + \frac {999 \left (1 - 2 x\right )^{\frac {5}{2}}}{1250} - \frac {4131 \left (1 - 2 x\right )^{\frac {3}{2}}}{2500} + \frac {24 \sqrt {1 - 2 x}}{3125} - \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}\: x \leq \frac {1}{2} \wedge x > - \frac {3}{5} \end {cases}\right )}{3125} + \frac {52 \left (\begin {cases} - \frac {\sqrt {55} \operatorname {acoth}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 < - \frac {11}{5} \\- \frac {\sqrt {55} \operatorname {atanh}{\left (\frac {\sqrt {55} \sqrt {1 - 2 x}}{11} \right )}}{55} & \text {for}\: 2 x - 1 > - \frac {11}{5} \end {cases}\right )}{625} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2+3*x)**4*(1-2*x)**(1/2)/(3+5*x)**2,x)

[Out]

-81*(1 - 2*x)**(7/2)/700 + 999*(1 - 2*x)**(5/2)/1250 - 4131*(1 - 2*x)**(3/2)/2500 + 24*sqrt(1 - 2*x)/3125 - 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*(sqr
t(55)*sqrt(1 - 2*x)/11 + 1)) - 1/(4*(sqrt(55)*sqrt(1 - 2*x)/11 - 1)))/605, (x <= 1/2) & (x > -3/5)))/3125 + 52
*Piecewise((-sqrt(55)*acoth(sqrt(55)*sqrt(1 - 2*x)/11)/55, 2*x - 1 < -11/5), (-sqrt(55)*atanh(sqrt(55)*sqrt(1
- 2*x)/11)/55, 2*x - 1 > -11/5))/625

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