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

Optimal. Leaf size=44 \[ -\frac {54 x^{5/2}}{5}+72 x^{3/2}-\frac {125 \sqrt {x}}{x+1}-450 \sqrt {x}+575 \tan ^{-1}\left (\sqrt {x}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 58, normalized size of antiderivative = 1.32, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {97, 153, 147, 63, 203} \begin {gather*} -\frac {\sqrt {x} (2-3 x)^3}{x+1}-\frac {21}{5} \sqrt {x} (2-3 x)^2-\frac {3}{5} (917-171 x) \sqrt {x}+575 \tan ^{-1}\left (\sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(917 - 171*x)*Sqrt[x])/5 - (21*(2 - 3*x)^2*Sqrt[x])/5 - ((2 - 3*x)^3*Sqrt[x])/(1 + x) + 575*ArcTan[Sqrt[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 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 203

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

Rubi steps

\begin {align*} \int \frac {(2-3 x)^3 \sqrt {x}}{(1+x)^2} \, dx &=-\frac {(2-3 x)^3 \sqrt {x}}{1+x}+\int \frac {\left (1-\frac {21 x}{2}\right ) (2-3 x)^2}{\sqrt {x} (1+x)} \, dx\\ &=-\frac {21}{5} (2-3 x)^2 \sqrt {x}-\frac {(2-3 x)^3 \sqrt {x}}{1+x}+\frac {2}{5} \int \frac {\left (\frac {31}{2}-\frac {513 x}{4}\right ) (2-3 x)}{\sqrt {x} (1+x)} \, dx\\ &=-\frac {3}{5} (917-171 x) \sqrt {x}-\frac {21}{5} (2-3 x)^2 \sqrt {x}-\frac {(2-3 x)^3 \sqrt {x}}{1+x}+\frac {575}{2} \int \frac {1}{\sqrt {x} (1+x)} \, dx\\ &=-\frac {3}{5} (917-171 x) \sqrt {x}-\frac {21}{5} (2-3 x)^2 \sqrt {x}-\frac {(2-3 x)^3 \sqrt {x}}{1+x}+575 \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\sqrt {x}\right )\\ &=-\frac {3}{5} (917-171 x) \sqrt {x}-\frac {21}{5} (2-3 x)^2 \sqrt {x}-\frac {(2-3 x)^3 \sqrt {x}}{1+x}+575 \tan ^{-1}\left (\sqrt {x}\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 38, normalized size = 0.86 \begin {gather*} 575 \tan ^{-1}\left (\sqrt {x}\right )-\frac {\sqrt {x} \left (54 x^3-306 x^2+1890 x+2875\right )}{5 (x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(Sqrt[x]*(2875 + 1890*x - 306*x^2 + 54*x^3))/(1 + x) + 575*ArcTan[Sqrt[x]]

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IntegrateAlgebraic [A]  time = 0.04, size = 38, normalized size = 0.86 \begin {gather*} 575 \tan ^{-1}\left (\sqrt {x}\right )-\frac {\sqrt {x} \left (54 x^3-306 x^2+1890 x+2875\right )}{5 (x+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/5*(Sqrt[x]*(2875 + 1890*x - 306*x^2 + 54*x^3))/(1 + x) + 575*ArcTan[Sqrt[x]]

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fricas [A]  time = 1.01, size = 37, normalized size = 0.84 \begin {gather*} \frac {2875 \, {\left (x + 1\right )} \arctan \left (\sqrt {x}\right ) - {\left (54 \, x^{3} - 306 \, x^{2} + 1890 \, x + 2875\right )} \sqrt {x}}{5 \, {\left (x + 1\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(2875*(x + 1)*arctan(sqrt(x)) - (54*x^3 - 306*x^2 + 1890*x + 2875)*sqrt(x))/(x + 1)

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giac [A]  time = 1.26, size = 32, normalized size = 0.73 \begin {gather*} -\frac {54}{5} \, x^{\frac {5}{2}} + 72 \, x^{\frac {3}{2}} - 450 \, \sqrt {x} - \frac {125 \, \sqrt {x}}{x + 1} + 575 \, \arctan \left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-54/5*x^(5/2) + 72*x^(3/2) - 450*sqrt(x) - 125*sqrt(x)/(x + 1) + 575*arctan(sqrt(x))

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maple [A]  time = 0.01, size = 33, normalized size = 0.75 \begin {gather*} -\frac {54 x^{\frac {5}{2}}}{5}+72 x^{\frac {3}{2}}+575 \arctan \left (\sqrt {x}\right )-450 \sqrt {x}-\frac {125 \sqrt {x}}{x +1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

72*x^(3/2)-54/5*x^(5/2)+575*arctan(x^(1/2))-450*x^(1/2)-125*x^(1/2)/(x+1)

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maxima [A]  time = 1.96, size = 32, normalized size = 0.73 \begin {gather*} -\frac {54}{5} \, x^{\frac {5}{2}} + 72 \, x^{\frac {3}{2}} - 450 \, \sqrt {x} - \frac {125 \, \sqrt {x}}{x + 1} + 575 \, \arctan \left (\sqrt {x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-54/5*x^(5/2) + 72*x^(3/2) - 450*sqrt(x) - 125*sqrt(x)/(x + 1) + 575*arctan(sqrt(x))

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mupad [B]  time = 0.36, size = 32, normalized size = 0.73 \begin {gather*} 575\,\mathrm {atan}\left (\sqrt {x}\right )-\frac {125\,\sqrt {x}}{x+1}-450\,\sqrt {x}+72\,x^{3/2}-\frac {54\,x^{5/2}}{5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

575*atan(x^(1/2)) - (125*x^(1/2))/(x + 1) - 450*x^(1/2) + 72*x^(3/2) - (54*x^(5/2))/5

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sympy [A]  time = 1.33, size = 75, normalized size = 1.70 \begin {gather*} - \frac {54 x^{\frac {7}{2}}}{5 x + 5} + \frac {306 x^{\frac {5}{2}}}{5 x + 5} - \frac {1890 x^{\frac {3}{2}}}{5 x + 5} - \frac {2875 \sqrt {x}}{5 x + 5} + \frac {2875 x \operatorname {atan}{\left (\sqrt {x} \right )}}{5 x + 5} + \frac {2875 \operatorname {atan}{\left (\sqrt {x} \right )}}{5 x + 5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-54*x**(7/2)/(5*x + 5) + 306*x**(5/2)/(5*x + 5) - 1890*x**(3/2)/(5*x + 5) - 2875*sqrt(x)/(5*x + 5) + 2875*x*at
an(sqrt(x))/(5*x + 5) + 2875*atan(sqrt(x))/(5*x + 5)

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