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

Optimal. Leaf size=100 \[ -\frac {\sqrt {1-2 x} (3 x+2)^3}{110 (5 x+3)^2}-\frac {84 \sqrt {1-2 x} (3 x+2)^2}{3025 (5 x+3)}-\frac {63 \sqrt {1-2 x} (75 x+352)}{30250}-\frac {2667 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15125 \sqrt {55}} \]

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Rubi [A]  time = 0.03, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {98, 149, 147, 63, 206} \begin {gather*} -\frac {\sqrt {1-2 x} (3 x+2)^3}{110 (5 x+3)^2}-\frac {84 \sqrt {1-2 x} (3 x+2)^2}{3025 (5 x+3)}-\frac {63 \sqrt {1-2 x} (75 x+352)}{30250}-\frac {2667 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15125 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[1 - 2*x]*(2 + 3*x)^3)/(110*(3 + 5*x)^2) - (84*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(3025*(3 + 5*x)) - (63*Sqrt[1
- 2*x]*(352 + 75*x))/30250 - (2667*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(15125*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 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (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 149

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] - Dist[1
/(b*(b*e - a*f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b*c*(f*g - e*h)*(m + 1) + (
b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; Free
Q[{a, b, c, d, e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 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 {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^3} \, dx &=-\frac {\sqrt {1-2 x} (2+3 x)^3}{110 (3+5 x)^2}-\frac {1}{110} \int \frac {(-147-189 x) (2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^2} \, dx\\ &=-\frac {\sqrt {1-2 x} (2+3 x)^3}{110 (3+5 x)^2}-\frac {84 \sqrt {1-2 x} (2+3 x)^2}{3025 (3+5 x)}-\frac {\int \frac {(-5502-4725 x) (2+3 x)}{\sqrt {1-2 x} (3+5 x)} \, dx}{6050}\\ &=-\frac {\sqrt {1-2 x} (2+3 x)^3}{110 (3+5 x)^2}-\frac {84 \sqrt {1-2 x} (2+3 x)^2}{3025 (3+5 x)}-\frac {63 \sqrt {1-2 x} (352+75 x)}{30250}+\frac {2667 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{30250}\\ &=-\frac {\sqrt {1-2 x} (2+3 x)^3}{110 (3+5 x)^2}-\frac {84 \sqrt {1-2 x} (2+3 x)^2}{3025 (3+5 x)}-\frac {63 \sqrt {1-2 x} (352+75 x)}{30250}-\frac {2667 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{30250}\\ &=-\frac {\sqrt {1-2 x} (2+3 x)^3}{110 (3+5 x)^2}-\frac {84 \sqrt {1-2 x} (2+3 x)^2}{3025 (3+5 x)}-\frac {63 \sqrt {1-2 x} (352+75 x)}{30250}-\frac {2667 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15125 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 63, normalized size = 0.63 \begin {gather*} \frac {-\frac {55 \sqrt {1-2 x} \left (163350 x^3+784080 x^2+764745 x+211864\right )}{(5 x+3)^2}-5334 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{1663750} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-55*Sqrt[1 - 2*x]*(211864 + 764745*x + 784080*x^2 + 163350*x^3))/(3 + 5*x)^2 - 5334*Sqrt[55]*ArcTanh[Sqrt[5/
11]*Sqrt[1 - 2*x]])/1663750

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IntegrateAlgebraic [A]  time = 0.18, size = 79, normalized size = 0.79 \begin {gather*} \frac {\left (81675 (1-2 x)^3-1029105 (1-2 x)^2+3342675 (1-2 x)-3242701\right ) \sqrt {1-2 x}}{30250 (5 (1-2 x)-11)^2}-\frac {2667 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{15125 \sqrt {55}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-3242701 + 3342675*(1 - 2*x) - 1029105*(1 - 2*x)^2 + 81675*(1 - 2*x)^3)*Sqrt[1 - 2*x])/(30250*(-11 + 5*(1 -
2*x))^2) - (2667*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(15125*Sqrt[55])

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fricas [A]  time = 1.55, size = 79, normalized size = 0.79 \begin {gather*} \frac {2667 \, \sqrt {55} {\left (25 \, x^{2} + 30 \, x + 9\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (163350 \, x^{3} + 784080 \, x^{2} + 764745 \, x + 211864\right )} \sqrt {-2 \, x + 1}}{1663750 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1663750*(2667*sqrt(55)*(25*x^2 + 30*x + 9)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(163350*x
^3 + 784080*x^2 + 764745*x + 211864)*sqrt(-2*x + 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 1.30, size = 86, normalized size = 0.86 \begin {gather*} \frac {27}{250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {2667}{1663750} \, \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 {1107}{1250} \, \sqrt {-2 \, x + 1} + \frac {1335 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2959 \, \sqrt {-2 \, x + 1}}{302500 \, {\left (5 \, x + 3\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/250*(-2*x + 1)^(3/2) + 2667/1663750*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqr
t(-2*x + 1))) - 1107/1250*sqrt(-2*x + 1) + 1/302500*(1335*(-2*x + 1)^(3/2) - 2959*sqrt(-2*x + 1))/(5*x + 3)^2

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maple [A]  time = 0.01, size = 66, normalized size = 0.66 \begin {gather*} -\frac {2667 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{831875}+\frac {27 \left (-2 x +1\right )^{\frac {3}{2}}}{250}-\frac {1107 \sqrt {-2 x +1}}{1250}+\frac {\frac {267 \left (-2 x +1\right )^{\frac {3}{2}}}{15125}-\frac {269 \sqrt {-2 x +1}}{6875}}{\left (-10 x -6\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

27/250*(-2*x+1)^(3/2)-1107/1250*(-2*x+1)^(1/2)+4/25*(267/2420*(-2*x+1)^(3/2)-269/1100*(-2*x+1)^(1/2))/(-10*x-6
)^2-2667/831875*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.34, size = 92, normalized size = 0.92 \begin {gather*} \frac {27}{250} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {2667}{1663750} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {1107}{1250} \, \sqrt {-2 \, x + 1} + \frac {1335 \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} - 2959 \, \sqrt {-2 \, x + 1}}{75625 \, {\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((2+3*x)^4/(3+5*x)^3/(1-2*x)^(1/2),x, algorithm="maxima")

[Out]

27/250*(-2*x + 1)^(3/2) + 2667/1663750*sqrt(55)*log(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1
))) - 1107/1250*sqrt(-2*x + 1) + 1/75625*(1335*(-2*x + 1)^(3/2) - 2959*sqrt(-2*x + 1))/(25*(2*x - 1)^2 + 220*x
+ 11)

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mupad [B]  time = 1.23, size = 74, normalized size = 0.74 \begin {gather*} \frac {27\,{\left (1-2\,x\right )}^{3/2}}{250}-\frac {1107\,\sqrt {1-2\,x}}{1250}-\frac {\frac {269\,\sqrt {1-2\,x}}{171875}-\frac {267\,{\left (1-2\,x\right )}^{3/2}}{378125}}{\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 )\,2667{}\mathrm {i}}{831875} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*2667i)/831875 - (1107*(1 - 2*x)^(1/2))/1250 + (27*(1 - 2*x)^(
3/2))/250 - ((269*(1 - 2*x)^(1/2))/171875 - (267*(1 - 2*x)^(3/2))/378125)/((44*x)/5 + (2*x - 1)^2 + 11/25)

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sympy [F(-1)]  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((2+3*x)**4/(3+5*x)**3/(1-2*x)**(1/2),x)

[Out]

Timed out

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