3.2048 \(\int \frac {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)^2} \, dx\)

Optimal. Leaf size=113 \[ -\frac {\sqrt {1-2 x} (3 x+2)^4}{55 (5 x+3)}-\frac {78 \sqrt {1-2 x} (3 x+2)^3}{1925}-\frac {1668 \sqrt {1-2 x} (3 x+2)^2}{6875}-\frac {6 \sqrt {1-2 x} (19875 x+59708)}{34375}-\frac {332 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{34375 \sqrt {55}} \]

[Out]

-332/1890625*arctanh(1/11*55^(1/2)*(1-2*x)^(1/2))*55^(1/2)-1668/6875*(2+3*x)^2*(1-2*x)^(1/2)-78/1925*(2+3*x)^3
*(1-2*x)^(1/2)-1/55*(2+3*x)^4*(1-2*x)^(1/2)/(3+5*x)-6/34375*(59708+19875*x)*(1-2*x)^(1/2)

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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 = {98, 153, 147, 63, 206} \[ -\frac {\sqrt {1-2 x} (3 x+2)^4}{55 (5 x+3)}-\frac {78 \sqrt {1-2 x} (3 x+2)^3}{1925}-\frac {1668 \sqrt {1-2 x} (3 x+2)^2}{6875}-\frac {6 \sqrt {1-2 x} (19875 x+59708)}{34375}-\frac {332 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{34375 \sqrt {55}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-1668*Sqrt[1 - 2*x]*(2 + 3*x)^2)/6875 - (78*Sqrt[1 - 2*x]*(2 + 3*x)^3)/1925 - (Sqrt[1 - 2*x]*(2 + 3*x)^4)/(55
*(3 + 5*x)) - (6*Sqrt[1 - 2*x]*(59708 + 19875*x))/34375 - (332*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/(34375*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 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 {(2+3 x)^5}{\sqrt {1-2 x} (3+5 x)^2} \, dx &=-\frac {\sqrt {1-2 x} (2+3 x)^4}{55 (3+5 x)}-\frac {1}{55} \int \frac {(-80-78 x) (2+3 x)^3}{\sqrt {1-2 x} (3+5 x)} \, dx\\ &=-\frac {78 \sqrt {1-2 x} (2+3 x)^3}{1925}-\frac {\sqrt {1-2 x} (2+3 x)^4}{55 (3+5 x)}+\frac {\int \frac {(2+3 x)^2 (7238+11676 x)}{\sqrt {1-2 x} (3+5 x)} \, dx}{1925}\\ &=-\frac {1668 \sqrt {1-2 x} (2+3 x)^2}{6875}-\frac {78 \sqrt {1-2 x} (2+3 x)^3}{1925}-\frac {\sqrt {1-2 x} (2+3 x)^4}{55 (3+5 x)}-\frac {\int \frac {(-502012-834750 x) (2+3 x)}{\sqrt {1-2 x} (3+5 x)} \, dx}{48125}\\ &=-\frac {1668 \sqrt {1-2 x} (2+3 x)^2}{6875}-\frac {78 \sqrt {1-2 x} (2+3 x)^3}{1925}-\frac {\sqrt {1-2 x} (2+3 x)^4}{55 (3+5 x)}-\frac {6 \sqrt {1-2 x} (59708+19875 x)}{34375}+\frac {166 \int \frac {1}{\sqrt {1-2 x} (3+5 x)} \, dx}{34375}\\ &=-\frac {1668 \sqrt {1-2 x} (2+3 x)^2}{6875}-\frac {78 \sqrt {1-2 x} (2+3 x)^3}{1925}-\frac {\sqrt {1-2 x} (2+3 x)^4}{55 (3+5 x)}-\frac {6 \sqrt {1-2 x} (59708+19875 x)}{34375}-\frac {166 \operatorname {Subst}\left (\int \frac {1}{\frac {11}{2}-\frac {5 x^2}{2}} \, dx,x,\sqrt {1-2 x}\right )}{34375}\\ &=-\frac {1668 \sqrt {1-2 x} (2+3 x)^2}{6875}-\frac {78 \sqrt {1-2 x} (2+3 x)^3}{1925}-\frac {\sqrt {1-2 x} (2+3 x)^4}{55 (3+5 x)}-\frac {6 \sqrt {1-2 x} (59708+19875 x)}{34375}-\frac {332 \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{34375 \sqrt {55}}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 68, normalized size = 0.60 \[ \frac {-\frac {55 \sqrt {1-2 x} \left (1670625 x^4+6994350 x^3+13532310 x^2+20175210 x+8527768\right )}{5 x+3}-2324 \sqrt {55} \tanh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {1-2 x}\right )}{13234375} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-55*Sqrt[1 - 2*x]*(8527768 + 20175210*x + 13532310*x^2 + 6994350*x^3 + 1670625*x^4))/(3 + 5*x) - 2324*Sqrt[5
5]*ArcTanh[Sqrt[5/11]*Sqrt[1 - 2*x]])/13234375

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fricas [A]  time = 1.02, size = 74, normalized size = 0.65 \[ \frac {1162 \, \sqrt {55} {\left (5 \, x + 3\right )} \log \left (\frac {5 \, x + \sqrt {55} \sqrt {-2 \, x + 1} - 8}{5 \, x + 3}\right ) - 55 \, {\left (1670625 \, x^{4} + 6994350 \, x^{3} + 13532310 \, x^{2} + 20175210 \, x + 8527768\right )} \sqrt {-2 \, x + 1}}{13234375 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/13234375*(1162*sqrt(55)*(5*x + 3)*log((5*x + sqrt(55)*sqrt(-2*x + 1) - 8)/(5*x + 3)) - 55*(1670625*x^4 + 699
4350*x^3 + 13532310*x^2 + 20175210*x + 8527768)*sqrt(-2*x + 1))/(5*x + 3)

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giac [A]  time = 1.25, size = 106, normalized size = 0.94 \[ -\frac {243}{1400} \, {\left (2 \, x - 1\right )}^{3} \sqrt {-2 \, x + 1} - \frac {8829}{5000} \, {\left (2 \, x - 1\right )}^{2} \sqrt {-2 \, x + 1} + \frac {35703}{5000} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {166}{1890625} \, \sqrt {55} \log \left (\frac {{\left | -2 \, \sqrt {55} + 10 \, \sqrt {-2 \, x + 1} \right |}}{2 \, {\left (\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}\right )}}\right ) - \frac {434043}{25000} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{34375 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-243/1400*(2*x - 1)^3*sqrt(-2*x + 1) - 8829/5000*(2*x - 1)^2*sqrt(-2*x + 1) + 35703/5000*(-2*x + 1)^(3/2) + 16
6/1890625*sqrt(55)*log(1/2*abs(-2*sqrt(55) + 10*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 434043/25000*
sqrt(-2*x + 1) - 1/34375*sqrt(-2*x + 1)/(5*x + 3)

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maple [A]  time = 0.01, size = 72, normalized size = 0.64 \[ -\frac {332 \sqrt {55}\, \arctanh \left (\frac {\sqrt {55}\, \sqrt {-2 x +1}}{11}\right )}{1890625}+\frac {243 \left (-2 x +1\right )^{\frac {7}{2}}}{1400}-\frac {8829 \left (-2 x +1\right )^{\frac {5}{2}}}{5000}+\frac {35703 \left (-2 x +1\right )^{\frac {3}{2}}}{5000}-\frac {434043 \sqrt {-2 x +1}}{25000}+\frac {2 \sqrt {-2 x +1}}{171875 \left (-2 x -\frac {6}{5}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

243/1400*(-2*x+1)^(7/2)-8829/5000*(-2*x+1)^(5/2)+35703/5000*(-2*x+1)^(3/2)-434043/25000*(-2*x+1)^(1/2)+2/17187
5*(-2*x+1)^(1/2)/(-2*x-6/5)-332/1890625*arctanh(1/11*55^(1/2)*(-2*x+1)^(1/2))*55^(1/2)

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maxima [A]  time = 1.15, size = 89, normalized size = 0.79 \[ \frac {243}{1400} \, {\left (-2 \, x + 1\right )}^{\frac {7}{2}} - \frac {8829}{5000} \, {\left (-2 \, x + 1\right )}^{\frac {5}{2}} + \frac {35703}{5000} \, {\left (-2 \, x + 1\right )}^{\frac {3}{2}} + \frac {166}{1890625} \, \sqrt {55} \log \left (-\frac {\sqrt {55} - 5 \, \sqrt {-2 \, x + 1}}{\sqrt {55} + 5 \, \sqrt {-2 \, x + 1}}\right ) - \frac {434043}{25000} \, \sqrt {-2 \, x + 1} - \frac {\sqrt {-2 \, x + 1}}{34375 \, {\left (5 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

243/1400*(-2*x + 1)^(7/2) - 8829/5000*(-2*x + 1)^(5/2) + 35703/5000*(-2*x + 1)^(3/2) + 166/1890625*sqrt(55)*lo
g(-(sqrt(55) - 5*sqrt(-2*x + 1))/(sqrt(55) + 5*sqrt(-2*x + 1))) - 434043/25000*sqrt(-2*x + 1) - 1/34375*sqrt(-
2*x + 1)/(5*x + 3)

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mupad [B]  time = 1.22, size = 73, normalized size = 0.65 \[ \frac {35703\,{\left (1-2\,x\right )}^{3/2}}{5000}-\frac {434043\,\sqrt {1-2\,x}}{25000}-\frac {2\,\sqrt {1-2\,x}}{171875\,\left (2\,x+\frac {6}{5}\right )}-\frac {8829\,{\left (1-2\,x\right )}^{5/2}}{5000}+\frac {243\,{\left (1-2\,x\right )}^{7/2}}{1400}+\frac {\sqrt {55}\,\mathrm {atan}\left (\frac {\sqrt {55}\,\sqrt {1-2\,x}\,1{}\mathrm {i}}{11}\right )\,332{}\mathrm {i}}{1890625} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(55^(1/2)*atan((55^(1/2)*(1 - 2*x)^(1/2)*1i)/11)*332i)/1890625 - (2*(1 - 2*x)^(1/2))/(171875*(2*x + 6/5)) - (4
34043*(1 - 2*x)^(1/2))/25000 + (35703*(1 - 2*x)^(3/2))/5000 - (8829*(1 - 2*x)^(5/2))/5000 + (243*(1 - 2*x)^(7/
2))/1400

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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