3.2581 \(\int \frac {(2+3 x)^3 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx\)

Optimal. Leaf size=113 \[ \frac {\sqrt {5 x+3} (3 x+2)^3}{3 (1-2 x)^{3/2}}-\frac {233 \sqrt {5 x+3} (3 x+2)^2}{66 \sqrt {1-2 x}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3} (69780 x+168157)}{3520}+\frac {126513 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{320 \sqrt {10}} \]

[Out]

126513/3200*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+1/3*(2+3*x)^3*(3+5*x)^(1/2)/(1-2*x)^(3/2)-233/66*(2+3
*x)^2*(3+5*x)^(1/2)/(1-2*x)^(1/2)-1/3520*(168157+69780*x)*(1-2*x)^(1/2)*(3+5*x)^(1/2)

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Rubi [A]  time = 0.03, antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {97, 150, 147, 54, 216} \[ \frac {\sqrt {5 x+3} (3 x+2)^3}{3 (1-2 x)^{3/2}}-\frac {233 \sqrt {5 x+3} (3 x+2)^2}{66 \sqrt {1-2 x}}-\frac {\sqrt {1-2 x} \sqrt {5 x+3} (69780 x+168157)}{3520}+\frac {126513 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{320 \sqrt {10}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-233*(2 + 3*x)^2*Sqrt[3 + 5*x])/(66*Sqrt[1 - 2*x]) + ((2 + 3*x)^3*Sqrt[3 + 5*x])/(3*(1 - 2*x)^(3/2)) - (Sqrt[
1 - 2*x]*Sqrt[3 + 5*x]*(168157 + 69780*x))/3520 + (126513*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(320*Sqrt[10])

Rule 54

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

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 150

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] && IntegersQ[2*m, 2*n, 2*p]

Rule 216

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

Rubi steps

\begin {align*} \int \frac {(2+3 x)^3 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx &=\frac {(2+3 x)^3 \sqrt {3+5 x}}{3 (1-2 x)^{3/2}}-\frac {1}{3} \int \frac {(2+3 x)^2 \left (32+\frac {105 x}{2}\right )}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx\\ &=-\frac {233 (2+3 x)^2 \sqrt {3+5 x}}{66 \sqrt {1-2 x}}+\frac {(2+3 x)^3 \sqrt {3+5 x}}{3 (1-2 x)^{3/2}}-\frac {1}{33} \int \frac {\left (-\frac {5349}{2}-\frac {17445 x}{4}\right ) (2+3 x)}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {233 (2+3 x)^2 \sqrt {3+5 x}}{66 \sqrt {1-2 x}}+\frac {(2+3 x)^3 \sqrt {3+5 x}}{3 (1-2 x)^{3/2}}-\frac {\sqrt {1-2 x} \sqrt {3+5 x} (168157+69780 x)}{3520}+\frac {126513}{640} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {233 (2+3 x)^2 \sqrt {3+5 x}}{66 \sqrt {1-2 x}}+\frac {(2+3 x)^3 \sqrt {3+5 x}}{3 (1-2 x)^{3/2}}-\frac {\sqrt {1-2 x} \sqrt {3+5 x} (168157+69780 x)}{3520}+\frac {126513 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{320 \sqrt {5}}\\ &=-\frac {233 (2+3 x)^2 \sqrt {3+5 x}}{66 \sqrt {1-2 x}}+\frac {(2+3 x)^3 \sqrt {3+5 x}}{3 (1-2 x)^{3/2}}-\frac {\sqrt {1-2 x} \sqrt {3+5 x} (168157+69780 x)}{3520}+\frac {126513 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{320 \sqrt {10}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 90, normalized size = 0.80 \[ \frac {10 \sqrt {2 x-1} \sqrt {5 x+3} \left (71280 x^3+431244 x^2-1786144 x+625431\right )+4174929 \sqrt {10} (1-2 x)^2 \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{105600 \sqrt {1-2 x} (2 x-1)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[-1 + 2*x]*Sqrt[3 + 5*x]*(625431 - 1786144*x + 431244*x^2 + 71280*x^3) + 4174929*Sqrt[10]*(1 - 2*x)^2*
ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]])/(105600*Sqrt[1 - 2*x]*(-1 + 2*x)^(3/2))

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fricas [A]  time = 1.29, size = 96, normalized size = 0.85 \[ -\frac {4174929 \, \sqrt {10} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (71280 \, x^{3} + 431244 \, x^{2} - 1786144 \, x + 625431\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{211200 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/211200*(4174929*sqrt(10)*(4*x^2 - 4*x + 1)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10
*x^2 + x - 3)) + 20*(71280*x^3 + 431244*x^2 - 1786144*x + 625431)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(4*x^2 - 4*x +
1)

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giac [A]  time = 1.23, size = 84, normalized size = 0.74 \[ \frac {126513}{3200} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {{\left (4 \, {\left (891 \, {\left (4 \, \sqrt {5} {\left (5 \, x + 3\right )} + 85 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 2783318 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 45924219 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{1320000 \, {\left (2 \, x - 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

126513/3200*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/1320000*(4*(891*(4*sqrt(5)*(5*x + 3) + 85*sqrt(5)
)*(5*x + 3) - 2783318*sqrt(5))*(5*x + 3) + 45924219*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2

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maple [A]  time = 0.02, size = 137, normalized size = 1.21 \[ \frac {\left (-1425600 \sqrt {-10 x^{2}-x +3}\, x^{3}[phone] \sqrt {10}\, x^{2} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-8624880 \sqrt {-10 x^{2}-x +3}\, x^{2}-16699716 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )[phone] \sqrt {-10 x^{2}-x +3}\, x +4174929 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-12508620 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}\, \sqrt {5 x +3}}{211200 \left (2 x -1\right )^{2} \sqrt {-10 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/211200*(16699716*10^(1/2)*x^2*arcsin(20/11*x+1/11)-1425600*(-10*x^2-x+3)^(1/2)*x^3-16699716*10^(1/2)*x*arcsi
n(20/11*x+1/11)-8624880*(-10*x^2-x+3)^(1/2)*x^2+4174929*10^(1/2)*arcsin(20/11*x+1/11)[phone]*(-10*x^2-x+3)^(
1/2)*x-12508620*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)*(5*x+3)^(1/2)/(2*x-1)^2/(-10*x^2-x+3)^(1/2)

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (3\,x+2\right )}^3\,\sqrt {5\,x+3}}{{\left (1-2\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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