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

Optimal. Leaf size=142 \[ -\frac {2 \sqrt {1-2 x} (3 x+2)^4}{15 (5 x+3)^{3/2}}-\frac {524 \sqrt {1-2 x} (3 x+2)^3}{825 \sqrt {5 x+3}}+\frac {623 \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^2}{1375}+\frac {7 \sqrt {1-2 x} \sqrt {5 x+3} (8940 x+2563)}{220000}+\frac {35511 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{20000 \sqrt {10}} \]

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Rubi [A]  time = 0.04, antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {97, 150, 153, 147, 54, 216} \begin {gather*} -\frac {2 \sqrt {1-2 x} (3 x+2)^4}{15 (5 x+3)^{3/2}}-\frac {524 \sqrt {1-2 x} (3 x+2)^3}{825 \sqrt {5 x+3}}+\frac {623 \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^2}{1375}+\frac {7 \sqrt {1-2 x} \sqrt {5 x+3} (8940 x+2563)}{220000}+\frac {35511 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{20000 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 - 2*x]*(2 + 3*x)^4)/(15*(3 + 5*x)^(3/2)) - (524*Sqrt[1 - 2*x]*(2 + 3*x)^3)/(825*Sqrt[3 + 5*x]) + (6
23*Sqrt[1 - 2*x]*(2 + 3*x)^2*Sqrt[3 + 5*x])/1375 + (7*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(2563 + 8940*x))/220000 + (3
5511*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(20000*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 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 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 {\sqrt {1-2 x} (2+3 x)^4}{(3+5 x)^{5/2}} \, dx &=-\frac {2 \sqrt {1-2 x} (2+3 x)^4}{15 (3+5 x)^{3/2}}+\frac {2}{15} \int \frac {(10-27 x) (2+3 x)^3}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx\\ &=-\frac {2 \sqrt {1-2 x} (2+3 x)^4}{15 (3+5 x)^{3/2}}-\frac {524 \sqrt {1-2 x} (2+3 x)^3}{825 \sqrt {3+5 x}}+\frac {4}{825} \int \frac {\left (882-\frac {5607 x}{2}\right ) (2+3 x)^2}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {2 \sqrt {1-2 x} (2+3 x)^4}{15 (3+5 x)^{3/2}}-\frac {524 \sqrt {1-2 x} (2+3 x)^3}{825 \sqrt {3+5 x}}+\frac {623 \sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}}{1375}-\frac {2 \int \frac {(2+3 x) \left (-\frac {10521}{2}+\frac {46935 x}{4}\right )}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx}{12375}\\ &=-\frac {2 \sqrt {1-2 x} (2+3 x)^4}{15 (3+5 x)^{3/2}}-\frac {524 \sqrt {1-2 x} (2+3 x)^3}{825 \sqrt {3+5 x}}+\frac {623 \sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}}{1375}+\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (2563+8940 x)}{220000}+\frac {35511 \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx}{40000}\\ &=-\frac {2 \sqrt {1-2 x} (2+3 x)^4}{15 (3+5 x)^{3/2}}-\frac {524 \sqrt {1-2 x} (2+3 x)^3}{825 \sqrt {3+5 x}}+\frac {623 \sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}}{1375}+\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (2563+8940 x)}{220000}+\frac {35511 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{20000 \sqrt {5}}\\ &=-\frac {2 \sqrt {1-2 x} (2+3 x)^4}{15 (3+5 x)^{3/2}}-\frac {524 \sqrt {1-2 x} (2+3 x)^3}{825 \sqrt {3+5 x}}+\frac {623 \sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}}{1375}+\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (2563+8940 x)}{220000}+\frac {35511 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{20000 \sqrt {10}}\\ \end {align*}

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Mathematica [A]  time = 0.11, size = 88, normalized size = 0.62 \begin {gather*} \frac {1171863 (5 x+3)^{3/2} \sqrt {20 x-10} \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )-10 \left (7128000 x^5+14434200 x^4+3768930 x^3-4392275 x^2-1433776 x+218953\right )}{6600000 \sqrt {1-2 x} (5 x+3)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-10*(218953 - 1433776*x - 4392275*x^2 + 3768930*x^3 + 14434200*x^4 + 7128000*x^5) + 1171863*(3 + 5*x)^(3/2)*S
qrt[-10 + 20*x]*ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]])/(6600000*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.19, size = 141, normalized size = 0.99 \begin {gather*} -\frac {\sqrt {1-2 x} \left (\frac {8000 (1-2 x)^4}{(5 x+3)^4}+\frac {643200 (1-2 x)^3}{(5 x+3)^3}+\frac {24041535 (1-2 x)^2}{(5 x+3)^2}+\frac {16770320 (1-2 x)}{5 x+3}-4687452\right )}{660000 \sqrt {5 x+3} \left (\frac {5 (1-2 x)}{5 x+3}+2\right )^3}-\frac {35511 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )}{20000 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/660000*(Sqrt[1 - 2*x]*(-4687452 + (8000*(1 - 2*x)^4)/(3 + 5*x)^4 + (643200*(1 - 2*x)^3)/(3 + 5*x)^3 + (2404
1535*(1 - 2*x)^2)/(3 + 5*x)^2 + (16770320*(1 - 2*x))/(3 + 5*x)))/(Sqrt[3 + 5*x]*(2 + (5*(1 - 2*x))/(3 + 5*x))^
3) - (35511*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/(20000*Sqrt[10])

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fricas [A]  time = 1.59, size = 101, normalized size = 0.71 \begin {gather*} -\frac {1171863 \, \sqrt {10} {\left (25 \, x^{2} + 30 \, x + 9\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 (3564000 \, x^{4} + 8999100 \, x^{3} + 6384015 \, x^{2} + 995870 \, x - 218953\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{13200000 \, {\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*(1-2*x)^(1/2)/(3+5*x)^(5/2),x, algorithm="fricas")

[Out]

-1/13200000*(1171863*sqrt(10)*(25*x^2 + 30*x + 9)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)
/(10*x^2 + x - 3)) - 20*(3564000*x^4 + 8999100*x^3 + 6384015*x^2 + 995870*x - 218953)*sqrt(5*x + 3)*sqrt(-2*x
+ 1))/(25*x^2 + 30*x + 9)

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giac [A]  time = 2.11, size = 184, normalized size = 1.30 \begin {gather*} \frac {27}{500000} \, {\left (4 \, {\left (8 \, \sqrt {5} {\left (5 \, x + 3\right )} + 5 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 475 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {1}{8250000} \, \sqrt {10} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{{\left (5 \, x + 3\right )}^{\frac {3}{2}}} + \frac {3156 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}}\right )} + \frac {35511}{200000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {\sqrt {10} {\left (5 \, x + 3\right )}^{\frac {3}{2}} {\left (\frac {789 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{515625 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}} \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)^(5/2),x, algorithm="giac")

[Out]

27/500000*(4*(8*sqrt(5)*(5*x + 3) + 5*sqrt(5))*(5*x + 3) - 475*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) - 1/8250
000*sqrt(10)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) + 3156*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22
))/sqrt(5*x + 3)) + 35511/200000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/515625*sqrt(10)*(5*x + 3)^(3
/2)*(789*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) + 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^3

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maple [A]  time = 0.02, size = 147, normalized size = 1.04 \begin {gather*} \frac {\left (71280000 \sqrt {-10 x^{2}-x +3}\, x^{4}[phone] \sqrt {-10 x^{2}-x +3}\, x^{3}[phone] \sqrt {10}\, x^{2} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )[phone] \sqrt {-10 x^{2}-x +3}\, x^{2}[phone] \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )[phone] \sqrt {-10 x^{2}-x +3}\, x [phone] \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-4379060 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}}{13200000 \sqrt {-10 x^{2}-x +3}\, \left (5 x +3\right )^{\frac {3}{2}}} \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)^(5/2),x)

[Out]

1/13200000*(71280000*(-10*x^2-x+3)^(1/2)*x^4+29296575*10^(1/2)*x^2*arcsin(20/11*x+1/11)[phone]*(-10*x^2-x+3
)^(1/2)*x^3+35155890*10^(1/2)*x*arcsin(20/11*x+1/11)[phone]*(-10*x^2-x+3)^(1/2)*x^2+10546767*10^(1/2)*arcsi
n(20/11*x+1/11)[phone]*(-10*x^2-x+3)^(1/2)*x-4379060*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)/(-10*x^2-x+3)^(1/2)
/(5*x+3)^(3/2)

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {1-2\,x}\,{\left (3\,x+2\right )}^4}{{\left (5\,x+3\right )}^{5/2}} \,d x \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)^(5/2),x)

[Out]

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

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

[Out]

Timed out

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