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

Optimal. Leaf size=150 \[ -\frac {1}{20} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{7/2}-\frac {\sqrt {1-2 x} (18960 x+37439) (5 x+3)^{7/2}}{32000}-\frac {2012291 \sqrt {1-2 x} (5 x+3)^{5/2}}{384000}-\frac {22135201 \sqrt {1-2 x} (5 x+3)^{3/2}}{614400}-\frac {243487211 \sqrt {1-2 x} \sqrt {5 x+3}}{819200}+\frac {2678359321 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{819200 \sqrt {10}} \]

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Rubi [A]  time = 0.04, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {100, 147, 50, 54, 216} \begin {gather*} -\frac {1}{20} \sqrt {1-2 x} (3 x+2)^2 (5 x+3)^{7/2}-\frac {\sqrt {1-2 x} (18960 x+37439) (5 x+3)^{7/2}}{32000}-\frac {2012291 \sqrt {1-2 x} (5 x+3)^{5/2}}{384000}-\frac {22135201 \sqrt {1-2 x} (5 x+3)^{3/2}}{614400}-\frac {243487211 \sqrt {1-2 x} \sqrt {5 x+3}}{819200}+\frac {2678359321 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{819200 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-243487211*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/819200 - (22135201*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2))/614400 - (2012291*S
qrt[1 - 2*x]*(3 + 5*x)^(5/2))/384000 - (Sqrt[1 - 2*x]*(2 + 3*x)^2*(3 + 5*x)^(7/2))/20 - (Sqrt[1 - 2*x]*(3 + 5*
x)^(7/2)*(37439 + 18960*x))/32000 + (2678359321*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(819200*Sqrt[10])

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

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 100

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m - 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d*f*(m + n + p + 1)), x] + Dist[1/(d*f*(m + n + p + 1)), I
nt[(a + b*x)^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1)
+ c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d,
e, f, n, p}, x] && GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegerQ[m]

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 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 (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx &=-\frac {1}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{7/2}-\frac {1}{60} \int \frac {\left (-381-\frac {1185 x}{2}\right ) (2+3 x) (3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx\\ &=-\frac {1}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{7/2}-\frac {\sqrt {1-2 x} (3+5 x)^{7/2} (37439+18960 x)}{32000}+\frac {2012291 \int \frac {(3+5 x)^{5/2}}{\sqrt {1-2 x}} \, dx}{64000}\\ &=-\frac {2012291 \sqrt {1-2 x} (3+5 x)^{5/2}}{384000}-\frac {1}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{7/2}-\frac {\sqrt {1-2 x} (3+5 x)^{7/2} (37439+18960 x)}{32000}+\frac {22135201 \int \frac {(3+5 x)^{3/2}}{\sqrt {1-2 x}} \, dx}{153600}\\ &=-\frac {22135201 \sqrt {1-2 x} (3+5 x)^{3/2}}{614400}-\frac {2012291 \sqrt {1-2 x} (3+5 x)^{5/2}}{384000}-\frac {1}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{7/2}-\frac {\sqrt {1-2 x} (3+5 x)^{7/2} (37439+18960 x)}{32000}+\frac {243487211 \int \frac {\sqrt {3+5 x}}{\sqrt {1-2 x}} \, dx}{409600}\\ &=-\frac {243487211 \sqrt {1-2 x} \sqrt {3+5 x}}{819200}-\frac {22135201 \sqrt {1-2 x} (3+5 x)^{3/2}}{614400}-\frac {2012291 \sqrt {1-2 x} (3+5 x)^{5/2}}{384000}-\frac {1}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{7/2}-\frac {\sqrt {1-2 x} (3+5 x)^{7/2} (37439+18960 x)}{32000}+\frac {2678359321 \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx}{1638400}\\ &=-\frac {243487211 \sqrt {1-2 x} \sqrt {3+5 x}}{819200}-\frac {22135201 \sqrt {1-2 x} (3+5 x)^{3/2}}{614400}-\frac {2012291 \sqrt {1-2 x} (3+5 x)^{5/2}}{384000}-\frac {1}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{7/2}-\frac {\sqrt {1-2 x} (3+5 x)^{7/2} (37439+18960 x)}{32000}+\frac {2678359321 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{819200 \sqrt {5}}\\ &=-\frac {243487211 \sqrt {1-2 x} \sqrt {3+5 x}}{819200}-\frac {22135201 \sqrt {1-2 x} (3+5 x)^{3/2}}{614400}-\frac {2012291 \sqrt {1-2 x} (3+5 x)^{5/2}}{384000}-\frac {1}{20} \sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{7/2}-\frac {\sqrt {1-2 x} (3+5 x)^{7/2} (37439+18960 x)}{32000}+\frac {2678359321 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{819200 \sqrt {10}}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 93, normalized size = 0.62 \begin {gather*} -\frac {\sqrt {1-2 x} \left (10 \sqrt {2 x-1} \sqrt {5 x+3} \left (138240000 x^5+615168000 x^4+1229558400 x^3+1505007200 x^2+1362715220 x+1202896557\right )[phone] \sqrt {10} \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )\right )}{24576000 \sqrt {2 x-1}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/24576000*(Sqrt[1 - 2*x]*(10*Sqrt[-1 + 2*x]*Sqrt[3 + 5*x]*(1202896557 + 1362715220*x + 1505007200*x^2 + 1229
558400*x^3 + 615168000*x^4 + 138240000*x^5) + 8035077963*Sqrt[10]*ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]]))/Sqrt[-1
+ 2*x]

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IntegrateAlgebraic [A]  time = 0.25, size = 157, normalized size = 1.05 \begin {gather*} -\frac {1331 \sqrt {1-2 x} \left (\frac {18865228125 (1-2 x)^5}{(5 x+3)^5}+\frac {42761183750 (1-2 x)^4}{(5 x+3)^4}+\frac {39843361800 (1-2 x)^3}{(5 x+3)^3}+\frac {19386846000 (1-2 x)^2}{(5 x+3)^2}+\frac {5110810640 (1-2 x)}{5 x+3}[phone]\right )}{2457600 \sqrt {5 x+3} \left (\frac {5 (1-2 x)}{5 x+3}+2\right )^6}-\frac {2678359321 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )}{819200 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-1331*Sqrt[1 - 2*x]*(649776864 + (18865228125*(1 - 2*x)^5)/(3 + 5*x)^5 + (42761183750*(1 - 2*x)^4)/(3 + 5*x)^
4 + (39843361800*(1 - 2*x)^3)/(3 + 5*x)^3 + (19386846000*(1 - 2*x)^2)/(3 + 5*x)^2 + (5110810640*(1 - 2*x))/(3
+ 5*x)))/(2457600*Sqrt[3 + 5*x]*(2 + (5*(1 - 2*x))/(3 + 5*x))^6) - (2678359321*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x]
)/Sqrt[3 + 5*x]])/(819200*Sqrt[10])

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fricas [A]  time = 1.14, size = 82, normalized size = 0.55 \begin {gather*} -\frac {1}{2457600} \, {\left (138240000 \, x^{5} + 615168000 \, x^{4} + 1229558400 \, x^{3} + 1505007200 \, x^{2} + 1362715220 \, x + 1202896557\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} - \frac {2678359321}{16384000} \, \sqrt {10} \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 ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2457600*(138240000*x^5 + 615168000*x^4 + 1229558400*x^3 + 1505007200*x^2 + 1362715220*x + 1202896557)*sqrt(
5*x + 3)*sqrt(-2*x + 1) - 2678359321/16384000*sqrt(10)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x
+ 1)/(10*x^2 + x - 3))

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giac [A]  time = 1.10, size = 81, normalized size = 0.54 \begin {gather*} -\frac {1}{122880000} \, \sqrt {5} {\left (2 \, {\left (4 \, {\left (8 \, {\left (108 \, {\left (16 \, {\left (20 \, x + 41\right )} {\left (5 \, x + 3\right )} + 2903\right )} {\left (5 \, x + 3\right )} + 2012291\right )} {\left (5 \, x + 3\right )} + 110676005\right )} {\left (5 \, x + 3\right )} + 3652308165\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - 40175389815 \, \sqrt {2} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/122880000*sqrt(5)*(2*(4*(8*(108*(16*(20*x + 41)*(5*x + 3) + 2903)*(5*x + 3) + 2012291)*(5*x + 3) + 11067600
5)*(5*x + 3) + 3652308165)*sqrt(5*x + 3)*sqrt(-10*x + 5) - 40175389815*sqrt(2)*arcsin(1/11*sqrt(22)*sqrt(5*x +
3)))

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maple [A]  time = 0.01, size = 138, normalized size = 0.92 \begin {gather*} \frac {\sqrt {5 x +3}\, \sqrt {-2 x +1}\, \left (-2764800000 \sqrt {-10 x^{2}-x +3}\, x^{5}-12303360000 \sqrt {-10 x^{2}-x +3}\, x^{4}-24591168000 \sqrt {-10 x^{2}-x +3}\, x^{3}-30100144000 \sqrt {-10 x^{2}-x +3}\, x^{2}-27254304400 \sqrt {-10 x^{2}-x +3}\, x [phone] \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-24057931140 \sqrt {-10 x^{2}-x +3}\right )}{49152000 \sqrt {-10 x^{2}-x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/49152000*(5*x+3)^(1/2)*(-2*x+1)^(1/2)*(-2764800000*(-10*x^2-x+3)^(1/2)*x^5-12303360000*(-10*x^2-x+3)^(1/2)*x
^4-24591168000*(-10*x^2-x+3)^(1/2)*x^3-30100144000*(-10*x^2-x+3)^(1/2)*x^2+8035077963*10^(1/2)*arcsin(20/11*x+
1/11)-27254304400*(-10*x^2-x+3)^(1/2)*x-24057931140*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)

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maxima [A]  time = 1.17, size = 109, normalized size = 0.73 \begin {gather*} -\frac {225}{4} \, \sqrt {-10 \, x^{2} - x + 3} x^{5} - \frac {4005}{16} \, \sqrt {-10 \, x^{2} - x + 3} x^{4} - \frac {128079}{256} \, \sqrt {-10 \, x^{2} - x + 3} x^{3} - \frac {1881259}{3072} \, \sqrt {-10 \, x^{2} - x + 3} x^{2} - \frac {68135761}{122880} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {2678359321}{16384000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) - \frac {400965519}{819200} \, \sqrt {-10 \, x^{2} - x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-225/4*sqrt(-10*x^2 - x + 3)*x^5 - 4005/16*sqrt(-10*x^2 - x + 3)*x^4 - 128079/256*sqrt(-10*x^2 - x + 3)*x^3 -
1881259/3072*sqrt(-10*x^2 - x + 3)*x^2 - 68135761/122880*sqrt(-10*x^2 - x + 3)*x - 2678359321/16384000*sqrt(10
)*arcsin(-20/11*x - 1/11) - 400965519/819200*sqrt(-10*x^2 - x + 3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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