3.26.8 \(\int \frac {(2+3 x)^4}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx\) [2508]

Optimal. Leaf size=113 \[ -\frac {2 \sqrt {1-2 x} (2+3 x)^3}{165 (3+5 x)^{3/2}}-\frac {602 \sqrt {1-2 x} (2+3 x)^2}{9075 \sqrt {3+5 x}}-\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (12199+1020 x)}{242000}+\frac {8127 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{2000 \sqrt {10}} \]

[Out]

8127/20000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-2/165*(2+3*x)^3*(1-2*x)^(1/2)/(3+5*x)^(3/2)-602/9075*(
2+3*x)^2*(1-2*x)^(1/2)/(3+5*x)^(1/2)-7/242000*(12199+1020*x)*(1-2*x)^(1/2)*(3+5*x)^(1/2)

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Rubi [A]
time = 0.02, 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 = {100, 155, 152, 56, 222} \begin {gather*} \frac {8127 \text {ArcSin}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{2000 \sqrt {10}}-\frac {2 \sqrt {1-2 x} (3 x+2)^3}{165 (5 x+3)^{3/2}}-\frac {602 \sqrt {1-2 x} (3 x+2)^2}{9075 \sqrt {5 x+3}}-\frac {7 \sqrt {1-2 x} \sqrt {5 x+3} (1020 x+12199)}{242000} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[1 - 2*x]*(2 + 3*x)^3)/(165*(3 + 5*x)^(3/2)) - (602*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(9075*Sqrt[3 + 5*x]) -
(7*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(12199 + 1020*x))/242000 + (8127*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(2000*Sqrt[1
0])

Rule 56

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*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 152

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 155

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 222

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)^4}{\sqrt {1-2 x} (3+5 x)^{5/2}} \, dx &=-\frac {2 \sqrt {1-2 x} (2+3 x)^3}{165 (3+5 x)^{3/2}}-\frac {2}{165} \int \frac {\left (-112-\frac {273 x}{2}\right ) (2+3 x)^2}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx\\ &=-\frac {2 \sqrt {1-2 x} (2+3 x)^3}{165 (3+5 x)^{3/2}}-\frac {602 \sqrt {1-2 x} (2+3 x)^2}{9075 \sqrt {3+5 x}}-\frac {4 \int \frac {\left (-\frac {4809}{2}-\frac {1785 x}{4}\right ) (2+3 x)}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx}{9075}\\ &=-\frac {2 \sqrt {1-2 x} (2+3 x)^3}{165 (3+5 x)^{3/2}}-\frac {602 \sqrt {1-2 x} (2+3 x)^2}{9075 \sqrt {3+5 x}}-\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (12199+1020 x)}{242000}+\frac {8127 \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx}{4000}\\ &=-\frac {2 \sqrt {1-2 x} (2+3 x)^3}{165 (3+5 x)^{3/2}}-\frac {602 \sqrt {1-2 x} (2+3 x)^2}{9075 \sqrt {3+5 x}}-\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (12199+1020 x)}{242000}+\frac {8127 \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{2000 \sqrt {5}}\\ &=-\frac {2 \sqrt {1-2 x} (2+3 x)^3}{165 (3+5 x)^{3/2}}-\frac {602 \sqrt {1-2 x} (2+3 x)^2}{9075 \sqrt {3+5 x}}-\frac {7 \sqrt {1-2 x} \sqrt {3+5 x} (12199+1020 x)}{242000}+\frac {8127 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{2000 \sqrt {10}}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 69, normalized size = 0.61 \begin {gather*} -\frac {\sqrt {1-2 x} \left (2953931+10891910 x+11712195 x^2+2940300 x^3\right )}{726000 (3+5 x)^{3/2}}-\frac {8127 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{2000 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/726000*(Sqrt[1 - 2*x]*(2953931 + 10891910*x + 11712195*x^2 + 2940300*x^3))/(3 + 5*x)^(3/2) - (8127*ArcTan[S
qrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(2000*Sqrt[10])

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Maple [A]
time = 0.08, size = 130, normalized size = 1.15

method	result	size
default	\(\frac {\left (73752525 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-58806000 x^{3} \sqrt {-10 x^{2}-x +3}[phone] \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -234243900 x^{2} \sqrt {-10 x^{2}-x +3}[phone] \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-217838200 x \sqrt {-10 x^{2}-x +3}-59078620 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}}{14520000 \sqrt {-10 x^{2}-x +3}\, \left (3+5 x \right )^{\frac {3}{2}}}\)	\(130\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/14520000*(73752525*10^(1/2)*arcsin(20/11*x+1/11)*x^2-58806000*x^3*(-10*x^2-x+3)^(1/2)[phone]*10^(1/2)*arcs
in(20/11*x+1/11)*x-234243900*x^2*(-10*x^2-x+3)^(1/2)[phone]*10^(1/2)*arcsin(20/11*x+1/11)-217838200*x*(-10*x
^2-x+3)^(1/2)-59078620*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)

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Maxima [A]
time = 0.60, size = 91, normalized size = 0.81 \begin {gather*} \frac {8127}{40000} \, \sqrt {5} \sqrt {2} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {81}{500} \, \sqrt {-10 \, x^{2} - x + 3} x - \frac {4509}{10000} \, \sqrt {-10 \, x^{2} - x + 3} - \frac {2 \, \sqrt {-10 \, x^{2} - x + 3}}{20625 \, {\left (25 \, x^{2} + 30 \, x + 9\right )}} - \frac {32 \, \sqrt {-10 \, x^{2} - x + 3}}{9075 \, {\left (5 \, x + 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

8127/40000*sqrt(5)*sqrt(2)*arcsin(20/11*x + 1/11) - 81/500*sqrt(-10*x^2 - x + 3)*x - 4509/10000*sqrt(-10*x^2 -
x + 3) - 2/20625*sqrt(-10*x^2 - x + 3)/(25*x^2 + 30*x + 9) - 32/9075*sqrt(-10*x^2 - x + 3)/(5*x + 3)

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Fricas [A]
time = 0.52, size = 96, normalized size = 0.85 \begin {gather*} -\frac {2950101 \, \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 (2940300 \, x^{3} + 11712195 \, x^{2} + 10891910 \, x + 2953931\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14520000 \, {\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)^(5/2)/(1-2*x)^(1/2),x, algorithm="fricas")

[Out]

-1/14520000*(2950101*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*(2940300*x^3 + 11712195*x^2 + 10891910*x + 2953931)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(25*
x^2 + 30*x + 9)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (3 x + 2\right )^{4}}{\sqrt {1 - 2 x} \left (5 x + 3\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((3*x + 2)**4/(sqrt(1 - 2*x)*(5*x + 3)**(5/2)), x)

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Giac [A]
time = 1.07, size = 171, normalized size = 1.51 \begin {gather*} -\frac {27}{50000} \, {\left (12 \, \sqrt {5} {\left (5 \, x + 3\right )} + 131 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} - \frac {1}{18150000} \, \sqrt {10} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{3}}{{\left (5 \, x + 3\right )}^{\frac {3}{2}}} + \frac {3204 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}}\right )} + \frac {8127}{20000} \, \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 {801 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} + 4\right )}}{1134375 \, {\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/(3+5*x)^(5/2)/(1-2*x)^(1/2),x, algorithm="giac")

[Out]

-27/50000*(12*sqrt(5)*(5*x + 3) + 131*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) - 1/18150000*sqrt(10)*((sqrt(2)*s
qrt(-10*x + 5) - sqrt(22))^3/(5*x + 3)^(3/2) + 3204*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3)) + 8127
/20000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/1134375*sqrt(10)*(5*x + 3)^(3/2)*(801*(sqrt(2)*sqrt(-1
0*x + 5) - sqrt(22))^2/(5*x + 3) + 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^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 )}^4}{\sqrt {1-2\,x}\,{\left (5\,x+3\right )}^{5/2}} \,d x \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)^(5/2)),x)

[Out]

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

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