3.26.62 \(\int \frac {(2+3 x)^4}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx\) [2562]

Optimal. Leaf size=113 \[ -\frac {37 \sqrt {1-2 x} (2+3 x)^2}{605 \sqrt {3+5 x}}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {3 \sqrt {1-2 x} \sqrt {3+5 x} (173063+72060 x)}{96800}-\frac {35451 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{800 \sqrt {10}} \]

[Out]

-35451/8000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+7/11*(2+3*x)^3/(1-2*x)^(1/2)/(3+5*x)^(1/2)-37/605*(2+
3*x)^2*(1-2*x)^(1/2)/(3+5*x)^(1/2)+3/96800*(173063+72060*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 {35451 \text {ArcSin}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{800 \sqrt {10}}+\frac {7 (3 x+2)^3}{11 \sqrt {1-2 x} \sqrt {5 x+3}}-\frac {37 \sqrt {1-2 x} (3 x+2)^2}{605 \sqrt {5 x+3}}+\frac {3 \sqrt {1-2 x} \sqrt {5 x+3} (72060 x+173063)}{96800} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-37*Sqrt[1 - 2*x]*(2 + 3*x)^2)/(605*Sqrt[3 + 5*x]) + (7*(2 + 3*x)^3)/(11*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) + (3*Sq
rt[1 - 2*x]*Sqrt[3 + 5*x]*(173063 + 72060*x))/96800 - (35451*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(800*Sqrt[10])

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}{(1-2 x)^{3/2} (3+5 x)^{3/2}} \, dx &=\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {1}{11} \int \frac {(2+3 x)^2 \left (152+\frac {519 x}{2}\right )}{\sqrt {1-2 x} (3+5 x)^{3/2}} \, dx\\ &=-\frac {37 \sqrt {1-2 x} (2+3 x)^2}{605 \sqrt {3+5 x}}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {2}{605} \int \frac {(2+3 x) \left (\frac {5487}{2}+\frac {18015 x}{4}\right )}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx\\ &=-\frac {37 \sqrt {1-2 x} (2+3 x)^2}{605 \sqrt {3+5 x}}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {3 \sqrt {1-2 x} \sqrt {3+5 x} (173063+72060 x)}{96800}-\frac {35451 \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx}{1600}\\ &=-\frac {37 \sqrt {1-2 x} (2+3 x)^2}{605 \sqrt {3+5 x}}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {3 \sqrt {1-2 x} \sqrt {3+5 x} (173063+72060 x)}{96800}-\frac {35451 \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{800 \sqrt {5}}\\ &=-\frac {37 \sqrt {1-2 x} (2+3 x)^2}{605 \sqrt {3+5 x}}+\frac {7 (2+3 x)^3}{11 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {3 \sqrt {1-2 x} \sqrt {3+5 x} (173063+72060 x)}{96800}-\frac {35451 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{800 \sqrt {10}}\\ \end {align*}

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Mathematica [A]
time = 0.16, size = 82, normalized size = 0.73 \begin {gather*} \frac {10 \left (2026687+2323271 x-1992870 x^2-392040 x^3\right )+4289571 \sqrt {10-20 x} \sqrt {3+5 x} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{968000 \sqrt {1-2 x} \sqrt {3+5 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*(2026687 + 2323271*x - 1992870*x^2 - 392040*x^3) + 4289571*Sqrt[10 - 20*x]*Sqrt[3 + 5*x]*ArcTan[Sqrt[5/2 -
5*x]/Sqrt[3 + 5*x]])/(968000*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])

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Maple [A]
time = 0.08, size = 137, normalized size = 1.21

method	result	size
default	\(-\frac {\sqrt {1-2 x}\, \left (42895710 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-7840800 x^{3} \sqrt {-10 x^{2}-x +3}+4289571 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -39857400 x^{2} \sqrt {-10 x^{2}-x +3}-12868713 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )[phone] x \sqrt {-10 x^{2}-x +3}[phone] \sqrt {-10 x^{2}-x +3}\right )}{1936000 \left (-1+2 x \right ) \sqrt {-10 x^{2}-x +3}\, \sqrt {3+5 x}}\)	\(137\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1936000*(1-2*x)^(1/2)*(42895710*10^(1/2)*arcsin(20/11*x+1/11)*x^2-7840800*x^3*(-10*x^2-x+3)^(1/2)+4289571*1
0^(1/2)*arcsin(20/11*x+1/11)*x-39857400*x^2*(-10*x^2-x+3)^(1/2)-12868713*10^(1/2)*arcsin(20/11*x+1/11)+4646542
0*x*(-10*x^2-x+3)^(1/2)[phone]*(-10*x^2-x+3)^(1/2))/(-1+2*x)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(1/2)

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Maxima [A]
time = 0.66, size = 75, normalized size = 0.66 \begin {gather*} -\frac {81 \, x^{3}}{20 \, \sqrt {-10 \, x^{2} - x + 3}} - \frac {1647 \, x^{2}}{80 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {35451}{16000} \, \sqrt {10} \arcsin \left (-\frac {20}{11} \, x - \frac {1}{11}\right ) + \frac {2323271 \, x}{96800 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {2026687}{96800 \, \sqrt {-10 \, x^{2} - x + 3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-81/20*x^3/sqrt(-10*x^2 - x + 3) - 1647/80*x^2/sqrt(-10*x^2 - x + 3) + 35451/16000*sqrt(10)*arcsin(-20/11*x -
1/11) + 2323271/96800*x/sqrt(-10*x^2 - x + 3) + 2026687/96800/sqrt(-10*x^2 - x + 3)

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Fricas [A]
time = 0.45, size = 92, normalized size = 0.81 \begin {gather*} \frac {4289571 \, \sqrt {10} {\left (10 \, x^{2} + x - 3\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 (392040 \, x^{3} + 1992870 \, x^{2} - 2323271 \, x - 2026687\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1936000 \, {\left (10 \, x^{2} + x - 3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1936000*(4289571*sqrt(10)*(10*x^2 + x - 3)*arctan(1/20*sqrt(10)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*
x^2 + x - 3)) + 20*(392040*x^3 + 1992870*x^2 - 2323271*x - 2026687)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(10*x^2 + x
- 3)

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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}}{\left (1 - 2 x\right )^{\frac {3}{2}} \left (5 x + 3\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.93, size = 131, normalized size = 1.16 \begin {gather*} -\frac {35451}{8000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) + \frac {{\left (6534 \, {\left (12 \, \sqrt {5} {\left (5 \, x + 3\right )} + 197 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 21456431 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{12100000 \, {\left (2 \, x - 1\right )}} - \frac {\sqrt {10} {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{151250 \, \sqrt {5 \, x + 3}} + \frac {2 \, \sqrt {10} \sqrt {5 \, x + 3}}{75625 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-35451/8000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) + 1/12100000*(6534*(12*sqrt(5)*(5*x + 3) + 197*sqrt(5
))*(5*x + 3) - 21456431*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1) - 1/151250*sqrt(10)*(sqrt(2)*sqrt(-10
*x + 5) - sqrt(22))/sqrt(5*x + 3) + 2/75625*sqrt(10)*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))

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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}{{\left (1-2\,x\right )}^{3/2}\,{\left (5\,x+3\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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