\(\int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx\) [2580]

Optimal result

Integrand size = 26, antiderivative size = 142 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=-\frac {697}{88} \sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}-\frac {299 (2+3 x)^3 \sqrt {3+5 x}}{66 \sqrt {1-2 x}}+\frac {(2+3 x)^4 \sqrt {3+5 x}}{3 (1-2 x)^{3/2}}-\frac {\sqrt {1-2 x} \sqrt {3+5 x} (17606479+7306140 x)}{70400}+\frac {13246251 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{6400 \sqrt {10}} \]

[Out]

13246251/64000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)+1/3*(2+3*x)^4*(3+5*x)^(1/2)/(1-2*x)^(3/2)-299/66*(
2+3*x)^3*(3+5*x)^(1/2)/(1-2*x)^(1/2)-697/88*(2+3*x)^2*(1-2*x)^(1/2)*(3+5*x)^(1/2)-1/70400*(17606479+7306140*x)
*(1-2*x)^(1/2)*(3+5*x)^(1/2)

Rubi [A] (verified)

Time = 0.03 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {99, 155, 158, 152, 56, 222} \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\frac {13246251 \arcsin \left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{6400 \sqrt {10}}+\frac {\sqrt {5 x+3} (3 x+2)^4}{3 (1-2 x)^{3/2}}-\frac {299 \sqrt {5 x+3} (3 x+2)^3}{66 \sqrt {1-2 x}}-\frac {697}{88} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^2-\frac {\sqrt {1-2 x} \sqrt {5 x+3} (7306140 x+17606479)}{70400} \]

[In]

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

[Out]

(-697*Sqrt[1 - 2*x]*(2 + 3*x)^2*Sqrt[3 + 5*x])/88 - (299*(2 + 3*x)^3*Sqrt[3 + 5*x])/(66*Sqrt[1 - 2*x]) + ((2 +
3*x)^4*Sqrt[3 + 5*x])/(3*(1 - 2*x)^(3/2)) - (Sqrt[1 - 2*x]*Sqrt[3 + 5*x]*(17606479 + 7306140*x))/70400 + (132
46251*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(6400*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 99

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 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 158

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 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*} \text {integral}& = \frac {(2+3 x)^4 \sqrt {3+5 x}}{3 (1-2 x)^{3/2}}-\frac {1}{3} \int \frac {(2+3 x)^3 \left (41+\frac {135 x}{2}\right )}{(1-2 x)^{3/2} \sqrt {3+5 x}} \, dx \\ & = -\frac {299 (2+3 x)^3 \sqrt {3+5 x}}{66 \sqrt {1-2 x}}+\frac {(2+3 x)^4 \sqrt {3+5 x}}{3 (1-2 x)^{3/2}}-\frac {1}{33} \int \frac {\left (-4779-\frac {31365 x}{4}\right ) (2+3 x)^2}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx \\ & = -\frac {697}{88} \sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}-\frac {299 (2+3 x)^3 \sqrt {3+5 x}}{66 \sqrt {1-2 x}}+\frac {(2+3 x)^4 \sqrt {3+5 x}}{3 (1-2 x)^{3/2}}+\frac {1}{990} \int \frac {(2+3 x) \left (\frac {1680165}{4}+\frac {5479605 x}{8}\right )}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx \\ & = -\frac {697}{88} \sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}-\frac {299 (2+3 x)^3 \sqrt {3+5 x}}{66 \sqrt {1-2 x}}+\frac {(2+3 x)^4 \sqrt {3+5 x}}{3 (1-2 x)^{3/2}}-\frac {\sqrt {1-2 x} \sqrt {3+5 x} (17606479+7306140 x)}{70400}+\frac {13246251 \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx}{12800} \\ & = -\frac {697}{88} \sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}-\frac {299 (2+3 x)^3 \sqrt {3+5 x}}{66 \sqrt {1-2 x}}+\frac {(2+3 x)^4 \sqrt {3+5 x}}{3 (1-2 x)^{3/2}}-\frac {\sqrt {1-2 x} \sqrt {3+5 x} (17606479+7306140 x)}{70400}+\frac {13246251 \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{6400 \sqrt {5}} \\ & = -\frac {697}{88} \sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}-\frac {299 (2+3 x)^3 \sqrt {3+5 x}}{66 \sqrt {1-2 x}}+\frac {(2+3 x)^4 \sqrt {3+5 x}}{3 (1-2 x)^{3/2}}-\frac {\sqrt {1-2 x} \sqrt {3+5 x} (17606479+7306140 x)}{70400}+\frac {13246251 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{6400 \sqrt {10}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.19 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.58 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\frac {-10 \sqrt {3+5 x} \left (66038637-183672928 x+52700868 x^2+15040080 x^3+2851200 x^4\right )[phone] \sqrt {10-20 x} (-1+2 x) \arctan \left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )}{2112000 (1-2 x)^{3/2}} \]

[In]

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

[Out]

(-10*Sqrt[3 + 5*x]*(66038637 - 183672928*x + 52700868*x^2 + 15040080*x^3 + 2851200*x^4) + 437126283*Sqrt[10 -
20*x]*(-1 + 2*x)*ArcTan[Sqrt[5/2 - 5*x]/Sqrt[3 + 5*x]])/(2112000*(1 - 2*x)^(3/2))

Maple [A] (verified)

Time = 4.06 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.08

method	result	size
default	\(\frac {\left (-57024000 x^{4} \sqrt {-10 x^{2}-x +3}[phone] \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x^{2}-300801600 x^{3} \sqrt {-10 x^{2}-x +3}-1748505132 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right ) x -1054017360 x^{2} \sqrt {-10 x^{2}-x +3}[phone] \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )[phone] x \sqrt {-10 x^{2}-x +3}-1320772740 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {1-2 x}\, \sqrt {3+5 x}}{4224000 \left (-1+2 x \right )^{2} \sqrt {-10 x^{2}-x +3}}\)	\(154\)

[In]

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

[Out]

1/4224000*(-57024000*x^4*(-10*x^2-x+3)^(1/2)[phone]*10^(1/2)*arcsin(20/11*x+1/11)*x^2-300801600*x^3*(-10*x
^2-x+3)^(1/2)-1748505132*10^(1/2)*arcsin(20/11*x+1/11)*x-1054017360*x^2*(-10*x^2-x+3)^(1/2)[phone]*10^(1/2)
*arcsin(20/11*x+1/11)[phone]*x*(-10*x^2-x+3)^(1/2)-1320772740*(-10*x^2-x+3)^(1/2))*(1-2*x)^(1/2)*(3+5*x)^(
1/2)/(-1+2*x)^2/(-10*x^2-x+3)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.71 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=-\frac {437126283 \, \sqrt {10} {\left (4 \, x^{2} - 4 \, x + 1\right )} \arctan \left (\frac {\sqrt {10} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 20 \, {\left (2851200 \, x^{4} + 15040080 \, x^{3} + 52700868 \, x^{2} - 183672928 \, x + 66038637\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{4224000 \, {\left (4 \, x^{2} - 4 \, x + 1\right )}} \]

[In]

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

[Out]

-1/4224000*(437126283*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*(2851200*x^4 + 15040080*x^3 + 52700868*x^2 - 183672928*x + 66038637)*sqrt(5*x + 3)*sqrt
(-2*x + 1))/(4*x^2 - 4*x + 1)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-1)]

Timed out. \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.68 \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\frac {13246251}{64000} \, \sqrt {10} \arcsin \left (\frac {1}{11} \, \sqrt {22} \sqrt {5 \, x + 3}\right ) - \frac {{\left (4 \, {\left (891 \, {\left (4 \, {\left (8 \, \sqrt {5} {\left (5 \, x + 3\right )} + 115 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 8919 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} - 291417650 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 4808389113 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{26400000 \, {\left (2 \, x - 1\right )}^{2}} \]

[In]

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

[Out]

13246251/64000*sqrt(10)*arcsin(1/11*sqrt(22)*sqrt(5*x + 3)) - 1/26400000*(4*(891*(4*(8*sqrt(5)*(5*x + 3) + 115
*sqrt(5))*(5*x + 3) + 8919*sqrt(5))*(5*x + 3) - 291417650*sqrt(5))*(5*x + 3) + 4808389113*sqrt(5))*sqrt(5*x +
3)*sqrt(-10*x + 5)/(2*x - 1)^2

Mupad [F(-1)]

Timed out. \[ \int \frac {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx=\int \frac {{\left (3\,x+2\right )}^4\,\sqrt {5\,x+3}}{{\left (1-2\,x\right )}^{5/2}} \,d x \]

[In]

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

[Out]

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