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

Optimal. Leaf size=142 \[ \frac {\sqrt {5 x+3} (3 x+2)^4}{3 (1-2 x)^{3/2}}-\frac {299 \sqrt {5 x+3} (3 x+2)^3}{66 \sqrt {1-2 x}}-\frac {697}{88} \sqrt {1-2 x} \sqrt {5 x+3} (3 x+2)^2-\frac {\sqrt {1-2 x} \sqrt {5 x+3} (7306140 x+17606479)}{70400}+\frac {13246251 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{6400 \sqrt {10}} \]

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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 {\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}+\frac {13246251 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{6400 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[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 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 {(2+3 x)^4 \sqrt {3+5 x}}{(1-2 x)^{5/2}} \, dx &=\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 \operatorname {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*}

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Mathematica [A]  time = 0.07, size = 95, normalized size = 0.67 \begin {gather*} \frac {10 \sqrt {2 x-1} \sqrt {5 x+3} \left (2851200 x^4+15040080 x^3+52700868 x^2-183672928 x+66038637\right )[phone] \sqrt {10} (1-2 x)^2 \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{2112000 \sqrt {1-2 x} (2 x-1)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[-1 + 2*x]*Sqrt[3 + 5*x]*(66038637 - 183672928*x + 52700868*x^2 + 15040080*x^3 + 2851200*x^4) + 437126
283*Sqrt[10]*(1 - 2*x)^2*ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]])/(2112000*Sqrt[1 - 2*x]*(-1 + 2*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.20, size = 141, normalized size = 0.99 \begin {gather*} \frac {(5 x+3)^{3/2} \left (-\frac {10928157075 (1-2 x)^4}{(5 x+3)^4}-\frac {11656698320 (1-2 x)^3}{(5 x+3)^3}-\frac {3846667668 (1-2 x)^2}{(5 x+3)^2}-\frac {319401600 (1-2 x)}{5 x+3}[phone]\right )}{211200 (1-2 x)^{3/2} \left (\frac {5 (1-2 x)}{5 x+3}+2\right )^3}-\frac {13246251 \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}} \sqrt {1-2 x}}{\sqrt {5 x+3}}\right )}{6400 \sqrt {10}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3 + 5*x)^(3/2)*(15366400 - (10928157075*(1 - 2*x)^4)/(3 + 5*x)^4 - (11656698320*(1 - 2*x)^3)/(3 + 5*x)^3 - (
3846667668*(1 - 2*x)^2)/(3 + 5*x)^2 - (319401600*(1 - 2*x))/(3 + 5*x)))/(211200*(1 - 2*x)^(3/2)*(2 + (5*(1 - 2
*x))/(3 + 5*x))^3) - (13246251*ArcTan[(Sqrt[5/2]*Sqrt[1 - 2*x])/Sqrt[3 + 5*x]])/(6400*Sqrt[10])

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fricas [A]  time = 1.31, size = 101, normalized size = 0.71 \begin {gather*} -\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [A]  time = 1.09, size = 97, normalized size = 0.68 \begin {gather*} \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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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maple [A]  time = 0.02, size = 154, normalized size = 1.08 \begin {gather*} \frac {\left (-57024000 \sqrt {-10 x^{2}-x +3}\, x^{4}-300801600 \sqrt {-10 x^{2}-x +3}\, x^{3}[phone] \sqrt {10}\, x^{2} \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-1054017360 \sqrt {-10 x^{2}-x +3}\, x^{2}-1748505132 \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 )-1320772740 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}\, \sqrt {5 x +3}}{4224000 \left (2 x -1\right )^{2} \sqrt {-10 x^{2}-x +3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4224000*(-57024000*(-10*x^2-x+3)^(1/2)*x^4+1748505132*10^(1/2)*x^2*arcsin(20/11*x+1/11)-300801600*(-10*x^2-x
+3)^(1/2)*x^3-1748505132*10^(1/2)*x*arcsin(20/11*x+1/11)-1054017360*(-10*x^2-x+3)^(1/2)*x^2+437126283*10^(1/2)
*arcsin(20/11*x+1/11)[phone]*(-10*x^2-x+3)^(1/2)*x-1320772740*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)*(5*x+3)^
(1/2)/(2*x-1)^2/(-10*x^2-x+3)^(1/2)

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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*(3+5*x)^(1/2)/(1-2*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 {{\left (3\,x+2\right )}^4\,\sqrt {5\,x+3}}{{\left (1-2\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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