3.24.92 \(\int \frac {(1-2 x)^{5/2} \sqrt {3+5 x}}{2+3 x} \, dx\) [2392]

Optimal. Leaf size=128 \[ \frac {6401 \sqrt {1-2 x} \sqrt {3+5 x}}{5400}+\frac {59}{180} (1-2 x)^{3/2} \sqrt {3+5 x}+\frac {1}{9} (1-2 x)^{5/2} \sqrt {3+5 x}+\frac {250433 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{16200 \sqrt {10}}+\frac {98}{81} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right ) \]

[Out]

98/81*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+250433/162000*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2)
)*10^(1/2)+59/180*(1-2*x)^(3/2)*(3+5*x)^(1/2)+1/9*(1-2*x)^(5/2)*(3+5*x)^(1/2)+6401/5400*(1-2*x)^(1/2)*(3+5*x)^
(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {103, 159, 163, 56, 222, 95, 210} \begin {gather*} \frac {250433 \text {ArcSin}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )}{16200 \sqrt {10}}+\frac {98}{81} \sqrt {7} \text {ArcTan}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )+\frac {1}{9} \sqrt {5 x+3} (1-2 x)^{5/2}+\frac {59}{180} \sqrt {5 x+3} (1-2 x)^{3/2}+\frac {6401 \sqrt {5 x+3} \sqrt {1-2 x}}{5400} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(6401*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/5400 + (59*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x])/180 + ((1 - 2*x)^(5/2)*Sqrt[3 + 5
*x])/9 + (250433*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/(16200*Sqrt[10]) + (98*Sqrt[7]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7
]*Sqrt[3 + 5*x])])/81

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 95

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 103

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

Rule 159

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] && IntegersQ[2*m, 2
*n, 2*p]

Rule 163

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 210

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

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 {(1-2 x)^{5/2} \sqrt {3+5 x}}{2+3 x} \, dx &=\frac {1}{9} (1-2 x)^{5/2} \sqrt {3+5 x}-\frac {1}{9} \int \frac {\left (-52-\frac {177 x}{2}\right ) (1-2 x)^{3/2}}{(2+3 x) \sqrt {3+5 x}} \, dx\\ &=\frac {59}{180} (1-2 x)^{3/2} \sqrt {3+5 x}+\frac {1}{9} (1-2 x)^{5/2} \sqrt {3+5 x}-\frac {1}{270} \int \frac {\left (-\frac {5421}{2}-\frac {19203 x}{4}\right ) \sqrt {1-2 x}}{(2+3 x) \sqrt {3+5 x}} \, dx\\ &=\frac {6401 \sqrt {1-2 x} \sqrt {3+5 x}}{5400}+\frac {59}{180} (1-2 x)^{3/2} \sqrt {3+5 x}+\frac {1}{9} (1-2 x)^{5/2} \sqrt {3+5 x}-\frac {\int \frac {-\frac {181833}{4}-\frac {751299 x}{8}}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{4050}\\ &=\frac {6401 \sqrt {1-2 x} \sqrt {3+5 x}}{5400}+\frac {59}{180} (1-2 x)^{3/2} \sqrt {3+5 x}+\frac {1}{9} (1-2 x)^{5/2} \sqrt {3+5 x}-\frac {343}{81} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx+\frac {250433 \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx}{32400}\\ &=\frac {6401 \sqrt {1-2 x} \sqrt {3+5 x}}{5400}+\frac {59}{180} (1-2 x)^{3/2} \sqrt {3+5 x}+\frac {1}{9} (1-2 x)^{5/2} \sqrt {3+5 x}-\frac {686}{81} \text {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )+\frac {250433 \text {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{16200 \sqrt {5}}\\ &=\frac {6401 \sqrt {1-2 x} \sqrt {3+5 x}}{5400}+\frac {59}{180} (1-2 x)^{3/2} \sqrt {3+5 x}+\frac {1}{9} (1-2 x)^{5/2} \sqrt {3+5 x}+\frac {250433 \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )}{16200 \sqrt {10}}+\frac {98}{81} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 114, normalized size = 0.89 \begin {gather*} \frac {30 \sqrt {1-2 x} \left (26313+26035 x-22500 x^2+12000 x^3\right )-250433 \sqrt {30+50 x} \tan ^{-1}\left (\frac {\sqrt {\frac {5}{2}-5 x}}{\sqrt {3+5 x}}\right )+196000 \sqrt {7} \sqrt {3+5 x} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{162000 \sqrt {3+5 x}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(30*Sqrt[1 - 2*x]*(26313 + 26035*x - 22500*x^2 + 12000*x^3) - 250433*Sqrt[30 + 50*x]*ArcTan[Sqrt[5/2 - 5*x]/Sq
rt[3 + 5*x]] + 196000*Sqrt[7]*Sqrt[3 + 5*x]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(162000*Sqrt[3 + 5*
x])

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Maple [A]
time = 0.12, size = 115, normalized size = 0.90

method	result	size
default	\(\frac {\sqrt {1-2 x}\, \sqrt {3+5 x}\, \left (144000 x^{2} \sqrt {-10 x^{2}-x +3}+250433 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )-196000 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-356400 x \sqrt {-10 x^{2}-x +3}+526260 \sqrt {-10 x^{2}-x +3}\right )}{324000 \sqrt {-10 x^{2}-x +3}}\)	\(115\)
risch	\(-\frac {\left (2400 x^{2}-5940 x +8771\right ) \sqrt {3+5 x}\, \left (-1+2 x \right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{5400 \sqrt {-\left (3+5 x \right ) \left (-1+2 x \right )}\, \sqrt {1-2 x}}-\frac {\left (-\frac {250433 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )}{324000}+\frac {49 \sqrt {7}\, \arctan \left (\frac {9 \left (\frac {20}{3}+\frac {37 x}{3}\right ) \sqrt {7}}{14 \sqrt {-90 \left (\frac {2}{3}+x \right )^{2}+67+111 x}}\right )}{81}\right ) \sqrt {\left (1-2 x \right ) \left (3+5 x \right )}}{\sqrt {1-2 x}\, \sqrt {3+5 x}}\)	\(131\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/324000*(1-2*x)^(1/2)*(3+5*x)^(1/2)*(144000*x^2*(-10*x^2-x+3)^(1/2)+250433*10^(1/2)*arcsin(20/11*x+1/11)-1960
00*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-356400*x*(-10*x^2-x+3)^(1/2)+526260*(-10*x^2-x+3
)^(1/2))/(-10*x^2-x+3)^(1/2)

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Maxima [A]
time = 0.65, size = 83, normalized size = 0.65 \begin {gather*} -\frac {2}{45} \, {\left (-10 \, x^{2} - x + 3\right )}^{\frac {3}{2}} - \frac {103}{90} \, \sqrt {-10 \, x^{2} - x + 3} x + \frac {250433}{324000} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) - \frac {49}{81} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {9491}{5400} \, \sqrt {-10 \, x^{2} - x + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/45*(-10*x^2 - x + 3)^(3/2) - 103/90*sqrt(-10*x^2 - x + 3)*x + 250433/324000*sqrt(10)*arcsin(20/11*x + 1/11)
- 49/81*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 9491/5400*sqrt(-10*x^2 - x + 3)

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Fricas [A]
time = 0.61, size = 107, normalized size = 0.84 \begin {gather*} \frac {1}{5400} \, {\left (2400 \, x^{2} - 5940 \, x + 8771\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1} + \frac {49}{81} \, \sqrt {7} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - \frac {250433}{324000} \, \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((1-2*x)^(5/2)*(3+5*x)^(1/2)/(2+3*x),x, algorithm="fricas")

[Out]

1/5400*(2400*x^2 - 5940*x + 8771)*sqrt(5*x + 3)*sqrt(-2*x + 1) + 49/81*sqrt(7)*arctan(1/14*sqrt(7)*(37*x + 20)
*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 250433/324000*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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (1 - 2 x\right )^{\frac {5}{2}} \sqrt {5 x + 3}}{3 x + 2}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (92) = 184\).
time = 0.61, size = 186, normalized size = 1.45 \begin {gather*} -\frac {49}{810} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {1}{27000} \, {\left (12 \, {\left (8 \, \sqrt {5} {\left (5 \, x + 3\right )} - 147 \, \sqrt {5}\right )} {\left (5 \, x + 3\right )} + 13199 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5} + \frac {250433}{324000} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-49/810*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2
/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) + 1/27000*(12*(8*sqrt(5)*(5*x + 3) - 147*sqrt(5))*(5*x
+ 3) + 13199*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5) + 250433/324000*sqrt(10)*(pi + 2*arctan(-1/4*sqrt(5*x + 3)
*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))^2/(5*x + 3) - 4)/(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 (1-2\,x\right )}^{5/2}\,\sqrt {5\,x+3}}{3\,x+2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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