3.2366 \(\int \frac {(1-2 x)^{3/2}}{(2+3 x)^2 \sqrt {3+5 x}} \, dx\)

Optimal. Leaf size=93 \[ \frac {7 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}+\frac {4}{9} \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {29}{9} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \]

[Out]

4/45*arcsin(1/11*22^(1/2)*(3+5*x)^(1/2))*10^(1/2)-29/9*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)
+7/3*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)

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Rubi [A]  time = 0.03, antiderivative size = 93, 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 = {98, 157, 54, 216, 93, 204} \[ \frac {7 \sqrt {1-2 x} \sqrt {5 x+3}}{3 (3 x+2)}+\frac {4}{9} \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {5 x+3}\right )-\frac {29}{9} \sqrt {7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(3*(2 + 3*x)) + (4*Sqrt[2/5]*ArcSin[Sqrt[2/11]*Sqrt[3 + 5*x]])/9 - (29*Sqrt[7]
*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/9

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 93

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 98

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 157

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 204

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

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 {(1-2 x)^{3/2}}{(2+3 x)^2 \sqrt {3+5 x}} \, dx &=\frac {7 \sqrt {1-2 x} \sqrt {3+5 x}}{3 (2+3 x)}+\frac {1}{3} \int \frac {\frac {73}{2}+4 x}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=\frac {7 \sqrt {1-2 x} \sqrt {3+5 x}}{3 (2+3 x)}+\frac {4}{9} \int \frac {1}{\sqrt {1-2 x} \sqrt {3+5 x}} \, dx+\frac {203}{18} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=\frac {7 \sqrt {1-2 x} \sqrt {3+5 x}}{3 (2+3 x)}+\frac {203}{9} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )+\frac {8 \operatorname {Subst}\left (\int \frac {1}{\sqrt {11-2 x^2}} \, dx,x,\sqrt {3+5 x}\right )}{9 \sqrt {5}}\\ &=\frac {7 \sqrt {1-2 x} \sqrt {3+5 x}}{3 (2+3 x)}+\frac {4}{9} \sqrt {\frac {2}{5}} \sin ^{-1}\left (\sqrt {\frac {2}{11}} \sqrt {3+5 x}\right )-\frac {29}{9} \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 = 0.14, size = 120, normalized size = 1.29 \[ \frac {105 \sqrt {-(2 x-1)^2} \sqrt {5 x+3}-145 (3 x+2) \sqrt {14 x-7} \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )-4 \sqrt {10-20 x} (3 x+2) \sinh ^{-1}\left (\sqrt {\frac {5}{11}} \sqrt {2 x-1}\right )}{45 \sqrt {2 x-1} (3 x+2)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(105*Sqrt[-(-1 + 2*x)^2]*Sqrt[3 + 5*x] - 4*Sqrt[10 - 20*x]*(2 + 3*x)*ArcSinh[Sqrt[5/11]*Sqrt[-1 + 2*x]] - 145*
(2 + 3*x)*Sqrt[-7 + 14*x]*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(45*Sqrt[-1 + 2*x]*(2 + 3*x))

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fricas [A]  time = 0.86, size = 122, normalized size = 1.31 \[ -\frac {4 \, \sqrt {5} \sqrt {2} {\left (3 \, x + 2\right )} \arctan \left (\frac {\sqrt {5} \sqrt {2} {\left (20 \, x + 1\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{20 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) + 145 \, \sqrt {7} {\left (3 \, x + 2\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 210 \, \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{90 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/90*(4*sqrt(5)*sqrt(2)*(3*x + 2)*arctan(1/20*sqrt(5)*sqrt(2)*(20*x + 1)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2
+ x - 3)) + 145*sqrt(7)*(3*x + 2)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x -
3)) - 210*sqrt(5*x + 3)*sqrt(-2*x + 1))/(3*x + 2)

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giac [B]  time = 1.65, size = 260, normalized size = 2.80 \[ \frac {29}{180} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {2}{45} \, \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{4 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} + \frac {154 \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{3 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

29/180*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)))) + 2/45*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)))) + 154/3*sqrt(10)*((sqr
t(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt
(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)

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maple [A]  time = 0.01, size = 131, normalized size = 1.41 \[ \frac {\sqrt {-2 x +1}\, \sqrt {5 x +3}\, \left (12 \sqrt {10}\, x \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+435 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+8 \sqrt {10}\, \arcsin \left (\frac {20 x}{11}+\frac {1}{11}\right )+290 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+210 \sqrt {-10 x^{2}-x +3}\right )}{90 \sqrt {-10 x^{2}-x +3}\, \left (3 x +2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/90*(-2*x+1)^(1/2)*(5*x+3)^(1/2)*(12*10^(1/2)*x*arcsin(20/11*x+1/11)+435*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1
/2)/(-10*x^2-x+3)^(1/2))+8*10^(1/2)*arcsin(20/11*x+1/11)+290*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+
3)^(1/2))+210*(-10*x^2-x+3)^(1/2))/(-10*x^2-x+3)^(1/2)/(3*x+2)

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maxima [A]  time = 1.38, size = 61, normalized size = 0.66 \[ \frac {2}{45} \, \sqrt {10} \arcsin \left (\frac {20}{11} \, x + \frac {1}{11}\right ) + \frac {29}{18} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) + \frac {7 \, \sqrt {-10 \, x^{2} - x + 3}}{3 \, {\left (3 \, x + 2\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/45*sqrt(10)*arcsin(20/11*x + 1/11) + 29/18*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) + 7/3*s
qrt(-10*x^2 - x + 3)/(3*x + 2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (1-2\,x\right )}^{3/2}}{{\left (3\,x+2\right )}^2\,\sqrt {5\,x+3}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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