3.2526 \(\int \frac {\sqrt {3+5 x}}{(1-2 x)^{3/2} (2+3 x)^3} \, dx\)

Optimal. Leaf size=122 \[ \frac {15 \sqrt {1-2 x} \sqrt {5 x+3}}{1372 (3 x+2)}-\frac {15 \sqrt {1-2 x} \sqrt {5 x+3}}{98 (3 x+2)^2}+\frac {2 \sqrt {5 x+3}}{7 \sqrt {1-2 x} (3 x+2)^2}-\frac {1585 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{1372 \sqrt {7}} \]

[Out]

-1585/9604*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+2/7*(3+5*x)^(1/2)/(2+3*x)^2/(1-2*x)^(1/2)-1
5/98*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)^2+15/1372*(1-2*x)^(1/2)*(3+5*x)^(1/2)/(2+3*x)

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Rubi [A]  time = 0.04, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {99, 151, 12, 93, 204} \[ \frac {15 \sqrt {1-2 x} \sqrt {5 x+3}}{1372 (3 x+2)}-\frac {15 \sqrt {1-2 x} \sqrt {5 x+3}}{98 (3 x+2)^2}+\frac {2 \sqrt {5 x+3}}{7 \sqrt {1-2 x} (3 x+2)^2}-\frac {1585 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{1372 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[3 + 5*x])/(7*Sqrt[1 - 2*x]*(2 + 3*x)^2) - (15*Sqrt[1 - 2*x]*Sqrt[3 + 5*x])/(98*(2 + 3*x)^2) + (15*Sqrt
[1 - 2*x]*Sqrt[3 + 5*x])/(1372*(2 + 3*x)) - (1585*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(1372*Sqrt[7]
)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 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 + 1))/((m + 1)*(b*e - a*f)), x] - Dist[1/((m + 1)*(b*e - a*f)), Int[(a +
b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[d*e*n + c*f*(m + p + 2) + d*f*(m + n + p + 2)*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 151

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 + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

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])

Rubi steps

\begin {align*} \int \frac {\sqrt {3+5 x}}{(1-2 x)^{3/2} (2+3 x)^3} \, dx &=\frac {2 \sqrt {3+5 x}}{7 \sqrt {1-2 x} (2+3 x)^2}-\frac {2}{7} \int \frac {-\frac {35}{2}-30 x}{\sqrt {1-2 x} (2+3 x)^3 \sqrt {3+5 x}} \, dx\\ &=\frac {2 \sqrt {3+5 x}}{7 \sqrt {1-2 x} (2+3 x)^2}-\frac {15 \sqrt {1-2 x} \sqrt {3+5 x}}{98 (2+3 x)^2}-\frac {1}{49} \int \frac {-\frac {205}{4}-75 x}{\sqrt {1-2 x} (2+3 x)^2 \sqrt {3+5 x}} \, dx\\ &=\frac {2 \sqrt {3+5 x}}{7 \sqrt {1-2 x} (2+3 x)^2}-\frac {15 \sqrt {1-2 x} \sqrt {3+5 x}}{98 (2+3 x)^2}+\frac {15 \sqrt {1-2 x} \sqrt {3+5 x}}{1372 (2+3 x)}-\frac {1}{343} \int -\frac {1585}{8 \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=\frac {2 \sqrt {3+5 x}}{7 \sqrt {1-2 x} (2+3 x)^2}-\frac {15 \sqrt {1-2 x} \sqrt {3+5 x}}{98 (2+3 x)^2}+\frac {15 \sqrt {1-2 x} \sqrt {3+5 x}}{1372 (2+3 x)}+\frac {1585 \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{2744}\\ &=\frac {2 \sqrt {3+5 x}}{7 \sqrt {1-2 x} (2+3 x)^2}-\frac {15 \sqrt {1-2 x} \sqrt {3+5 x}}{98 (2+3 x)^2}+\frac {15 \sqrt {1-2 x} \sqrt {3+5 x}}{1372 (2+3 x)}+\frac {1585 \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )}{1372}\\ &=\frac {2 \sqrt {3+5 x}}{7 \sqrt {1-2 x} (2+3 x)^2}-\frac {15 \sqrt {1-2 x} \sqrt {3+5 x}}{98 (2+3 x)^2}+\frac {15 \sqrt {1-2 x} \sqrt {3+5 x}}{1372 (2+3 x)}-\frac {1585 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{1372 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 85, normalized size = 0.70 \[ \frac {7 \sqrt {5 x+3} \left (-90 x^2+405 x+212\right )-1585 \sqrt {7-14 x} (3 x+2)^2 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{9604 \sqrt {1-2 x} (3 x+2)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Sqrt[3 + 5*x]*(212 + 405*x - 90*x^2) - 1585*Sqrt[7 - 14*x]*(2 + 3*x)^2*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3
+ 5*x])])/(9604*Sqrt[1 - 2*x]*(2 + 3*x)^2)

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fricas [A]  time = 0.81, size = 101, normalized size = 0.83 \[ -\frac {1585 \, \sqrt {7} {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (90 \, x^{2} - 405 \, x - 212\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{19208 \, {\left (18 \, x^{3} + 15 \, x^{2} - 4 \, x - 4\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/19208*(1585*sqrt(7)*(18*x^3 + 15*x^2 - 4*x - 4)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x + 1
)/(10*x^2 + x - 3)) - 14*(90*x^2 - 405*x - 212)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(18*x^3 + 15*x^2 - 4*x - 4)

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giac [B]  time = 2.37, size = 278, normalized size = 2.28 \[ \frac {317}{38416} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {8 \, \sqrt {5} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{1715 \, {\left (2 \, x - 1\right )}} - \frac {33 \, \sqrt {10} {\left (7 \, {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{3} - \frac {680 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} + \frac {2720 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}}{98 \, {\left ({\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )}^{2} + 280\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

317/38416*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)))) - 8/1715*sqrt(5)*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x -
1) - 33/98*sqrt(10)*(7*((sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-1
0*x + 5) - sqrt(22)))^3 - 680*(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) + 2720*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)^2

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maple [B]  time = 0.02, size = 209, normalized size = 1.71 \[ \frac {\left (28530 \sqrt {7}\, x^{3} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+23775 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+1260 \sqrt {-10 x^{2}-x +3}\, x^{2}-6340 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-5670 \sqrt {-10 x^{2}-x +3}\, x -6340 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-2968 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}\, \sqrt {5 x +3}}{19208 \left (3 x +2\right )^{2} \left (2 x -1\right ) \sqrt {-10 x^{2}-x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/19208*(28530*7^(1/2)*x^3*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+23775*7^(1/2)*x^2*arctan(1/14*(3
7*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-6340*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+1260*(-
10*x^2-x+3)^(1/2)*x^2-6340*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-5670*(-10*x^2-x+3)^(1/2)
*x-2968*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)*(5*x+3)^(1/2)/(3*x+2)^2/(2*x-1)/(-10*x^2-x+3)^(1/2)

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maxima [A]  time = 1.27, size = 143, normalized size = 1.17 \[ \frac {1585}{19208} \, \sqrt {7} \arcsin \left (\frac {37 \, x}{11 \, {\left | 3 \, x + 2 \right |}} + \frac {20}{11 \, {\left | 3 \, x + 2 \right |}}\right ) - \frac {25 \, x}{686 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {785}{4116 \, \sqrt {-10 \, x^{2} - x + 3}} + \frac {1}{42 \, {\left (9 \, \sqrt {-10 \, x^{2} - x + 3} x^{2} + 12 \, \sqrt {-10 \, x^{2} - x + 3} x + 4 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} - \frac {95}{588 \, {\left (3 \, \sqrt {-10 \, x^{2} - x + 3} x + 2 \, \sqrt {-10 \, x^{2} - x + 3}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1585/19208*sqrt(7)*arcsin(37/11*x/abs(3*x + 2) + 20/11/abs(3*x + 2)) - 25/686*x/sqrt(-10*x^2 - x + 3) + 785/41
16/sqrt(-10*x^2 - x + 3) + 1/42/(9*sqrt(-10*x^2 - x + 3)*x^2 + 12*sqrt(-10*x^2 - x + 3)*x + 4*sqrt(-10*x^2 - x
+ 3)) - 95/588/(3*sqrt(-10*x^2 - x + 3)*x + 2*sqrt(-10*x^2 - x + 3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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