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

Optimal. Leaf size=108 \[ \frac {84235 \sqrt {1-2 x}}{2541 \sqrt {5 x+3}}-\frac {845 \sqrt {1-2 x}}{231 (5 x+3)^{3/2}}+\frac {3 \sqrt {1-2 x}}{7 (3 x+2) (5 x+3)^{3/2}}-\frac {1593 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}} \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 108, 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 = {103, 152, 12, 93, 204} \begin {gather*} \frac {84235 \sqrt {1-2 x}}{2541 \sqrt {5 x+3}}-\frac {845 \sqrt {1-2 x}}{231 (5 x+3)^{3/2}}+\frac {3 \sqrt {1-2 x}}{7 (3 x+2) (5 x+3)^{3/2}}-\frac {1593 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-845*Sqrt[1 - 2*x])/(231*(3 + 5*x)^(3/2)) + (3*Sqrt[1 - 2*x])/(7*(2 + 3*x)*(3 + 5*x)^(3/2)) + (84235*Sqrt[1 -
2*x])/(2541*Sqrt[3 + 5*x]) - (1593*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*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 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(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*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 152

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] && IntegersQ[2*m, 2*n, 2*p]

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 {1}{\sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{5/2}} \, dx &=\frac {3 \sqrt {1-2 x}}{7 (2+3 x) (3+5 x)^{3/2}}+\frac {1}{7} \int \frac {\frac {97}{2}-60 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{5/2}} \, dx\\ &=-\frac {845 \sqrt {1-2 x}}{231 (3+5 x)^{3/2}}+\frac {3 \sqrt {1-2 x}}{7 (2+3 x) (3+5 x)^{3/2}}-\frac {2}{231} \int \frac {\frac {10763}{4}-2535 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{3/2}} \, dx\\ &=-\frac {845 \sqrt {1-2 x}}{231 (3+5 x)^{3/2}}+\frac {3 \sqrt {1-2 x}}{7 (2+3 x) (3+5 x)^{3/2}}+\frac {84235 \sqrt {1-2 x}}{2541 \sqrt {3+5 x}}+\frac {4 \int \frac {578259}{8 \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{2541}\\ &=-\frac {845 \sqrt {1-2 x}}{231 (3+5 x)^{3/2}}+\frac {3 \sqrt {1-2 x}}{7 (2+3 x) (3+5 x)^{3/2}}+\frac {84235 \sqrt {1-2 x}}{2541 \sqrt {3+5 x}}+\frac {1593}{14} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {845 \sqrt {1-2 x}}{231 (3+5 x)^{3/2}}+\frac {3 \sqrt {1-2 x}}{7 (2+3 x) (3+5 x)^{3/2}}+\frac {84235 \sqrt {1-2 x}}{2541 \sqrt {3+5 x}}+\frac {1593}{7} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )\\ &=-\frac {845 \sqrt {1-2 x}}{231 (3+5 x)^{3/2}}+\frac {3 \sqrt {1-2 x}}{7 (2+3 x) (3+5 x)^{3/2}}+\frac {84235 \sqrt {1-2 x}}{2541 \sqrt {3+5 x}}-\frac {1593 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{7 \sqrt {7}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.04, size = 93, normalized size = 0.86 \begin {gather*} \frac {7 \sqrt {1-2 x} \left (1263525 x^2+1572580 x+487909\right )-578259 \sqrt {7} \sqrt {5 x+3} \left (15 x^2+19 x+6\right ) \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{17787 (3 x+2) (5 x+3)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Sqrt[1 - 2*x]*(487909 + 1572580*x + 1263525*x^2) - 578259*Sqrt[7]*Sqrt[3 + 5*x]*(6 + 19*x + 15*x^2)*ArcTan[
Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(17787*(2 + 3*x)*(3 + 5*x)^(3/2))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.14, size = 117, normalized size = 1.08 \begin {gather*} \frac {-\frac {1750 (1-2 x)^{5/2}}{(5 x+3)^{5/2}}+\frac {54950 (1-2 x)^{3/2}}{(5 x+3)^{3/2}}+\frac {578211 \sqrt {1-2 x}}{\sqrt {5 x+3}}}{2541 \left (\frac {1-2 x}{5 x+3}+7\right )}-\frac {1593 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{7 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1750*(1 - 2*x)^(5/2))/(3 + 5*x)^(5/2) + (54950*(1 - 2*x)^(3/2))/(3 + 5*x)^(3/2) + (578211*Sqrt[1 - 2*x])/Sq
rt[3 + 5*x])/(2541*(7 + (1 - 2*x)/(3 + 5*x))) - (1593*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*Sqrt[7
])

________________________________________________________________________________________

fricas [A]  time = 0.91, size = 101, normalized size = 0.94 \begin {gather*} -\frac {578259 \, \sqrt {7} {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\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 (1263525 \, x^{2} + 1572580 \, x + 487909\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{35574 \, {\left (75 \, x^{3} + 140 \, x^{2} + 87 \, x + 18\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/35574*(578259*sqrt(7)*(75*x^3 + 140*x^2 + 87*x + 18)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*
x + 1)/(10*x^2 + x - 3)) - 14*(1263525*x^2 + 1572580*x + 487909)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(75*x^3 + 140*x
^2 + 87*x + 18)

________________________________________________________________________________________

giac [B]  time = 1.61, size = 309, normalized size = 2.86 \begin {gather*} \frac {1593}{980} \, \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 {5}{5808} \, \sqrt {10} {\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 )}^{3} - \frac {1536 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} + \frac {6144 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} + \frac {594 \, \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 )}}{7 \, {\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 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1593/980*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)))) - 5/5808*sqrt(10)*(((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)))^3 - 1536*(sqrt(2)*sqrt(-10*x + 5) - s
qrt(22))/sqrt(5*x + 3) + 6144*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))) + 594/7*sqrt(10)*((sqrt(2)*s
qrt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))/(((sqrt(2)*sq
rt(-10*x + 5) - sqrt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^2 + 280)

________________________________________________________________________________________

maple [B]  time = 0.02, size = 202, normalized size = 1.87 \begin {gather*} \frac {\left (43369425 \sqrt {7}\, x^{3} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )[phone] \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )[phone] \sqrt {-10 x^{2}-x +3}\, x^{2}[phone] \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )[phone] \sqrt {-10 x^{2}-x +3}\, x [phone] \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+6830726 \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}}{35574 \left (3 x +2\right ) \sqrt {-10 x^{2}-x +3}\, \left (5 x +3\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35574*(43369425*7^(1/2)*x^3*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))[phone]*7^(1/2)*x^2*arctan(1
/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))[phone]*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2
))[phone]*(-10*x^2-x+3)^(1/2)*x^2+10408662*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))+220161
20*(-10*x^2-x+3)^(1/2)*x+6830726*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)/(3*x+2)/(-10*x^2-x+3)^(1/2)/(5*x+3)^(3/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{{\left (5 \, x + 3\right )}^{\frac {5}{2}} {\left (3 \, x + 2\right )}^{2} \sqrt {-2 \, x + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________