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

Optimal. Leaf size=137 \[ \frac {20743985 \sqrt {1-2 x}}{71148 \sqrt {5 x+3}}-\frac {207895 \sqrt {1-2 x}}{6468 (5 x+3)^{3/2}}+\frac {753 \sqrt {1-2 x}}{196 (3 x+2) (5 x+3)^{3/2}}+\frac {3 \sqrt {1-2 x}}{14 (3 x+2)^2 (5 x+3)^{3/2}}-\frac {392283 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{196 \sqrt {7}} \]

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Rubi [A]  time = 0.04, antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {103, 151, 152, 12, 93, 204} \begin {gather*} \frac {20743985 \sqrt {1-2 x}}{71148 \sqrt {5 x+3}}-\frac {207895 \sqrt {1-2 x}}{6468 (5 x+3)^{3/2}}+\frac {753 \sqrt {1-2 x}}{196 (3 x+2) (5 x+3)^{3/2}}+\frac {3 \sqrt {1-2 x}}{14 (3 x+2)^2 (5 x+3)^{3/2}}-\frac {392283 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{196 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-207895*Sqrt[1 - 2*x])/(6468*(3 + 5*x)^(3/2)) + (3*Sqrt[1 - 2*x])/(14*(2 + 3*x)^2*(3 + 5*x)^(3/2)) + (753*Sqr
t[1 - 2*x])/(196*(2 + 3*x)*(3 + 5*x)^(3/2)) + (20743985*Sqrt[1 - 2*x])/(71148*Sqrt[3 + 5*x]) - (392283*ArcTan[
Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(196*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 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 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)^3 (3+5 x)^{5/2}} \, dx &=\frac {3 \sqrt {1-2 x}}{14 (2+3 x)^2 (3+5 x)^{3/2}}+\frac {1}{14} \int \frac {\frac {131}{2}-90 x}{\sqrt {1-2 x} (2+3 x)^2 (3+5 x)^{5/2}} \, dx\\ &=\frac {3 \sqrt {1-2 x}}{14 (2+3 x)^2 (3+5 x)^{3/2}}+\frac {753 \sqrt {1-2 x}}{196 (2+3 x) (3+5 x)^{3/2}}+\frac {1}{98} \int \frac {\frac {23507}{4}-7530 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{5/2}} \, dx\\ &=-\frac {207895 \sqrt {1-2 x}}{6468 (3+5 x)^{3/2}}+\frac {3 \sqrt {1-2 x}}{14 (2+3 x)^2 (3+5 x)^{3/2}}+\frac {753 \sqrt {1-2 x}}{196 (2+3 x) (3+5 x)^{3/2}}-\frac {\int \frac {\frac {2651953}{8}-\frac {623685 x}{2}}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{3/2}} \, dx}{1617}\\ &=-\frac {207895 \sqrt {1-2 x}}{6468 (3+5 x)^{3/2}}+\frac {3 \sqrt {1-2 x}}{14 (2+3 x)^2 (3+5 x)^{3/2}}+\frac {753 \sqrt {1-2 x}}{196 (2+3 x) (3+5 x)^{3/2}}+\frac {20743985 \sqrt {1-2 x}}{71148 \sqrt {3+5 x}}+\frac {2 \int \frac {142398729}{16 \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{17787}\\ &=-\frac {207895 \sqrt {1-2 x}}{6468 (3+5 x)^{3/2}}+\frac {3 \sqrt {1-2 x}}{14 (2+3 x)^2 (3+5 x)^{3/2}}+\frac {753 \sqrt {1-2 x}}{196 (2+3 x) (3+5 x)^{3/2}}+\frac {20743985 \sqrt {1-2 x}}{71148 \sqrt {3+5 x}}+\frac {392283}{392} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=-\frac {207895 \sqrt {1-2 x}}{6468 (3+5 x)^{3/2}}+\frac {3 \sqrt {1-2 x}}{14 (2+3 x)^2 (3+5 x)^{3/2}}+\frac {753 \sqrt {1-2 x}}{196 (2+3 x) (3+5 x)^{3/2}}+\frac {20743985 \sqrt {1-2 x}}{71148 \sqrt {3+5 x}}+\frac {392283}{196} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )\\ &=-\frac {207895 \sqrt {1-2 x}}{6468 (3+5 x)^{3/2}}+\frac {3 \sqrt {1-2 x}}{14 (2+3 x)^2 (3+5 x)^{3/2}}+\frac {753 \sqrt {1-2 x}}{196 (2+3 x) (3+5 x)^{3/2}}+\frac {20743985 \sqrt {1-2 x}}{71148 \sqrt {3+5 x}}-\frac {392283 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{196 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 79, normalized size = 0.58 \begin {gather*} \frac {\sqrt {1-2 x} \left (933479325 x^3+1784145090 x^2+1135041037 x+240342364\right )}{71148 (3 x+2)^2 (5 x+3)^{3/2}}-\frac {392283 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{196 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[1 - 2*x]*(240342364 + 1135041037*x + 1784145090*x^2 + 933479325*x^3))/(71148*(2 + 3*x)^2*(3 + 5*x)^(3/2)
) - (392283*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(196*Sqrt[7])

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IntegrateAlgebraic [A]  time = 0.21, size = 137, normalized size = 1.00 \begin {gather*} \frac {-\frac {245000 (1-2 x)^{7/2}}{(5 x+3)^{7/2}}+\frac {10829000 (1-2 x)^{5/2}}{(5 x+3)^{5/2}}+\frac {237321871 (1-2 x)^{3/2}}{(5 x+3)^{3/2}}+\frac {996788415 \sqrt {1-2 x}}{\sqrt {5 x+3}}}{71148 \left (\frac {1-2 x}{5 x+3}+7\right )^2}-\frac {392283 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{196 \sqrt {7}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-245000*(1 - 2*x)^(7/2))/(3 + 5*x)^(7/2) + (10829000*(1 - 2*x)^(5/2))/(3 + 5*x)^(5/2) + (237321871*(1 - 2*x)
^(3/2))/(3 + 5*x)^(3/2) + (996788415*Sqrt[1 - 2*x])/Sqrt[3 + 5*x])/(71148*(7 + (1 - 2*x)/(3 + 5*x))^2) - (3922
83*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(196*Sqrt[7])

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fricas [A]  time = 1.25, size = 116, normalized size = 0.85 \begin {gather*} -\frac {142398729 \, \sqrt {7} {\left (225 \, x^{4} + 570 \, x^{3} + 541 \, x^{2} + 228 \, x + 36\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 (933479325 \, x^{3} + 1784145090 \, x^{2} + 1135041037 \, x + 240342364\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{996072 \, {\left (225 \, x^{4} + 570 \, x^{3} + 541 \, x^{2} + 228 \, x + 36\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/996072*(142398729*sqrt(7)*(225*x^4 + 570*x^3 + 541*x^2 + 228*x + 36)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5
*x + 3)*sqrt(-2*x + 1)/(10*x^2 + x - 3)) - 14*(933479325*x^3 + 1784145090*x^2 + 1135041037*x + 240342364)*sqrt
(5*x + 3)*sqrt(-2*x + 1))/(225*x^4 + 570*x^3 + 541*x^2 + 228*x + 36)

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giac [B]  time = 1.84, size = 373, normalized size = 2.72 \begin {gather*} \frac {392283}{27440} \, \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 {25}{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 {2328 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}{\sqrt {5 \, x + 3}} + \frac {9312 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} + \frac {297 \, {\left (461 \, \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} + 110600 \, \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 )}\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}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

392283/27440*sqrt(70)*sqrt(10)*(pi + 2*arctan(-1/140*sqrt(70)*sqrt(5*x + 3)*((sqrt(2)*sqrt(-10*x + 5) - sqrt(2
2))^2/(5*x + 3) - 4)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))) - 25/5808*sqrt(10)*(((sqrt(2)*sqrt(-10*x + 5) - sq
rt(22))/sqrt(5*x + 3) - 4*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22)))^3 - 2328*(sqrt(2)*sqrt(-10*x + 5
) - sqrt(22))/sqrt(5*x + 3) + 9312*sqrt(5*x + 3)/(sqrt(2)*sqrt(-10*x + 5) - sqrt(22))) + 297/98*(461*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
+ 110600*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))))/(((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 = 250, normalized size = 1.82 \begin {gather*} \frac {\left (32039714025 \sqrt {7}\, x^{4} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )[phone] \sqrt {7}\, x^{3} \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^{3}[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 )[phone] \sqrt {-10 x^{2}-x +3}\right ) \sqrt {-2 x +1}}{996072 \left (3 x +2\right )^{2} \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)^3/(5*x+3)^(5/2)/(-2*x+1)^(1/2),x)

[Out]

1/996072*(32039714025*7^(1/2)*x^4*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))[phone]*7^(1/2)*x^3*a
rctan(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]*(-10*x^2-x+3)^(1/2)*x^3+32466910212*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+5126354244*7^(1/2)*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+3364793096*(-10*x^2-x+3)^(1/2))*(-2*x+1)^(1/2)/(3*x+2)^2/(-10*x^2-
x+3)^(1/2)/(5*x+3)^(3/2)

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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 )}^{3} \sqrt {-2 \, x + 1}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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 )}^3\,{\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)^3*(5*x + 3)^(5/2)),x)

[Out]

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

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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 )^{3} \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)**3/(3+5*x)**(5/2)/(1-2*x)**(1/2),x)

[Out]

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

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