3.2624 \(\int \frac {1}{(1-2 x)^{5/2} (2+3 x) (3+5 x)^{3/2}} \, dx\)

Optimal. Leaf size=101 \[ -\frac {42230 \sqrt {1-2 x}}{195657 \sqrt {5 x+3}}+\frac {956}{17787 \sqrt {1-2 x} \sqrt {5 x+3}}+\frac {4}{231 (1-2 x)^{3/2} \sqrt {5 x+3}}+\frac {54 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{49 \sqrt {7}} \]

[Out]

54/343*arctan(1/7*(1-2*x)^(1/2)*7^(1/2)/(3+5*x)^(1/2))*7^(1/2)+4/231/(1-2*x)^(3/2)/(3+5*x)^(1/2)+956/17787/(1-
2*x)^(1/2)/(3+5*x)^(1/2)-42230/195657*(1-2*x)^(1/2)/(3+5*x)^(1/2)

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Rubi [A]  time = 0.04, antiderivative size = 101, 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 = {104, 152, 12, 93, 204} \[ -\frac {42230 \sqrt {1-2 x}}{195657 \sqrt {5 x+3}}+\frac {956}{17787 \sqrt {1-2 x} \sqrt {5 x+3}}+\frac {4}{231 (1-2 x)^{3/2} \sqrt {5 x+3}}+\frac {54 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{49 \sqrt {7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

4/(231*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x]) + 956/(17787*Sqrt[1 - 2*x]*Sqrt[3 + 5*x]) - (42230*Sqrt[1 - 2*x])/(19565
7*Sqrt[3 + 5*x]) + (54*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(49*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 104

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] &&
IntegersQ[2*m, 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}{(1-2 x)^{5/2} (2+3 x) (3+5 x)^{3/2}} \, dx &=\frac {4}{231 (1-2 x)^{3/2} \sqrt {3+5 x}}-\frac {2}{231} \int \frac {-\frac {179}{2}-60 x}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^{3/2}} \, dx\\ &=\frac {4}{231 (1-2 x)^{3/2} \sqrt {3+5 x}}+\frac {956}{17787 \sqrt {1-2 x} \sqrt {3+5 x}}+\frac {4 \int \frac {\frac {12827}{4}+3585 x}{\sqrt {1-2 x} (2+3 x) (3+5 x)^{3/2}} \, dx}{17787}\\ &=\frac {4}{231 (1-2 x)^{3/2} \sqrt {3+5 x}}+\frac {956}{17787 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {42230 \sqrt {1-2 x}}{195657 \sqrt {3+5 x}}-\frac {8 \int \frac {107811}{8 \sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx}{195657}\\ &=\frac {4}{231 (1-2 x)^{3/2} \sqrt {3+5 x}}+\frac {956}{17787 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {42230 \sqrt {1-2 x}}{195657 \sqrt {3+5 x}}-\frac {27}{49} \int \frac {1}{\sqrt {1-2 x} (2+3 x) \sqrt {3+5 x}} \, dx\\ &=\frac {4}{231 (1-2 x)^{3/2} \sqrt {3+5 x}}+\frac {956}{17787 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {42230 \sqrt {1-2 x}}{195657 \sqrt {3+5 x}}-\frac {54}{49} \operatorname {Subst}\left (\int \frac {1}{-7-x^2} \, dx,x,\frac {\sqrt {1-2 x}}{\sqrt {3+5 x}}\right )\\ &=\frac {4}{231 (1-2 x)^{3/2} \sqrt {3+5 x}}+\frac {956}{17787 \sqrt {1-2 x} \sqrt {3+5 x}}-\frac {42230 \sqrt {1-2 x}}{195657 \sqrt {3+5 x}}+\frac {54 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {3+5 x}}\right )}{49 \sqrt {7}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 67, normalized size = 0.66 \[ \frac {54 \tan ^{-1}\left (\frac {\sqrt {1-2 x}}{\sqrt {7} \sqrt {5 x+3}}\right )}{49 \sqrt {7}}-\frac {2 \left (84460 x^2-73944 x+14163\right )}{195657 (1-2 x)^{3/2} \sqrt {5 x+3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(14163 - 73944*x + 84460*x^2))/(195657*(1 - 2*x)^(3/2)*Sqrt[3 + 5*x]) + (54*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*
Sqrt[3 + 5*x])])/(49*Sqrt[7])

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fricas [A]  time = 1.05, size = 101, normalized size = 1.00 \[ \frac {107811 \, \sqrt {7} {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )} \arctan \left (\frac {\sqrt {7} {\left (37 \, x + 20\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{14 \, {\left (10 \, x^{2} + x - 3\right )}}\right ) - 14 \, {\left (84460 \, x^{2} - 73944 \, x + 14163\right )} \sqrt {5 \, x + 3} \sqrt {-2 \, x + 1}}{1369599 \, {\left (20 \, x^{3} - 8 \, x^{2} - 7 \, x + 3\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1369599*(107811*sqrt(7)*(20*x^3 - 8*x^2 - 7*x + 3)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x +
1)/(10*x^2 + x - 3)) - 14*(84460*x^2 - 73944*x + 14163)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(20*x^3 - 8*x^2 - 7*x +
3)

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giac [B]  time = 1.38, size = 172, normalized size = 1.70 \[ -\frac {27}{3430} \, \sqrt {70} \sqrt {10} {\left (\pi + 2 \, \arctan \left (-\frac {\sqrt {70} \sqrt {5 \, x + 3} {\left (\frac {{\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}^{2}}{5 \, x + 3} - 4\right )}}{140 \, {\left (\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}\right )}}\right )\right )} - \frac {25}{2662} \, \sqrt {10} {\left (\frac {\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}{\sqrt {5 \, x + 3}} - \frac {4 \, \sqrt {5 \, x + 3}}{\sqrt {2} \sqrt {-10 \, x + 5} - \sqrt {22}}\right )} - \frac {8 \, {\left (548 \, \sqrt {5} {\left (5 \, x + 3\right )} - 3399 \, \sqrt {5}\right )} \sqrt {5 \, x + 3} \sqrt {-10 \, x + 5}}{4891425 \, {\left (2 \, x - 1\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-27/3430*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)))) - 25/2662*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))) - 8/4891425*(548*sqrt(5)*(5*x + 3) -
3399*sqrt(5))*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)^2

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maple [B]  time = 0.02, size = 202, normalized size = 2.00 \[ -\frac {\sqrt {-2 x +1}\, \left (2156220 \sqrt {7}\, x^{3} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-862488 \sqrt {7}\, x^{2} \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+1182440 \sqrt {-10 x^{2}-x +3}\, x^{2}-754677 \sqrt {7}\, x \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )-1035216 \sqrt {-10 x^{2}-x +3}\, x +323433 \sqrt {7}\, \arctan \left (\frac {\left (37 x +20\right ) \sqrt {7}}{14 \sqrt {-10 x^{2}-x +3}}\right )+198282 \sqrt {-10 x^{2}-x +3}\right )}{1369599 \left (2 x -1\right )^{2} \sqrt {-10 x^{2}-x +3}\, \sqrt {5 x +3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/1369599*(-2*x+1)^(1/2)*(2156220*7^(1/2)*x^3*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-862488*7^(1/
2)*x^2*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))-754677*7^(1/2)*x*arctan(1/14*(37*x+20)*7^(1/2)/(-10*
x^2-x+3)^(1/2))+1182440*(-10*x^2-x+3)^(1/2)*x^2+323433*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/
2))-1035216*(-10*x^2-x+3)^(1/2)*x+198282*(-10*x^2-x+3)^(1/2))/(2*x-1)^2/(-10*x^2-x+3)^(1/2)/(5*x+3)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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