3.2577 \(\int \frac{1}{(1-2 x)^{3/2} (2+3 x) (3+5 x)^{5/2}} \, dx\)

Optimal. Leaf size=101 \[ \frac{31030 \sqrt{1-2 x}}{27951 \sqrt{5 x+3}}-\frac{410 \sqrt{1-2 x}}{2541 (5 x+3)^{3/2}}+\frac{4}{77 \sqrt{1-2 x} (5 x+3)^{3/2}}-\frac{54 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{7 \sqrt{7}} \]

[Out]

4/(77*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) - (410*Sqrt[1 - 2*x])/(2541*(3 + 5*x)^(3/2)) + (31030*Sqrt[1 - 2*x])/(279
51*Sqrt[3 + 5*x]) - (54*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*Sqrt[7])

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Rubi [A]  time = 0.036159, antiderivative size = 101, normalized size of antiderivative = 1., 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} \[ \frac{31030 \sqrt{1-2 x}}{27951 \sqrt{5 x+3}}-\frac{410 \sqrt{1-2 x}}{2541 (5 x+3)^{3/2}}+\frac{4}{77 \sqrt{1-2 x} (5 x+3)^{3/2}}-\frac{54 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{7 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

4/(77*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) - (410*Sqrt[1 - 2*x])/(2541*(3 + 5*x)^(3/2)) + (31030*Sqrt[1 - 2*x])/(279
51*Sqrt[3 + 5*x]) - (54*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]*Sqrt[3 + 5*x])])/(7*Sqrt[7])

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 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 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)^{3/2} (2+3 x) (3+5 x)^{5/2}} \, dx &=\frac{4}{77 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{2}{77} \int \frac{-\frac{113}{2}-60 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)^{5/2}} \, dx\\ &=\frac{4}{77 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{410 \sqrt{1-2 x}}{2541 (3+5 x)^{3/2}}+\frac{4 \int \frac{-\frac{1627}{4}+615 x}{\sqrt{1-2 x} (2+3 x) (3+5 x)^{3/2}} \, dx}{2541}\\ &=\frac{4}{77 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{410 \sqrt{1-2 x}}{2541 (3+5 x)^{3/2}}+\frac{31030 \sqrt{1-2 x}}{27951 \sqrt{3+5 x}}-\frac{8 \int -\frac{107811}{8 \sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx}{27951}\\ &=\frac{4}{77 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{410 \sqrt{1-2 x}}{2541 (3+5 x)^{3/2}}+\frac{31030 \sqrt{1-2 x}}{27951 \sqrt{3+5 x}}+\frac{27}{7} \int \frac{1}{\sqrt{1-2 x} (2+3 x) \sqrt{3+5 x}} \, dx\\ &=\frac{4}{77 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{410 \sqrt{1-2 x}}{2541 (3+5 x)^{3/2}}+\frac{31030 \sqrt{1-2 x}}{27951 \sqrt{3+5 x}}+\frac{54}{7} \operatorname{Subst}\left (\int \frac{1}{-7-x^2} \, dx,x,\frac{\sqrt{1-2 x}}{\sqrt{3+5 x}}\right )\\ &=\frac{4}{77 \sqrt{1-2 x} (3+5 x)^{3/2}}-\frac{410 \sqrt{1-2 x}}{2541 (3+5 x)^{3/2}}+\frac{31030 \sqrt{1-2 x}}{27951 \sqrt{3+5 x}}-\frac{54 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{3+5 x}}\right )}{7 \sqrt{7}}\\ \end{align*}

Mathematica [A]  time = 0.0645453, size = 67, normalized size = 0.66 \[ -\frac{2 \left (155150 x^2+11005 x-45016\right )}{27951 \sqrt{1-2 x} (5 x+3)^{3/2}}-\frac{54 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{7 \sqrt{7}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(-45016 + 11005*x + 155150*x^2))/(27951*Sqrt[1 - 2*x]*(3 + 5*x)^(3/2)) - (54*ArcTan[Sqrt[1 - 2*x]/(Sqrt[7]
*Sqrt[3 + 5*x])])/(7*Sqrt[7])

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Maple [B]  time = 0.013, size = 202, normalized size = 2. \begin{align*}{\frac{1}{391314\,x-195657}\sqrt{1-2\,x} \left ( 5390550\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{3}+3773385\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}-1293732\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x+2172100\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}-970299\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) +154070\,x\sqrt{-10\,{x}^{2}-x+3}-630224\,\sqrt{-10\,{x}^{2}-x+3} \right ){\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}} \left ( 3+5\,x \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/195657*(1-2*x)^(1/2)*(5390550*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^3+3773385*7^(1/2)
*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2))*x^2-1293732*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2
-x+3)^(1/2))*x+2172100*x^2*(-10*x^2-x+3)^(1/2)-970299*7^(1/2)*arctan(1/14*(37*x+20)*7^(1/2)/(-10*x^2-x+3)^(1/2
))+154070*x*(-10*x^2-x+3)^(1/2)-630224*(-10*x^2-x+3)^(1/2))/(2*x-1)/(-10*x^2-x+3)^(1/2)/(3+5*x)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.56953, size = 311, normalized size = 3.08 \begin{align*} -\frac{107811 \, \sqrt{7}{\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\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 (155150 \, x^{2} + 11005 \, x - 45016\right )} \sqrt{5 \, x + 3} \sqrt{-2 \, x + 1}}{195657 \,{\left (50 \, x^{3} + 35 \, x^{2} - 12 \, x - 9\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/195657*(107811*sqrt(7)*(50*x^3 + 35*x^2 - 12*x - 9)*arctan(1/14*sqrt(7)*(37*x + 20)*sqrt(5*x + 3)*sqrt(-2*x
+ 1)/(10*x^2 + x - 3)) - 14*(155150*x^2 + 11005*x - 45016)*sqrt(5*x + 3)*sqrt(-2*x + 1))/(50*x^3 + 35*x^2 - 1
2*x - 9)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 2.77023, size = 297, normalized size = 2.94 \begin{align*} -\frac{5}{63888} \, \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} + \frac{27}{490} \, \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{145}{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{16 \, \sqrt{5} \sqrt{5 \, x + 3} \sqrt{-10 \, x + 5}}{46585 \,{\left (2 \, x - 1\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-5/63888*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 + 27/490*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)))) + 145/2662*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))) - 16/46585*sqrt(5)
*sqrt(5*x + 3)*sqrt(-10*x + 5)/(2*x - 1)