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]

(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]
)

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Rubi [A]  time = 0.0381462, antiderivative size = 122, 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 = {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 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 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{\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*}

Mathematica [A]  time = 0.0455029, size = 85, normalized size = 0.7 \[ \frac{7 \sqrt{5 x+3} \left (-90 x^2+405 x+212\right )-1585 \sqrt{7-14 x} (3 x+2)^2 \tan ^{-1}\left (\frac{\sqrt{1-2 x}}{\sqrt{7} \sqrt{5 x+3}}\right )}{9604 \sqrt{1-2 x} (3 x+2)^2} \]

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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Maple [B]  time = 0.013, size = 209, normalized size = 1.7 \begin{align*}{\frac{1}{19208\, \left ( 2+3\,x \right ) ^{2} \left ( 2\,x-1 \right ) } \left ( 28530\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{3}+23775\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ){x}^{2}-6340\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) x+1260\,{x}^{2}\sqrt{-10\,{x}^{2}-x+3}-6340\,\sqrt{7}\arctan \left ( 1/14\,{\frac{ \left ( 37\,x+20 \right ) \sqrt{7}}{\sqrt{-10\,{x}^{2}-x+3}}} \right ) -5670\,x\sqrt{-10\,{x}^{2}-x+3}-2968\,\sqrt{-10\,{x}^{2}-x+3} \right ) \sqrt{1-2\,x}\sqrt{3+5\,x}{\frac{1}{\sqrt{-10\,{x}^{2}-x+3}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 3.59048, size = 193, normalized size = 1.58 \begin{align*} \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 )}} \end{align*}

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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Fricas [A]  time = 1.76944, size = 293, normalized size = 2.4 \begin{align*} -\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 )}} \end{align*}

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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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Giac [B]  time = 2.74585, size = 382, normalized size = 3.13 \begin{align*} \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 \,{\left (7 \, \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} - 680 \, \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{align*}

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*(7*sqrt(10)*((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(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