3.1365 \(\int \frac{(5-x) \sqrt{2+3 x^2}}{(3+2 x)^5} \, dx\)

Optimal. Leaf size=104 \[ -\frac{89 \left (3 x^2+2\right )^{3/2}}{2940 (2 x+3)^3}-\frac{13 \left (3 x^2+2\right )^{3/2}}{140 (2 x+3)^4}-\frac{33 (4-9 x) \sqrt{3 x^2+2}}{8575 (2 x+3)^2}-\frac{198 \tanh ^{-1}\left (\frac{4-9 x}{\sqrt{35} \sqrt{3 x^2+2}}\right )}{8575 \sqrt{35}} \]

[Out]

(-33*(4 - 9*x)*Sqrt[2 + 3*x^2])/(8575*(3 + 2*x)^2) - (13*(2 + 3*x^2)^(3/2))/(140*(3 + 2*x)^4) - (89*(2 + 3*x^2
)^(3/2))/(2940*(3 + 2*x)^3) - (198*ArcTanh[(4 - 9*x)/(Sqrt[35]*Sqrt[2 + 3*x^2])])/(8575*Sqrt[35])

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Rubi [A]  time = 0.0523357, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {835, 807, 721, 725, 206} \[ -\frac{89 \left (3 x^2+2\right )^{3/2}}{2940 (2 x+3)^3}-\frac{13 \left (3 x^2+2\right )^{3/2}}{140 (2 x+3)^4}-\frac{33 (4-9 x) \sqrt{3 x^2+2}}{8575 (2 x+3)^2}-\frac{198 \tanh ^{-1}\left (\frac{4-9 x}{\sqrt{35} \sqrt{3 x^2+2}}\right )}{8575 \sqrt{35}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-33*(4 - 9*x)*Sqrt[2 + 3*x^2])/(8575*(3 + 2*x)^2) - (13*(2 + 3*x^2)^(3/2))/(140*(3 + 2*x)^4) - (89*(2 + 3*x^2
)^(3/2))/(2940*(3 + 2*x)^3) - (198*ArcTanh[(4 - 9*x)/(Sqrt[35]*Sqrt[2 + 3*x^2])])/(8575*Sqrt[35])

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 721

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(m + 1)*(-2*a*e + (2*c*
d)*x)*(a + c*x^2)^p)/(2*(m + 1)*(c*d^2 + a*e^2)), x] - Dist[(4*a*c*p)/(2*(m + 1)*(c*d^2 + a*e^2)), Int[(d + e*
x)^(m + 2)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && EqQ[m + 2*p + 2,
0] && GtQ[p, 0]

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{(5-x) \sqrt{2+3 x^2}}{(3+2 x)^5} \, dx &=-\frac{13 \left (2+3 x^2\right )^{3/2}}{140 (3+2 x)^4}-\frac{1}{140} \int \frac{(-164+39 x) \sqrt{2+3 x^2}}{(3+2 x)^4} \, dx\\ &=-\frac{13 \left (2+3 x^2\right )^{3/2}}{140 (3+2 x)^4}-\frac{89 \left (2+3 x^2\right )^{3/2}}{2940 (3+2 x)^3}+\frac{66}{245} \int \frac{\sqrt{2+3 x^2}}{(3+2 x)^3} \, dx\\ &=-\frac{33 (4-9 x) \sqrt{2+3 x^2}}{8575 (3+2 x)^2}-\frac{13 \left (2+3 x^2\right )^{3/2}}{140 (3+2 x)^4}-\frac{89 \left (2+3 x^2\right )^{3/2}}{2940 (3+2 x)^3}+\frac{198 \int \frac{1}{(3+2 x) \sqrt{2+3 x^2}} \, dx}{8575}\\ &=-\frac{33 (4-9 x) \sqrt{2+3 x^2}}{8575 (3+2 x)^2}-\frac{13 \left (2+3 x^2\right )^{3/2}}{140 (3+2 x)^4}-\frac{89 \left (2+3 x^2\right )^{3/2}}{2940 (3+2 x)^3}-\frac{198 \operatorname{Subst}\left (\int \frac{1}{35-x^2} \, dx,x,\frac{4-9 x}{\sqrt{2+3 x^2}}\right )}{8575}\\ &=-\frac{33 (4-9 x) \sqrt{2+3 x^2}}{8575 (3+2 x)^2}-\frac{13 \left (2+3 x^2\right )^{3/2}}{140 (3+2 x)^4}-\frac{89 \left (2+3 x^2\right )^{3/2}}{2940 (3+2 x)^3}-\frac{198 \tanh ^{-1}\left (\frac{4-9 x}{\sqrt{35} \sqrt{2+3 x^2}}\right )}{8575 \sqrt{35}}\\ \end{align*}

Mathematica [A]  time = 0.0812768, size = 70, normalized size = 0.67 \[ -\frac{\sqrt{3 x^2+2} \left (2217 x^3+10134 x^2-304 x+26028\right )}{51450 (2 x+3)^4}-\frac{198 \tanh ^{-1}\left (\frac{4-9 x}{\sqrt{35} \sqrt{3 x^2+2}}\right )}{8575 \sqrt{35}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[2 + 3*x^2]*(26028 - 304*x + 10134*x^2 + 2217*x^3))/(51450*(3 + 2*x)^4) - (198*ArcTanh[(4 - 9*x)/(Sqrt[3
5]*Sqrt[2 + 3*x^2])])/(8575*Sqrt[35])

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Maple [A]  time = 0.011, size = 149, normalized size = 1.4 \begin{align*} -{\frac{89}{23520} \left ( 3\, \left ( x+3/2 \right ) ^{2}-9\,x-{\frac{19}{4}} \right ) ^{{\frac{3}{2}}} \left ( x+{\frac{3}{2}} \right ) ^{-3}}-{\frac{33}{17150} \left ( 3\, \left ( x+3/2 \right ) ^{2}-9\,x-{\frac{19}{4}} \right ) ^{{\frac{3}{2}}} \left ( x+{\frac{3}{2}} \right ) ^{-2}}-{\frac{297}{300125} \left ( 3\, \left ( x+3/2 \right ) ^{2}-9\,x-{\frac{19}{4}} \right ) ^{{\frac{3}{2}}} \left ( x+{\frac{3}{2}} \right ) ^{-1}}+{\frac{198}{300125}\sqrt{12\, \left ( x+3/2 \right ) ^{2}-36\,x-19}}-{\frac{198\,\sqrt{35}}{300125}{\it Artanh} \left ({\frac{ \left ( 8-18\,x \right ) \sqrt{35}}{35}{\frac{1}{\sqrt{12\, \left ( x+3/2 \right ) ^{2}-36\,x-19}}}} \right ) }+{\frac{891\,x}{300125}\sqrt{3\, \left ( x+3/2 \right ) ^{2}-9\,x-{\frac{19}{4}}}}-{\frac{13}{2240} \left ( 3\, \left ( x+3/2 \right ) ^{2}-9\,x-{\frac{19}{4}} \right ) ^{{\frac{3}{2}}} \left ( x+{\frac{3}{2}} \right ) ^{-4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-89/23520/(x+3/2)^3*(3*(x+3/2)^2-9*x-19/4)^(3/2)-33/17150/(x+3/2)^2*(3*(x+3/2)^2-9*x-19/4)^(3/2)-297/300125/(x
+3/2)*(3*(x+3/2)^2-9*x-19/4)^(3/2)+198/300125*(12*(x+3/2)^2-36*x-19)^(1/2)-198/300125*35^(1/2)*arctanh(2/35*(4
-9*x)*35^(1/2)/(12*(x+3/2)^2-36*x-19)^(1/2))+891/300125*x*(3*(x+3/2)^2-9*x-19/4)^(1/2)-13/2240/(x+3/2)^4*(3*(x
+3/2)^2-9*x-19/4)^(3/2)

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Maxima [A]  time = 1.50529, size = 200, normalized size = 1.92 \begin{align*} \frac{198}{300125} \, \sqrt{35} \operatorname{arsinh}\left (\frac{3 \, \sqrt{6} x}{2 \,{\left | 2 \, x + 3 \right |}} - \frac{2 \, \sqrt{6}}{3 \,{\left | 2 \, x + 3 \right |}}\right ) + \frac{99}{17150} \, \sqrt{3 \, x^{2} + 2} - \frac{13 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}}{140 \,{\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )}} - \frac{89 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}}{2940 \,{\left (8 \, x^{3} + 36 \, x^{2} + 54 \, x + 27\right )}} - \frac{66 \,{\left (3 \, x^{2} + 2\right )}^{\frac{3}{2}}}{8575 \,{\left (4 \, x^{2} + 12 \, x + 9\right )}} - \frac{297 \, \sqrt{3 \, x^{2} + 2}}{17150 \,{\left (2 \, x + 3\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

198/300125*sqrt(35)*arcsinh(3/2*sqrt(6)*x/abs(2*x + 3) - 2/3*sqrt(6)/abs(2*x + 3)) + 99/17150*sqrt(3*x^2 + 2)
- 13/140*(3*x^2 + 2)^(3/2)/(16*x^4 + 96*x^3 + 216*x^2 + 216*x + 81) - 89/2940*(3*x^2 + 2)^(3/2)/(8*x^3 + 36*x^
2 + 54*x + 27) - 66/8575*(3*x^2 + 2)^(3/2)/(4*x^2 + 12*x + 9) - 297/17150*sqrt(3*x^2 + 2)/(2*x + 3)

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Fricas [A]  time = 2.26299, size = 339, normalized size = 3.26 \begin{align*} \frac{594 \, \sqrt{35}{\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )} \log \left (-\frac{\sqrt{35} \sqrt{3 \, x^{2} + 2}{\left (9 \, x - 4\right )} + 93 \, x^{2} - 36 \, x + 43}{4 \, x^{2} + 12 \, x + 9}\right ) - 35 \,{\left (2217 \, x^{3} + 10134 \, x^{2} - 304 \, x + 26028\right )} \sqrt{3 \, x^{2} + 2}}{1800750 \,{\left (16 \, x^{4} + 96 \, x^{3} + 216 \, x^{2} + 216 \, x + 81\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/1800750*(594*sqrt(35)*(16*x^4 + 96*x^3 + 216*x^2 + 216*x + 81)*log(-(sqrt(35)*sqrt(3*x^2 + 2)*(9*x - 4) + 93
*x^2 - 36*x + 43)/(4*x^2 + 12*x + 9)) - 35*(2217*x^3 + 10134*x^2 - 304*x + 26028)*sqrt(3*x^2 + 2))/(16*x^4 + 9
6*x^3 + 216*x^2 + 216*x + 81)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.17641, size = 244, normalized size = 2.35 \begin{align*} \frac{1}{9604000} \, \sqrt{35}{\left (739 \, \sqrt{35} \sqrt{3} + 6336 \, \log \left (\sqrt{35} \sqrt{3} - 9\right )\right )} \mathrm{sgn}\left (\frac{1}{2 \, x + 3}\right ) - \frac{198}{300125} \, \sqrt{35} \log \left (\sqrt{35}{\left (\sqrt{-\frac{18}{2 \, x + 3} + \frac{35}{{\left (2 \, x + 3\right )}^{2}} + 3} + \frac{\sqrt{35}}{2 \, x + 3}\right )} - 9\right ) \mathrm{sgn}\left (\frac{1}{2 \, x + 3}\right ) - \frac{1}{823200} \,{\left (\frac{35 \,{\left (\frac{7 \,{\left (\frac{1365 \, \mathrm{sgn}\left (\frac{1}{2 \, x + 3}\right )}{2 \, x + 3} - 257 \, \mathrm{sgn}\left (\frac{1}{2 \, x + 3}\right )\right )}}{2 \, x + 3} + 9 \, \mathrm{sgn}\left (\frac{1}{2 \, x + 3}\right )\right )}}{2 \, x + 3} + 2217 \, \mathrm{sgn}\left (\frac{1}{2 \, x + 3}\right )\right )} \sqrt{-\frac{18}{2 \, x + 3} + \frac{35}{{\left (2 \, x + 3\right )}^{2}} + 3} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/9604000*sqrt(35)*(739*sqrt(35)*sqrt(3) + 6336*log(sqrt(35)*sqrt(3) - 9))*sgn(1/(2*x + 3)) - 198/300125*sqrt(
35)*log(sqrt(35)*(sqrt(-18/(2*x + 3) + 35/(2*x + 3)^2 + 3) + sqrt(35)/(2*x + 3)) - 9)*sgn(1/(2*x + 3)) - 1/823
200*(35*(7*(1365*sgn(1/(2*x + 3))/(2*x + 3) - 257*sgn(1/(2*x + 3)))/(2*x + 3) + 9*sgn(1/(2*x + 3)))/(2*x + 3)
+ 2217*sgn(1/(2*x + 3)))*sqrt(-18/(2*x + 3) + 35/(2*x + 3)^2 + 3)