3.1388 \(\int \frac {(5-x) (2+3 x^2)^{5/2}}{(3+2 x)^3} \, dx\)

Optimal. Leaf size=126 \[ -\frac {(2 x+29) \left (3 x^2+2\right )^{5/2}}{16 (2 x+3)^2}+\frac {5 (29 x+178) \left (3 x^2+2\right )^{3/2}}{32 (2 x+3)}+\frac {15}{64} (859-267 x) \sqrt {3 x^2+2}-\frac {12885}{128} \sqrt {35} \tanh ^{-1}\left (\frac {4-9 x}{\sqrt {35} \sqrt {3 x^2+2}}\right )-\frac {43995}{128} \sqrt {3} \sinh ^{-1}\left (\sqrt {\frac {3}{2}} x\right ) \]

[Out]

5/32*(178+29*x)*(3*x^2+2)^(3/2)/(3+2*x)-1/16*(29+2*x)*(3*x^2+2)^(5/2)/(3+2*x)^2-43995/128*arcsinh(1/2*x*6^(1/2
))*3^(1/2)-12885/128*arctanh(1/35*(4-9*x)*35^(1/2)/(3*x^2+2)^(1/2))*35^(1/2)+15/64*(859-267*x)*(3*x^2+2)^(1/2)

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Rubi [A]  time = 0.08, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {813, 815, 844, 215, 725, 206} \[ -\frac {(2 x+29) \left (3 x^2+2\right )^{5/2}}{16 (2 x+3)^2}+\frac {5 (29 x+178) \left (3 x^2+2\right )^{3/2}}{32 (2 x+3)}+\frac {15}{64} (859-267 x) \sqrt {3 x^2+2}-\frac {12885}{128} \sqrt {35} \tanh ^{-1}\left (\frac {4-9 x}{\sqrt {35} \sqrt {3 x^2+2}}\right )-\frac {43995}{128} \sqrt {3} \sinh ^{-1}\left (\sqrt {\frac {3}{2}} x\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(859 - 267*x)*Sqrt[2 + 3*x^2])/64 + (5*(178 + 29*x)*(2 + 3*x^2)^(3/2))/(32*(3 + 2*x)) - ((29 + 2*x)*(2 + 3
*x^2)^(5/2))/(16*(3 + 2*x)^2) - (43995*Sqrt[3]*ArcSinh[Sqrt[3/2]*x])/128 - (12885*Sqrt[35]*ArcTanh[(4 - 9*x)/(
Sqrt[35]*Sqrt[2 + 3*x^2])])/128

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])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

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 813

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(e*f*(m + 2*p + 2) - d*g*(2*p + 1) + e*g*(m + 1)*x)*(a + c*x^2)^p)/(e^2*(m + 1)*(m + 2*p + 2)), x] + Di
st[p/(e^2*(m + 1)*(m + 2*p + 2)), Int[(d + e*x)^(m + 1)*(a + c*x^2)^(p - 1)*Simp[g*(2*a*e + 2*a*e*m) + (g*(2*c
*d + 4*c*d*p) - 2*c*e*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2,
0] && RationalQ[p] && p > 0 && (LtQ[m, -1] || EqQ[p, 1] || (IntegerQ[p] &&  !RationalQ[m])) && NeQ[m, -1] &&
!ILtQ[m + 2*p + 1, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 815

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((d + e*x)^(
m + 1)*(c*e*f*(m + 2*p + 2) - g*c*d*(2*p + 1) + g*c*e*(m + 2*p + 1)*x)*(a + c*x^2)^p)/(c*e^2*(m + 2*p + 1)*(m
+ 2*p + 2)), x] + Dist[(2*p)/(c*e^2*(m + 2*p + 1)*(m + 2*p + 2)), Int[(d + e*x)^m*(a + c*x^2)^(p - 1)*Simp[f*a
*c*e^2*(m + 2*p + 2) + a*c*d*e*g*m - (c^2*f*d*e*(m + 2*p + 2) - g*(c^2*d^2*(2*p + 1) + a*c*e^2*(m + 2*p + 1)))
*x, x], x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[p, 0] && (IntegerQ[p] ||  !R
ationalQ[m] || (GeQ[m, -1] && LtQ[m, 0])) &&  !ILtQ[m + 2*p, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*
m, 2*p])

Rule 844

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[g/e, Int[(d
+ e*x)^(m + 1)*(a + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(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] &&  !IGtQ[m, 0]

Rubi steps

\begin {align*} \int \frac {(5-x) \left (2+3 x^2\right )^{5/2}}{(3+2 x)^3} \, dx &=-\frac {(29+2 x) \left (2+3 x^2\right )^{5/2}}{16 (3+2 x)^2}-\frac {5}{64} \int \frac {(16-348 x) \left (2+3 x^2\right )^{3/2}}{(3+2 x)^2} \, dx\\ &=\frac {5 (178+29 x) \left (2+3 x^2\right )^{3/2}}{32 (3+2 x)}-\frac {(29+2 x) \left (2+3 x^2\right )^{5/2}}{16 (3+2 x)^2}+\frac {5}{512} \int \frac {(2784-25632 x) \sqrt {2+3 x^2}}{3+2 x} \, dx\\ &=\frac {15}{64} (859-267 x) \sqrt {2+3 x^2}+\frac {5 (178+29 x) \left (2+3 x^2\right )^{3/2}}{32 (3+2 x)}-\frac {(29+2 x) \left (2+3 x^2\right )^{5/2}}{16 (3+2 x)^2}+\frac {5 \int \frac {1056384-5068224 x}{(3+2 x) \sqrt {2+3 x^2}} \, dx}{12288}\\ &=\frac {15}{64} (859-267 x) \sqrt {2+3 x^2}+\frac {5 (178+29 x) \left (2+3 x^2\right )^{3/2}}{32 (3+2 x)}-\frac {(29+2 x) \left (2+3 x^2\right )^{5/2}}{16 (3+2 x)^2}-\frac {131985}{128} \int \frac {1}{\sqrt {2+3 x^2}} \, dx+\frac {450975}{128} \int \frac {1}{(3+2 x) \sqrt {2+3 x^2}} \, dx\\ &=\frac {15}{64} (859-267 x) \sqrt {2+3 x^2}+\frac {5 (178+29 x) \left (2+3 x^2\right )^{3/2}}{32 (3+2 x)}-\frac {(29+2 x) \left (2+3 x^2\right )^{5/2}}{16 (3+2 x)^2}-\frac {43995}{128} \sqrt {3} \sinh ^{-1}\left (\sqrt {\frac {3}{2}} x\right )-\frac {450975}{128} \operatorname {Subst}\left (\int \frac {1}{35-x^2} \, dx,x,\frac {4-9 x}{\sqrt {2+3 x^2}}\right )\\ &=\frac {15}{64} (859-267 x) \sqrt {2+3 x^2}+\frac {5 (178+29 x) \left (2+3 x^2\right )^{3/2}}{32 (3+2 x)}-\frac {(29+2 x) \left (2+3 x^2\right )^{5/2}}{16 (3+2 x)^2}-\frac {43995}{128} \sqrt {3} \sinh ^{-1}\left (\sqrt {\frac {3}{2}} x\right )-\frac {12885}{128} \sqrt {35} \tanh ^{-1}\left (\frac {4-9 x}{\sqrt {35} \sqrt {2+3 x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 97, normalized size = 0.77 \[ \frac {1}{128} \left (-12885 \sqrt {35} \tanh ^{-1}\left (\frac {4-9 x}{\sqrt {35} \sqrt {3 x^2+2}}\right )-\frac {2 \sqrt {3 x^2+2} \left (72 x^5-696 x^4+2826 x^3-19268 x^2-127403 x-126181\right )}{(2 x+3)^2}-43995 \sqrt {3} \sinh ^{-1}\left (\sqrt {\frac {3}{2}} x\right )\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-2*Sqrt[2 + 3*x^2]*(-126181 - 127403*x - 19268*x^2 + 2826*x^3 - 696*x^4 + 72*x^5))/(3 + 2*x)^2 - 43995*Sqrt[
3]*ArcSinh[Sqrt[3/2]*x] - 12885*Sqrt[35]*ArcTanh[(4 - 9*x)/(Sqrt[35]*Sqrt[2 + 3*x^2])])/128

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fricas [A]  time = 0.63, size = 146, normalized size = 1.16 \[ \frac {43995 \, \sqrt {3} {\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (\sqrt {3} \sqrt {3 \, x^{2} + 2} x - 3 \, x^{2} - 1\right ) + 12885 \, \sqrt {35} {\left (4 \, x^{2} + 12 \, x + 9\right )} \log \left (-\frac {\sqrt {35} \sqrt {3 \, x^{2} + 2} {\left (9 \, x - 4\right )} + 93 \, x^{2} - 36 \, x + 43}{4 \, x^{2} + 12 \, x + 9}\right ) - 4 \, {\left (72 \, x^{5} - 696 \, x^{4} + 2826 \, x^{3} - 19268 \, x^{2} - 127403 \, x - 126181\right )} \sqrt {3 \, x^{2} + 2}}{256 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*(43995*sqrt(3)*(4*x^2 + 12*x + 9)*log(sqrt(3)*sqrt(3*x^2 + 2)*x - 3*x^2 - 1) + 12885*sqrt(35)*(4*x^2 + 1
2*x + 9)*log(-(sqrt(35)*sqrt(3*x^2 + 2)*(9*x - 4) + 93*x^2 - 36*x + 43)/(4*x^2 + 12*x + 9)) - 4*(72*x^5 - 696*
x^4 + 2826*x^3 - 19268*x^2 - 127403*x - 126181)*sqrt(3*x^2 + 2))/(4*x^2 + 12*x + 9)

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giac [B]  time = 0.29, size = 230, normalized size = 1.83 \[ -\frac {1}{32} \, {\left (3 \, {\left ({\left (3 \, x - 38\right )} x + 225\right )} x - 4177\right )} \sqrt {3 \, x^{2} + 2} + \frac {43995}{128} \, \sqrt {3} \log \left (-\sqrt {3} x + \sqrt {3 \, x^{2} + 2}\right ) + \frac {12885}{128} \, \sqrt {35} \log \left (-\frac {{\left | -2 \, \sqrt {3} x - \sqrt {35} - 3 \, \sqrt {3} + 2 \, \sqrt {3 \, x^{2} + 2} \right |}}{2 \, \sqrt {3} x - \sqrt {35} + 3 \, \sqrt {3} - 2 \, \sqrt {3 \, x^{2} + 2}}\right ) + \frac {35 \, {\left (11472 \, {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )}^{3} + 25829 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )}^{2} - 57912 \, \sqrt {3} x + 8984 \, \sqrt {3} + 57912 \, \sqrt {3 \, x^{2} + 2}\right )}}{256 \, {\left ({\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )}^{2} + 3 \, \sqrt {3} {\left (\sqrt {3} x - \sqrt {3 \, x^{2} + 2}\right )} - 2\right )}^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(3*((3*x - 38)*x + 225)*x - 4177)*sqrt(3*x^2 + 2) + 43995/128*sqrt(3)*log(-sqrt(3)*x + sqrt(3*x^2 + 2))
+ 12885/128*sqrt(35)*log(-abs(-2*sqrt(3)*x - sqrt(35) - 3*sqrt(3) + 2*sqrt(3*x^2 + 2))/(2*sqrt(3)*x - sqrt(35)
+ 3*sqrt(3) - 2*sqrt(3*x^2 + 2))) + 35/256*(11472*(sqrt(3)*x - sqrt(3*x^2 + 2))^3 + 25829*sqrt(3)*(sqrt(3)*x
- sqrt(3*x^2 + 2))^2 - 57912*sqrt(3)*x + 8984*sqrt(3) + 57912*sqrt(3*x^2 + 2))/((sqrt(3)*x - sqrt(3*x^2 + 2))^
2 + 3*sqrt(3)*(sqrt(3)*x - sqrt(3*x^2 + 2)) - 2)^2

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maple [A]  time = 0.06, size = 185, normalized size = 1.47 \[ -\frac {807 \left (-9 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {3}{2}} x}{224}-\frac {4005 \sqrt {-9 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}}\, x}{64}-\frac {1263 \left (-9 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {5}{2}} x}{4900}-\frac {43995 \sqrt {3}\, \arcsinh \left (\frac {\sqrt {6}\, x}{2}\right )}{128}-\frac {12885 \sqrt {35}\, \arctanh \left (\frac {2 \left (-9 x +4\right ) \sqrt {35}}{35 \sqrt {-36 x +12 \left (x +\frac {3}{2}\right )^{2}-19}}\right )}{128}-\frac {13 \left (-9 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {7}{2}}}{280 \left (x +\frac {3}{2}\right )^{2}}+\frac {421 \left (-9 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {7}{2}}}{4900 \left (x +\frac {3}{2}\right )}+\frac {2577 \left (-9 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {5}{2}}}{4900}+\frac {859 \left (-9 x +3 \left (x +\frac {3}{2}\right )^{2}-\frac {19}{4}\right )^{\frac {3}{2}}}{112}+\frac {12885 \sqrt {-36 x +12 \left (x +\frac {3}{2}\right )^{2}-19}}{128} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-13/280/(x+3/2)^2*(-9*x+3*(x+3/2)^2-19/4)^(7/2)+421/4900/(x+3/2)*(-9*x+3*(x+3/2)^2-19/4)^(7/2)+2577/4900*(-9*x
+3*(x+3/2)^2-19/4)^(5/2)-807/224*(-9*x+3*(x+3/2)^2-19/4)^(3/2)*x-4005/64*(-9*x+3*(x+3/2)^2-19/4)^(1/2)*x-43995
/128*arcsinh(1/2*6^(1/2)*x)*3^(1/2)+859/112*(-9*x+3*(x+3/2)^2-19/4)^(3/2)+12885/128*(-36*x+12*(x+3/2)^2-19)^(1
/2)-12885/128*35^(1/2)*arctanh(2/35*(-9*x+4)*35^(1/2)/(-36*x+12*(x+3/2)^2-19)^(1/2))-1263/4900*(-9*x+3*(x+3/2)
^2-19/4)^(5/2)*x

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maxima [A]  time = 1.32, size = 145, normalized size = 1.15 \[ \frac {39}{280} \, {\left (3 \, x^{2} + 2\right )}^{\frac {5}{2}} - \frac {13 \, {\left (3 \, x^{2} + 2\right )}^{\frac {7}{2}}}{70 \, {\left (4 \, x^{2} + 12 \, x + 9\right )}} - \frac {807}{224} \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}} x + \frac {859}{112} \, {\left (3 \, x^{2} + 2\right )}^{\frac {3}{2}} + \frac {421 \, {\left (3 \, x^{2} + 2\right )}^{\frac {5}{2}}}{280 \, {\left (2 \, x + 3\right )}} - \frac {4005}{64} \, \sqrt {3 \, x^{2} + 2} x - \frac {43995}{128} \, \sqrt {3} \operatorname {arsinh}\left (\frac {1}{2} \, \sqrt {6} x\right ) + \frac {12885}{128} \, \sqrt {35} \operatorname {arsinh}\left (\frac {3 \, \sqrt {6} x}{2 \, {\left | 2 \, x + 3 \right |}} - \frac {2 \, \sqrt {6}}{3 \, {\left | 2 \, x + 3 \right |}}\right ) + \frac {12885}{64} \, \sqrt {3 \, x^{2} + 2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

39/280*(3*x^2 + 2)^(5/2) - 13/70*(3*x^2 + 2)^(7/2)/(4*x^2 + 12*x + 9) - 807/224*(3*x^2 + 2)^(3/2)*x + 859/112*
(3*x^2 + 2)^(3/2) + 421/280*(3*x^2 + 2)^(5/2)/(2*x + 3) - 4005/64*sqrt(3*x^2 + 2)*x - 43995/128*sqrt(3)*arcsin
h(1/2*sqrt(6)*x) + 12885/128*sqrt(35)*arcsinh(3/2*sqrt(6)*x/abs(2*x + 3) - 2/3*sqrt(6)/abs(2*x + 3)) + 12885/6
4*sqrt(3*x^2 + 2)

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mupad [B]  time = 0.13, size = 147, normalized size = 1.17 \[ \frac {12885\,\sqrt {35}\,\ln \left (x+\frac {3}{2}\right )}{128}+\frac {4177\,\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{32}-\frac {43995\,\sqrt {3}\,\mathrm {asinh}\left (\frac {\sqrt {2}\,\sqrt {3}\,x}{2}\right )}{128}-\frac {12885\,\sqrt {35}\,\ln \left (x-\frac {\sqrt {3}\,\sqrt {35}\,\sqrt {x^2+\frac {2}{3}}}{9}-\frac {4}{9}\right )}{128}+\frac {57\,\sqrt {3}\,x^2\,\sqrt {x^2+\frac {2}{3}}}{16}-\frac {9\,\sqrt {3}\,x^3\,\sqrt {x^2+\frac {2}{3}}}{32}+\frac {39305\,\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{256\,\left (x+\frac {3}{2}\right )}-\frac {15925\,\sqrt {3}\,\sqrt {x^2+\frac {2}{3}}}{512\,\left (x^2+3\,x+\frac {9}{4}\right )}-\frac {675\,\sqrt {3}\,x\,\sqrt {x^2+\frac {2}{3}}}{32} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(12885*35^(1/2)*log(x + 3/2))/128 + (4177*3^(1/2)*(x^2 + 2/3)^(1/2))/32 - (43995*3^(1/2)*asinh((2^(1/2)*3^(1/2
)*x)/2))/128 - (12885*35^(1/2)*log(x - (3^(1/2)*35^(1/2)*(x^2 + 2/3)^(1/2))/9 - 4/9))/128 + (57*3^(1/2)*x^2*(x
^2 + 2/3)^(1/2))/16 - (9*3^(1/2)*x^3*(x^2 + 2/3)^(1/2))/32 + (39305*3^(1/2)*(x^2 + 2/3)^(1/2))/(256*(x + 3/2))
- (15925*3^(1/2)*(x^2 + 2/3)^(1/2))/(512*(3*x + x^2 + 9/4)) - (675*3^(1/2)*x*(x^2 + 2/3)^(1/2))/32

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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