3.1052 \(\int \frac {(2-5 x) (2+5 x+3 x^2)^{3/2}}{x^{11/2}} \, dx\)

Optimal. Leaf size=210 \[ \frac {5438 \sqrt {3 x^2+5 x+2}}{315 \sqrt {x}}-\frac {5438 \sqrt {x} (3 x+2)}{315 \sqrt {3 x^2+5 x+2}}-\frac {899 (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{21 \sqrt {2} \sqrt {3 x^2+5 x+2}}+\frac {5438 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{315 \sqrt {3 x^2+5 x+2}}-\frac {4 (7-15 x) \left (3 x^2+5 x+2\right )^{3/2}}{63 x^{9/2}}+\frac {(4055 x+1446) \sqrt {3 x^2+5 x+2}}{315 x^{5/2}} \]

[Out]

-4/63*(7-15*x)*(3*x^2+5*x+2)^(3/2)/x^(9/2)-5438/315*(2+3*x)*x^(1/2)/(3*x^2+5*x+2)^(1/2)-899/42*(1+x)^(3/2)*(1/
(1+x))^(1/2)*EllipticF(x^(1/2)/(1+x)^(1/2),1/2*I*2^(1/2))*2^(1/2)*((2+3*x)/(1+x))^(1/2)/(3*x^2+5*x+2)^(1/2)+54
38/315*(1+x)^(3/2)*(1/(1+x))^(1/2)*EllipticE(x^(1/2)/(1+x)^(1/2),1/2*I*2^(1/2))*2^(1/2)*((2+3*x)/(1+x))^(1/2)/
(3*x^2+5*x+2)^(1/2)+1/315*(1446+4055*x)*(3*x^2+5*x+2)^(1/2)/x^(5/2)+5438/315*(3*x^2+5*x+2)^(1/2)/x^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {810, 834, 839, 1189, 1100, 1136} \[ -\frac {4 (7-15 x) \left (3 x^2+5 x+2\right )^{3/2}}{63 x^{9/2}}+\frac {(4055 x+1446) \sqrt {3 x^2+5 x+2}}{315 x^{5/2}}+\frac {5438 \sqrt {3 x^2+5 x+2}}{315 \sqrt {x}}-\frac {5438 \sqrt {x} (3 x+2)}{315 \sqrt {3 x^2+5 x+2}}-\frac {899 (x+1) \sqrt {\frac {3 x+2}{x+1}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{21 \sqrt {2} \sqrt {3 x^2+5 x+2}}+\frac {5438 \sqrt {2} (x+1) \sqrt {\frac {3 x+2}{x+1}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{315 \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

Int[((2 - 5*x)*(2 + 5*x + 3*x^2)^(3/2))/x^(11/2),x]

[Out]

(-5438*Sqrt[x]*(2 + 3*x))/(315*Sqrt[2 + 5*x + 3*x^2]) + (5438*Sqrt[2 + 5*x + 3*x^2])/(315*Sqrt[x]) + ((1446 +
4055*x)*Sqrt[2 + 5*x + 3*x^2])/(315*x^(5/2)) - (4*(7 - 15*x)*(2 + 5*x + 3*x^2)^(3/2))/(63*x^(9/2)) + (5438*Sqr
t[2]*(1 + x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticE[ArcTan[Sqrt[x]], -1/2])/(315*Sqrt[2 + 5*x + 3*x^2]) - (899*(1 +
x)*Sqrt[(2 + 3*x)/(1 + x)]*EllipticF[ArcTan[Sqrt[x]], -1/2])/(21*Sqrt[2]*Sqrt[2 + 5*x + 3*x^2])

Rule 810

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*((d*g - e*f*(m + 2))*(c*d^2 - b*d*e + a*e^2) - d*p*(2*c*d - b*e)*(e*
f - d*g) - e*(g*(m + 1)*(c*d^2 - b*d*e + a*e^2) + p*(2*c*d - b*e)*(e*f - d*g))*x))/(e^2*(m + 1)*(m + 2)*(c*d^2
- b*d*e + a*e^2)), x] - Dist[p/(e^2*(m + 1)*(m + 2)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 2)*(a + b*x
+ c*x^2)^(p - 1)*Simp[2*a*c*e*(e*f - d*g)*(m + 2) + b^2*e*(d*g*(p + 1) - e*f*(m + p + 2)) + b*(a*e^2*g*(m + 1)
- c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2))) - c*(2*c*d*(d*g*(2*p + 1) - e*f*(m + 2*p + 2)) - e*(2*a*e*g*(m + 1
) - b*(d*g*(m - 2*p) + e*f*(m + 2*p + 2))))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 - 4*a*
c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && GtQ[p, 0] && LtQ[m, -2] && LtQ[m + 2*p, 0] &&  !ILtQ[m + 2*p + 3, 0]

Rule 834

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Sim
p[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[1/((m
+ 1)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p*Simp[(c*d*f - f*b*e + a*e*g)*(m + 1)
+ b*(d*g - e*f)*(p + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&
NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ
[2*m, 2*p])

Rule 839

Int[((f_) + (g_.)*(x_))/(Sqrt[x_]*Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2, Subst[Int[(f +
g*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x, Sqrt[x]], x] /; FreeQ[{a, b, c, f, g}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 1100

Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[((2*a + (b -
q)*x^2)*Sqrt[(2*a + (b + q)*x^2)/(2*a + (b - q)*x^2)]*EllipticF[ArcTan[Rt[(b - q)/(2*a), 2]*x], (-2*q)/(b - q)
])/(2*a*Rt[(b - q)/(2*a), 2]*Sqrt[a + b*x^2 + c*x^4]), x] /; PosQ[(b - q)/a]] /; FreeQ[{a, b, c}, x] && GtQ[b^
2 - 4*a*c, 0]

Rule 1136

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Simp[(x*(b -
q + 2*c*x^2))/(2*c*Sqrt[a + b*x^2 + c*x^4]), x] - Simp[(Rt[(b - q)/(2*a), 2]*(2*a + (b - q)*x^2)*Sqrt[(2*a + (
b + q)*x^2)/(2*a + (b - q)*x^2)]*EllipticE[ArcTan[Rt[(b - q)/(2*a), 2]*x], (-2*q)/(b - q)])/(2*c*Sqrt[a + b*x^
2 + c*x^4]), x] /; PosQ[(b - q)/a]] /; FreeQ[{a, b, c}, x] && GtQ[b^2 - 4*a*c, 0]

Rule 1189

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}
, Dist[d, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Dist[e, Int[x^2/Sqrt[a + b*x^2 + c*x^4], x], x] /; PosQ[(b +
q)/a] || PosQ[(b - q)/a]] /; FreeQ[{a, b, c, d, e}, x] && GtQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {(2-5 x) \left (2+5 x+3 x^2\right )^{3/2}}{x^{11/2}} \, dx &=-\frac {4 (7-15 x) \left (2+5 x+3 x^2\right )^{3/2}}{63 x^{9/2}}-\frac {1}{21} \int \frac {(241+285 x) \sqrt {2+5 x+3 x^2}}{x^{7/2}} \, dx\\ &=\frac {(1446+4055 x) \sqrt {2+5 x+3 x^2}}{315 x^{5/2}}-\frac {4 (7-15 x) \left (2+5 x+3 x^2\right )^{3/2}}{63 x^{9/2}}+\frac {1}{315} \int \frac {-5438-\frac {13485 x}{2}}{x^{3/2} \sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {5438 \sqrt {2+5 x+3 x^2}}{315 \sqrt {x}}+\frac {(1446+4055 x) \sqrt {2+5 x+3 x^2}}{315 x^{5/2}}-\frac {4 (7-15 x) \left (2+5 x+3 x^2\right )^{3/2}}{63 x^{9/2}}-\frac {1}{315} \int \frac {\frac {13485}{2}+8157 x}{\sqrt {x} \sqrt {2+5 x+3 x^2}} \, dx\\ &=\frac {5438 \sqrt {2+5 x+3 x^2}}{315 \sqrt {x}}+\frac {(1446+4055 x) \sqrt {2+5 x+3 x^2}}{315 x^{5/2}}-\frac {4 (7-15 x) \left (2+5 x+3 x^2\right )^{3/2}}{63 x^{9/2}}-\frac {2}{315} \operatorname {Subst}\left (\int \frac {\frac {13485}{2}+8157 x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=\frac {5438 \sqrt {2+5 x+3 x^2}}{315 \sqrt {x}}+\frac {(1446+4055 x) \sqrt {2+5 x+3 x^2}}{315 x^{5/2}}-\frac {4 (7-15 x) \left (2+5 x+3 x^2\right )^{3/2}}{63 x^{9/2}}-\frac {899}{21} \operatorname {Subst}\left (\int \frac {1}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )-\frac {5438}{105} \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {2+5 x^2+3 x^4}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {5438 \sqrt {x} (2+3 x)}{315 \sqrt {2+5 x+3 x^2}}+\frac {5438 \sqrt {2+5 x+3 x^2}}{315 \sqrt {x}}+\frac {(1446+4055 x) \sqrt {2+5 x+3 x^2}}{315 x^{5/2}}-\frac {4 (7-15 x) \left (2+5 x+3 x^2\right )^{3/2}}{63 x^{9/2}}+\frac {5438 \sqrt {2} (1+x) \sqrt {\frac {2+3 x}{1+x}} E\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{315 \sqrt {2+5 x+3 x^2}}-\frac {899 (1+x) \sqrt {\frac {2+3 x}{1+x}} F\left (\tan ^{-1}\left (\sqrt {x}\right )|-\frac {1}{2}\right )}{21 \sqrt {2} \sqrt {2+5 x+3 x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.17, size = 160, normalized size = 0.76 \[ \frac {-2609 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{11/2} F\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )-10876 i \sqrt {2} \sqrt {\frac {1}{x}+1} \sqrt {\frac {2}{x}+3} x^{11/2} E\left (i \sinh ^{-1}\left (\frac {\sqrt {\frac {2}{3}}}{\sqrt {x}}\right )|\frac {3}{2}\right )+29730 x^5+64706 x^4+44480 x^3+7424 x^2-3200 x-1120}{630 x^{9/2} \sqrt {3 x^2+5 x+2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((2 - 5*x)*(2 + 5*x + 3*x^2)^(3/2))/x^(11/2),x]

[Out]

(-1120 - 3200*x + 7424*x^2 + 44480*x^3 + 64706*x^4 + 29730*x^5 - (10876*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2
/x]*x^(11/2)*EllipticE[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2] - (2609*I)*Sqrt[2]*Sqrt[1 + x^(-1)]*Sqrt[3 + 2/x]*x^
(11/2)*EllipticF[I*ArcSinh[Sqrt[2/3]/Sqrt[x]], 3/2])/(630*x^(9/2)*Sqrt[2 + 5*x + 3*x^2])

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fricas [F]  time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {{\left (15 \, x^{3} + 19 \, x^{2} - 4\right )} \sqrt {3 \, x^{2} + 5 \, x + 2}}{x^{\frac {11}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*(3*x^2+5*x+2)^(3/2)/x^(11/2),x, algorithm="fricas")

[Out]

integral(-(15*x^3 + 19*x^2 - 4)*sqrt(3*x^2 + 5*x + 2)/x^(11/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} {\left (5 \, x - 2\right )}}{x^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*(3*x^2+5*x+2)^(3/2)/x^(11/2),x, algorithm="giac")

[Out]

integrate(-(3*x^2 + 5*x + 2)^(3/2)*(5*x - 2)/x^(11/2), x)

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maple [A]  time = 0.08, size = 134, normalized size = 0.64 \[ \frac {97884 x^{6}+252330 x^{5}-5438 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x^{4} \EllipticE \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+2829 \sqrt {6 x +4}\, \sqrt {3 x +3}\, \sqrt {6}\, \sqrt {-x}\, x^{4} \EllipticF \left (\frac {\sqrt {6 x +4}}{2}, i \sqrt {2}\right )+259374 x^{4}+133440 x^{3}+22272 x^{2}-9600 x -3360}{1890 \sqrt {3 x^{2}+5 x +2}\, x^{\frac {9}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2-5*x)*(3*x^2+5*x+2)^(3/2)/x^(11/2),x)

[Out]

1/1890*(2829*(6*x+4)^(1/2)*(3*x+3)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticF(1/2*(6*x+4)^(1/2),I*2^(1/2))*x^4-5438*(6
*x+4)^(1/2)*(3*x+3)^(1/2)*6^(1/2)*(-x)^(1/2)*EllipticE(1/2*(6*x+4)^(1/2),I*2^(1/2))*x^4+97884*x^6+252330*x^5+2
59374*x^4+133440*x^3+22272*x^2-9600*x-3360)/(3*x^2+5*x+2)^(1/2)/x^(9/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {{\left (3 \, x^{2} + 5 \, x + 2\right )}^{\frac {3}{2}} {\left (5 \, x - 2\right )}}{x^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*(3*x^2+5*x+2)^(3/2)/x^(11/2),x, algorithm="maxima")

[Out]

-integrate((3*x^2 + 5*x + 2)^(3/2)*(5*x - 2)/x^(11/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int -\frac {\left (5\,x-2\right )\,{\left (3\,x^2+5\,x+2\right )}^{3/2}}{x^{11/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-((5*x - 2)*(5*x + 3*x^2 + 2)^(3/2))/x^(11/2),x)

[Out]

int(-((5*x - 2)*(5*x + 3*x^2 + 2)^(3/2))/x^(11/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \left (- \frac {4 \sqrt {3 x^{2} + 5 x + 2}}{x^{\frac {11}{2}}}\right )\, dx - \int \frac {19 \sqrt {3 x^{2} + 5 x + 2}}{x^{\frac {7}{2}}}\, dx - \int \frac {15 \sqrt {3 x^{2} + 5 x + 2}}{x^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((2-5*x)*(3*x**2+5*x+2)**(3/2)/x**(11/2),x)

[Out]

-Integral(-4*sqrt(3*x**2 + 5*x + 2)/x**(11/2), x) - Integral(19*sqrt(3*x**2 + 5*x + 2)/x**(7/2), x) - Integral
(15*sqrt(3*x**2 + 5*x + 2)/x**(5/2), x)

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