3.2372 \(\int \frac {x^5}{\sqrt {a+b x+c x^2}} \, dx\)

Optimal. Leaf size=202 \[ -\frac {b \left (240 a^2 c^2-280 a b^2 c+63 b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{11/2}}+\frac {\left (1024 a^2 c^2-14 b c x \left (45 b^2-92 a c\right )-2940 a b^2 c+945 b^4\right ) \sqrt {a+b x+c x^2}}{1920 c^5}+\frac {x^2 \left (63 b^2-64 a c\right ) \sqrt {a+b x+c x^2}}{240 c^3}-\frac {9 b x^3 \sqrt {a+b x+c x^2}}{40 c^2}+\frac {x^4 \sqrt {a+b x+c x^2}}{5 c} \]

[Out]

-1/256*b*(240*a^2*c^2-280*a*b^2*c+63*b^4)*arctanh(1/2*(2*c*x+b)/c^(1/2)/(c*x^2+b*x+a)^(1/2))/c^(11/2)+1/240*(-
64*a*c+63*b^2)*x^2*(c*x^2+b*x+a)^(1/2)/c^3-9/40*b*x^3*(c*x^2+b*x+a)^(1/2)/c^2+1/5*x^4*(c*x^2+b*x+a)^(1/2)/c+1/
1920*(945*b^4-2940*a*b^2*c+1024*a^2*c^2-14*b*c*(-92*a*c+45*b^2)*x)*(c*x^2+b*x+a)^(1/2)/c^5

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Rubi [A]  time = 0.22, antiderivative size = 202, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {742, 832, 779, 621, 206} \[ \frac {\left (1024 a^2 c^2-14 b c x \left (45 b^2-92 a c\right )-2940 a b^2 c+945 b^4\right ) \sqrt {a+b x+c x^2}}{1920 c^5}-\frac {b \left (240 a^2 c^2-280 a b^2 c+63 b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{11/2}}+\frac {x^2 \left (63 b^2-64 a c\right ) \sqrt {a+b x+c x^2}}{240 c^3}-\frac {9 b x^3 \sqrt {a+b x+c x^2}}{40 c^2}+\frac {x^4 \sqrt {a+b x+c x^2}}{5 c} \]

Antiderivative was successfully verified.

[In]

Int[x^5/Sqrt[a + b*x + c*x^2],x]

[Out]

((63*b^2 - 64*a*c)*x^2*Sqrt[a + b*x + c*x^2])/(240*c^3) - (9*b*x^3*Sqrt[a + b*x + c*x^2])/(40*c^2) + (x^4*Sqrt
[a + b*x + c*x^2])/(5*c) + ((945*b^4 - 2940*a*b^2*c + 1024*a^2*c^2 - 14*b*c*(45*b^2 - 92*a*c)*x)*Sqrt[a + b*x
+ c*x^2])/(1920*c^5) - (b*(63*b^4 - 280*a*b^2*c + 240*a^2*c^2)*ArcTanh[(b + 2*c*x)/(2*Sqrt[c]*Sqrt[a + b*x + c
*x^2])])/(256*c^(11/2))

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 621

Int[1/Sqrt[(a_) + (b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Dist[2, Subst[Int[1/(4*c - x^2), x], x, (b + 2*c*x)
/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 742

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

Rule 779

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

Rule 832

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

Rubi steps

\begin {align*} \int \frac {x^5}{\sqrt {a+b x+c x^2}} \, dx &=\frac {x^4 \sqrt {a+b x+c x^2}}{5 c}+\frac {\int \frac {x^3 \left (-4 a-\frac {9 b x}{2}\right )}{\sqrt {a+b x+c x^2}} \, dx}{5 c}\\ &=-\frac {9 b x^3 \sqrt {a+b x+c x^2}}{40 c^2}+\frac {x^4 \sqrt {a+b x+c x^2}}{5 c}+\frac {\int \frac {x^2 \left (\frac {27 a b}{2}+\frac {1}{4} \left (63 b^2-64 a c\right ) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{20 c^2}\\ &=\frac {\left (63 b^2-64 a c\right ) x^2 \sqrt {a+b x+c x^2}}{240 c^3}-\frac {9 b x^3 \sqrt {a+b x+c x^2}}{40 c^2}+\frac {x^4 \sqrt {a+b x+c x^2}}{5 c}+\frac {\int \frac {x \left (-\frac {1}{2} a \left (63 b^2-64 a c\right )-\frac {7}{8} b \left (45 b^2-92 a c\right ) x\right )}{\sqrt {a+b x+c x^2}} \, dx}{60 c^3}\\ &=\frac {\left (63 b^2-64 a c\right ) x^2 \sqrt {a+b x+c x^2}}{240 c^3}-\frac {9 b x^3 \sqrt {a+b x+c x^2}}{40 c^2}+\frac {x^4 \sqrt {a+b x+c x^2}}{5 c}+\frac {\left (945 b^4-2940 a b^2 c+1024 a^2 c^2-14 b c \left (45 b^2-92 a c\right ) x\right ) \sqrt {a+b x+c x^2}}{1920 c^5}-\frac {\left (b \left (63 b^4-280 a b^2 c+240 a^2 c^2\right )\right ) \int \frac {1}{\sqrt {a+b x+c x^2}} \, dx}{256 c^5}\\ &=\frac {\left (63 b^2-64 a c\right ) x^2 \sqrt {a+b x+c x^2}}{240 c^3}-\frac {9 b x^3 \sqrt {a+b x+c x^2}}{40 c^2}+\frac {x^4 \sqrt {a+b x+c x^2}}{5 c}+\frac {\left (945 b^4-2940 a b^2 c+1024 a^2 c^2-14 b c \left (45 b^2-92 a c\right ) x\right ) \sqrt {a+b x+c x^2}}{1920 c^5}-\frac {\left (b \left (63 b^4-280 a b^2 c+240 a^2 c^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{4 c-x^2} \, dx,x,\frac {b+2 c x}{\sqrt {a+b x+c x^2}}\right )}{128 c^5}\\ &=\frac {\left (63 b^2-64 a c\right ) x^2 \sqrt {a+b x+c x^2}}{240 c^3}-\frac {9 b x^3 \sqrt {a+b x+c x^2}}{40 c^2}+\frac {x^4 \sqrt {a+b x+c x^2}}{5 c}+\frac {\left (945 b^4-2940 a b^2 c+1024 a^2 c^2-14 b c \left (45 b^2-92 a c\right ) x\right ) \sqrt {a+b x+c x^2}}{1920 c^5}-\frac {b \left (63 b^4-280 a b^2 c+240 a^2 c^2\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+b x+c x^2}}\right )}{256 c^{11/2}}\\ \end {align*}

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Mathematica [A]  time = 0.20, size = 213, normalized size = 1.05 \[ \frac {1024 a^3 c^2+4 a^2 c \left (-735 b^2+578 b c x+128 c^2 x^2\right )+a \left (945 b^4-3570 b^3 c x-1148 b^2 c^2 x^2+344 b c^3 x^3-128 c^4 x^4\right )+3 x \left (315 b^5+105 b^4 c x-42 b^3 c^2 x^2+24 b^2 c^3 x^3-16 b c^4 x^4+128 c^5 x^5\right )}{1920 c^5 \sqrt {a+x (b+c x)}}-\frac {b \left (240 a^2 c^2-280 a b^2 c+63 b^4\right ) \tanh ^{-1}\left (\frac {b+2 c x}{2 \sqrt {c} \sqrt {a+x (b+c x)}}\right )}{256 c^{11/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5/Sqrt[a + b*x + c*x^2],x]

[Out]

(1024*a^3*c^2 + 4*a^2*c*(-735*b^2 + 578*b*c*x + 128*c^2*x^2) + a*(945*b^4 - 3570*b^3*c*x - 1148*b^2*c^2*x^2 +
344*b*c^3*x^3 - 128*c^4*x^4) + 3*x*(315*b^5 + 105*b^4*c*x - 42*b^3*c^2*x^2 + 24*b^2*c^3*x^3 - 16*b*c^4*x^4 + 1
28*c^5*x^5))/(1920*c^5*Sqrt[a + x*(b + c*x)]) - (b*(63*b^4 - 280*a*b^2*c + 240*a^2*c^2)*ArcTanh[(b + 2*c*x)/(2
*Sqrt[c]*Sqrt[a + x*(b + c*x)])])/(256*c^(11/2))

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fricas [A]  time = 0.98, size = 347, normalized size = 1.72 \[ \left [\frac {15 \, {\left (63 \, b^{5} - 280 \, a b^{3} c + 240 \, a^{2} b c^{2}\right )} \sqrt {c} \log \left (-8 \, c^{2} x^{2} - 8 \, b c x - b^{2} + 4 \, \sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {c} - 4 \, a c\right ) + 4 \, {\left (384 \, c^{5} x^{4} - 432 \, b c^{4} x^{3} + 945 \, b^{4} c - 2940 \, a b^{2} c^{2} + 1024 \, a^{2} c^{3} + 8 \, {\left (63 \, b^{2} c^{3} - 64 \, a c^{4}\right )} x^{2} - 14 \, {\left (45 \, b^{3} c^{2} - 92 \, a b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{7680 \, c^{6}}, \frac {15 \, {\left (63 \, b^{5} - 280 \, a b^{3} c + 240 \, a^{2} b c^{2}\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x + a} {\left (2 \, c x + b\right )} \sqrt {-c}}{2 \, {\left (c^{2} x^{2} + b c x + a c\right )}}\right ) + 2 \, {\left (384 \, c^{5} x^{4} - 432 \, b c^{4} x^{3} + 945 \, b^{4} c - 2940 \, a b^{2} c^{2} + 1024 \, a^{2} c^{3} + 8 \, {\left (63 \, b^{2} c^{3} - 64 \, a c^{4}\right )} x^{2} - 14 \, {\left (45 \, b^{3} c^{2} - 92 \, a b c^{3}\right )} x\right )} \sqrt {c x^{2} + b x + a}}{3840 \, c^{6}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/7680*(15*(63*b^5 - 280*a*b^3*c + 240*a^2*b*c^2)*sqrt(c)*log(-8*c^2*x^2 - 8*b*c*x - b^2 + 4*sqrt(c*x^2 + b*x
 + a)*(2*c*x + b)*sqrt(c) - 4*a*c) + 4*(384*c^5*x^4 - 432*b*c^4*x^3 + 945*b^4*c - 2940*a*b^2*c^2 + 1024*a^2*c^
3 + 8*(63*b^2*c^3 - 64*a*c^4)*x^2 - 14*(45*b^3*c^2 - 92*a*b*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^6, 1/3840*(15*(63
*b^5 - 280*a*b^3*c + 240*a^2*b*c^2)*sqrt(-c)*arctan(1/2*sqrt(c*x^2 + b*x + a)*(2*c*x + b)*sqrt(-c)/(c^2*x^2 +
b*c*x + a*c)) + 2*(384*c^5*x^4 - 432*b*c^4*x^3 + 945*b^4*c - 2940*a*b^2*c^2 + 1024*a^2*c^3 + 8*(63*b^2*c^3 - 6
4*a*c^4)*x^2 - 14*(45*b^3*c^2 - 92*a*b*c^3)*x)*sqrt(c*x^2 + b*x + a))/c^6]

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giac [A]  time = 0.23, size = 161, normalized size = 0.80 \[ \frac {1}{1920} \, \sqrt {c x^{2} + b x + a} {\left (2 \, {\left (4 \, {\left (6 \, x {\left (\frac {8 \, x}{c} - \frac {9 \, b}{c^{2}}\right )} + \frac {63 \, b^{2} c^{2} - 64 \, a c^{3}}{c^{5}}\right )} x - \frac {7 \, {\left (45 \, b^{3} c - 92 \, a b c^{2}\right )}}{c^{5}}\right )} x + \frac {945 \, b^{4} - 2940 \, a b^{2} c + 1024 \, a^{2} c^{2}}{c^{5}}\right )} + \frac {{\left (63 \, b^{5} - 280 \, a b^{3} c + 240 \, a^{2} b c^{2}\right )} \log \left ({\left | -2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x + a}\right )} \sqrt {c} - b \right |}\right )}{256 \, c^{\frac {11}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

1/1920*sqrt(c*x^2 + b*x + a)*(2*(4*(6*x*(8*x/c - 9*b/c^2) + (63*b^2*c^2 - 64*a*c^3)/c^5)*x - 7*(45*b^3*c - 92*
a*b*c^2)/c^5)*x + (945*b^4 - 2940*a*b^2*c + 1024*a^2*c^2)/c^5) + 1/256*(63*b^5 - 280*a*b^3*c + 240*a^2*b*c^2)*
log(abs(-2*(sqrt(c)*x - sqrt(c*x^2 + b*x + a))*sqrt(c) - b))/c^(11/2)

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maple [A]  time = 0.06, size = 290, normalized size = 1.44 \[ \frac {\sqrt {c \,x^{2}+b x +a}\, x^{4}}{5 c}-\frac {9 \sqrt {c \,x^{2}+b x +a}\, b \,x^{3}}{40 c^{2}}-\frac {4 \sqrt {c \,x^{2}+b x +a}\, a \,x^{2}}{15 c^{2}}+\frac {21 \sqrt {c \,x^{2}+b x +a}\, b^{2} x^{2}}{80 c^{3}}-\frac {15 a^{2} b \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{16 c^{\frac {7}{2}}}+\frac {35 a \,b^{3} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{32 c^{\frac {9}{2}}}-\frac {63 b^{5} \ln \left (\frac {c x +\frac {b}{2}}{\sqrt {c}}+\sqrt {c \,x^{2}+b x +a}\right )}{256 c^{\frac {11}{2}}}+\frac {161 \sqrt {c \,x^{2}+b x +a}\, a b x}{240 c^{3}}-\frac {21 \sqrt {c \,x^{2}+b x +a}\, b^{3} x}{64 c^{4}}+\frac {8 \sqrt {c \,x^{2}+b x +a}\, a^{2}}{15 c^{3}}-\frac {49 \sqrt {c \,x^{2}+b x +a}\, a \,b^{2}}{32 c^{4}}+\frac {63 \sqrt {c \,x^{2}+b x +a}\, b^{4}}{128 c^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(c*x^2+b*x+a)^(1/2),x)

[Out]

1/5*x^4*(c*x^2+b*x+a)^(1/2)/c-9/40*b*x^3*(c*x^2+b*x+a)^(1/2)/c^2+21/80*b^2/c^3*x^2*(c*x^2+b*x+a)^(1/2)-21/64*b
^3/c^4*x*(c*x^2+b*x+a)^(1/2)+63/128*b^4/c^5*(c*x^2+b*x+a)^(1/2)-63/256*b^5/c^(11/2)*ln((c*x+1/2*b)/c^(1/2)+(c*
x^2+b*x+a)^(1/2))+35/32*b^3/c^(9/2)*a*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-49/32*b^2/c^4*a*(c*x^2+b*x+a
)^(1/2)+161/240*b/c^3*a*x*(c*x^2+b*x+a)^(1/2)-15/16*b/c^(7/2)*a^2*ln((c*x+1/2*b)/c^(1/2)+(c*x^2+b*x+a)^(1/2))-
4/15*a/c^2*x^2*(c*x^2+b*x+a)^(1/2)+8/15*a^2/c^3*(c*x^2+b*x+a)^(1/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a*c-b^2>0)', see `assume?` f
or more details)Is 4*a*c-b^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {x^5}{\sqrt {c\,x^2+b\,x+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5/(a + b*x + c*x^2)^(1/2),x)

[Out]

int(x^5/(a + b*x + c*x^2)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{5}}{\sqrt {a + b x + c x^{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5/(c*x**2+b*x+a)**(1/2),x)

[Out]

Integral(x**5/sqrt(a + b*x + c*x**2), x)

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