3.2.42 \(\int \frac {c+d x^2+e x^4+f x^6}{x \sqrt {a+b x^2}} \, dx\)

Optimal. Leaf size=103 \[ \frac {\sqrt {a+b x^2} \left (a^2 f-a b e+b^2 d\right )}{b^3}+\frac {\left (a+b x^2\right )^{3/2} (b e-2 a f)}{3 b^3}+\frac {f \left (a+b x^2\right )^{5/2}}{5 b^3}-\frac {c \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}} \]

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Rubi [A]  time = 0.14, antiderivative size = 103, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 32, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {1799, 1620, 63, 208} \begin {gather*} \frac {\sqrt {a+b x^2} \left (a^2 f-a b e+b^2 d\right )}{b^3}+\frac {\left (a+b x^2\right )^{3/2} (b e-2 a f)}{3 b^3}+\frac {f \left (a+b x^2\right )^{5/2}}{5 b^3}-\frac {c \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2 + e*x^4 + f*x^6)/(x*Sqrt[a + b*x^2]),x]

[Out]

((b^2*d - a*b*e + a^2*f)*Sqrt[a + b*x^2])/b^3 + ((b*e - 2*a*f)*(a + b*x^2)^(3/2))/(3*b^3) + (f*(a + b*x^2)^(5/
2))/(5*b^3) - (c*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 1799

Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m - 1)/2)*SubstFor[x^2,
Pq, x]*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x^2] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \frac {c+d x^2+e x^4+f x^6}{x \sqrt {a+b x^2}} \, dx &=\frac {1}{2} \operatorname {Subst}\left (\int \frac {c+d x+e x^2+f x^3}{x \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=\frac {1}{2} \operatorname {Subst}\left (\int \left (\frac {b^2 d-a b e+a^2 f}{b^2 \sqrt {a+b x}}+\frac {c}{x \sqrt {a+b x}}+\frac {(b e-2 a f) \sqrt {a+b x}}{b^2}+\frac {f (a+b x)^{3/2}}{b^2}\right ) \, dx,x,x^2\right )\\ &=\frac {\left (b^2 d-a b e+a^2 f\right ) \sqrt {a+b x^2}}{b^3}+\frac {(b e-2 a f) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {f \left (a+b x^2\right )^{5/2}}{5 b^3}+\frac {1}{2} c \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+b x}} \, dx,x,x^2\right )\\ &=\frac {\left (b^2 d-a b e+a^2 f\right ) \sqrt {a+b x^2}}{b^3}+\frac {(b e-2 a f) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {f \left (a+b x^2\right )^{5/2}}{5 b^3}+\frac {c \operatorname {Subst}\left (\int \frac {1}{-\frac {a}{b}+\frac {x^2}{b}} \, dx,x,\sqrt {a+b x^2}\right )}{b}\\ &=\frac {\left (b^2 d-a b e+a^2 f\right ) \sqrt {a+b x^2}}{b^3}+\frac {(b e-2 a f) \left (a+b x^2\right )^{3/2}}{3 b^3}+\frac {f \left (a+b x^2\right )^{5/2}}{5 b^3}-\frac {c \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 86, normalized size = 0.83 \begin {gather*} \frac {\sqrt {a+b x^2} \left (8 a^2 f-2 a b \left (5 e+2 f x^2\right )+b^2 \left (15 d+5 e x^2+3 f x^4\right )\right )}{15 b^3}-\frac {c \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2 + e*x^4 + f*x^6)/(x*Sqrt[a + b*x^2]),x]

[Out]

(Sqrt[a + b*x^2]*(8*a^2*f - 2*a*b*(5*e + 2*f*x^2) + b^2*(15*d + 5*e*x^2 + 3*f*x^4)))/(15*b^3) - (c*ArcTanh[Sqr
t[a + b*x^2]/Sqrt[a]])/Sqrt[a]

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IntegrateAlgebraic [A]  time = 0.09, size = 89, normalized size = 0.86 \begin {gather*} \frac {\sqrt {a+b x^2} \left (8 a^2 f-10 a b e-4 a b f x^2+15 b^2 d+5 b^2 e x^2+3 b^2 f x^4\right )}{15 b^3}-\frac {c \tanh ^{-1}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{\sqrt {a}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(c + d*x^2 + e*x^4 + f*x^6)/(x*Sqrt[a + b*x^2]),x]

[Out]

(Sqrt[a + b*x^2]*(15*b^2*d - 10*a*b*e + 8*a^2*f + 5*b^2*e*x^2 - 4*a*b*f*x^2 + 3*b^2*f*x^4))/(15*b^3) - (c*ArcT
anh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a]

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fricas [A]  time = 1.03, size = 205, normalized size = 1.99 \begin {gather*} \left [\frac {15 \, \sqrt {a} b^{3} c \log \left (-\frac {b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) + 2 \, {\left (3 \, a b^{2} f x^{4} + 15 \, a b^{2} d - 10 \, a^{2} b e + 8 \, a^{3} f + {\left (5 \, a b^{2} e - 4 \, a^{2} b f\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{30 \, a b^{3}}, \frac {15 \, \sqrt {-a} b^{3} c \arctan \left (\frac {\sqrt {-a}}{\sqrt {b x^{2} + a}}\right ) + {\left (3 \, a b^{2} f x^{4} + 15 \, a b^{2} d - 10 \, a^{2} b e + 8 \, a^{3} f + {\left (5 \, a b^{2} e - 4 \, a^{2} b f\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, a b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/30*(15*sqrt(a)*b^3*c*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(3*a*b^2*f*x^4 + 15*a*b^2*d -
10*a^2*b*e + 8*a^3*f + (5*a*b^2*e - 4*a^2*b*f)*x^2)*sqrt(b*x^2 + a))/(a*b^3), 1/15*(15*sqrt(-a)*b^3*c*arctan(s
qrt(-a)/sqrt(b*x^2 + a)) + (3*a*b^2*f*x^4 + 15*a*b^2*d - 10*a^2*b*e + 8*a^3*f + (5*a*b^2*e - 4*a^2*b*f)*x^2)*s
qrt(b*x^2 + a))/(a*b^3)]

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giac [A]  time = 0.40, size = 127, normalized size = 1.23 \begin {gather*} \frac {c \arctan \left (\frac {\sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} + \frac {15 \, \sqrt {b x^{2} + a} b^{14} d + 3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} b^{12} f - 10 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a b^{12} f + 15 \, \sqrt {b x^{2} + a} a^{2} b^{12} f + 5 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{13} e - 15 \, \sqrt {b x^{2} + a} a b^{13} e}{15 \, b^{15}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

c*arctan(sqrt(b*x^2 + a)/sqrt(-a))/sqrt(-a) + 1/15*(15*sqrt(b*x^2 + a)*b^14*d + 3*(b*x^2 + a)^(5/2)*b^12*f - 1
0*(b*x^2 + a)^(3/2)*a*b^12*f + 15*sqrt(b*x^2 + a)*a^2*b^12*f + 5*(b*x^2 + a)^(3/2)*b^13*e - 15*sqrt(b*x^2 + a)
*a*b^13*e)/b^15

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maple [A]  time = 0.01, size = 134, normalized size = 1.30 \begin {gather*} \frac {\sqrt {b \,x^{2}+a}\, f \,x^{4}}{5 b}-\frac {4 \sqrt {b \,x^{2}+a}\, a f \,x^{2}}{15 b^{2}}+\frac {\sqrt {b \,x^{2}+a}\, e \,x^{2}}{3 b}-\frac {c \ln \left (\frac {2 a +2 \sqrt {b \,x^{2}+a}\, \sqrt {a}}{x}\right )}{\sqrt {a}}+\frac {8 \sqrt {b \,x^{2}+a}\, a^{2} f}{15 b^{3}}-\frac {2 \sqrt {b \,x^{2}+a}\, a e}{3 b^{2}}+\frac {\sqrt {b \,x^{2}+a}\, d}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((f*x^6+e*x^4+d*x^2+c)/x/(b*x^2+a)^(1/2),x)

[Out]

1/5*f*x^4/b*(b*x^2+a)^(1/2)-4/15*f*a/b^2*x^2*(b*x^2+a)^(1/2)+8/15*f*a^2/b^3*(b*x^2+a)^(1/2)+1/3*e*x^2/b*(b*x^2
+a)^(1/2)-2/3*e*a/b^2*(b*x^2+a)^(1/2)+d/b*(b*x^2+a)^(1/2)-c/a^(1/2)*ln((2*a+2*(b*x^2+a)^(1/2)*a^(1/2))/x)

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maxima [A]  time = 1.35, size = 122, normalized size = 1.18 \begin {gather*} \frac {\sqrt {b x^{2} + a} f x^{4}}{5 \, b} + \frac {\sqrt {b x^{2} + a} e x^{2}}{3 \, b} - \frac {4 \, \sqrt {b x^{2} + a} a f x^{2}}{15 \, b^{2}} - \frac {c \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{\sqrt {a}} + \frac {\sqrt {b x^{2} + a} d}{b} - \frac {2 \, \sqrt {b x^{2} + a} a e}{3 \, b^{2}} + \frac {8 \, \sqrt {b x^{2} + a} a^{2} f}{15 \, b^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x^6+e*x^4+d*x^2+c)/x/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

1/5*sqrt(b*x^2 + a)*f*x^4/b + 1/3*sqrt(b*x^2 + a)*e*x^2/b - 4/15*sqrt(b*x^2 + a)*a*f*x^2/b^2 - c*arcsinh(a/(sq
rt(a*b)*abs(x)))/sqrt(a) + sqrt(b*x^2 + a)*d/b - 2/3*sqrt(b*x^2 + a)*a*e/b^2 + 8/15*sqrt(b*x^2 + a)*a^2*f/b^3

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mupad [B]  time = 1.81, size = 99, normalized size = 0.96 \begin {gather*} \sqrt {b\,x^2+a}\,\left (\frac {8\,a^2\,f}{15\,b^3}+\frac {f\,x^4}{5\,b}-\frac {4\,a\,f\,x^2}{15\,b^2}\right )+\frac {d\,\sqrt {b\,x^2+a}}{b}-\frac {c\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {e\,\sqrt {b\,x^2+a}\,\left (2\,a-b\,x^2\right )}{3\,b^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2 + e*x^4 + f*x^6)/(x*(a + b*x^2)^(1/2)),x)

[Out]

(a + b*x^2)^(1/2)*((8*a^2*f)/(15*b^3) + (f*x^4)/(5*b) - (4*a*f*x^2)/(15*b^2)) + (d*(a + b*x^2)^(1/2))/b - (c*a
tanh((a + b*x^2)^(1/2)/a^(1/2)))/a^(1/2) - (e*(a + b*x^2)^(1/2)*(2*a - b*x^2))/(3*b^2)

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sympy [A]  time = 37.87, size = 102, normalized size = 0.99 \begin {gather*} \frac {f \left (a + b x^{2}\right )^{\frac {5}{2}}}{5 b^{3}} - \frac {\left (a + b x^{2}\right )^{\frac {3}{2}} \left (2 a f - b e\right )}{3 b^{3}} + \frac {\sqrt {a + b x^{2}} \left (a^{2} f - a b e + b^{2} d\right )}{b^{3}} + \frac {c \operatorname {atan}{\left (\frac {1}{\sqrt {- \frac {1}{a}} \sqrt {a + b x^{2}}} \right )}}{a \sqrt {- \frac {1}{a}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x**6+e*x**4+d*x**2+c)/x/(b*x**2+a)**(1/2),x)

[Out]

f*(a + b*x**2)**(5/2)/(5*b**3) - (a + b*x**2)**(3/2)*(2*a*f - b*e)/(3*b**3) + sqrt(a + b*x**2)*(a**2*f - a*b*e
+ b**2*d)/b**3 + c*atan(1/(sqrt(-1/a)*sqrt(a + b*x**2)))/(a*sqrt(-1/a))

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