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

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

[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]

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Rubi [A]  time = 0.140266, antiderivative size = 103, normalized size of antiderivative = 1., 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} \[ \frac{\sqrt{a+b x^2} \left (a^2 f-a b e+b^2 d\right )}{b^3}+\frac{\left (a+b x^2\right )^{3/2} (b e-2 a f)}{3 b^3}+\frac{f \left (a+b x^2\right )^{5/2}}{5 b^3}-\frac{c \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{\sqrt{a}} \]

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 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]

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 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]

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*}

Mathematica [A]  time = 0.124084, size = 86, normalized size = 0.83 \[ \frac{\sqrt{a+b x^2} \left (8 a^2 f-2 a b \left (5 e+2 f x^2\right )+b^2 \left (15 d+5 e x^2+3 f x^4\right )\right )}{15 b^3}-\frac{c \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{\sqrt{a}} \]

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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Maple [A]  time = 0.008, size = 134, normalized size = 1.3 \begin{align*}{\frac{f{x}^{4}}{5\,b}\sqrt{b{x}^{2}+a}}-{\frac{4\,af{x}^{2}}{15\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{8\,{a}^{2}f}{15\,{b}^{3}}\sqrt{b{x}^{2}+a}}+{\frac{e{x}^{2}}{3\,b}\sqrt{b{x}^{2}+a}}-{\frac{2\,ae}{3\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{d}{b}\sqrt{b{x}^{2}+a}}-{c\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}} \end{align*}

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/b^2*a*x^2*(b*x^2+a)^(1/2)+8/15*f/b^3*a^2*(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*a^(1/2)*(b*x^2+a)^(1/2))/x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 1.39378, size = 483, normalized size = 4.69 \begin{align*} \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{align*}

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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Sympy [A]  time = 24.9387, size = 102, normalized size = 0.99 \begin{align*} \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{align*}

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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Giac [A]  time = 1.22864, size = 171, normalized size = 1.66 \begin{align*} \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{align*}

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