3.147 \(\int \frac{c+d x^2+e x^4+f x^6}{x^3 \sqrt{a+b x^2}} \, dx\)

Optimal. Leaf size=100 \[ \frac{(b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{3/2}}+\frac{\sqrt{a+b x^2} (b e-a f)}{b^2}+\frac{f \left (a+b x^2\right )^{3/2}}{3 b^2}-\frac{c \sqrt{a+b x^2}}{2 a x^2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.202996, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {1799, 1621, 897, 1153, 208} \[ \frac{(b c-2 a d) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{2 a^{3/2}}+\frac{\sqrt{a+b x^2} (b e-a f)}{b^2}+\frac{f \left (a+b x^2\right )^{3/2}}{3 b^2}-\frac{c \sqrt{a+b x^2}}{2 a x^2} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2 + e*x^4 + f*x^6)/(x^3*Sqrt[a + b*x^2]),x]

[Out]

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

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]

Rule 1621

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> With[{Qx = PolynomialQuotient[Px,
a + b*x, x], R = PolynomialRemainder[Px, a + b*x, x]}, Simp[(R*(a + b*x)^(m + 1)*(c + d*x)^(n + 1))/((m + 1)*
(b*c - a*d)), x] + Dist[1/((m + 1)*(b*c - a*d)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*ExpandToSum[(m + 1)*(b*c -
a*d)*Qx - d*R*(m + n + 2), x], x], x]] /; FreeQ[{a, b, c, d, n}, x] && PolyQ[Px, x] && ILtQ[m, -1] && GtQ[Expo
n[Px, x], 2]

Rule 897

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> With[{q = Denominator[m]}, Dist[q/e, Subst[Int[x^(q*(m + 1) - 1)*((e*f - d*g)/e + (g*x^q)/e)^n*((c*d^2 - b*d
*e + a*e^2)/e^2 - ((2*c*d - b*e)*x^q)/e^2 + (c*x^(2*q))/e^2)^p, x], x, (d + e*x)^(1/q)], x]] /; FreeQ[{a, b, c
, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegersQ[n,
p] && FractionQ[m]

Rule 1153

Int[((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(
d + e*x^2)^q*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 -
b*d*e + a*e^2, 0] && IGtQ[p, 0] && IGtQ[q, -2]

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

Mathematica [A]  time = 0.382209, size = 131, normalized size = 1.31 \[ \frac{3 b^3 c x^2 \sqrt{\frac{b x^2}{a}+1} \tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )-\left (a+b x^2\right ) \left (4 a^2 f x^2-2 a b x^2 \left (3 e+f x^2\right )+3 b^2 c\right )}{6 a b^2 x^2 \sqrt{a+b x^2}}-\frac{d \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{\sqrt{a}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2 + e*x^4 + f*x^6)/(x^3*Sqrt[a + b*x^2]),x]

[Out]

-((d*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a]) + (-((a + b*x^2)*(3*b^2*c + 4*a^2*f*x^2 - 2*a*b*x^2*(3*e + f*x
^2))) + 3*b^3*c*x^2*Sqrt[1 + (b*x^2)/a]*ArcTanh[Sqrt[1 + (b*x^2)/a]])/(6*a*b^2*x^2*Sqrt[a + b*x^2])

________________________________________________________________________________________

Maple [A]  time = 0.009, size = 127, normalized size = 1.3 \begin{align*}{\frac{f{x}^{2}}{3\,b}\sqrt{b{x}^{2}+a}}-{\frac{2\,af}{3\,{b}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{e}{b}\sqrt{b{x}^{2}+a}}-{d\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}-{\frac{c}{2\,a{x}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{bc}{2}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{3}{2}}}} \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^3/(b*x^2+a)^(1/2),x)

[Out]

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

________________________________________________________________________________________

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^3/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.46024, size = 482, normalized size = 4.82 \begin{align*} \left [-\frac{3 \,{\left (b^{3} c - 2 \, a b^{2} d\right )} \sqrt{a} x^{2} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) - 2 \,{\left (2 \, a^{2} b f x^{4} - 3 \, a b^{2} c + 2 \,{\left (3 \, a^{2} b e - 2 \, a^{3} f\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{12 \, a^{2} b^{2} x^{2}}, -\frac{3 \,{\left (b^{3} c - 2 \, a b^{2} d\right )} \sqrt{-a} x^{2} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) -{\left (2 \, a^{2} b f x^{4} - 3 \, a b^{2} c + 2 \,{\left (3 \, a^{2} b e - 2 \, a^{3} f\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{6 \, a^{2} b^{2} x^{2}}\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^3/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/12*(3*(b^3*c - 2*a*b^2*d)*sqrt(a)*x^2*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) - 2*(2*a^2*b*f*x
^4 - 3*a*b^2*c + 2*(3*a^2*b*e - 2*a^3*f)*x^2)*sqrt(b*x^2 + a))/(a^2*b^2*x^2), -1/6*(3*(b^3*c - 2*a*b^2*d)*sqrt
(-a)*x^2*arctan(sqrt(-a)/sqrt(b*x^2 + a)) - (2*a^2*b*f*x^4 - 3*a*b^2*c + 2*(3*a^2*b*e - 2*a^3*f)*x^2)*sqrt(b*x
^2 + a))/(a^2*b^2*x^2)]

________________________________________________________________________________________

Sympy [A]  time = 40.1251, size = 138, normalized size = 1.38 \begin{align*} e \left (\begin{cases} \frac{x^{2}}{2 \sqrt{a}} & \text{for}\: b = 0 \\\frac{\sqrt{a + b x^{2}}}{b} & \text{otherwise} \end{cases}\right ) + f \left (\begin{cases} - \frac{2 a \sqrt{a + b x^{2}}}{3 b^{2}} + \frac{x^{2} \sqrt{a + b x^{2}}}{3 b} & \text{for}\: b \neq 0 \\\frac{x^{4}}{4 \sqrt{a}} & \text{otherwise} \end{cases}\right ) - \frac{\sqrt{b} c \sqrt{\frac{a}{b x^{2}} + 1}}{2 a x} - \frac{d \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{\sqrt{a}} + \frac{b c \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2 a^{\frac{3}{2}}} \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**3/(b*x**2+a)**(1/2),x)

[Out]

e*Piecewise((x**2/(2*sqrt(a)), Eq(b, 0)), (sqrt(a + b*x**2)/b, True)) + f*Piecewise((-2*a*sqrt(a + b*x**2)/(3*
b**2) + x**2*sqrt(a + b*x**2)/(3*b), Ne(b, 0)), (x**4/(4*sqrt(a)), True)) - sqrt(b)*c*sqrt(a/(b*x**2) + 1)/(2*
a*x) - d*asinh(sqrt(a)/(sqrt(b)*x))/sqrt(a) + b*c*asinh(sqrt(a)/(sqrt(b)*x))/(2*a**(3/2))

________________________________________________________________________________________

Giac [A]  time = 1.23002, size = 154, normalized size = 1.54 \begin{align*} -\frac{\frac{3 \,{\left (b^{2} c - 2 \, a b d\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a} + \frac{3 \, \sqrt{b x^{2} + a} b c}{a x^{2}} - \frac{2 \,{\left ({\left (b x^{2} + a\right )}^{\frac{3}{2}} b^{2} f - 3 \, \sqrt{b x^{2} + a} a b^{2} f + 3 \, \sqrt{b x^{2} + a} b^{3} e\right )}}{b^{3}}}{6 \, b} \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^3/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-1/6*(3*(b^2*c - 2*a*b*d)*arctan(sqrt(b*x^2 + a)/sqrt(-a))/(sqrt(-a)*a) + 3*sqrt(b*x^2 + a)*b*c/(a*x^2) - 2*((
b*x^2 + a)^(3/2)*b^2*f - 3*sqrt(b*x^2 + a)*a*b^2*f + 3*sqrt(b*x^2 + a)*b^3*e)/b^3)/b