3.148 \(\int \frac{c+d x^2+e x^4+f x^6}{x^5 \sqrt{a+b x^2}} \, dx\)

Optimal. Leaf size=114 \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right ) \left (8 a^2 e-4 a b d+3 b^2 c\right )}{8 a^{5/2}}+\frac{\sqrt{a+b x^2} (3 b c-4 a d)}{8 a^2 x^2}-\frac{c \sqrt{a+b x^2}}{4 a x^4}+\frac{f \sqrt{a+b x^2}}{b} \]

[Out]

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

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Rubi [A]  time = 0.232688, antiderivative size = 114, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.188, Rules used = {1799, 1621, 897, 1157, 388, 208} \[ -\frac{\tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right ) \left (8 a^2 e-4 a b d+3 b^2 c\right )}{8 a^{5/2}}+\frac{\sqrt{a+b x^2} (3 b c-4 a d)}{8 a^2 x^2}-\frac{c \sqrt{a+b x^2}}{4 a x^4}+\frac{f \sqrt{a+b x^2}}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x^2 + e*x^4 + f*x^6)/(x^5*Sqrt[a + b*x^2]),x]

[Out]

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

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 1157

Int[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> With[{Qx = PolynomialQ
uotient[(a + b*x^2 + c*x^4)^p, d + e*x^2, x], R = Coeff[PolynomialRemainder[(a + b*x^2 + c*x^4)^p, d + e*x^2,
x], x, 0]}, -Simp[(R*x*(d + e*x^2)^(q + 1))/(2*d*(q + 1)), x] + Dist[1/(2*d*(q + 1)), Int[(d + e*x^2)^(q + 1)*
ExpandToSum[2*d*(q + 1)*Qx + R*(2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && N
eQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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^5 \sqrt{a+b x^2}} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{c+d x+e x^2+f x^3}{x^3 \sqrt{a+b x}} \, dx,x,x^2\right )\\ &=-\frac{c \sqrt{a+b x^2}}{4 a x^4}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (3 b c-4 a d)-2 a e x-2 a f x^2}{x^2 \sqrt{a+b x}} \, dx,x,x^2\right )}{4 a}\\ &=-\frac{c \sqrt{a+b x^2}}{4 a x^4}-\frac{\operatorname{Subst}\left (\int \frac{\frac{\frac{1}{2} b^2 (3 b c-4 a d)+2 a^2 b e-2 a^3 f}{b^2}-\frac{\left (2 a b e-4 a^2 f\right ) x^2}{b^2}-\frac{2 a f x^4}{b^2}}{\left (-\frac{a}{b}+\frac{x^2}{b}\right )^2} \, dx,x,\sqrt{a+b x^2}\right )}{2 a b}\\ &=-\frac{c \sqrt{a+b x^2}}{4 a x^4}+\frac{(3 b c-4 a d) \sqrt{a+b x^2}}{8 a^2 x^2}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} \left (-3 b c+4 a d-\frac{8 a^2 e}{b}+\frac{8 a^3 f}{b^2}\right )-\frac{4 a^2 f x^2}{b^2}}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{4 a^2}\\ &=\frac{f \sqrt{a+b x^2}}{b}-\frac{c \sqrt{a+b x^2}}{4 a x^4}+\frac{(3 b c-4 a d) \sqrt{a+b x^2}}{8 a^2 x^2}+\frac{\left (3 b c-4 a d+\frac{8 a^2 e}{b}\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b x^2}\right )}{8 a^2}\\ &=\frac{f \sqrt{a+b x^2}}{b}-\frac{c \sqrt{a+b x^2}}{4 a x^4}+\frac{(3 b c-4 a d) \sqrt{a+b x^2}}{8 a^2 x^2}-\frac{\left (3 b^2 c-4 a b d+8 a^2 e\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{8 a^{5/2}}\\ \end{align*}

Mathematica [C]  time = 0.345416, size = 141, normalized size = 1.24 \[ -\frac{b^2 c \sqrt{a+b x^2} \, _2F_1\left (\frac{1}{2},3;\frac{3}{2};\frac{b x^2}{a}+1\right )}{a^3}-\frac{b d \sqrt{a+b x^2} \left (\frac{a}{b x^2}-\frac{\tanh ^{-1}\left (\sqrt{\frac{b x^2}{a}+1}\right )}{\sqrt{\frac{b x^2}{a}+1}}\right )}{2 a^2}-\frac{e \tanh ^{-1}\left (\frac{\sqrt{a+b x^2}}{\sqrt{a}}\right )}{\sqrt{a}}+\frac{f \sqrt{a+b x^2}}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x^2 + e*x^4 + f*x^6)/(x^5*Sqrt[a + b*x^2]),x]

[Out]

(f*Sqrt[a + b*x^2])/b - (e*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a] - (b*d*Sqrt[a + b*x^2]*(a/(b*x^2) - ArcTa
nh[Sqrt[1 + (b*x^2)/a]]/Sqrt[1 + (b*x^2)/a]))/(2*a^2) - (b^2*c*Sqrt[a + b*x^2]*Hypergeometric2F1[1/2, 3, 3/2,
1 + (b*x^2)/a])/a^3

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Maple [A]  time = 0.01, size = 162, normalized size = 1.4 \begin{align*}{\frac{f}{b}\sqrt{b{x}^{2}+a}}-{e\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){\frac{1}{\sqrt{a}}}}-{\frac{c}{4\,a{x}^{4}}\sqrt{b{x}^{2}+a}}+{\frac{3\,bc}{8\,{a}^{2}{x}^{2}}\sqrt{b{x}^{2}+a}}-{\frac{3\,{b}^{2}c}{8}\ln \left ({\frac{1}{x} \left ( 2\,a+2\,\sqrt{a}\sqrt{b{x}^{2}+a} \right ) } \right ){a}^{-{\frac{5}{2}}}}-{\frac{d}{2\,a{x}^{2}}\sqrt{b{x}^{2}+a}}+{\frac{bd}{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^5/(b*x^2+a)^(1/2),x)

[Out]

f*(b*x^2+a)^(1/2)/b-e/a^(1/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)-1/4*c*(b*x^2+a)^(1/2)/a/x^4+3/8*c*b/a^2/x^
2*(b*x^2+a)^(1/2)-3/8*c*b^2/a^(5/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x)-1/2*d/a/x^2*(b*x^2+a)^(1/2)+1/2*d*b/
a^(3/2)*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^5/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.47811, size = 501, normalized size = 4.39 \begin{align*} \left [\frac{{\left (3 \, b^{3} c - 4 \, a b^{2} d + 8 \, a^{2} b e\right )} \sqrt{a} x^{4} \log \left (-\frac{b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{a} + 2 \, a}{x^{2}}\right ) + 2 \,{\left (8 \, a^{3} f x^{4} - 2 \, a^{2} b c +{\left (3 \, a b^{2} c - 4 \, a^{2} b d\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{16 \, a^{3} b x^{4}}, \frac{{\left (3 \, b^{3} c - 4 \, a b^{2} d + 8 \, a^{2} b e\right )} \sqrt{-a} x^{4} \arctan \left (\frac{\sqrt{-a}}{\sqrt{b x^{2} + a}}\right ) +{\left (8 \, a^{3} f x^{4} - 2 \, a^{2} b c +{\left (3 \, a b^{2} c - 4 \, a^{2} b d\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{8 \, a^{3} b x^{4}}\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^5/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

[1/16*((3*b^3*c - 4*a*b^2*d + 8*a^2*b*e)*sqrt(a)*x^4*log(-(b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/x^2) + 2*(
8*a^3*f*x^4 - 2*a^2*b*c + (3*a*b^2*c - 4*a^2*b*d)*x^2)*sqrt(b*x^2 + a))/(a^3*b*x^4), 1/8*((3*b^3*c - 4*a*b^2*d
+ 8*a^2*b*e)*sqrt(-a)*x^4*arctan(sqrt(-a)/sqrt(b*x^2 + a)) + (8*a^3*f*x^4 - 2*a^2*b*c + (3*a*b^2*c - 4*a^2*b*
d)*x^2)*sqrt(b*x^2 + a))/(a^3*b*x^4)]

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Sympy [A]  time = 78.6774, size = 194, normalized size = 1.7 \begin{align*} f \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 ) - \frac{c}{4 \sqrt{b} x^{5} \sqrt{\frac{a}{b x^{2}} + 1}} + \frac{\sqrt{b} c}{8 a x^{3} \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{\sqrt{b} d \sqrt{\frac{a}{b x^{2}} + 1}}{2 a x} + \frac{3 b^{\frac{3}{2}} c}{8 a^{2} x \sqrt{\frac{a}{b x^{2}} + 1}} - \frac{e \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{\sqrt{a}} + \frac{b d \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{2 a^{\frac{3}{2}}} - \frac{3 b^{2} c \operatorname{asinh}{\left (\frac{\sqrt{a}}{\sqrt{b} x} \right )}}{8 a^{\frac{5}{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**5/(b*x**2+a)**(1/2),x)

[Out]

f*Piecewise((x**2/(2*sqrt(a)), Eq(b, 0)), (sqrt(a + b*x**2)/b, True)) - c/(4*sqrt(b)*x**5*sqrt(a/(b*x**2) + 1)
) + sqrt(b)*c/(8*a*x**3*sqrt(a/(b*x**2) + 1)) - sqrt(b)*d*sqrt(a/(b*x**2) + 1)/(2*a*x) + 3*b**(3/2)*c/(8*a**2*
x*sqrt(a/(b*x**2) + 1)) - e*asinh(sqrt(a)/(sqrt(b)*x))/sqrt(a) + b*d*asinh(sqrt(a)/(sqrt(b)*x))/(2*a**(3/2)) -
3*b**2*c*asinh(sqrt(a)/(sqrt(b)*x))/(8*a**(5/2))

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Giac [A]  time = 1.16988, size = 190, normalized size = 1.67 \begin{align*} \frac{8 \, \sqrt{b x^{2} + a} f + \frac{{\left (3 \, b^{3} c - 4 \, a b^{2} d + 8 \, a^{2} b e\right )} \arctan \left (\frac{\sqrt{b x^{2} + a}}{\sqrt{-a}}\right )}{\sqrt{-a} a^{2}} + \frac{3 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} b^{3} c - 5 \, \sqrt{b x^{2} + a} a b^{3} c - 4 \,{\left (b x^{2} + a\right )}^{\frac{3}{2}} a b^{2} d + 4 \, \sqrt{b x^{2} + a} a^{2} b^{2} d}{a^{2} b^{2} x^{4}}}{8 \, 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^5/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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