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

Optimal. Leaf size=118 \[ -\frac{\sqrt{a+b x^2} \left (15 a^2 e-10 a b d+8 b^2 c\right )}{15 a^3 x}+\frac{\sqrt{a+b x^2} (4 b c-5 a d)}{15 a^2 x^3}-\frac{c \sqrt{a+b x^2}}{5 a x^5}+\frac{f \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{\sqrt{b}} \]

[Out]

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

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Rubi [A]  time = 0.133001, antiderivative size = 118, 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 = {1807, 1585, 1265, 451, 217, 206} \[ -\frac{\sqrt{a+b x^2} \left (15 a^2 e-10 a b d+8 b^2 c\right )}{15 a^3 x}+\frac{\sqrt{a+b x^2} (4 b c-5 a d)}{15 a^2 x^3}-\frac{c \sqrt{a+b x^2}}{5 a x^5}+\frac{f \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{\sqrt{b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1807

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuotient[Pq, c*x, x],
R = PolynomialRemainder[Pq, c*x, x]}, Simp[(R*(c*x)^(m + 1)*(a + b*x^2)^(p + 1))/(a*c*(m + 1)), x] + Dist[1/(
a*c*(m + 1)), Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*(m + 2*p + 3)*x, x], x], x]] /;
FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && LtQ[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])

Rule 1585

Int[(u_.)*(x_)^(m_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.) + (c_.)*(x_)^(r_.))^(n_.), x_Symbol] :> Int[u*x^(m +
n*p)*(a + b*x^(q - p) + c*x^(r - p))^n, x] /; FreeQ[{a, b, c, m, p, q, r}, x] && IntegerQ[n] && PosQ[q - p] &
& PosQ[r - p]

Rule 1265

Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Wit
h[{Qx = PolynomialQuotient[(a + b*x^2 + c*x^4)^p, f*x, x], R = PolynomialRemainder[(a + b*x^2 + c*x^4)^p, f*x,
x]}, Simp[(R*(f*x)^(m + 1)*(d + e*x^2)^(q + 1))/(d*f*(m + 1)), x] + Dist[1/(d*f^2*(m + 1)), Int[(f*x)^(m + 2)
*(d + e*x^2)^q*ExpandToSum[(d*f*(m + 1)*Qx)/x - e*R*(m + 2*q + 3), x], x], x]] /; FreeQ[{a, b, c, d, e, f, q},
x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -1]

Rule 451

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

Rule 217

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

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])

Rubi steps

\begin{align*} \int \frac{c+d x^2+e x^4+f x^6}{x^6 \sqrt{a+b x^2}} \, dx &=-\frac{c \sqrt{a+b x^2}}{5 a x^5}-\frac{\int \frac{(4 b c-5 a d) x-5 a e x^3-5 a f x^5}{x^5 \sqrt{a+b x^2}} \, dx}{5 a}\\ &=-\frac{c \sqrt{a+b x^2}}{5 a x^5}-\frac{\int \frac{4 b c-5 a d-5 a e x^2-5 a f x^4}{x^4 \sqrt{a+b x^2}} \, dx}{5 a}\\ &=-\frac{c \sqrt{a+b x^2}}{5 a x^5}+\frac{(4 b c-5 a d) \sqrt{a+b x^2}}{15 a^2 x^3}+\frac{\int \frac{8 b^2 c-10 a b d+15 a^2 e+15 a^2 f x^2}{x^2 \sqrt{a+b x^2}} \, dx}{15 a^2}\\ &=-\frac{c \sqrt{a+b x^2}}{5 a x^5}+\frac{(4 b c-5 a d) \sqrt{a+b x^2}}{15 a^2 x^3}-\frac{\left (8 b^2 c-10 a b d+15 a^2 e\right ) \sqrt{a+b x^2}}{15 a^3 x}+f \int \frac{1}{\sqrt{a+b x^2}} \, dx\\ &=-\frac{c \sqrt{a+b x^2}}{5 a x^5}+\frac{(4 b c-5 a d) \sqrt{a+b x^2}}{15 a^2 x^3}-\frac{\left (8 b^2 c-10 a b d+15 a^2 e\right ) \sqrt{a+b x^2}}{15 a^3 x}+f \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{x}{\sqrt{a+b x^2}}\right )\\ &=-\frac{c \sqrt{a+b x^2}}{5 a x^5}+\frac{(4 b c-5 a d) \sqrt{a+b x^2}}{15 a^2 x^3}-\frac{\left (8 b^2 c-10 a b d+15 a^2 e\right ) \sqrt{a+b x^2}}{15 a^3 x}+\frac{f \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{\sqrt{b}}\\ \end{align*}

Mathematica [A]  time = 0.106839, size = 95, normalized size = 0.81 \[ \frac{f \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{\sqrt{b}}-\frac{\sqrt{a+b x^2} \left (a^2 \left (3 c+5 d x^2+15 e x^4\right )-2 a b x^2 \left (2 c+5 d x^2\right )+8 b^2 c x^4\right )}{15 a^3 x^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.007, size = 136, normalized size = 1.2 \begin{align*}{f\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){\frac{1}{\sqrt{b}}}}-{\frac{d}{3\,a{x}^{3}}\sqrt{b{x}^{2}+a}}+{\frac{2\,bd}{3\,{a}^{2}x}\sqrt{b{x}^{2}+a}}-{\frac{c}{5\,a{x}^{5}}\sqrt{b{x}^{2}+a}}+{\frac{4\,bc}{15\,{x}^{3}{a}^{2}}\sqrt{b{x}^{2}+a}}-{\frac{8\,{b}^{2}c}{15\,{a}^{3}x}\sqrt{b{x}^{2}+a}}-{\frac{e}{ax}\sqrt{b{x}^{2}+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^6/(b*x^2+a)^(1/2),x)

[Out]

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

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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^6/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.40456, size = 509, normalized size = 4.31 \begin{align*} \left [\frac{15 \, a^{3} \sqrt{b} f x^{5} \log \left (-2 \, b x^{2} - 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left ({\left (8 \, b^{3} c - 10 \, a b^{2} d + 15 \, a^{2} b e\right )} x^{4} + 3 \, a^{2} b c -{\left (4 \, a b^{2} c - 5 \, a^{2} b d\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{30 \, a^{3} b x^{5}}, -\frac{15 \, a^{3} \sqrt{-b} f x^{5} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) +{\left ({\left (8 \, b^{3} c - 10 \, a b^{2} d + 15 \, a^{2} b e\right )} x^{4} + 3 \, a^{2} b c -{\left (4 \, a b^{2} c - 5 \, a^{2} b d\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{15 \, a^{3} b x^{5}}\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^6/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.20046, size = 456, normalized size = 3.86 \begin{align*} - \frac{3 a^{4} b^{\frac{9}{2}} c \sqrt{\frac{a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac{2 a^{3} b^{\frac{11}{2}} c x^{2} \sqrt{\frac{a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac{3 a^{2} b^{\frac{13}{2}} c x^{4} \sqrt{\frac{a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac{12 a b^{\frac{15}{2}} c x^{6} \sqrt{\frac{a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac{8 b^{\frac{17}{2}} c x^{8} \sqrt{\frac{a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} + f \left (\begin{cases} \frac{\sqrt{- \frac{a}{b}} \operatorname{asin}{\left (x \sqrt{- \frac{b}{a}} \right )}}{\sqrt{a}} & \text{for}\: a > 0 \wedge b < 0 \\\frac{\sqrt{\frac{a}{b}} \operatorname{asinh}{\left (x \sqrt{\frac{b}{a}} \right )}}{\sqrt{a}} & \text{for}\: a > 0 \wedge b > 0 \\\frac{\sqrt{- \frac{a}{b}} \operatorname{acosh}{\left (x \sqrt{- \frac{b}{a}} \right )}}{\sqrt{- a}} & \text{for}\: b > 0 \wedge a < 0 \end{cases}\right ) - \frac{\sqrt{b} d \sqrt{\frac{a}{b x^{2}} + 1}}{3 a x^{2}} - \frac{\sqrt{b} e \sqrt{\frac{a}{b x^{2}} + 1}}{a} + \frac{2 b^{\frac{3}{2}} d \sqrt{\frac{a}{b x^{2}} + 1}}{3 a^{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**6/(b*x**2+a)**(1/2),x)

[Out]

-3*a**4*b**(9/2)*c*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 2*a**3*b
**(11/2)*c*x**2*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 3*a**2*b**(
13/2)*c*x**4*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 12*a*b**(15/2)
*c*x**6*sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) - 8*b**(17/2)*c*x**8*
sqrt(a/(b*x**2) + 1)/(15*a**5*b**4*x**4 + 30*a**4*b**5*x**6 + 15*a**3*b**6*x**8) + f*Piecewise((sqrt(-a/b)*asi
n(x*sqrt(-b/a))/sqrt(a), (a > 0) & (b < 0)), (sqrt(a/b)*asinh(x*sqrt(b/a))/sqrt(a), (a > 0) & (b > 0)), (sqrt(
-a/b)*acosh(x*sqrt(-b/a))/sqrt(-a), (b > 0) & (a < 0))) - sqrt(b)*d*sqrt(a/(b*x**2) + 1)/(3*a*x**2) - sqrt(b)*
e*sqrt(a/(b*x**2) + 1)/a + 2*b**(3/2)*d*sqrt(a/(b*x**2) + 1)/(3*a**2)

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Giac [B]  time = 1.29502, size = 437, normalized size = 3.7 \begin{align*} -\frac{f \log \left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2}\right )}{2 \, \sqrt{b}} + \frac{2 \,{\left (15 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{8} \sqrt{b} e + 30 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{6} b^{\frac{3}{2}} d - 60 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{6} a \sqrt{b} e + 80 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} b^{\frac{5}{2}} c - 70 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} a b^{\frac{3}{2}} d + 90 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} a^{2} \sqrt{b} e - 40 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a b^{\frac{5}{2}} c + 50 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a^{2} b^{\frac{3}{2}} d - 60 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a^{3} \sqrt{b} e + 8 \, a^{2} b^{\frac{5}{2}} c - 10 \, a^{3} b^{\frac{3}{2}} d + 15 \, a^{4} \sqrt{b} e\right )}}{15 \,{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )}^{5}} \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^6/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

-1/2*f*log((sqrt(b)*x - sqrt(b*x^2 + a))^2)/sqrt(b) + 2/15*(15*(sqrt(b)*x - sqrt(b*x^2 + a))^8*sqrt(b)*e + 30*
(sqrt(b)*x - sqrt(b*x^2 + a))^6*b^(3/2)*d - 60*(sqrt(b)*x - sqrt(b*x^2 + a))^6*a*sqrt(b)*e + 80*(sqrt(b)*x - s
qrt(b*x^2 + a))^4*b^(5/2)*c - 70*(sqrt(b)*x - sqrt(b*x^2 + a))^4*a*b^(3/2)*d + 90*(sqrt(b)*x - sqrt(b*x^2 + a)
)^4*a^2*sqrt(b)*e - 40*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a*b^(5/2)*c + 50*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^2*b^
(3/2)*d - 60*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a^3*sqrt(b)*e + 8*a^2*b^(5/2)*c - 10*a^3*b^(3/2)*d + 15*a^4*sqrt(
b)*e)/((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^5