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

Optimal. Leaf size=110 \[ \frac{\sqrt{a+b x^2} (2 b c-3 a d)}{3 a^2 x}+\frac{(2 b e-a f) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{3/2}}-\frac{c \sqrt{a+b x^2}}{3 a x^3}+\frac{f x \sqrt{a+b x^2}}{2 b} \]

[Out]

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

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Rubi [A]  time = 0.127105, antiderivative size = 110, 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, 388, 217, 206} \[ \frac{\sqrt{a+b x^2} (2 b c-3 a d)}{3 a^2 x}+\frac{(2 b e-a f) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{3/2}}-\frac{c \sqrt{a+b x^2}}{3 a x^3}+\frac{f x \sqrt{a+b x^2}}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.105422, size = 93, normalized size = 0.85 \[ \frac{\sqrt{a+b x^2} \left (3 a^2 f x^4-2 a b \left (c+3 d x^2\right )+4 b^2 c x^2\right )}{6 a^2 b x^3}+\frac{(2 b e-a f) \tanh ^{-1}\left (\frac{\sqrt{b} x}{\sqrt{a+b x^2}}\right )}{2 b^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.009, size = 117, normalized size = 1.1 \begin{align*}{\frac{fx}{2\,b}\sqrt{b{x}^{2}+a}}-{\frac{af}{2}\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){b}^{-{\frac{3}{2}}}}+{e\ln \left ( x\sqrt{b}+\sqrt{b{x}^{2}+a} \right ){\frac{1}{\sqrt{b}}}}-{\frac{c}{3\,a{x}^{3}}\sqrt{b{x}^{2}+a}}+{\frac{2\,bc}{3\,{a}^{2}x}\sqrt{b{x}^{2}+a}}-{\frac{d}{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^4/(b*x^2+a)^(1/2),x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.46964, size = 479, normalized size = 4.35 \begin{align*} \left [-\frac{3 \,{\left (2 \, a^{2} b e - a^{3} f\right )} \sqrt{b} x^{3} \log \left (-2 \, b x^{2} + 2 \, \sqrt{b x^{2} + a} \sqrt{b} x - a\right ) - 2 \,{\left (3 \, a^{2} b f x^{4} - 2 \, a b^{2} c + 2 \,{\left (2 \, b^{3} c - 3 \, a b^{2} d\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{12 \, a^{2} b^{2} x^{3}}, -\frac{3 \,{\left (2 \, a^{2} b e - a^{3} f\right )} \sqrt{-b} x^{3} \arctan \left (\frac{\sqrt{-b} x}{\sqrt{b x^{2} + a}}\right ) -{\left (3 \, a^{2} b f x^{4} - 2 \, a b^{2} c + 2 \,{\left (2 \, b^{3} c - 3 \, a b^{2} d\right )} x^{2}\right )} \sqrt{b x^{2} + a}}{6 \, a^{2} b^{2} x^{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^4/(b*x^2+a)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 3.97899, size = 197, normalized size = 1.79 \begin{align*} \frac{\sqrt{a} f x \sqrt{1 + \frac{b x^{2}}{a}}}{2 b} - \frac{a f \operatorname{asinh}{\left (\frac{\sqrt{b} x}{\sqrt{a}} \right )}}{2 b^{\frac{3}{2}}} + e \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} c \sqrt{\frac{a}{b x^{2}} + 1}}{3 a x^{2}} - \frac{\sqrt{b} d \sqrt{\frac{a}{b x^{2}} + 1}}{a} + \frac{2 b^{\frac{3}{2}} c \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**4/(b*x**2+a)**(1/2),x)

[Out]

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

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Giac [A]  time = 1.20394, size = 238, normalized size = 2.16 \begin{align*} \frac{\sqrt{b x^{2} + a} f x}{2 \, b} + \frac{{\left (a \sqrt{b} f - 2 \, b^{\frac{3}{2}} e\right )} \log \left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2}\right )}{4 \, b^{2}} + \frac{2 \,{\left (3 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{4} \sqrt{b} d + 6 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} b^{\frac{3}{2}} c - 6 \,{\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} a \sqrt{b} d - 2 \, a b^{\frac{3}{2}} c + 3 \, a^{2} \sqrt{b} d\right )}}{3 \,{\left ({\left (\sqrt{b} x - \sqrt{b x^{2} + a}\right )}^{2} - a\right )}^{3}} \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^4/(b*x^2+a)^(1/2),x, algorithm="giac")

[Out]

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