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

[Out]

f*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(1/2)-1/5*c*(b*x^2+a)^(1/2)/a/x^5+1/15*(-5*a*d+4*b*c)*(b*x^2+a)^(1/2)/a
^2/x^3-1/15*(15*a^2*e-10*a*b*d+8*b^2*c)*(b*x^2+a)^(1/2)/a^3/x

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Rubi [A]  time = 0.13, antiderivative size = 118, normalized size of antiderivative = 1.00, 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 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])

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 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 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 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 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])

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*}

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Mathematica [A]  time = 0.11, 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]

-1/15*(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)))/(a^3*x^5) +
(f*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b]

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fricas [A]  time = 0.79, size = 221, normalized size = 1.87 \[ \left [\frac {15 \, a^{3} \sqrt {b} f x^{5} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left ({\left (8 \, b^{3} c - 10 \, a b^{2} d + 15 \, a^{2} b e\right )} x^{4} + 3 \, a^{2} b c - {\left (4 \, a b^{2} c - 5 \, a^{2} b d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{30 \, a^{3} b x^{5}}, -\frac {15 \, a^{3} \sqrt {-b} f x^{5} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left ({\left (8 \, b^{3} c - 10 \, a b^{2} d + 15 \, a^{2} b e\right )} x^{4} + 3 \, a^{2} b c - {\left (4 \, a b^{2} c - 5 \, a^{2} b d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{15 \, a^{3} b x^{5}}\right ] \]

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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giac [B]  time = 0.60, size = 324, normalized size = 2.75 \[ -\frac {f \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right )}{2 \, \sqrt {b}} + \frac {2 \, {\left (15 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{8} \sqrt {b} e + 30 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} b^{\frac {3}{2}} d - 60 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{6} a \sqrt {b} e + 80 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} b^{\frac {5}{2}} c - 70 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a b^{\frac {3}{2}} d + 90 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} a^{2} \sqrt {b} e - 40 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a b^{\frac {5}{2}} c + 50 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{2} b^{\frac {3}{2}} d - 60 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a^{3} \sqrt {b} e + 8 \, a^{2} b^{\frac {5}{2}} c - 10 \, a^{3} b^{\frac {3}{2}} d + 15 \, a^{4} \sqrt {b} e\right )}}{15 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{5}} \]

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

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maple [A]  time = 0.01, size = 136, normalized size = 1.15 \[ \frac {f \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}-\frac {\sqrt {b \,x^{2}+a}\, e}{a x}+\frac {2 \sqrt {b \,x^{2}+a}\, b d}{3 a^{2} x}-\frac {8 \sqrt {b \,x^{2}+a}\, b^{2} c}{15 a^{3} x}-\frac {\sqrt {b \,x^{2}+a}\, d}{3 a \,x^{3}}+\frac {4 \sqrt {b \,x^{2}+a}\, b c}{15 a^{2} x^{3}}-\frac {\sqrt {b \,x^{2}+a}\, c}{5 a \,x^{5}} \]

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

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maxima [A]  time = 1.33, size = 128, normalized size = 1.08 \[ \frac {f \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {8 \, \sqrt {b x^{2} + a} b^{2} c}{15 \, a^{3} x} + \frac {2 \, \sqrt {b x^{2} + a} b d}{3 \, a^{2} x} - \frac {\sqrt {b x^{2} + a} e}{a x} + \frac {4 \, \sqrt {b x^{2} + a} b c}{15 \, a^{2} x^{3}} - \frac {\sqrt {b x^{2} + a} d}{3 \, a x^{3}} - \frac {\sqrt {b x^{2} + a} c}{5 \, a x^{5}} \]

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]

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

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mupad [B]  time = 1.72, size = 105, normalized size = 0.89 \[ \frac {f\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{\sqrt {b}}-\frac {e\,\sqrt {b\,x^2+a}}{a\,x}-\frac {d\,\sqrt {b\,x^2+a}\,\left (a-2\,b\,x^2\right )}{3\,a^2\,x^3}-\frac {c\,\sqrt {b\,x^2+a}\,\left (3\,a^2-4\,a\,b\,x^2+8\,b^2\,x^4\right )}{15\,a^3\,x^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2 + e*x^4 + f*x^6)/(x^6*(a + b*x^2)^(1/2)),x)

[Out]

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

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sympy [A]  time = 6.21, size = 456, normalized size = 3.86 \[ - \frac {3 a^{4} b^{\frac {9}{2}} c \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {2 a^{3} b^{\frac {11}{2}} c x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {3 a^{2} b^{\frac {13}{2}} c x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {12 a b^{\frac {15}{2}} c x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} - \frac {8 b^{\frac {17}{2}} c x^{8} \sqrt {\frac {a}{b x^{2}} + 1}}{15 a^{5} b^{4} x^{4} + 30 a^{4} b^{5} x^{6} + 15 a^{3} b^{6} x^{8}} + f \left (\begin {cases} \frac {\sqrt {- \frac {a}{b}} \operatorname {asin}{\left (x \sqrt {- \frac {b}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge b < 0 \\\frac {\sqrt {\frac {a}{b}} \operatorname {asinh}{\left (x \sqrt {\frac {b}{a}} \right )}}{\sqrt {a}} & \text {for}\: a > 0 \wedge b > 0 \\\frac {\sqrt {- \frac {a}{b}} \operatorname {acosh}{\left (x \sqrt {- \frac {b}{a}} \right )}}{\sqrt {- a}} & \text {for}\: b > 0 \wedge a < 0 \end {cases}\right ) - \frac {\sqrt {b} d \sqrt {\frac {a}{b x^{2}} + 1}}{3 a x^{2}} - \frac {\sqrt {b} e \sqrt {\frac {a}{b x^{2}} + 1}}{a} + \frac {2 b^{\frac {3}{2}} d \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{2}} \]

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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