3.2.48 \(\int \frac {c+d x^2+e x^4+f x^6}{x^5 \sqrt {a+b x^2}} \, dx\) [148]

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

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 114, 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 = {1813, 1635, 911, 1171, 396, 214} \begin {gather*} \frac {\sqrt {a+b x^2} (3 b c-4 a d)}{8 a^2 x^2}-\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 {c \sqrt {a+b x^2}}{4 a x^4}+\frac {f \sqrt {a+b x^2}}{b} \end {gather*}

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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 396

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 911

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 1171

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] &&
NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && LtQ[q, -1]

Rule 1635

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 1813

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]

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} \text {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 {\text {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 {\text {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 {\text {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 ) \text {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*}

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Mathematica [A]
time = 0.21, size = 102, normalized size = 0.89 \begin {gather*} \frac {\sqrt {a+b x^2} \left (-2 a b c+3 b^2 c x^2-4 a b d x^2+8 a^2 f x^4\right )}{8 a^2 b x^4}+\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 {gather*}

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]

(Sqrt[a + b*x^2]*(-2*a*b*c + 3*b^2*c*x^2 - 4*a*b*d*x^2 + 8*a^2*f*x^4))/(8*a^2*b*x^4) + ((-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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Maple [A]
time = 0.13, size = 167, normalized size = 1.46

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (4 a d \,x^{2}-3 c \,x^{2} b +2 a c \right )}{8 a^{2} x^{4}}+\frac {f \sqrt {b \,x^{2}+a}}{b}-\frac {e \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{\sqrt {a}}+\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) b d}{2 a^{\frac {3}{2}}}-\frac {3 \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right ) b^{2} c}{8 a^{\frac {5}{2}}}\)	\(143\)
default	\(\frac {f \sqrt {b \,x^{2}+a}}{b}+c \left (-\frac {\sqrt {b \,x^{2}+a}}{4 a \,x^{4}}-\frac {3 b \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )}{4 a}\right )+d \left (-\frac {\sqrt {b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )-\frac {e \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{\sqrt {a}}\)	\(167\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.29, size = 129, normalized size = 1.13 \begin {gather*} -\frac {3 \, b^{2} c \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{8 \, a^{\frac {5}{2}}} + \frac {b d \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {3}{2}}} - \frac {\operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right ) e}{\sqrt {a}} + \frac {\sqrt {b x^{2} + a} f}{b} + \frac {3 \, \sqrt {b x^{2} + a} b c}{8 \, a^{2} x^{2}} - \frac {\sqrt {b x^{2} + a} d}{2 \, a x^{2}} - \frac {\sqrt {b x^{2} + a} c}{4 \, a x^{4}} \end {gather*}

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]

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

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Fricas [A]
time = 1.24, size = 233, normalized size = 2.04 \begin {gather*} \left [\frac {{\left (8 \, a^{2} b x^{4} e + {\left (3 \, b^{3} c - 4 \, a b^{2} d\right )} x^{4}\right )} \sqrt {a} \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 (8 \, a^{2} b x^{4} e + {\left (3 \, b^{3} c - 4 \, a b^{2} d\right )} x^{4}\right )} \sqrt {-a} \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 {gather*}

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*((8*a^2*b*x^4*e + (3*b^3*c - 4*a*b^2*d)*x^4)*sqrt(a)*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*((8*a^2*b*x^4*e
+ (3*b^3*c - 4*a*b^2*d)*x^4)*sqrt(-a)*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 = 60.98, size = 194, normalized size = 1.70 \begin {gather*} 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 {gather*}

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 = 2.10, size = 141, normalized size = 1.24 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 2.19, size = 133, normalized size = 1.17 \begin {gather*} \frac {f\,\sqrt {b\,x^2+a}}{b}-\frac {e\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{\sqrt {a}}-\frac {5\,c\,\sqrt {b\,x^2+a}}{8\,a\,x^4}+\frac {3\,c\,{\left (b\,x^2+a\right )}^{3/2}}{8\,a^2\,x^4}-\frac {d\,\sqrt {b\,x^2+a}}{2\,a\,x^2}+\frac {b\,d\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{2\,a^{3/2}}-\frac {3\,b^2\,c\,\mathrm {atanh}\left (\frac {\sqrt {b\,x^2+a}}{\sqrt {a}}\right )}{8\,a^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x^2 + e*x^4 + f*x^6)/(x^5*(a + b*x^2)^(1/2)),x)

[Out]

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

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