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

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

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Rubi [A]  time = 0.13, antiderivative size = 110, 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, 388, 217, 206} \begin {gather*} \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} \end {gather*}

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 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 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 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^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*}

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Mathematica [A]  time = 0.11, size = 93, normalized size = 0.85 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [A]  time = 0.24, size = 95, normalized size = 0.86 \begin {gather*} \frac {\sqrt {a+b x^2} \left (3 a^2 f x^4-2 a b c-6 a b d x^2+4 b^2 c x^2\right )}{6 a^2 b x^3}+\frac {(a f-2 b e) \log \left (\sqrt {a+b x^2}-\sqrt {b} x\right )}{2 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*x^2]*(-2*a*b*c + 4*b^2*c*x^2 - 6*a*b*d*x^2 + 3*a^2*f*x^4))/(6*a^2*b*x^3) + ((-2*b*e + a*f)*Log[-(S
qrt[b]*x) + Sqrt[a + b*x^2]])/(2*b^(3/2))

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fricas [A]  time = 0.74, size = 210, normalized size = 1.91 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.56, size = 176, normalized size = 1.60 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.01, size = 117, normalized size = 1.06 \begin {gather*} -\frac {a f \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}+\frac {e \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+\frac {\sqrt {b \,x^{2}+a}\, f x}{2 b}-\frac {\sqrt {b \,x^{2}+a}\, d}{a x}+\frac {2 \sqrt {b \,x^{2}+a}\, b c}{3 a^{2} x}-\frac {\sqrt {b \,x^{2}+a}\, c}{3 a \,x^{3}} \end {gather*}

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(b^(1/2)*x+(b*x^2+a)^(1/2))+e*ln(b^(1/2)*x+(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 [A]  time = 1.33, size = 102, normalized size = 0.93 \begin {gather*} \frac {\sqrt {b x^{2} + a} f x}{2 \, b} + \frac {e \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {a f \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {2 \, \sqrt {b x^{2} + a} b c}{3 \, a^{2} x} - \frac {\sqrt {b x^{2} + a} d}{a x} - \frac {\sqrt {b x^{2} + a} c}{3 \, a x^{3}} \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^4/(b*x^2+a)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(b*x^2 + a)*f*x/b + e*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 1/2*a*f*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 2/3*sq
rt(b*x^2 + a)*b*c/(a^2*x) - sqrt(b*x^2 + a)*d/(a*x) - 1/3*sqrt(b*x^2 + a)*c/(a*x^3)

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mupad [B]  time = 2.20, size = 143, normalized size = 1.30 \begin {gather*} \left \{\begin {array}{cl} -\frac {-f\,x^6-3\,e\,x^4+3\,d\,x^2+c}{3\,\sqrt {a}\,x^3} & \text {\ if\ \ }b=0\\ \frac {e\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{\sqrt {b}}-\frac {d\,\sqrt {b\,x^2+a}}{a\,x}-\frac {a\,f\,\ln \left (2\,\sqrt {b}\,x+2\,\sqrt {b\,x^2+a}\right )}{2\,b^{3/2}}+\frac {f\,x\,\sqrt {b\,x^2+a}}{2\,b}-\frac {c\,\sqrt {b\,x^2+a}\,\left (a-2\,b\,x^2\right )}{3\,a^2\,x^3} & \text {\ if\ \ }b\neq 0 \end {array}\right . \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^4*(a + b*x^2)^(1/2)),x)

[Out]

piecewise(b == 0, -(c + 3*d*x^2 - 3*e*x^4 - f*x^6)/(3*a^(1/2)*x^3), b ~= 0, (e*log(b^(1/2)*x + (a + b*x^2)^(1/
2)))/b^(1/2) - (d*(a + b*x^2)^(1/2))/(a*x) - (a*f*log(2*b^(1/2)*x + 2*(a + b*x^2)^(1/2)))/(2*b^(3/2)) + (f*x*(
a + b*x^2)^(1/2))/(2*b) - (c*(a + b*x^2)^(1/2)*(a - 2*b*x^2))/(3*a^2*x^3))

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sympy [A]  time = 4.73, size = 197, normalized size = 1.79 \begin {gather*} \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 {gather*}

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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