Integrand size = 32, antiderivative size = 110 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \sqrt {a+b x^2}} \, dx=-\frac {c \sqrt {a+b x^2}}{3 a x^3}+\frac {(2 b c-3 a d) \sqrt {a+b x^2}}{3 a^2 x}+\frac {f x \sqrt {a+b x^2}}{2 b}+\frac {(2 b e-a f) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}} \]
1/2*(-a*f+2*b*e)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(3/2)-1/3*c*(b*x^2+a )^(1/2)/a/x^3+1/3*(-3*a*d+2*b*c)*(b*x^2+a)^(1/2)/a^2/x+1/2*f*x*(b*x^2+a)^( 1/2)/b
Time = 0.19 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.86 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \sqrt {a+b x^2}} \, dx=\frac {\sqrt {a+b x^2} \left (-2 a b c+4 b^2 c x^2-6 a b d x^2+3 a^2 f x^4\right )}{6 a^2 b x^3}+\frac {(-2 b e+a f) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 b^{3/2}} \]
(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[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(2*b^(3/ 2))
Time = 0.34 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {2338, 9, 1588, 27, 299, 224, 219}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \frac {c+d x^2+e x^4+f x^6}{x^4 \sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 2338 |
\(\displaystyle -\frac {\int \frac {-3 a f x^5-3 a e x^3+(2 b c-3 a d) x}{x^3 \sqrt {b x^2+a}}dx}{3 a}-\frac {c \sqrt {a+b x^2}}{3 a x^3}\) |
\(\Big \downarrow \) 9 |
\(\displaystyle -\frac {\int \frac {-3 a f x^4-3 a e x^2+2 b c-3 a d}{x^2 \sqrt {b x^2+a}}dx}{3 a}-\frac {c \sqrt {a+b x^2}}{3 a x^3}\) |
\(\Big \downarrow \) 1588 |
\(\displaystyle -\frac {-\frac {\int \frac {3 a^2 \left (f x^2+e\right )}{\sqrt {b x^2+a}}dx}{a}-\frac {\sqrt {a+b x^2} (2 b c-3 a d)}{a x}}{3 a}-\frac {c \sqrt {a+b x^2}}{3 a x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {-3 a \int \frac {f x^2+e}{\sqrt {b x^2+a}}dx-\frac {\sqrt {a+b x^2} (2 b c-3 a d)}{a x}}{3 a}-\frac {c \sqrt {a+b x^2}}{3 a x^3}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle -\frac {-3 a \left (\frac {(2 b e-a f) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}+\frac {f x \sqrt {a+b x^2}}{2 b}\right )-\frac {\sqrt {a+b x^2} (2 b c-3 a d)}{a x}}{3 a}-\frac {c \sqrt {a+b x^2}}{3 a x^3}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle -\frac {-3 a \left (\frac {(2 b e-a f) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}+\frac {f x \sqrt {a+b x^2}}{2 b}\right )-\frac {\sqrt {a+b x^2} (2 b c-3 a d)}{a x}}{3 a}-\frac {c \sqrt {a+b x^2}}{3 a x^3}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {-3 a \left (\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) (2 b e-a f)}{2 b^{3/2}}+\frac {f x \sqrt {a+b x^2}}{2 b}\right )-\frac {\sqrt {a+b x^2} (2 b c-3 a d)}{a x}}{3 a}-\frac {c \sqrt {a+b x^2}}{3 a x^3}\) |
-1/3*(c*Sqrt[a + b*x^2])/(a*x^3) - (-(((2*b*c - 3*a*d)*Sqrt[a + b*x^2])/(a *x)) - 3*a*((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))))/(3*a)
3.2.55.3.1 Defintions of rubi rules used
Int[(u_.)*(Px_)^(p_.)*((e_.)*(x_))^(m_.), x_Symbol] :> With[{r = Expon[Px, x, Min]}, Simp[1/e^(p*r) Int[u*(e*x)^(m + p*r)*ExpandToSum[Px/x^r, x]^p, x], x] /; IGtQ[r, 0]] /; FreeQ[{e, m}, x] && PolyQ[Px, x] && IntegerQ[p] && !MonomialQ[Px, x]
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
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]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x *((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 *p + 3)) Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && NeQ[2*p + 3, 0]
Int[((f_.)*(x_))^(m_)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c _.)*(x_)^4)^(p_.), x_Symbol] :> With[{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] + Simp[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] && Ne Q[b^2 - 4*a*c, 0] && IGtQ[p, 0] && LtQ[m, -1]
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]}, S imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[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] && Lt Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
Time = 3.60 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.75
method | result | size |
risch | \(\frac {\sqrt {b \,x^{2}+a}\, \left (3 a^{2} f \,x^{4}-6 x^{2} a b d +4 b^{2} c \,x^{2}-2 a b c \right )}{6 b \,a^{2} x^{3}}-\frac {\left (a f -2 b e \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\) | \(83\) |
pseudoelliptic | \(-\frac {x^{3} a^{2} \left (a f -2 b e \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-\left (-\frac {2 a \left (3 d \,x^{2}+c \right ) b^{\frac {3}{2}}}{3}+x^{2} \left (\sqrt {b}\, a^{2} f \,x^{2}+\frac {4 b^{\frac {5}{2}} c}{3}\right )\right ) \sqrt {b \,x^{2}+a}}{2 b^{\frac {3}{2}} x^{3} a^{2}}\) | \(93\) |
default | \(\frac {e \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+f \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )+c \left (-\frac {\sqrt {b \,x^{2}+a}}{3 a \,x^{3}}+\frac {2 b \sqrt {b \,x^{2}+a}}{3 a^{2} x}\right )-\frac {d \sqrt {b \,x^{2}+a}}{a x}\) | \(119\) |
1/6*(b*x^2+a)^(1/2)*(3*a^2*f*x^4-6*a*b*d*x^2+4*b^2*c*x^2-2*a*b*c)/b/a^2/x^ 3-1/2*(a*f-2*b*e)/b^(3/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))
Time = 0.28 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.91 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \sqrt {a+b x^2}} \, dx=\left [-\frac {3 \, {\left (2 \, a^{2} b e - a^{3} f\right )} \sqrt {b} x^{3} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (3 \, a^{2} b f x^{4} - 2 \, a b^{2} c + 2 \, {\left (2 \, b^{3} c - 3 \, a b^{2} d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{12 \, a^{2} b^{2} x^{3}}, -\frac {3 \, {\left (2 \, a^{2} b e - a^{3} f\right )} \sqrt {-b} x^{3} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (3 \, a^{2} b f x^{4} - 2 \, a b^{2} c + 2 \, {\left (2 \, b^{3} c - 3 \, a b^{2} d\right )} x^{2}\right )} \sqrt {b x^{2} + a}}{6 \, a^{2} b^{2} x^{3}}\right ] \]
[-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)]
Time = 1.09 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.78 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \sqrt {a+b x^2}} \, dx=e \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \wedge b \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {for}\: b \neq 0 \\\frac {x}{\sqrt {a}} & \text {otherwise} \end {cases}\right ) + f \left (\begin {cases} - \frac {a \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{2 b} + \frac {x \sqrt {a + b x^{2}}}{2 b} & \text {for}\: b \neq 0 \\\frac {x^{3}}{3 \sqrt {a}} & \text {otherwise} \end {cases}\right ) - \frac {\sqrt {b} c \sqrt {\frac {a}{b x^{2}} + 1}}{3 a x^{2}} - \frac {\sqrt {b} d \sqrt {\frac {a}{b x^{2}} + 1}}{a} + \frac {2 b^{\frac {3}{2}} c \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{2}} \]
e*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0) & N e(b, 0)), (x*log(x)/sqrt(b*x**2), Ne(b, 0)), (x/sqrt(a), True)) + f*Piecew ise((-a*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/(2*b) + x*sqrt(a + b*x**2)/(2*b), Ne(b , 0)), (x**3/(3*sqrt(a)), True)) - 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)
Time = 0.20 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.93 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} f x}{2 \, b} + \frac {e \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {a f \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {2 \, \sqrt {b x^{2} + a} b c}{3 \, a^{2} x} - \frac {\sqrt {b x^{2} + a} d}{a x} - \frac {\sqrt {b x^{2} + a} c}{3 \, a x^{3}} \]
1/2*sqrt(b*x^2 + a)*f*x/b + e*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 1/2*a*f*arc sinh(b*x/sqrt(a*b))/b^(3/2) + 2/3*sqrt(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)
Time = 0.36 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.55 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} f x}{2 \, b} - \frac {{\left (2 \, b e - a f\right )} \log \left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2}\right )}{4 \, b^{\frac {3}{2}}} + \frac {2 \, {\left (3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} \sqrt {b} d + 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} b^{\frac {3}{2}} c - 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a \sqrt {b} d - 2 \, a b^{\frac {3}{2}} c + 3 \, a^{2} \sqrt {b} d\right )}}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{3}} \]
1/2*sqrt(b*x^2 + a)*f*x/b - 1/4*(2*b*e - a*f)*log((sqrt(b)*x - sqrt(b*x^2 + a))^2)/b^(3/2) + 2/3*(3*(sqrt(b)*x - sqrt(b*x^2 + a))^4*sqrt(b)*d + 6*(s qrt(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
Time = 7.20 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.30 \[ \int \frac {c+d x^2+e x^4+f x^6}{x^4 \sqrt {a+b x^2}} \, dx=\left \{\begin {array}{cl} -\frac {-f\,x^6-3\,e\,x^4+3\,d\,x^2+c}{3\,\sqrt {a}\,x^3} & \text {\ if\ \ }b=0\\ \frac {e\,\ln \left (\sqrt {b}\,x+\sqrt {b\,x^2+a}\right )}{\sqrt {b}}-\frac {d\,\sqrt {b\,x^2+a}}{a\,x}-\frac {a\,f\,\ln \left (2\,\sqrt {b}\,x+2\,\sqrt {b\,x^2+a}\right )}{2\,b^{3/2}}+\frac {f\,x\,\sqrt {b\,x^2+a}}{2\,b}-\frac {c\,\sqrt {b\,x^2+a}\,\left (a-2\,b\,x^2\right )}{3\,a^2\,x^3} & \text {\ if\ \ }b\neq 0 \end {array}\right . \]
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))