Integrand size = 29, antiderivative size = 145 \[ \int \frac {c+d x^2+e x^4+f x^6}{\sqrt {a+b x^2}} \, dx=\frac {\left (8 b^2 d-6 a b e+5 a^2 f\right ) x \sqrt {a+b x^2}}{16 b^3}+\frac {(6 b e-5 a f) x^3 \sqrt {a+b x^2}}{24 b^2}+\frac {f x^5 \sqrt {a+b x^2}}{6 b}+\frac {\left (16 b^3 c-8 a b^2 d+6 a^2 b e-5 a^3 f\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{7/2}} \]
1/16*(-5*a^3*f+6*a^2*b*e-8*a*b^2*d+16*b^3*c)*arctanh(x*b^(1/2)/(b*x^2+a)^( 1/2))/b^(7/2)+1/16*(5*a^2*f-6*a*b*e+8*b^2*d)*x*(b*x^2+a)^(1/2)/b^3+1/24*(- 5*a*f+6*b*e)*x^3*(b*x^2+a)^(1/2)/b^2+1/6*f*x^5*(b*x^2+a)^(1/2)/b
Time = 0.34 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.89 \[ \int \frac {c+d x^2+e x^4+f x^6}{\sqrt {a+b x^2}} \, dx=\frac {x \sqrt {a+b x^2} \left (24 b^2 d-18 a b e+15 a^2 f+12 b^2 e x^2-10 a b f x^2+8 b^2 f x^4\right )}{48 b^3}+\frac {\left (16 b^3 c-8 a b^2 d+6 a^2 b e-5 a^3 f\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{8 b^{7/2}} \]
(x*Sqrt[a + b*x^2]*(24*b^2*d - 18*a*b*e + 15*a^2*f + 12*b^2*e*x^2 - 10*a*b *f*x^2 + 8*b^2*f*x^4))/(48*b^3) + ((16*b^3*c - 8*a*b^2*d + 6*a^2*b*e - 5*a ^3*f)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2])])/(8*b^(7/2))
Time = 0.33 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {2346, 1473, 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.
| \(\displaystyle \int \frac {c+d x^2+e x^4+f x^6}{\sqrt {a+b x^2}} \, dx\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {\int \frac {(6 b e-5 a f) x^4+6 b d x^2+6 b c}{\sqrt {b x^2+a}}dx}{6 b}+\frac {f x^5 \sqrt {a+b x^2}}{6 b}\) |
| \(\Big \downarrow \) 1473 |
| \(\displaystyle \frac {\frac {\int \frac {3 \left (8 c b^2+\left (5 f a^2-6 b e a+8 b^2 d\right ) x^2\right )}{\sqrt {b x^2+a}}dx}{4 b}+\frac {x^3 \sqrt {a+b x^2} (6 b e-5 a f)}{4 b}}{6 b}+\frac {f x^5 \sqrt {a+b x^2}}{6 b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {3 \int \frac {8 c b^2+\left (5 f a^2-6 b e a+8 b^2 d\right ) x^2}{\sqrt {b x^2+a}}dx}{4 b}+\frac {x^3 \sqrt {a+b x^2} (6 b e-5 a f)}{4 b}}{6 b}+\frac {f x^5 \sqrt {a+b x^2}}{6 b}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\left (-5 a^3 f+6 a^2 b e-8 a b^2 d+16 b^3 c\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}+\frac {x \sqrt {a+b x^2} \left (5 a^2 f-6 a b e+8 b^2 d\right )}{2 b}\right )}{4 b}+\frac {x^3 \sqrt {a+b x^2} (6 b e-5 a f)}{4 b}}{6 b}+\frac {f x^5 \sqrt {a+b x^2}}{6 b}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\left (-5 a^3 f+6 a^2 b e-8 a b^2 d+16 b^3 c\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}+\frac {x \sqrt {a+b x^2} \left (5 a^2 f-6 a b e+8 b^2 d\right )}{2 b}\right )}{4 b}+\frac {x^3 \sqrt {a+b x^2} (6 b e-5 a f)}{4 b}}{6 b}+\frac {f x^5 \sqrt {a+b x^2}}{6 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {3 \left (\frac {x \sqrt {a+b x^2} \left (5 a^2 f-6 a b e+8 b^2 d\right )}{2 b}+\frac {\text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (-5 a^3 f+6 a^2 b e-8 a b^2 d+16 b^3 c\right )}{2 b^{3/2}}\right )}{4 b}+\frac {x^3 \sqrt {a+b x^2} (6 b e-5 a f)}{4 b}}{6 b}+\frac {f x^5 \sqrt {a+b x^2}}{6 b}\) |
(f*x^5*Sqrt[a + b*x^2])/(6*b) + (((6*b*e - 5*a*f)*x^3*Sqrt[a + b*x^2])/(4* b) + (3*(((8*b^2*d - 6*a*b*e + 5*a^2*f)*x*Sqrt[a + b*x^2])/(2*b) + ((16*b^ 3*c - 8*a*b^2*d + 6*a^2*b*e - 5*a^3*f)*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2] ])/(2*b^(3/2))))/(4*b))/(6*b)
3.2.53.3.1 Defintions of rubi rules used
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[((d_) + (e_.)*(x_)^2)^(q_)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[c^p*x^(4*p - 1)*((d + e*x^2)^(q + 1)/(e*(4*p + 2*q + 1))) , x] + Simp[1/(e*(4*p + 2*q + 1)) Int[(d + e*x^2)^q*ExpandToSum[e*(4*p + 2*q + 1)*(a + b*x^2 + c*x^4)^p - d*c^p*(4*p - 1)*x^(4*p - 2) - e*c^p*(4*p + 2*q + 1)*x^(4*p), x], x], x] /; FreeQ[{a, b, c, d, e, q}, 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]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq, x], e = Coeff[Pq, x, Expon[Pq, x]]}, Simp[e*x^(q - 1)*((a + b*x^2)^(p + 1)/(b*( q + 2*p + 1))), x] + Simp[1/(b*(q + 2*p + 1)) Int[(a + b*x^2)^p*ExpandToS um[b*(q + 2*p + 1)*Pq - a*e*(q - 1)*x^(q - 2) - b*e*(q + 2*p + 1)*x^q, x], x], x]] /; FreeQ[{a, b, p}, x] && PolyQ[Pq, x] && !LeQ[p, -1]
Time = 3.68 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.74
| method | result | size |
| pseudoelliptic | \(-\frac {5 \left (\left (f \,a^{3}-\frac {6}{5} a^{2} b e +\frac {8}{5} a \,b^{2} d -\frac {16}{5} b^{3} c \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-\left (\frac {4 \left (\frac {2}{3} f \,x^{4}+e \,x^{2}+2 d \right ) b^{\frac {5}{2}}}{5}+\left (2 \left (-\frac {f \,x^{2}}{3}-\frac {3 e}{5}\right ) b^{\frac {3}{2}}+a f \sqrt {b}\right ) a \right ) x \sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {7}{2}}}\) | \(108\) |
| risch | \(\frac {x \left (8 b^{2} f \,x^{4}-10 a b f \,x^{2}+12 b^{2} e \,x^{2}+15 a^{2} f -18 a e b +24 b^{2} d \right ) \sqrt {b \,x^{2}+a}}{48 b^{3}}-\frac {\left (5 f \,a^{3}-6 a^{2} b e +8 a \,b^{2} d -16 b^{3} c \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {7}{2}}}\) | \(109\) |
| default | \(\frac {c \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{\sqrt {b}}+f \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )+e \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+d \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )\) | \(215\) |
-5/16*((f*a^3-6/5*a^2*b*e+8/5*a*b^2*d-16/5*b^3*c)*arctanh((b*x^2+a)^(1/2)/ x/b^(1/2))-(4/5*(2/3*f*x^4+e*x^2+2*d)*b^(5/2)+(2*(-1/3*f*x^2-3/5*e)*b^(3/2 )+a*f*b^(1/2))*a)*x*(b*x^2+a)^(1/2))/b^(7/2)
Time = 0.30 (sec) , antiderivative size = 250, normalized size of antiderivative = 1.72 \[ \int \frac {c+d x^2+e x^4+f x^6}{\sqrt {a+b x^2}} \, dx=\left [-\frac {3 \, {\left (16 \, b^{3} c - 8 \, a b^{2} d + 6 \, a^{2} b e - 5 \, a^{3} f\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, b^{3} f x^{5} + 2 \, {\left (6 \, b^{3} e - 5 \, a b^{2} f\right )} x^{3} + 3 \, {\left (8 \, b^{3} d - 6 \, a b^{2} e + 5 \, a^{2} b f\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{4}}, -\frac {3 \, {\left (16 \, b^{3} c - 8 \, a b^{2} d + 6 \, a^{2} b e - 5 \, a^{3} f\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, b^{3} f x^{5} + 2 \, {\left (6 \, b^{3} e - 5 \, a b^{2} f\right )} x^{3} + 3 \, {\left (8 \, b^{3} d - 6 \, a b^{2} e + 5 \, a^{2} b f\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{4}}\right ] \]
[-1/96*(3*(16*b^3*c - 8*a*b^2*d + 6*a^2*b*e - 5*a^3*f)*sqrt(b)*log(-2*b*x^ 2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(8*b^3*f*x^5 + 2*(6*b^3*e - 5*a*b ^2*f)*x^3 + 3*(8*b^3*d - 6*a*b^2*e + 5*a^2*b*f)*x)*sqrt(b*x^2 + a))/b^4, - 1/48*(3*(16*b^3*c - 8*a*b^2*d + 6*a^2*b*e - 5*a^3*f)*sqrt(-b)*arctan(sqrt( -b)*x/sqrt(b*x^2 + a)) - (8*b^3*f*x^5 + 2*(6*b^3*e - 5*a*b^2*f)*x^3 + 3*(8 *b^3*d - 6*a*b^2*e + 5*a^2*b*f)*x)*sqrt(b*x^2 + a))/b^4]
Time = 0.41 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.07 \[ \int \frac {c+d x^2+e x^4+f x^6}{\sqrt {a+b x^2}} \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {f x^{5}}{6 b} + \frac {x^{3} \left (- \frac {5 a f}{6 b} + e\right )}{4 b} + \frac {x \left (- \frac {3 a \left (- \frac {5 a f}{6 b} + e\right )}{4 b} + d\right )}{2 b}\right ) + \left (- \frac {a \left (- \frac {3 a \left (- \frac {5 a f}{6 b} + e\right )}{4 b} + d\right )}{2 b} + c\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\\frac {c x + \frac {d x^{3}}{3} + \frac {e x^{5}}{5} + \frac {f x^{7}}{7}}{\sqrt {a}} & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a + b*x**2)*(f*x**5/(6*b) + x**3*(-5*a*f/(6*b) + e)/(4*b) + x*(-3*a*(-5*a*f/(6*b) + e)/(4*b) + d)/(2*b)) + (-a*(-3*a*(-5*a*f/(6*b) + e)/(4*b) + d)/(2*b) + c)*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)), Ne(b, 0)), ((c*x + d*x**3/3 + e*x**5/5 + f*x**7/7)/sqrt(a), True))
Time = 0.20 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.20 \[ \int \frac {c+d x^2+e x^4+f x^6}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} f x^{5}}{6 \, b} + \frac {\sqrt {b x^{2} + a} e x^{3}}{4 \, b} - \frac {5 \, \sqrt {b x^{2} + a} a f x^{3}}{24 \, b^{2}} + \frac {\sqrt {b x^{2} + a} d x}{2 \, b} - \frac {3 \, \sqrt {b x^{2} + a} a e x}{8 \, b^{2}} + \frac {5 \, \sqrt {b x^{2} + a} a^{2} f x}{16 \, b^{3}} + \frac {c \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {b}} - \frac {a d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {3}{2}}} + \frac {3 \, a^{2} e \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {5 \, a^{3} f \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {7}{2}}} \]
1/6*sqrt(b*x^2 + a)*f*x^5/b + 1/4*sqrt(b*x^2 + a)*e*x^3/b - 5/24*sqrt(b*x^ 2 + a)*a*f*x^3/b^2 + 1/2*sqrt(b*x^2 + a)*d*x/b - 3/8*sqrt(b*x^2 + a)*a*e*x /b^2 + 5/16*sqrt(b*x^2 + a)*a^2*f*x/b^3 + c*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 1/2*a*d*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 3/8*a^2*e*arcsinh(b*x/sqrt(a*b ))/b^(5/2) - 5/16*a^3*f*arcsinh(b*x/sqrt(a*b))/b^(7/2)
Time = 0.31 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.86 \[ \int \frac {c+d x^2+e x^4+f x^6}{\sqrt {a+b x^2}} \, dx=\frac {1}{48} \, {\left (2 \, {\left (\frac {4 \, f x^{2}}{b} + \frac {6 \, b^{4} e - 5 \, a b^{3} f}{b^{5}}\right )} x^{2} + \frac {3 \, {\left (8 \, b^{4} d - 6 \, a b^{3} e + 5 \, a^{2} b^{2} f\right )}}{b^{5}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (16 \, b^{3} c - 8 \, a b^{2} d + 6 \, a^{2} b e - 5 \, a^{3} f\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {7}{2}}} \]
1/48*(2*(4*f*x^2/b + (6*b^4*e - 5*a*b^3*f)/b^5)*x^2 + 3*(8*b^4*d - 6*a*b^3 *e + 5*a^2*b^2*f)/b^5)*sqrt(b*x^2 + a)*x - 1/16*(16*b^3*c - 8*a*b^2*d + 6* a^2*b*e - 5*a^3*f)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
Timed out. \[ \int \frac {c+d x^2+e x^4+f x^6}{\sqrt {a+b x^2}} \, dx=\int \frac {f\,x^6+e\,x^4+d\,x^2+c}{\sqrt {b\,x^2+a}} \,d x \]