Integrand size = 32, antiderivative size = 245 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\sqrt {a+b x^2}} \, dx=-\frac {a \left (96 b^3 c-80 a b^2 d+70 a^2 b e-63 a^3 f\right ) x \sqrt {a+b x^2}}{256 b^5}+\frac {\left (96 b^3 c-80 a b^2 d+70 a^2 b e-63 a^3 f\right ) x^3 \sqrt {a+b x^2}}{384 b^4}+\frac {\left (80 b^2 d-70 a b e+63 a^2 f\right ) x^5 \sqrt {a+b x^2}}{480 b^3}+\frac {(10 b e-9 a f) x^7 \sqrt {a+b x^2}}{80 b^2}+\frac {f x^9 \sqrt {a+b x^2}}{10 b}+\frac {a^2 \left (96 b^3 c-80 a b^2 d+70 a^2 b e-63 a^3 f\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{11/2}} \]
1/256*a^2*(-63*a^3*f+70*a^2*b*e-80*a*b^2*d+96*b^3*c)*arctanh(x*b^(1/2)/(b* x^2+a)^(1/2))/b^(11/2)-1/256*a*(-63*a^3*f+70*a^2*b*e-80*a*b^2*d+96*b^3*c)* x*(b*x^2+a)^(1/2)/b^5+1/384*(-63*a^3*f+70*a^2*b*e-80*a*b^2*d+96*b^3*c)*x^3 *(b*x^2+a)^(1/2)/b^4+1/480*(63*a^2*f-70*a*b*e+80*b^2*d)*x^5*(b*x^2+a)^(1/2 )/b^3+1/80*(-9*a*f+10*b*e)*x^7*(b*x^2+a)^(1/2)/b^2+1/10*f*x^9*(b*x^2+a)^(1 /2)/b
Time = 0.98 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.89 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\sqrt {a+b x^2}} \, dx=\frac {x \sqrt {a+b x^2} \left (-1440 a b^3 c+1200 a^2 b^2 d-1050 a^3 b e+945 a^4 f+960 b^4 c x^2-800 a b^3 d x^2+700 a^2 b^2 e x^2-630 a^3 b f x^2+640 b^4 d x^4-560 a b^3 e x^4+504 a^2 b^2 f x^4+480 b^4 e x^6-432 a b^3 f x^6+384 b^4 f x^8\right )}{3840 b^5}-\frac {a^2 \left (-96 b^3 c+80 a b^2 d-70 a^2 b e+63 a^3 f\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{128 b^{11/2}} \]
(x*Sqrt[a + b*x^2]*(-1440*a*b^3*c + 1200*a^2*b^2*d - 1050*a^3*b*e + 945*a^ 4*f + 960*b^4*c*x^2 - 800*a*b^3*d*x^2 + 700*a^2*b^2*e*x^2 - 630*a^3*b*f*x^ 2 + 640*b^4*d*x^4 - 560*a*b^3*e*x^4 + 504*a^2*b^2*f*x^4 + 480*b^4*e*x^6 - 432*a*b^3*f*x^6 + 384*b^4*f*x^8))/(3840*b^5) - (a^2*(-96*b^3*c + 80*a*b^2* d - 70*a^2*b*e + 63*a^3*f)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2] )])/(128*b^(11/2))
Time = 0.44 (sec) , antiderivative size = 221, normalized size of antiderivative = 0.90, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {2340, 1590, 363, 262, 262, 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.
\(\displaystyle \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 2340 |
\(\displaystyle \frac {\int \frac {x^4 \left ((10 b e-9 a f) x^4+10 b d x^2+10 b c\right )}{\sqrt {b x^2+a}}dx}{10 b}+\frac {f x^9 \sqrt {a+b x^2}}{10 b}\) |
\(\Big \downarrow \) 1590 |
\(\displaystyle \frac {\frac {\int \frac {x^4 \left (80 c b^2+\left (63 f a^2-70 b e a+80 b^2 d\right ) x^2\right )}{\sqrt {b x^2+a}}dx}{8 b}+\frac {x^7 \sqrt {a+b x^2} (10 b e-9 a f)}{8 b}}{10 b}+\frac {f x^9 \sqrt {a+b x^2}}{10 b}\) |
\(\Big \downarrow \) 363 |
\(\displaystyle \frac {\frac {\frac {5 \left (-63 a^3 f+70 a^2 b e-80 a b^2 d+96 b^3 c\right ) \int \frac {x^4}{\sqrt {b x^2+a}}dx}{6 b}+\frac {x^5 \sqrt {a+b x^2} \left (63 a^2 f-70 a b e+80 b^2 d\right )}{6 b}}{8 b}+\frac {x^7 \sqrt {a+b x^2} (10 b e-9 a f)}{8 b}}{10 b}+\frac {f x^9 \sqrt {a+b x^2}}{10 b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {\frac {5 \left (-63 a^3 f+70 a^2 b e-80 a b^2 d+96 b^3 c\right ) \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^2+a}}dx}{4 b}\right )}{6 b}+\frac {x^5 \sqrt {a+b x^2} \left (63 a^2 f-70 a b e+80 b^2 d\right )}{6 b}}{8 b}+\frac {x^7 \sqrt {a+b x^2} (10 b e-9 a f)}{8 b}}{10 b}+\frac {f x^9 \sqrt {a+b x^2}}{10 b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {\frac {5 \left (-63 a^3 f+70 a^2 b e-80 a b^2 d+96 b^3 c\right ) \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )}{4 b}\right )}{6 b}+\frac {x^5 \sqrt {a+b x^2} \left (63 a^2 f-70 a b e+80 b^2 d\right )}{6 b}}{8 b}+\frac {x^7 \sqrt {a+b x^2} (10 b e-9 a f)}{8 b}}{10 b}+\frac {f x^9 \sqrt {a+b x^2}}{10 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {\frac {5 \left (-63 a^3 f+70 a^2 b e-80 a b^2 d+96 b^3 c\right ) \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )}{4 b}\right )}{6 b}+\frac {x^5 \sqrt {a+b x^2} \left (63 a^2 f-70 a b e+80 b^2 d\right )}{6 b}}{8 b}+\frac {x^7 \sqrt {a+b x^2} (10 b e-9 a f)}{8 b}}{10 b}+\frac {f x^9 \sqrt {a+b x^2}}{10 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\frac {x^5 \sqrt {a+b x^2} \left (63 a^2 f-70 a b e+80 b^2 d\right )}{6 b}+\frac {5 \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}}\right )}{4 b}\right ) \left (-63 a^3 f+70 a^2 b e-80 a b^2 d+96 b^3 c\right )}{6 b}}{8 b}+\frac {x^7 \sqrt {a+b x^2} (10 b e-9 a f)}{8 b}}{10 b}+\frac {f x^9 \sqrt {a+b x^2}}{10 b}\) |
(f*x^9*Sqrt[a + b*x^2])/(10*b) + (((10*b*e - 9*a*f)*x^7*Sqrt[a + b*x^2])/( 8*b) + (((80*b^2*d - 70*a*b*e + 63*a^2*f)*x^5*Sqrt[a + b*x^2])/(6*b) + (5* (96*b^3*c - 80*a*b^2*d + 70*a^2*b*e - 63*a^3*f)*((x^3*Sqrt[a + b*x^2])/(4* b) - (3*a*((x*Sqrt[a + b*x^2])/(2*b) - (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x ^2]])/(2*b^(3/2))))/(4*b)))/(6*b))/(8*b))/(10*b)
3.2.51.3.1 Defintions of rubi rules used
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Simp[c^p*(f*x)^(m + 4*p - 1)*((d + e*x^2)^ (q + 1)/(e*f^(4*p - 1)*(m + 4*p + 2*q + 1))), x] + Simp[1/(e*(m + 4*p + 2*q + 1)) Int[(f*x)^m*(d + e*x^2)^q*ExpandToSum[e*(m + 4*p + 2*q + 1)*((a + b*x^2 + c*x^4)^p - c^p*x^(4*p)) - d*c^p*(m + 4*p - 1)*x^(4*p - 2), x], x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[b^2 - 4*a*c, 0] && IGtQ[p, 0] && !IntegerQ[q] && NeQ[m + 4*p + 2*q + 1, 0]
Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ {q = Expon[Pq, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[f*(c*x)^(m + q - 1 )*((a + b*x^2)^(p + 1)/(b*c^(q - 1)*(m + q + 2*p + 1))), x] + Simp[1/(b*(m + q + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^p*ExpandToSum[b*(m + q + 2*p + 1) *Pq - b*f*(m + q + 2*p + 1)*x^q - a*f*(m + q - 1)*x^(q - 2), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p + 1, 0]] /; FreeQ[{a, b, c, m, p}, x] && PolyQ [Pq, x] && ( !IGtQ[m, 0] || IGtQ[p + 1/2, -1])
Time = 3.70 (sec) , antiderivative size = 169, normalized size of antiderivative = 0.69
method | result | size |
pseudoelliptic | \(-\frac {63 \left (a^{2} \left (f \,a^{3}-\frac {10}{9} a^{2} b e +\frac {80}{63} a \,b^{2} d -\frac {32}{21} b^{3} c \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-\left (-\frac {32 \left (\frac {3}{10} f \,x^{6}+\frac {7}{18} e \,x^{4}+\frac {5}{9} d \,x^{2}+c \right ) a \,b^{\frac {7}{2}}}{21}+\frac {64 x^{2} \left (\frac {2}{5} f \,x^{6}+\frac {1}{2} e \,x^{4}+\frac {2}{3} d \,x^{2}+c \right ) b^{\frac {9}{2}}}{63}+\left (\left (\frac {8}{15} f \,x^{4}+\frac {20}{27} e \,x^{2}+\frac {80}{63} d \right ) b^{\frac {5}{2}}+\left (\left (-\frac {2 f \,x^{2}}{3}-\frac {10 e}{9}\right ) b^{\frac {3}{2}}+a f \sqrt {b}\right ) a \right ) a^{2}\right ) \sqrt {b \,x^{2}+a}\, x \right )}{256 b^{\frac {11}{2}}}\) | \(169\) |
risch | \(\frac {x \left (384 f \,x^{8} b^{4}-432 a \,b^{3} f \,x^{6}+480 b^{4} e \,x^{6}+504 a^{2} b^{2} f \,x^{4}-560 a \,b^{3} e \,x^{4}+640 b^{4} d \,x^{4}-630 a^{3} b f \,x^{2}+700 a^{2} b^{2} e \,x^{2}-800 a \,b^{3} d \,x^{2}+960 b^{4} c \,x^{2}+945 a^{4} f -1050 a^{3} b e +1200 a^{2} b^{2} d -1440 a \,b^{3} c \right ) \sqrt {b \,x^{2}+a}}{3840 b^{5}}-\frac {a^{2} \left (63 f \,a^{3}-70 a^{2} b e +80 a \,b^{2} d -96 b^{3} c \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{256 b^{\frac {11}{2}}}\) | \(198\) |
default | \(e \left (\frac {x^{7} \sqrt {b \,x^{2}+a}}{8 b}-\frac {7 a \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )}{8 b}\right )+d \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )+c \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+f \left (\frac {x^{9} \sqrt {b \,x^{2}+a}}{10 b}-\frac {9 a \left (\frac {x^{7} \sqrt {b \,x^{2}+a}}{8 b}-\frac {7 a \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )}{8 b}\right )}{10 b}\right )\) | \(402\) |
-63/256/b^(11/2)*(a^2*(f*a^3-10/9*a^2*b*e+80/63*a*b^2*d-32/21*b^3*c)*arcta nh((b*x^2+a)^(1/2)/x/b^(1/2))-(-32/21*(3/10*f*x^6+7/18*e*x^4+5/9*d*x^2+c)* a*b^(7/2)+64/63*x^2*(2/5*f*x^6+1/2*e*x^4+2/3*d*x^2+c)*b^(9/2)+((8/15*f*x^4 +20/27*e*x^2+80/63*d)*b^(5/2)+((-2/3*f*x^2-10/9*e)*b^(3/2)+a*f*b^(1/2))*a) *a^2)*(b*x^2+a)^(1/2)*x)
Time = 0.36 (sec) , antiderivative size = 414, normalized size of antiderivative = 1.69 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\sqrt {a+b x^2}} \, dx=\left [-\frac {15 \, {\left (96 \, a^{2} b^{3} c - 80 \, a^{3} b^{2} d + 70 \, a^{4} b e - 63 \, a^{5} f\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (384 \, b^{5} f x^{9} + 48 \, {\left (10 \, b^{5} e - 9 \, a b^{4} f\right )} x^{7} + 8 \, {\left (80 \, b^{5} d - 70 \, a b^{4} e + 63 \, a^{2} b^{3} f\right )} x^{5} + 10 \, {\left (96 \, b^{5} c - 80 \, a b^{4} d + 70 \, a^{2} b^{3} e - 63 \, a^{3} b^{2} f\right )} x^{3} - 15 \, {\left (96 \, a b^{4} c - 80 \, a^{2} b^{3} d + 70 \, a^{3} b^{2} e - 63 \, a^{4} b f\right )} x\right )} \sqrt {b x^{2} + a}}{7680 \, b^{6}}, -\frac {15 \, {\left (96 \, a^{2} b^{3} c - 80 \, a^{3} b^{2} d + 70 \, a^{4} b e - 63 \, a^{5} f\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (384 \, b^{5} f x^{9} + 48 \, {\left (10 \, b^{5} e - 9 \, a b^{4} f\right )} x^{7} + 8 \, {\left (80 \, b^{5} d - 70 \, a b^{4} e + 63 \, a^{2} b^{3} f\right )} x^{5} + 10 \, {\left (96 \, b^{5} c - 80 \, a b^{4} d + 70 \, a^{2} b^{3} e - 63 \, a^{3} b^{2} f\right )} x^{3} - 15 \, {\left (96 \, a b^{4} c - 80 \, a^{2} b^{3} d + 70 \, a^{3} b^{2} e - 63 \, a^{4} b f\right )} x\right )} \sqrt {b x^{2} + a}}{3840 \, b^{6}}\right ] \]
[-1/7680*(15*(96*a^2*b^3*c - 80*a^3*b^2*d + 70*a^4*b*e - 63*a^5*f)*sqrt(b) *log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(384*b^5*f*x^9 + 48*( 10*b^5*e - 9*a*b^4*f)*x^7 + 8*(80*b^5*d - 70*a*b^4*e + 63*a^2*b^3*f)*x^5 + 10*(96*b^5*c - 80*a*b^4*d + 70*a^2*b^3*e - 63*a^3*b^2*f)*x^3 - 15*(96*a*b ^4*c - 80*a^2*b^3*d + 70*a^3*b^2*e - 63*a^4*b*f)*x)*sqrt(b*x^2 + a))/b^6, -1/3840*(15*(96*a^2*b^3*c - 80*a^3*b^2*d + 70*a^4*b*e - 63*a^5*f)*sqrt(-b) *arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (384*b^5*f*x^9 + 48*(10*b^5*e - 9*a* b^4*f)*x^7 + 8*(80*b^5*d - 70*a*b^4*e + 63*a^2*b^3*f)*x^5 + 10*(96*b^5*c - 80*a*b^4*d + 70*a^2*b^3*e - 63*a^3*b^2*f)*x^3 - 15*(96*a*b^4*c - 80*a^2*b ^3*d + 70*a^3*b^2*e - 63*a^4*b*f)*x)*sqrt(b*x^2 + a))/b^6]
Time = 0.61 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.99 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} \frac {3 a^{2} \left (- \frac {5 a \left (- \frac {7 a \left (- \frac {9 a f}{10 b} + e\right )}{8 b} + d\right )}{6 b} + c\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} + \sqrt {a + b x^{2}} \left (- \frac {3 a x \left (- \frac {5 a \left (- \frac {7 a \left (- \frac {9 a f}{10 b} + e\right )}{8 b} + d\right )}{6 b} + c\right )}{8 b^{2}} + \frac {f x^{9}}{10 b} + \frac {x^{7} \left (- \frac {9 a f}{10 b} + e\right )}{8 b} + \frac {x^{5} \left (- \frac {7 a \left (- \frac {9 a f}{10 b} + e\right )}{8 b} + d\right )}{6 b} + \frac {x^{3} \left (- \frac {5 a \left (- \frac {7 a \left (- \frac {9 a f}{10 b} + e\right )}{8 b} + d\right )}{6 b} + c\right )}{4 b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {c x^{5}}{5} + \frac {d x^{7}}{7} + \frac {e x^{9}}{9} + \frac {f x^{11}}{11}}{\sqrt {a}} & \text {otherwise} \end {cases} \]
Piecewise((3*a**2*(-5*a*(-7*a*(-9*a*f/(10*b) + e)/(8*b) + d)/(6*b) + c)*Pi ecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqrt(b), Ne(a, 0)), (x*lo g(x)/sqrt(b*x**2), True))/(8*b**2) + sqrt(a + b*x**2)*(-3*a*x*(-5*a*(-7*a* (-9*a*f/(10*b) + e)/(8*b) + d)/(6*b) + c)/(8*b**2) + f*x**9/(10*b) + x**7* (-9*a*f/(10*b) + e)/(8*b) + x**5*(-7*a*(-9*a*f/(10*b) + e)/(8*b) + d)/(6*b ) + x**3*(-5*a*(-7*a*(-9*a*f/(10*b) + e)/(8*b) + d)/(6*b) + c)/(4*b)), Ne( b, 0)), ((c*x**5/5 + d*x**7/7 + e*x**9/9 + f*x**11/11)/sqrt(a), True))
Time = 0.20 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.38 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} f x^{9}}{10 \, b} + \frac {\sqrt {b x^{2} + a} e x^{7}}{8 \, b} - \frac {9 \, \sqrt {b x^{2} + a} a f x^{7}}{80 \, b^{2}} + \frac {\sqrt {b x^{2} + a} d x^{5}}{6 \, b} - \frac {7 \, \sqrt {b x^{2} + a} a e x^{5}}{48 \, b^{2}} + \frac {21 \, \sqrt {b x^{2} + a} a^{2} f x^{5}}{160 \, b^{3}} + \frac {\sqrt {b x^{2} + a} c x^{3}}{4 \, b} - \frac {5 \, \sqrt {b x^{2} + a} a d x^{3}}{24 \, b^{2}} + \frac {35 \, \sqrt {b x^{2} + a} a^{2} e x^{3}}{192 \, b^{3}} - \frac {21 \, \sqrt {b x^{2} + a} a^{3} f x^{3}}{128 \, b^{4}} - \frac {3 \, \sqrt {b x^{2} + a} a c x}{8 \, b^{2}} + \frac {5 \, \sqrt {b x^{2} + a} a^{2} d x}{16 \, b^{3}} - \frac {35 \, \sqrt {b x^{2} + a} a^{3} e x}{128 \, b^{4}} + \frac {63 \, \sqrt {b x^{2} + a} a^{4} f x}{256 \, b^{5}} + \frac {3 \, a^{2} c \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} - \frac {5 \, a^{3} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {7}{2}}} + \frac {35 \, a^{4} e \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {9}{2}}} - \frac {63 \, a^{5} f \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{256 \, b^{\frac {11}{2}}} \]
1/10*sqrt(b*x^2 + a)*f*x^9/b + 1/8*sqrt(b*x^2 + a)*e*x^7/b - 9/80*sqrt(b*x ^2 + a)*a*f*x^7/b^2 + 1/6*sqrt(b*x^2 + a)*d*x^5/b - 7/48*sqrt(b*x^2 + a)*a *e*x^5/b^2 + 21/160*sqrt(b*x^2 + a)*a^2*f*x^5/b^3 + 1/4*sqrt(b*x^2 + a)*c* x^3/b - 5/24*sqrt(b*x^2 + a)*a*d*x^3/b^2 + 35/192*sqrt(b*x^2 + a)*a^2*e*x^ 3/b^3 - 21/128*sqrt(b*x^2 + a)*a^3*f*x^3/b^4 - 3/8*sqrt(b*x^2 + a)*a*c*x/b ^2 + 5/16*sqrt(b*x^2 + a)*a^2*d*x/b^3 - 35/128*sqrt(b*x^2 + a)*a^3*e*x/b^4 + 63/256*sqrt(b*x^2 + a)*a^4*f*x/b^5 + 3/8*a^2*c*arcsinh(b*x/sqrt(a*b))/b ^(5/2) - 5/16*a^3*d*arcsinh(b*x/sqrt(a*b))/b^(7/2) + 35/128*a^4*e*arcsinh( b*x/sqrt(a*b))/b^(9/2) - 63/256*a^5*f*arcsinh(b*x/sqrt(a*b))/b^(11/2)
Time = 0.32 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.89 \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\sqrt {a+b x^2}} \, dx=\frac {1}{3840} \, {\left (2 \, {\left (4 \, {\left (6 \, {\left (\frac {8 \, f x^{2}}{b} + \frac {10 \, b^{8} e - 9 \, a b^{7} f}{b^{9}}\right )} x^{2} + \frac {80 \, b^{8} d - 70 \, a b^{7} e + 63 \, a^{2} b^{6} f}{b^{9}}\right )} x^{2} + \frac {5 \, {\left (96 \, b^{8} c - 80 \, a b^{7} d + 70 \, a^{2} b^{6} e - 63 \, a^{3} b^{5} f\right )}}{b^{9}}\right )} x^{2} - \frac {15 \, {\left (96 \, a b^{7} c - 80 \, a^{2} b^{6} d + 70 \, a^{3} b^{5} e - 63 \, a^{4} b^{4} f\right )}}{b^{9}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (96 \, a^{2} b^{3} c - 80 \, a^{3} b^{2} d + 70 \, a^{4} b e - 63 \, a^{5} f\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{256 \, b^{\frac {11}{2}}} \]
1/3840*(2*(4*(6*(8*f*x^2/b + (10*b^8*e - 9*a*b^7*f)/b^9)*x^2 + (80*b^8*d - 70*a*b^7*e + 63*a^2*b^6*f)/b^9)*x^2 + 5*(96*b^8*c - 80*a*b^7*d + 70*a^2*b ^6*e - 63*a^3*b^5*f)/b^9)*x^2 - 15*(96*a*b^7*c - 80*a^2*b^6*d + 70*a^3*b^5 *e - 63*a^4*b^4*f)/b^9)*sqrt(b*x^2 + a)*x - 1/256*(96*a^2*b^3*c - 80*a^3*b ^2*d + 70*a^4*b*e - 63*a^5*f)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(11 /2)
Timed out. \[ \int \frac {x^4 \left (c+d x^2+e x^4+f x^6\right )}{\sqrt {a+b x^2}} \, dx=\int \frac {x^4\,\left (f\,x^6+e\,x^4+d\,x^2+c\right )}{\sqrt {b\,x^2+a}} \,d x \]