Integrand size = 32, antiderivative size = 248 \[ \int \frac {x^4 \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=-\frac {a \left (96 A b^3-a \left (80 b^2 B-70 a b C+63 a^2 D\right )\right ) x \sqrt {a+b x^2}}{256 b^5}+\frac {\left (96 A b^3-a \left (80 b^2 B-70 a b C+63 a^2 D\right )\right ) x^3 \sqrt {a+b x^2}}{384 b^4}+\frac {\left (80 b^2 B-70 a b C+63 a^2 D\right ) x^5 \sqrt {a+b x^2}}{480 b^3}+\frac {(10 b C-9 a D) x^7 \sqrt {a+b x^2}}{80 b^2}+\frac {D x^9 \sqrt {a+b x^2}}{10 b}+\frac {a^2 \left (96 A b^3-a \left (80 b^2 B-70 a b C+63 a^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{11/2}} \] Output:
-1/256*a*(96*A*b^3-a*(80*B*b^2-70*C*a*b+63*D*a^2))*x*(b*x^2+a)^(1/2)/b^5+1 /384*(96*A*b^3-a*(80*B*b^2-70*C*a*b+63*D*a^2))*x^3*(b*x^2+a)^(1/2)/b^4+1/4 80*(80*B*b^2-70*C*a*b+63*D*a^2)*x^5*(b*x^2+a)^(1/2)/b^3+1/80*(10*C*b-9*D*a )*x^7*(b*x^2+a)^(1/2)/b^2+1/10*D*x^9*(b*x^2+a)^(1/2)/b+1/256*a^2*(96*A*b^3 -a*(80*B*b^2-70*C*a*b+63*D*a^2))*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(11/ 2)
Time = 1.55 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.88 \[ \int \frac {x^4 \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\frac {x \sqrt {a+b x^2} \left (-1440 a A b^3+1200 a^2 b^2 B-1050 a^3 b C+945 a^4 D+960 A b^4 x^2-800 a b^3 B x^2+700 a^2 b^2 C x^2-630 a^3 b D x^2+640 b^4 B x^4-560 a b^3 C x^4+504 a^2 b^2 D x^4+480 b^4 C x^6-432 a b^3 D x^6+384 b^4 D x^8\right )}{3840 b^5}-\frac {a^2 \left (-96 A b^3+80 a b^2 B-70 a^2 b C+63 a^3 D\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{-\sqrt {a}+\sqrt {a+b x^2}}\right )}{128 b^{11/2}} \] Input:
Integrate[(x^4*(A + B*x^2 + C*x^4 + D*x^6))/Sqrt[a + b*x^2],x]
Output:
(x*Sqrt[a + b*x^2]*(-1440*a*A*b^3 + 1200*a^2*b^2*B - 1050*a^3*b*C + 945*a^ 4*D + 960*A*b^4*x^2 - 800*a*b^3*B*x^2 + 700*a^2*b^2*C*x^2 - 630*a^3*b*D*x^ 2 + 640*b^4*B*x^4 - 560*a*b^3*C*x^4 + 504*a^2*b^2*D*x^4 + 480*b^4*C*x^6 - 432*a*b^3*D*x^6 + 384*b^4*D*x^8))/(3840*b^5) - (a^2*(-96*A*b^3 + 80*a*b^2* B - 70*a^2*b*C + 63*a^3*D)*ArcTanh[(Sqrt[b]*x)/(-Sqrt[a] + Sqrt[a + b*x^2] )])/(128*b^(11/2))
Time = 0.48 (sec) , antiderivative size = 222, 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.
\(\displaystyle \int \frac {x^4 \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 2340 |
\(\displaystyle \frac {\int \frac {x^4 \left ((10 b C-9 a D) x^4+10 b B x^2+10 A b\right )}{\sqrt {b x^2+a}}dx}{10 b}+\frac {D x^9 \sqrt {a+b x^2}}{10 b}\) |
\(\Big \downarrow \) 1590 |
\(\displaystyle \frac {\frac {\int \frac {x^4 \left (80 A b^2+\left (63 D a^2-70 b C a+80 b^2 B\right ) x^2\right )}{\sqrt {b x^2+a}}dx}{8 b}+\frac {x^7 \sqrt {a+b x^2} (10 b C-9 a D)}{8 b}}{10 b}+\frac {D x^9 \sqrt {a+b x^2}}{10 b}\) |
\(\Big \downarrow \) 363 |
\(\displaystyle \frac {\frac {\frac {5 \left (96 A b^3-a \left (63 a^2 D-70 a b C+80 b^2 B\right )\right ) \int \frac {x^4}{\sqrt {b x^2+a}}dx}{6 b}+\frac {x^5 \sqrt {a+b x^2} \left (63 a^2 D-70 a b C+80 b^2 B\right )}{6 b}}{8 b}+\frac {x^7 \sqrt {a+b x^2} (10 b C-9 a D)}{8 b}}{10 b}+\frac {D x^9 \sqrt {a+b x^2}}{10 b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {\frac {5 \left (96 A b^3-a \left (63 a^2 D-70 a b C+80 b^2 B\right )\right ) \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^2+a}}dx}{4 b}\right )}{6 b}+\frac {x^5 \sqrt {a+b x^2} \left (63 a^2 D-70 a b C+80 b^2 B\right )}{6 b}}{8 b}+\frac {x^7 \sqrt {a+b x^2} (10 b C-9 a D)}{8 b}}{10 b}+\frac {D x^9 \sqrt {a+b x^2}}{10 b}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {\frac {\frac {5 \left (96 A b^3-a \left (63 a^2 D-70 a b C+80 b^2 B\right )\right ) \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}\right )}{4 b}\right )}{6 b}+\frac {x^5 \sqrt {a+b x^2} \left (63 a^2 D-70 a b C+80 b^2 B\right )}{6 b}}{8 b}+\frac {x^7 \sqrt {a+b x^2} (10 b C-9 a D)}{8 b}}{10 b}+\frac {D x^9 \sqrt {a+b x^2}}{10 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\frac {\frac {5 \left (96 A b^3-a \left (63 a^2 D-70 a b C+80 b^2 B\right )\right ) \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}\right )}{4 b}\right )}{6 b}+\frac {x^5 \sqrt {a+b x^2} \left (63 a^2 D-70 a b C+80 b^2 B\right )}{6 b}}{8 b}+\frac {x^7 \sqrt {a+b x^2} (10 b C-9 a D)}{8 b}}{10 b}+\frac {D x^9 \sqrt {a+b x^2}}{10 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\frac {\frac {5 \left (\frac {x^3 \sqrt {a+b x^2}}{4 b}-\frac {3 a \left (\frac {x \sqrt {a+b x^2}}{2 b}-\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{3/2}}\right )}{4 b}\right ) \left (96 A b^3-a \left (63 a^2 D-70 a b C+80 b^2 B\right )\right )}{6 b}+\frac {x^5 \sqrt {a+b x^2} \left (63 a^2 D-70 a b C+80 b^2 B\right )}{6 b}}{8 b}+\frac {x^7 \sqrt {a+b x^2} (10 b C-9 a D)}{8 b}}{10 b}+\frac {D x^9 \sqrt {a+b x^2}}{10 b}\) |
Input:
Int[(x^4*(A + B*x^2 + C*x^4 + D*x^6))/Sqrt[a + b*x^2],x]
Output:
(D*x^9*Sqrt[a + b*x^2])/(10*b) + (((10*b*C - 9*a*D)*x^7*Sqrt[a + b*x^2])/( 8*b) + (((80*b^2*B - 70*a*b*C + 63*a^2*D)*x^5*Sqrt[a + b*x^2])/(6*b) + (5* (96*A*b^3 - a*(80*b^2*B - 70*a*b*C + 63*a^2*D))*((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)
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 = 0.59 (sec) , antiderivative size = 168, normalized size of antiderivative = 0.68
method | result | size |
pseudoelliptic | \(-\frac {3 \left (-\left (b^{3} A -\frac {5}{6} a \,b^{2} B +\frac {35}{48} a^{2} b C -\frac {21}{32} a^{3} D\right ) a^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\sqrt {b \,x^{2}+a}\, x \left (a \left (\frac {3}{10} D x^{6}+\frac {7}{18} C \,x^{4}+\frac {5}{9} x^{2} B +A \right ) b^{\frac {7}{2}}-\frac {2 \left (\frac {2}{5} D x^{6}+\frac {1}{2} C \,x^{4}+\frac {2}{3} x^{2} B +A \right ) x^{2} b^{\frac {9}{2}}}{3}+\frac {35 \left (\left (-\frac {12}{25} D x^{4}-\frac {2}{3} C \,x^{2}-\frac {8}{7} B \right ) b^{\frac {5}{2}}+\left (\left (\frac {3 D x^{2}}{5}+C \right ) b^{\frac {3}{2}}-\frac {9 D a \sqrt {b}}{10}\right ) a \right ) a^{2}}{48}\right )\right )}{8 b^{\frac {11}{2}}}\) | \(168\) |
default | \(A \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )+B \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )+C \left (\frac {x^{7} \sqrt {b \,x^{2}+a}}{8 b}-\frac {7 a \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )}{8 b}\right )+D \left (\frac {x^{9} \sqrt {b \,x^{2}+a}}{10 b}-\frac {9 a \left (\frac {x^{7} \sqrt {b \,x^{2}+a}}{8 b}-\frac {7 a \left (\frac {x^{5} \sqrt {b \,x^{2}+a}}{6 b}-\frac {5 a \left (\frac {x^{3} \sqrt {b \,x^{2}+a}}{4 b}-\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2 b}-\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 b^{\frac {3}{2}}}\right )}{4 b}\right )}{6 b}\right )}{8 b}\right )}{10 b}\right )\) | \(402\) |
Input:
int(x^4*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-3/8/b^(11/2)*(-(b^3*A-5/6*a*b^2*B+35/48*a^2*b*C-21/32*a^3*D)*a^2*arctanh( (b*x^2+a)^(1/2)/x/b^(1/2))+(b*x^2+a)^(1/2)*x*(a*(3/10*D*x^6+7/18*C*x^4+5/9 *x^2*B+A)*b^(7/2)-2/3*(2/5*D*x^6+1/2*C*x^4+2/3*x^2*B+A)*x^2*b^(9/2)+35/48* ((-12/25*D*x^4-2/3*C*x^2-8/7*B)*b^(5/2)+((3/5*D*x^2+C)*b^(3/2)-9/10*D*a*b^ (1/2))*a)*a^2))
Time = 0.14 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.67 \[ \int \frac {x^4 \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\left [-\frac {15 \, {\left (63 \, D a^{5} - 70 \, C a^{4} b + 80 \, B a^{3} b^{2} - 96 \, A a^{2} b^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (384 \, D b^{5} x^{9} - 48 \, {\left (9 \, D a b^{4} - 10 \, C b^{5}\right )} x^{7} + 8 \, {\left (63 \, D a^{2} b^{3} - 70 \, C a b^{4} + 80 \, B b^{5}\right )} x^{5} - 10 \, {\left (63 \, D a^{3} b^{2} - 70 \, C a^{2} b^{3} + 80 \, B a b^{4} - 96 \, A b^{5}\right )} x^{3} + 15 \, {\left (63 \, D a^{4} b - 70 \, C a^{3} b^{2} + 80 \, B a^{2} b^{3} - 96 \, A a b^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{7680 \, b^{6}}, \frac {15 \, {\left (63 \, D a^{5} - 70 \, C a^{4} b + 80 \, B a^{3} b^{2} - 96 \, A a^{2} b^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (384 \, D b^{5} x^{9} - 48 \, {\left (9 \, D a b^{4} - 10 \, C b^{5}\right )} x^{7} + 8 \, {\left (63 \, D a^{2} b^{3} - 70 \, C a b^{4} + 80 \, B b^{5}\right )} x^{5} - 10 \, {\left (63 \, D a^{3} b^{2} - 70 \, C a^{2} b^{3} + 80 \, B a b^{4} - 96 \, A b^{5}\right )} x^{3} + 15 \, {\left (63 \, D a^{4} b - 70 \, C a^{3} b^{2} + 80 \, B a^{2} b^{3} - 96 \, A a b^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{3840 \, b^{6}}\right ] \] Input:
integrate(x^4*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
[-1/7680*(15*(63*D*a^5 - 70*C*a^4*b + 80*B*a^3*b^2 - 96*A*a^2*b^3)*sqrt(b) *log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(384*D*b^5*x^9 - 48*( 9*D*a*b^4 - 10*C*b^5)*x^7 + 8*(63*D*a^2*b^3 - 70*C*a*b^4 + 80*B*b^5)*x^5 - 10*(63*D*a^3*b^2 - 70*C*a^2*b^3 + 80*B*a*b^4 - 96*A*b^5)*x^3 + 15*(63*D*a ^4*b - 70*C*a^3*b^2 + 80*B*a^2*b^3 - 96*A*a*b^4)*x)*sqrt(b*x^2 + a))/b^6, 1/3840*(15*(63*D*a^5 - 70*C*a^4*b + 80*B*a^3*b^2 - 96*A*a^2*b^3)*sqrt(-b)* arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (384*D*b^5*x^9 - 48*(9*D*a*b^4 - 10*C *b^5)*x^7 + 8*(63*D*a^2*b^3 - 70*C*a*b^4 + 80*B*b^5)*x^5 - 10*(63*D*a^3*b^ 2 - 70*C*a^2*b^3 + 80*B*a*b^4 - 96*A*b^5)*x^3 + 15*(63*D*a^4*b - 70*C*a^3* b^2 + 80*B*a^2*b^3 - 96*A*a*b^4)*x)*sqrt(b*x^2 + a))/b^6]
Time = 0.67 (sec) , antiderivative size = 243, normalized size of antiderivative = 0.98 \[ \int \frac {x^4 \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\begin {cases} \frac {3 a^{2} \left (A - \frac {5 a \left (B - \frac {7 a \left (C - \frac {9 D a}{10 b}\right )}{8 b}\right )}{6 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{8 b^{2}} + \sqrt {a + b x^{2}} \left (\frac {D x^{9}}{10 b} - \frac {3 a x \left (A - \frac {5 a \left (B - \frac {7 a \left (C - \frac {9 D a}{10 b}\right )}{8 b}\right )}{6 b}\right )}{8 b^{2}} + \frac {x^{7} \left (C - \frac {9 D a}{10 b}\right )}{8 b} + \frac {x^{5} \left (B - \frac {7 a \left (C - \frac {9 D a}{10 b}\right )}{8 b}\right )}{6 b} + \frac {x^{3} \left (A - \frac {5 a \left (B - \frac {7 a \left (C - \frac {9 D a}{10 b}\right )}{8 b}\right )}{6 b}\right )}{4 b}\right ) & \text {for}\: b \neq 0 \\\frac {\frac {A x^{5}}{5} + \frac {B x^{7}}{7} + \frac {C x^{9}}{9} + \frac {D x^{11}}{11}}{\sqrt {a}} & \text {otherwise} \end {cases} \] Input:
integrate(x**4*(D*x**6+C*x**4+B*x**2+A)/(b*x**2+a)**(1/2),x)
Output:
Piecewise((3*a**2*(A - 5*a*(B - 7*a*(C - 9*D*a/(10*b))/(8*b))/(6*b))*Piece wise((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))/(8*b**2) + sqrt(a + b*x**2)*(D*x**9/(10*b) - 3*a*x* (A - 5*a*(B - 7*a*(C - 9*D*a/(10*b))/(8*b))/(6*b))/(8*b**2) + x**7*(C - 9* D*a/(10*b))/(8*b) + x**5*(B - 7*a*(C - 9*D*a/(10*b))/(8*b))/(6*b) + x**3*( A - 5*a*(B - 7*a*(C - 9*D*a/(10*b))/(8*b))/(6*b))/(4*b)), Ne(b, 0)), ((A*x **5/5 + B*x**7/7 + C*x**9/9 + D*x**11/11)/sqrt(a), True))
Time = 0.04 (sec) , antiderivative size = 339, normalized size of antiderivative = 1.37 \[ \int \frac {x^4 \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\frac {\sqrt {b x^{2} + a} D x^{9}}{10 \, b} - \frac {9 \, \sqrt {b x^{2} + a} D a x^{7}}{80 \, b^{2}} + \frac {\sqrt {b x^{2} + a} C x^{7}}{8 \, b} + \frac {21 \, \sqrt {b x^{2} + a} D a^{2} x^{5}}{160 \, b^{3}} - \frac {7 \, \sqrt {b x^{2} + a} C a x^{5}}{48 \, b^{2}} + \frac {\sqrt {b x^{2} + a} B x^{5}}{6 \, b} - \frac {21 \, \sqrt {b x^{2} + a} D a^{3} x^{3}}{128 \, b^{4}} + \frac {35 \, \sqrt {b x^{2} + a} C a^{2} x^{3}}{192 \, b^{3}} - \frac {5 \, \sqrt {b x^{2} + a} B a x^{3}}{24 \, b^{2}} + \frac {\sqrt {b x^{2} + a} A x^{3}}{4 \, b} + \frac {63 \, \sqrt {b x^{2} + a} D a^{4} x}{256 \, b^{5}} - \frac {35 \, \sqrt {b x^{2} + a} C a^{3} x}{128 \, b^{4}} + \frac {5 \, \sqrt {b x^{2} + a} B a^{2} x}{16 \, b^{3}} - \frac {3 \, \sqrt {b x^{2} + a} A a x}{8 \, b^{2}} - \frac {63 \, D a^{5} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{256 \, b^{\frac {11}{2}}} + \frac {35 \, C a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {9}{2}}} - \frac {5 \, B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {7}{2}}} + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {5}{2}}} \] Input:
integrate(x^4*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="maxima")
Output:
1/10*sqrt(b*x^2 + a)*D*x^9/b - 9/80*sqrt(b*x^2 + a)*D*a*x^7/b^2 + 1/8*sqrt (b*x^2 + a)*C*x^7/b + 21/160*sqrt(b*x^2 + a)*D*a^2*x^5/b^3 - 7/48*sqrt(b*x ^2 + a)*C*a*x^5/b^2 + 1/6*sqrt(b*x^2 + a)*B*x^5/b - 21/128*sqrt(b*x^2 + a) *D*a^3*x^3/b^4 + 35/192*sqrt(b*x^2 + a)*C*a^2*x^3/b^3 - 5/24*sqrt(b*x^2 + a)*B*a*x^3/b^2 + 1/4*sqrt(b*x^2 + a)*A*x^3/b + 63/256*sqrt(b*x^2 + a)*D*a^ 4*x/b^5 - 35/128*sqrt(b*x^2 + a)*C*a^3*x/b^4 + 5/16*sqrt(b*x^2 + a)*B*a^2* x/b^3 - 3/8*sqrt(b*x^2 + a)*A*a*x/b^2 - 63/256*D*a^5*arcsinh(b*x/sqrt(a*b) )/b^(11/2) + 35/128*C*a^4*arcsinh(b*x/sqrt(a*b))/b^(9/2) - 5/16*B*a^3*arcs inh(b*x/sqrt(a*b))/b^(7/2) + 3/8*A*a^2*arcsinh(b*x/sqrt(a*b))/b^(5/2)
Time = 0.13 (sec) , antiderivative size = 219, normalized size of antiderivative = 0.88 \[ \int \frac {x^4 \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\frac {1}{3840} \, {\left (2 \, {\left (4 \, {\left (6 \, {\left (\frac {8 \, D x^{2}}{b} - \frac {9 \, D a b^{7} - 10 \, C b^{8}}{b^{9}}\right )} x^{2} + \frac {63 \, D a^{2} b^{6} - 70 \, C a b^{7} + 80 \, B b^{8}}{b^{9}}\right )} x^{2} - \frac {5 \, {\left (63 \, D a^{3} b^{5} - 70 \, C a^{2} b^{6} + 80 \, B a b^{7} - 96 \, A b^{8}\right )}}{b^{9}}\right )} x^{2} + \frac {15 \, {\left (63 \, D a^{4} b^{4} - 70 \, C a^{3} b^{5} + 80 \, B a^{2} b^{6} - 96 \, A a b^{7}\right )}}{b^{9}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (63 \, D a^{5} - 70 \, C a^{4} b + 80 \, B a^{3} b^{2} - 96 \, A a^{2} b^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{256 \, b^{\frac {11}{2}}} \] Input:
integrate(x^4*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x, algorithm="giac")
Output:
1/3840*(2*(4*(6*(8*D*x^2/b - (9*D*a*b^7 - 10*C*b^8)/b^9)*x^2 + (63*D*a^2*b ^6 - 70*C*a*b^7 + 80*B*b^8)/b^9)*x^2 - 5*(63*D*a^3*b^5 - 70*C*a^2*b^6 + 80 *B*a*b^7 - 96*A*b^8)/b^9)*x^2 + 15*(63*D*a^4*b^4 - 70*C*a^3*b^5 + 80*B*a^2 *b^6 - 96*A*a*b^7)/b^9)*sqrt(b*x^2 + a)*x + 1/256*(63*D*a^5 - 70*C*a^4*b + 80*B*a^3*b^2 - 96*A*a^2*b^3)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(11 /2)
Timed out. \[ \int \frac {x^4 \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\int \frac {x^4\,\left (A+B\,x^2+C\,x^4+x^6\,D\right )}{\sqrt {b\,x^2+a}} \,d x \] Input:
int((x^4*(A + B*x^2 + C*x^4 + x^6*D))/(a + b*x^2)^(1/2),x)
Output:
int((x^4*(A + B*x^2 + C*x^4 + x^6*D))/(a + b*x^2)^(1/2), x)
Time = 0.18 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.23 \[ \int \frac {x^4 \left (A+B x^2+C x^4+D x^6\right )}{\sqrt {a+b x^2}} \, dx=\frac {945 \sqrt {b \,x^{2}+a}\, a^{4} b d x -1050 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} c x -630 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} d \,x^{3}-240 \sqrt {b \,x^{2}+a}\, a^{2} b^{4} x +700 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} c \,x^{3}+504 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} d \,x^{5}+160 \sqrt {b \,x^{2}+a}\, a \,b^{5} x^{3}-560 \sqrt {b \,x^{2}+a}\, a \,b^{4} c \,x^{5}-432 \sqrt {b \,x^{2}+a}\, a \,b^{4} d \,x^{7}+640 \sqrt {b \,x^{2}+a}\, b^{6} x^{5}+480 \sqrt {b \,x^{2}+a}\, b^{5} c \,x^{7}+384 \sqrt {b \,x^{2}+a}\, b^{5} d \,x^{9}-945 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{5} d +1050 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4} b c +240 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b^{3}}{3840 b^{6}} \] Input:
int(x^4*(D*x^6+C*x^4+B*x^2+A)/(b*x^2+a)^(1/2),x)
Output:
(945*sqrt(a + b*x**2)*a**4*b*d*x - 1050*sqrt(a + b*x**2)*a**3*b**2*c*x - 6 30*sqrt(a + b*x**2)*a**3*b**2*d*x**3 - 240*sqrt(a + b*x**2)*a**2*b**4*x + 700*sqrt(a + b*x**2)*a**2*b**3*c*x**3 + 504*sqrt(a + b*x**2)*a**2*b**3*d*x **5 + 160*sqrt(a + b*x**2)*a*b**5*x**3 - 560*sqrt(a + b*x**2)*a*b**4*c*x** 5 - 432*sqrt(a + b*x**2)*a*b**4*d*x**7 + 640*sqrt(a + b*x**2)*b**6*x**5 + 480*sqrt(a + b*x**2)*b**5*c*x**7 + 384*sqrt(a + b*x**2)*b**5*d*x**9 - 945* sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**5*d + 1050*sqrt(b)* log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**4*b*c + 240*sqrt(b)*log((sq rt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b**3)/(3840*b**6)