Integrand size = 29, antiderivative size = 241 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {a \left (96 A b^3-a \left (16 b^2 B-6 a b C+3 a^2 D\right )\right ) x \sqrt {a+b x^2}}{256 b^3}+\frac {1}{384} \left (96 A-\frac {a \left (16 b^2 B-6 a b C+3 a^2 D\right )}{b^3}\right ) x \left (a+b x^2\right )^{3/2}+\frac {\left (16 b^2 B-6 a b C+3 a^2 D\right ) x \left (a+b x^2\right )^{5/2}}{96 b^3}+\frac {(2 b C-a D) x^3 \left (a+b x^2\right )^{5/2}}{16 b^2}+\frac {D x^5 \left (a+b x^2\right )^{5/2}}{10 b}+\frac {a^2 \left (96 A b^3-a \left (16 b^2 B-6 a b C+3 a^2 D\right )\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{7/2}} \] Output:
1/256*a*(96*A*b^3-a*(16*B*b^2-6*C*a*b+3*D*a^2))*x*(b*x^2+a)^(1/2)/b^3+1/38 4*(96*A-a*(16*B*b^2-6*C*a*b+3*D*a^2)/b^3)*x*(b*x^2+a)^(3/2)+1/96*(16*B*b^2 -6*C*a*b+3*D*a^2)*x*(b*x^2+a)^(5/2)/b^3+1/16*(2*C*b-D*a)*x^3*(b*x^2+a)^(5/ 2)/b^2+1/10*D*x^5*(b*x^2+a)^(5/2)/b+1/256*a^2*(96*A*b^3-a*(16*B*b^2-6*C*a* b+3*D*a^2))*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^(7/2)
Time = 0.34 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.77 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {\sqrt {b} x \sqrt {a+b x^2} \left (45 a^4 D-30 a^3 b \left (3 C+D x^2\right )+12 a^2 b^2 \left (20 B+5 C x^2+2 D x^4\right )+32 b^4 x^2 \left (30 A+20 B x^2+15 C x^4+12 D x^6\right )+16 a b^3 \left (150 A+70 B x^2+45 C x^4+33 D x^6\right )\right )+15 a^2 \left (-96 A b^3+a \left (16 b^2 B-6 a b C+3 a^2 D\right )\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{3840 b^{7/2}} \] Input:
Integrate[(a + b*x^2)^(3/2)*(A + B*x^2 + C*x^4 + D*x^6),x]
Output:
(Sqrt[b]*x*Sqrt[a + b*x^2]*(45*a^4*D - 30*a^3*b*(3*C + D*x^2) + 12*a^2*b^2 *(20*B + 5*C*x^2 + 2*D*x^4) + 32*b^4*x^2*(30*A + 20*B*x^2 + 15*C*x^4 + 12* D*x^6) + 16*a*b^3*(150*A + 70*B*x^2 + 45*C*x^4 + 33*D*x^6)) + 15*a^2*(-96* A*b^3 + a*(16*b^2*B - 6*a*b*C + 3*a^2*D))*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^ 2]])/(3840*b^(7/2))
Time = 0.36 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.87, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {2346, 27, 1473, 299, 211, 211, 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.
| \(\displaystyle \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right ) \, dx\) |
| \(\Big \downarrow \) 2346 |
| \(\displaystyle \frac {\int 5 \left (b x^2+a\right )^{3/2} \left ((2 b C-a D) x^4+2 b B x^2+2 A b\right )dx}{10 b}+\frac {D x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\int \left (b x^2+a\right )^{3/2} \left ((2 b C-a D) x^4+2 b B x^2+2 A b\right )dx}{2 b}+\frac {D x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
| \(\Big \downarrow \) 1473 |
| \(\displaystyle \frac {\frac {\int \left (b x^2+a\right )^{3/2} \left (16 A b^2+\left (3 D a^2-6 b C a+16 b^2 B\right ) x^2\right )dx}{8 b}+\frac {x^3 \left (a+b x^2\right )^{5/2} (2 b C-a D)}{8 b}}{2 b}+\frac {D x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {\frac {\left (96 A b^3-a \left (3 a^2 D-6 a b C+16 b^2 B\right )\right ) \int \left (b x^2+a\right )^{3/2}dx}{6 b}+\frac {x \left (a+b x^2\right )^{5/2} \left (3 a^2 D-6 a b C+16 b^2 B\right )}{6 b}}{8 b}+\frac {x^3 \left (a+b x^2\right )^{5/2} (2 b C-a D)}{8 b}}{2 b}+\frac {D x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\frac {\left (96 A b^3-a \left (3 a^2 D-6 a b C+16 b^2 B\right )\right ) \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {x \left (a+b x^2\right )^{5/2} \left (3 a^2 D-6 a b C+16 b^2 B\right )}{6 b}}{8 b}+\frac {x^3 \left (a+b x^2\right )^{5/2} (2 b C-a D)}{8 b}}{2 b}+\frac {D x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\frac {\left (96 A b^3-a \left (3 a^2 D-6 a b C+16 b^2 B\right )\right ) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {x \left (a+b x^2\right )^{5/2} \left (3 a^2 D-6 a b C+16 b^2 B\right )}{6 b}}{8 b}+\frac {x^3 \left (a+b x^2\right )^{5/2} (2 b C-a D)}{8 b}}{2 b}+\frac {D x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\frac {\left (96 A b^3-a \left (3 a^2 D-6 a b C+16 b^2 B\right )\right ) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {x \left (a+b x^2\right )^{5/2} \left (3 a^2 D-6 a b C+16 b^2 B\right )}{6 b}}{8 b}+\frac {x^3 \left (a+b x^2\right )^{5/2} (2 b C-a D)}{8 b}}{2 b}+\frac {D x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {\left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right ) \left (96 A b^3-a \left (3 a^2 D-6 a b C+16 b^2 B\right )\right )}{6 b}+\frac {x \left (a+b x^2\right )^{5/2} \left (3 a^2 D-6 a b C+16 b^2 B\right )}{6 b}}{8 b}+\frac {x^3 \left (a+b x^2\right )^{5/2} (2 b C-a D)}{8 b}}{2 b}+\frac {D x^5 \left (a+b x^2\right )^{5/2}}{10 b}\) |
Input:
Int[(a + b*x^2)^(3/2)*(A + B*x^2 + C*x^4 + D*x^6),x]
Output:
(D*x^5*(a + b*x^2)^(5/2))/(10*b) + (((2*b*C - a*D)*x^3*(a + b*x^2)^(5/2))/ (8*b) + (((16*b^2*B - 6*a*b*C + 3*a^2*D)*x*(a + b*x^2)^(5/2))/(6*b) + ((96 *A*b^3 - a*(16*b^2*B - 6*a*b*C + 3*a^2*D))*((x*(a + b*x^2)^(3/2))/4 + (3*a *((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt [b])))/4))/(6*b))/(8*b))/(2*b)
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)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 )), x] + Simp[2*a*(p/(2*p + 1)) Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ {a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
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 = 0.59 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.71
| method | result | size |
| pseudoelliptic | \(\frac {\frac {3 \left (b^{3} A -\frac {1}{6} a \,b^{2} B +\frac {1}{16} a^{2} b C -\frac {1}{32} a^{3} D\right ) a^{2} \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{8}+\frac {5 \sqrt {b \,x^{2}+a}\, x \left (a \left (\frac {11}{50} D x^{6}+\frac {3}{10} C \,x^{4}+\frac {7}{15} x^{2} B +A \right ) b^{\frac {7}{2}}+\frac {2 \left (\frac {2}{5} D x^{6}+\frac {1}{2} C \,x^{4}+\frac {2}{3} x^{2} B +A \right ) x^{2} b^{\frac {9}{2}}}{5}+\frac {3 \left (\frac {4 \left (\frac {2}{5} D x^{4}+C \,x^{2}+4 B \right ) b^{\frac {5}{2}}}{3}+\left (2 \left (-\frac {D x^{2}}{3}-C \right ) b^{\frac {3}{2}}+D a \sqrt {b}\right ) a \right ) a^{2}}{160}\right )}{8}}{b^{\frac {7}{2}}}\) | \(170\) |
| default | \(A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+C \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )+D \left (\frac {x^{5} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{10 b}-\frac {a \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )}{2 b}\right )+B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )\) | \(352\) |
Input:
int((b*x^2+a)^(3/2)*(D*x^6+C*x^4+B*x^2+A),x,method=_RETURNVERBOSE)
Output:
5/8/b^(7/2)*(3/5*(b^3*A-1/6*a*b^2*B+1/16*a^2*b*C-1/32*a^3*D)*a^2*arctanh(( b*x^2+a)^(1/2)/x/b^(1/2))+(b*x^2+a)^(1/2)*x*(a*(11/50*D*x^6+3/10*C*x^4+7/1 5*x^2*B+A)*b^(7/2)+2/5*(2/5*D*x^6+1/2*C*x^4+2/3*x^2*B+A)*x^2*b^(9/2)+3/160 *(4/3*(2/5*D*x^4+C*x^2+4*B)*b^(5/2)+(2*(-1/3*D*x^2-C)*b^(3/2)+D*a*b^(1/2)) *a)*a^2))
Time = 0.14 (sec) , antiderivative size = 413, normalized size of antiderivative = 1.71 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\left [-\frac {15 \, {\left (3 \, D a^{5} - 6 \, C a^{4} b + 16 \, B a^{3} b^{2} - 96 \, A a^{2} b^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (384 \, D b^{5} x^{9} + 48 \, {\left (11 \, D a b^{4} + 10 \, C b^{5}\right )} x^{7} + 8 \, {\left (3 \, D a^{2} b^{3} + 90 \, C a b^{4} + 80 \, B b^{5}\right )} x^{5} - 10 \, {\left (3 \, D a^{3} b^{2} - 6 \, C a^{2} b^{3} - 112 \, B a b^{4} - 96 \, A b^{5}\right )} x^{3} + 15 \, {\left (3 \, D a^{4} b - 6 \, C a^{3} b^{2} + 16 \, B a^{2} b^{3} + 160 \, A a b^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{7680 \, b^{4}}, \frac {15 \, {\left (3 \, D a^{5} - 6 \, C a^{4} b + 16 \, B a^{3} b^{2} - 96 \, A a^{2} b^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (384 \, D b^{5} x^{9} + 48 \, {\left (11 \, D a b^{4} + 10 \, C b^{5}\right )} x^{7} + 8 \, {\left (3 \, D a^{2} b^{3} + 90 \, C a b^{4} + 80 \, B b^{5}\right )} x^{5} - 10 \, {\left (3 \, D a^{3} b^{2} - 6 \, C a^{2} b^{3} - 112 \, B a b^{4} - 96 \, A b^{5}\right )} x^{3} + 15 \, {\left (3 \, D a^{4} b - 6 \, C a^{3} b^{2} + 16 \, B a^{2} b^{3} + 160 \, A a b^{4}\right )} x\right )} \sqrt {b x^{2} + a}}{3840 \, b^{4}}\right ] \] Input:
integrate((b*x^2+a)^(3/2)*(D*x^6+C*x^4+B*x^2+A),x, algorithm="fricas")
Output:
[-1/7680*(15*(3*D*a^5 - 6*C*a^4*b + 16*B*a^3*b^2 - 96*A*a^2*b^3)*sqrt(b)*l og(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(384*D*b^5*x^9 + 48*(11 *D*a*b^4 + 10*C*b^5)*x^7 + 8*(3*D*a^2*b^3 + 90*C*a*b^4 + 80*B*b^5)*x^5 - 1 0*(3*D*a^3*b^2 - 6*C*a^2*b^3 - 112*B*a*b^4 - 96*A*b^5)*x^3 + 15*(3*D*a^4*b - 6*C*a^3*b^2 + 16*B*a^2*b^3 + 160*A*a*b^4)*x)*sqrt(b*x^2 + a))/b^4, 1/38 40*(15*(3*D*a^5 - 6*C*a^4*b + 16*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*(11*D*a*b^4 + 10*C*b^5) *x^7 + 8*(3*D*a^2*b^3 + 90*C*a*b^4 + 80*B*b^5)*x^5 - 10*(3*D*a^3*b^2 - 6*C *a^2*b^3 - 112*B*a*b^4 - 96*A*b^5)*x^3 + 15*(3*D*a^4*b - 6*C*a^3*b^2 + 16* B*a^2*b^3 + 160*A*a*b^4)*x)*sqrt(b*x^2 + a))/b^4]
Time = 0.52 (sec) , antiderivative size = 396, normalized size of antiderivative = 1.64 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {D b x^{9}}{10} + \frac {x^{7} \left (C b^{2} + \frac {11 D a b}{10}\right )}{8 b} + \frac {x^{5} \left (B b^{2} + 2 C a b + D a^{2} - \frac {7 a \left (C b^{2} + \frac {11 D a b}{10}\right )}{8 b}\right )}{6 b} + \frac {x^{3} \left (A b^{2} + 2 B a b + C a^{2} - \frac {5 a \left (B b^{2} + 2 C a b + D a^{2} - \frac {7 a \left (C b^{2} + \frac {11 D a b}{10}\right )}{8 b}\right )}{6 b}\right )}{4 b} + \frac {x \left (2 A a b + B a^{2} - \frac {3 a \left (A b^{2} + 2 B a b + C a^{2} - \frac {5 a \left (B b^{2} + 2 C a b + D a^{2} - \frac {7 a \left (C b^{2} + \frac {11 D a b}{10}\right )}{8 b}\right )}{6 b}\right )}{4 b}\right )}{2 b}\right ) + \left (A a^{2} - \frac {a \left (2 A a b + B a^{2} - \frac {3 a \left (A b^{2} + 2 B a b + C a^{2} - \frac {5 a \left (B b^{2} + 2 C a b + D a^{2} - \frac {7 a \left (C b^{2} + \frac {11 D a b}{10}\right )}{8 b}\right )}{6 b}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (A x + \frac {B x^{3}}{3} + \frac {C x^{5}}{5} + \frac {D x^{7}}{7}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((b*x**2+a)**(3/2)*(D*x**6+C*x**4+B*x**2+A),x)
Output:
Piecewise((sqrt(a + b*x**2)*(D*b*x**9/10 + x**7*(C*b**2 + 11*D*a*b/10)/(8* b) + x**5*(B*b**2 + 2*C*a*b + D*a**2 - 7*a*(C*b**2 + 11*D*a*b/10)/(8*b))/( 6*b) + x**3*(A*b**2 + 2*B*a*b + C*a**2 - 5*a*(B*b**2 + 2*C*a*b + D*a**2 - 7*a*(C*b**2 + 11*D*a*b/10)/(8*b))/(6*b))/(4*b) + x*(2*A*a*b + B*a**2 - 3*a *(A*b**2 + 2*B*a*b + C*a**2 - 5*a*(B*b**2 + 2*C*a*b + D*a**2 - 7*a*(C*b**2 + 11*D*a*b/10)/(8*b))/(6*b))/(4*b))/(2*b)) + (A*a**2 - a*(2*A*a*b + B*a** 2 - 3*a*(A*b**2 + 2*B*a*b + C*a**2 - 5*a*(B*b**2 + 2*C*a*b + D*a**2 - 7*a* (C*b**2 + 11*D*a*b/10)/(8*b))/(6*b))/(4*b))/(2*b))*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), Tr ue)), Ne(b, 0)), (a**(3/2)*(A*x + B*x**3/3 + C*x**5/5 + D*x**7/7), True))
Time = 0.04 (sec) , antiderivative size = 319, normalized size of antiderivative = 1.32 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} D x^{5}}{10 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} D a x^{3}}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} C x^{3}}{8 \, b} + \frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A x + \frac {3}{8} \, \sqrt {b x^{2} + a} A a x + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} D a^{2} x}{32 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} D a^{3} x}{128 \, b^{3}} - \frac {3 \, \sqrt {b x^{2} + a} D a^{4} x}{256 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} C a x}{16 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} C a^{2} x}{64 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} C a^{3} x}{128 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x}{24 \, b} - \frac {\sqrt {b x^{2} + a} B a^{2} x}{16 \, b} - \frac {3 \, D a^{5} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{256 \, b^{\frac {7}{2}}} + \frac {3 \, C a^{4} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} - \frac {B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} \] Input:
integrate((b*x^2+a)^(3/2)*(D*x^6+C*x^4+B*x^2+A),x, algorithm="maxima")
Output:
1/10*(b*x^2 + a)^(5/2)*D*x^5/b - 1/16*(b*x^2 + a)^(5/2)*D*a*x^3/b^2 + 1/8* (b*x^2 + a)^(5/2)*C*x^3/b + 1/4*(b*x^2 + a)^(3/2)*A*x + 3/8*sqrt(b*x^2 + a )*A*a*x + 1/32*(b*x^2 + a)^(5/2)*D*a^2*x/b^3 - 1/128*(b*x^2 + a)^(3/2)*D*a ^3*x/b^3 - 3/256*sqrt(b*x^2 + a)*D*a^4*x/b^3 - 1/16*(b*x^2 + a)^(5/2)*C*a* x/b^2 + 1/64*(b*x^2 + a)^(3/2)*C*a^2*x/b^2 + 3/128*sqrt(b*x^2 + a)*C*a^3*x /b^2 + 1/6*(b*x^2 + a)^(5/2)*B*x/b - 1/24*(b*x^2 + a)^(3/2)*B*a*x/b - 1/16 *sqrt(b*x^2 + a)*B*a^2*x/b - 3/256*D*a^5*arcsinh(b*x/sqrt(a*b))/b^(7/2) + 3/128*C*a^4*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 1/16*B*a^3*arcsinh(b*x/sqrt(a *b))/b^(3/2) + 3/8*A*a^2*arcsinh(b*x/sqrt(a*b))/sqrt(b)
Time = 0.14 (sec) , antiderivative size = 216, normalized size of antiderivative = 0.90 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {1}{3840} \, {\left (2 \, {\left (4 \, {\left (6 \, {\left (8 \, D b x^{2} + \frac {11 \, D a b^{8} + 10 \, C b^{9}}{b^{8}}\right )} x^{2} + \frac {3 \, D a^{2} b^{7} + 90 \, C a b^{8} + 80 \, B b^{9}}{b^{8}}\right )} x^{2} - \frac {5 \, {\left (3 \, D a^{3} b^{6} - 6 \, C a^{2} b^{7} - 112 \, B a b^{8} - 96 \, A b^{9}\right )}}{b^{8}}\right )} x^{2} + \frac {15 \, {\left (3 \, D a^{4} b^{5} - 6 \, C a^{3} b^{6} + 16 \, B a^{2} b^{7} + 160 \, A a b^{8}\right )}}{b^{8}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (3 \, D a^{5} - 6 \, C a^{4} b + 16 \, B a^{3} b^{2} - 96 \, A a^{2} b^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{256 \, b^{\frac {7}{2}}} \] Input:
integrate((b*x^2+a)^(3/2)*(D*x^6+C*x^4+B*x^2+A),x, algorithm="giac")
Output:
1/3840*(2*(4*(6*(8*D*b*x^2 + (11*D*a*b^8 + 10*C*b^9)/b^8)*x^2 + (3*D*a^2*b ^7 + 90*C*a*b^8 + 80*B*b^9)/b^8)*x^2 - 5*(3*D*a^3*b^6 - 6*C*a^2*b^7 - 112* B*a*b^8 - 96*A*b^9)/b^8)*x^2 + 15*(3*D*a^4*b^5 - 6*C*a^3*b^6 + 16*B*a^2*b^ 7 + 160*A*a*b^8)/b^8)*sqrt(b*x^2 + a)*x + 1/256*(3*D*a^5 - 6*C*a^4*b + 16* B*a^3*b^2 - 96*A*a^2*b^3)*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
Timed out. \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\int {\left (b\,x^2+a\right )}^{3/2}\,\left (A+B\,x^2+C\,x^4+x^6\,D\right ) \,d x \] Input:
int((a + b*x^2)^(3/2)*(A + B*x^2 + C*x^4 + x^6*D),x)
Output:
int((a + b*x^2)^(3/2)*(A + B*x^2 + C*x^4 + x^6*D), x)
Time = 0.17 (sec) , antiderivative size = 304, normalized size of antiderivative = 1.26 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right ) \, dx=\frac {45 \sqrt {b \,x^{2}+a}\, a^{4} b d x -90 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} c x -30 \sqrt {b \,x^{2}+a}\, a^{3} b^{2} d \,x^{3}+2640 \sqrt {b \,x^{2}+a}\, a^{2} b^{4} x +60 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} c \,x^{3}+24 \sqrt {b \,x^{2}+a}\, a^{2} b^{3} d \,x^{5}+2080 \sqrt {b \,x^{2}+a}\, a \,b^{5} x^{3}+720 \sqrt {b \,x^{2}+a}\, a \,b^{4} c \,x^{5}+528 \sqrt {b \,x^{2}+a}\, a \,b^{4} d \,x^{7}+640 \sqrt {b \,x^{2}+a}\, b^{6} x^{5}+480 \sqrt {b \,x^{2}+a}\, b^{5} c \,x^{7}+384 \sqrt {b \,x^{2}+a}\, b^{5} d \,x^{9}-45 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{5} d +90 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{4} b c +1200 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} b^{3}}{3840 b^{4}} \] Input:
int((b*x^2+a)^(3/2)*(D*x^6+C*x^4+B*x^2+A),x)
Output:
(45*sqrt(a + b*x**2)*a**4*b*d*x - 90*sqrt(a + b*x**2)*a**3*b**2*c*x - 30*s qrt(a + b*x**2)*a**3*b**2*d*x**3 + 2640*sqrt(a + b*x**2)*a**2*b**4*x + 60* sqrt(a + b*x**2)*a**2*b**3*c*x**3 + 24*sqrt(a + b*x**2)*a**2*b**3*d*x**5 + 2080*sqrt(a + b*x**2)*a*b**5*x**3 + 720*sqrt(a + b*x**2)*a*b**4*c*x**5 + 528*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 - 45*sqrt( b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**5*d + 90*sqrt(b)*log((sq rt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**4*b*c + 1200*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*b**3)/(3840*b**4)