Integrand size = 21, antiderivative size = 272 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3 \, dx=\frac {3 a (4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right ) x \sqrt {a+b x^2}}{256 b^3}+\frac {(4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right ) x \left (a+b x^2\right )^{3/2}}{128 b^3}+\frac {d \left (36 b^2 c^2-20 a b c d+5 a^2 d^2\right ) x \left (a+b x^2\right )^{5/2}}{160 b^3}+\frac {d (14 b c-5 a d) x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )}{80 b^2}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}+\frac {3 a^2 (4 b c-a d) \left (8 b^2 c^2-2 a b c d+a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{7/2}} \]
1/128*(-a*d+4*b*c)*(a^2*d^2-2*a*b*c*d+8*b^2*c^2)*x*(b*x^2+a)^(3/2)/b^3+1/1 60*d*(5*a^2*d^2-20*a*b*c*d+36*b^2*c^2)*x*(b*x^2+a)^(5/2)/b^3+1/80*d*(-5*a* d+14*b*c)*x*(b*x^2+a)^(5/2)*(d*x^2+c)/b^2+1/10*d*x*(b*x^2+a)^(5/2)*(d*x^2+ c)^2/b+3/256*a^2*(-a*d+4*b*c)*(a^2*d^2-2*a*b*c*d+8*b^2*c^2)*arctanh(x*b^(1 /2)/(b*x^2+a)^(1/2))/b^(7/2)+3/256*a*(-a*d+4*b*c)*(a^2*d^2-2*a*b*c*d+8*b^2 *c^2)*x*(b*x^2+a)^(1/2)/b^3
Time = 0.35 (sec) , antiderivative size = 225, normalized size of antiderivative = 0.83 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3 \, dx=\frac {\sqrt {b} x \sqrt {a+b x^2} \left (15 a^4 d^3-10 a^3 b d^2 \left (9 c+d x^2\right )+4 a^2 b^2 d \left (60 c^2+15 c d x^2+2 d^2 x^4\right )+32 b^4 x^2 \left (10 c^3+20 c^2 d x^2+15 c d^2 x^4+4 d^3 x^6\right )+16 a b^3 \left (50 c^3+70 c^2 d x^2+45 c d^2 x^4+11 d^3 x^6\right )\right )+15 a^2 \left (-32 b^3 c^3+16 a b^2 c^2 d-6 a^2 b c d^2+a^3 d^3\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{1280 b^{7/2}} \]
(Sqrt[b]*x*Sqrt[a + b*x^2]*(15*a^4*d^3 - 10*a^3*b*d^2*(9*c + d*x^2) + 4*a^ 2*b^2*d*(60*c^2 + 15*c*d*x^2 + 2*d^2*x^4) + 32*b^4*x^2*(10*c^3 + 20*c^2*d* x^2 + 15*c*d^2*x^4 + 4*d^3*x^6) + 16*a*b^3*(50*c^3 + 70*c^2*d*x^2 + 45*c*d ^2*x^4 + 11*d^3*x^6)) + 15*a^2*(-32*b^3*c^3 + 16*a*b^2*c^2*d - 6*a^2*b*c*d ^2 + a^3*d^3)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(1280*b^(7/2))
Time = 0.35 (sec) , antiderivative size = 231, normalized size of antiderivative = 0.85, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {318, 403, 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.
| \(\displaystyle \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3 \, dx\) |
| \(\Big \downarrow \) 318 |
| \(\displaystyle \frac {\int \left (b x^2+a\right )^{3/2} \left (d x^2+c\right ) \left (d (14 b c-5 a d) x^2+c (10 b c-a d)\right )dx}{10 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}\) |
| \(\Big \downarrow \) 403 |
| \(\displaystyle \frac {\frac {\int \left (b x^2+a\right )^{3/2} \left (3 d \left (36 b^2 c^2-20 a b d c+5 a^2 d^2\right ) x^2+c \left (80 b^2 c^2-22 a b d c+5 a^2 d^2\right )\right )dx}{8 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right ) (14 b c-5 a d)}{8 b}}{10 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {\frac {\frac {5 (4 b c-a d) \left (a^2 d^2-2 a b c d+8 b^2 c^2\right ) \int \left (b x^2+a\right )^{3/2}dx}{2 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (5 a^2 d^2-20 a b c d+36 b^2 c^2\right )}{2 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right ) (14 b c-5 a d)}{8 b}}{10 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\frac {5 (4 b c-a d) \left (a^2 d^2-2 a b c d+8 b^2 c^2\right ) \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{2 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (5 a^2 d^2-20 a b c d+36 b^2 c^2\right )}{2 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right ) (14 b c-5 a d)}{8 b}}{10 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\frac {\frac {5 (4 b c-a d) \left (a^2 d^2-2 a b c d+8 b^2 c^2\right ) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{2 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (5 a^2 d^2-20 a b c d+36 b^2 c^2\right )}{2 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right ) (14 b c-5 a d)}{8 b}}{10 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\frac {5 (4 b c-a d) \left (a^2 d^2-2 a b c d+8 b^2 c^2\right ) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{2 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (5 a^2 d^2-20 a b c d+36 b^2 c^2\right )}{2 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right ) (14 b c-5 a d)}{8 b}}{10 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\frac {5 \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right ) (4 b c-a d) \left (a^2 d^2-2 a b c d+8 b^2 c^2\right )}{2 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (5 a^2 d^2-20 a b c d+36 b^2 c^2\right )}{2 b}}{8 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right ) (14 b c-5 a d)}{8 b}}{10 b}+\frac {d x \left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2}{10 b}\) |
(d*x*(a + b*x^2)^(5/2)*(c + d*x^2)^2)/(10*b) + ((d*(14*b*c - 5*a*d)*x*(a + b*x^2)^(5/2)*(c + d*x^2))/(8*b) + ((d*(36*b^2*c^2 - 20*a*b*c*d + 5*a^2*d^ 2)*x*(a + b*x^2)^(5/2))/(2*b) + (5*(4*b*c - a*d)*(8*b^2*c^2 - 2*a*b*c*d + a^2*d^2)*((x*(a + b*x^2)^(3/2))/4 + (3*a*((x*Sqrt[a + b*x^2])/2 + (a*ArcTa nh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b])))/4))/(2*b))/(8*b))/(10*b)
3.1.53.3.1 Defintions of rubi rules used
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
Time = 2.48 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.76
| method | result | size |
| pseudoelliptic | \(-\frac {3 \left (\left (a^{5} d^{3}-6 a^{4} b c \,d^{2}+16 a^{3} b^{2} c^{2} d -32 a^{2} b^{3} c^{3}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-x \left (\frac {160 \left (\frac {11}{50} d^{3} x^{6}+\frac {9}{10} c \,d^{2} x^{4}+\frac {7}{5} c^{2} d \,x^{2}+c^{3}\right ) a \,b^{\frac {7}{2}}}{3}+\frac {64 x^{2} \left (\frac {2}{5} d^{3} x^{6}+\frac {3}{2} c \,d^{2} x^{4}+2 c^{2} d \,x^{2}+c^{3}\right ) b^{\frac {9}{2}}}{3}+\left (\left (\frac {8}{15} d^{2} x^{4}+4 c d \,x^{2}+16 c^{2}\right ) b^{\frac {5}{2}}+\left (\left (-\frac {2 d \,x^{2}}{3}-6 c \right ) b^{\frac {3}{2}}+a d \sqrt {b}\right ) d a \right ) d \,a^{2}\right ) \sqrt {b \,x^{2}+a}\right )}{256 b^{\frac {7}{2}}}\) | \(208\) |
| risch | \(\frac {x \left (128 b^{4} d^{3} x^{8}+176 a \,b^{3} d^{3} x^{6}+480 b^{4} c \,d^{2} x^{6}+8 a^{2} b^{2} d^{3} x^{4}+720 a \,b^{3} c \,d^{2} x^{4}+640 b^{4} c^{2} d \,x^{4}-10 a^{3} b \,d^{3} x^{2}+60 a^{2} b^{2} c \,d^{2} x^{2}+1120 a \,b^{3} c^{2} d \,x^{2}+320 b^{4} c^{3} x^{2}+15 a^{4} d^{3}-90 a^{3} b c \,d^{2}+240 a^{2} b^{2} c^{2} d +800 a \,b^{3} c^{3}\right ) \sqrt {b \,x^{2}+a}}{1280 b^{3}}-\frac {3 a^{2} \left (a^{3} d^{3}-6 a^{2} b c \,d^{2}+16 a \,b^{2} c^{2} d -32 b^{3} c^{3}\right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{256 b^{\frac {7}{2}}}\) | \(242\) |
| default | \(c^{3} \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+d^{3} \left (\frac {x^{5} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{10 b}-\frac {a \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )}{2 b}\right )+3 c \,d^{2} \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{8 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )}{8 b}\right )+3 c^{2} d \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )\) | \(364\) |
-3/256/b^(7/2)*((a^5*d^3-6*a^4*b*c*d^2+16*a^3*b^2*c^2*d-32*a^2*b^3*c^3)*ar ctanh((b*x^2+a)^(1/2)/x/b^(1/2))-x*(160/3*(11/50*d^3*x^6+9/10*c*d^2*x^4+7/ 5*c^2*d*x^2+c^3)*a*b^(7/2)+64/3*x^2*(2/5*d^3*x^6+3/2*c*d^2*x^4+2*c^2*d*x^2 +c^3)*b^(9/2)+((8/15*d^2*x^4+4*c*d*x^2+16*c^2)*b^(5/2)+((-2/3*d*x^2-6*c)*b ^(3/2)+a*d*b^(1/2))*d*a)*d*a^2)*(b*x^2+a)^(1/2))
Time = 0.42 (sec) , antiderivative size = 502, normalized size of antiderivative = 1.85 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3 \, dx=\left [-\frac {15 \, {\left (32 \, a^{2} b^{3} c^{3} - 16 \, a^{3} b^{2} c^{2} d + 6 \, a^{4} b c d^{2} - a^{5} d^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (128 \, b^{5} d^{3} x^{9} + 16 \, {\left (30 \, b^{5} c d^{2} + 11 \, a b^{4} d^{3}\right )} x^{7} + 8 \, {\left (80 \, b^{5} c^{2} d + 90 \, a b^{4} c d^{2} + a^{2} b^{3} d^{3}\right )} x^{5} + 10 \, {\left (32 \, b^{5} c^{3} + 112 \, a b^{4} c^{2} d + 6 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )} x^{3} + 5 \, {\left (160 \, a b^{4} c^{3} + 48 \, a^{2} b^{3} c^{2} d - 18 \, a^{3} b^{2} c d^{2} + 3 \, a^{4} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{2560 \, b^{4}}, -\frac {15 \, {\left (32 \, a^{2} b^{3} c^{3} - 16 \, a^{3} b^{2} c^{2} d + 6 \, a^{4} b c d^{2} - a^{5} d^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (128 \, b^{5} d^{3} x^{9} + 16 \, {\left (30 \, b^{5} c d^{2} + 11 \, a b^{4} d^{3}\right )} x^{7} + 8 \, {\left (80 \, b^{5} c^{2} d + 90 \, a b^{4} c d^{2} + a^{2} b^{3} d^{3}\right )} x^{5} + 10 \, {\left (32 \, b^{5} c^{3} + 112 \, a b^{4} c^{2} d + 6 \, a^{2} b^{3} c d^{2} - a^{3} b^{2} d^{3}\right )} x^{3} + 5 \, {\left (160 \, a b^{4} c^{3} + 48 \, a^{2} b^{3} c^{2} d - 18 \, a^{3} b^{2} c d^{2} + 3 \, a^{4} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{1280 \, b^{4}}\right ] \]
[-1/2560*(15*(32*a^2*b^3*c^3 - 16*a^3*b^2*c^2*d + 6*a^4*b*c*d^2 - a^5*d^3) *sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(128*b^5*d^3* x^9 + 16*(30*b^5*c*d^2 + 11*a*b^4*d^3)*x^7 + 8*(80*b^5*c^2*d + 90*a*b^4*c* d^2 + a^2*b^3*d^3)*x^5 + 10*(32*b^5*c^3 + 112*a*b^4*c^2*d + 6*a^2*b^3*c*d^ 2 - a^3*b^2*d^3)*x^3 + 5*(160*a*b^4*c^3 + 48*a^2*b^3*c^2*d - 18*a^3*b^2*c* d^2 + 3*a^4*b*d^3)*x)*sqrt(b*x^2 + a))/b^4, -1/1280*(15*(32*a^2*b^3*c^3 - 16*a^3*b^2*c^2*d + 6*a^4*b*c*d^2 - a^5*d^3)*sqrt(-b)*arctan(sqrt(-b)*x/sqr t(b*x^2 + a)) - (128*b^5*d^3*x^9 + 16*(30*b^5*c*d^2 + 11*a*b^4*d^3)*x^7 + 8*(80*b^5*c^2*d + 90*a*b^4*c*d^2 + a^2*b^3*d^3)*x^5 + 10*(32*b^5*c^3 + 112 *a*b^4*c^2*d + 6*a^2*b^3*c*d^2 - a^3*b^2*d^3)*x^3 + 5*(160*a*b^4*c^3 + 48* a^2*b^3*c^2*d - 18*a^3*b^2*c*d^2 + 3*a^4*b*d^3)*x)*sqrt(b*x^2 + a))/b^4]
Time = 0.49 (sec) , antiderivative size = 529, normalized size of antiderivative = 1.94 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3 \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {b d^{3} x^{9}}{10} + \frac {x^{7} \cdot \left (\frac {11 a b d^{3}}{10} + 3 b^{2} c d^{2}\right )}{8 b} + \frac {x^{5} \left (a^{2} d^{3} + 6 a b c d^{2} - \frac {7 a \left (\frac {11 a b d^{3}}{10} + 3 b^{2} c d^{2}\right )}{8 b} + 3 b^{2} c^{2} d\right )}{6 b} + \frac {x^{3} \cdot \left (3 a^{2} c d^{2} + 6 a b c^{2} d - \frac {5 a \left (a^{2} d^{3} + 6 a b c d^{2} - \frac {7 a \left (\frac {11 a b d^{3}}{10} + 3 b^{2} c d^{2}\right )}{8 b} + 3 b^{2} c^{2} d\right )}{6 b} + b^{2} c^{3}\right )}{4 b} + \frac {x \left (3 a^{2} c^{2} d + 2 a b c^{3} - \frac {3 a \left (3 a^{2} c d^{2} + 6 a b c^{2} d - \frac {5 a \left (a^{2} d^{3} + 6 a b c d^{2} - \frac {7 a \left (\frac {11 a b d^{3}}{10} + 3 b^{2} c d^{2}\right )}{8 b} + 3 b^{2} c^{2} d\right )}{6 b} + b^{2} c^{3}\right )}{4 b}\right )}{2 b}\right ) + \left (a^{2} c^{3} - \frac {a \left (3 a^{2} c^{2} d + 2 a b c^{3} - \frac {3 a \left (3 a^{2} c d^{2} + 6 a b c^{2} d - \frac {5 a \left (a^{2} d^{3} + 6 a b c d^{2} - \frac {7 a \left (\frac {11 a b d^{3}}{10} + 3 b^{2} c d^{2}\right )}{8 b} + 3 b^{2} c^{2} d\right )}{6 b} + b^{2} c^{3}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (c^{3} x + c^{2} d x^{3} + \frac {3 c d^{2} x^{5}}{5} + \frac {d^{3} x^{7}}{7}\right ) & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a + b*x**2)*(b*d**3*x**9/10 + x**7*(11*a*b*d**3/10 + 3*b** 2*c*d**2)/(8*b) + x**5*(a**2*d**3 + 6*a*b*c*d**2 - 7*a*(11*a*b*d**3/10 + 3 *b**2*c*d**2)/(8*b) + 3*b**2*c**2*d)/(6*b) + x**3*(3*a**2*c*d**2 + 6*a*b*c **2*d - 5*a*(a**2*d**3 + 6*a*b*c*d**2 - 7*a*(11*a*b*d**3/10 + 3*b**2*c*d** 2)/(8*b) + 3*b**2*c**2*d)/(6*b) + b**2*c**3)/(4*b) + x*(3*a**2*c**2*d + 2* a*b*c**3 - 3*a*(3*a**2*c*d**2 + 6*a*b*c**2*d - 5*a*(a**2*d**3 + 6*a*b*c*d* *2 - 7*a*(11*a*b*d**3/10 + 3*b**2*c*d**2)/(8*b) + 3*b**2*c**2*d)/(6*b) + b **2*c**3)/(4*b))/(2*b)) + (a**2*c**3 - a*(3*a**2*c**2*d + 2*a*b*c**3 - 3*a *(3*a**2*c*d**2 + 6*a*b*c**2*d - 5*a*(a**2*d**3 + 6*a*b*c*d**2 - 7*a*(11*a *b*d**3/10 + 3*b**2*c*d**2)/(8*b) + 3*b**2*c**2*d)/(6*b) + b**2*c**3)/(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), True)), Ne(b, 0)), (a**(3/2)*(c**3*x + c**2 *d*x**3 + 3*c*d**2*x**5/5 + d**3*x**7/7), True))
Time = 0.20 (sec) , antiderivative size = 364, normalized size of antiderivative = 1.34 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3 \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} d^{3} x^{5}}{10 \, b} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} c d^{2} x^{3}}{8 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} a d^{3} x^{3}}{16 \, b^{2}} + \frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} c^{3} x + \frac {3}{8} \, \sqrt {b x^{2} + a} a c^{3} x + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} c^{2} d x}{2 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a c^{2} d x}{8 \, b} - \frac {3 \, \sqrt {b x^{2} + a} a^{2} c^{2} d x}{16 \, b} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} a c d^{2} x}{16 \, b^{2}} + \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} c d^{2} x}{64 \, b^{2}} + \frac {9 \, \sqrt {b x^{2} + a} a^{3} c d^{2} x}{128 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} a^{2} d^{3} x}{32 \, b^{3}} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{3} d^{3} x}{128 \, b^{3}} - \frac {3 \, \sqrt {b x^{2} + a} a^{4} d^{3} x}{256 \, b^{3}} + \frac {3 \, a^{2} c^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} - \frac {3 \, a^{3} c^{2} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} + \frac {9 \, a^{4} c d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{128 \, b^{\frac {5}{2}}} - \frac {3 \, a^{5} d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{256 \, b^{\frac {7}{2}}} \]
1/10*(b*x^2 + a)^(5/2)*d^3*x^5/b + 3/8*(b*x^2 + a)^(5/2)*c*d^2*x^3/b - 1/1 6*(b*x^2 + a)^(5/2)*a*d^3*x^3/b^2 + 1/4*(b*x^2 + a)^(3/2)*c^3*x + 3/8*sqrt (b*x^2 + a)*a*c^3*x + 1/2*(b*x^2 + a)^(5/2)*c^2*d*x/b - 1/8*(b*x^2 + a)^(3 /2)*a*c^2*d*x/b - 3/16*sqrt(b*x^2 + a)*a^2*c^2*d*x/b - 3/16*(b*x^2 + a)^(5 /2)*a*c*d^2*x/b^2 + 3/64*(b*x^2 + a)^(3/2)*a^2*c*d^2*x/b^2 + 9/128*sqrt(b* x^2 + a)*a^3*c*d^2*x/b^2 + 1/32*(b*x^2 + a)^(5/2)*a^2*d^3*x/b^3 - 1/128*(b *x^2 + a)^(3/2)*a^3*d^3*x/b^3 - 3/256*sqrt(b*x^2 + a)*a^4*d^3*x/b^3 + 3/8* a^2*c^3*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 3/16*a^3*c^2*d*arcsinh(b*x/sqrt(a *b))/b^(3/2) + 9/128*a^4*c*d^2*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 3/256*a^5* d^3*arcsinh(b*x/sqrt(a*b))/b^(7/2)
Time = 0.31 (sec) , antiderivative size = 260, normalized size of antiderivative = 0.96 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3 \, dx=\frac {1}{1280} \, {\left (2 \, {\left (4 \, {\left (2 \, {\left (8 \, b d^{3} x^{2} + \frac {30 \, b^{9} c d^{2} + 11 \, a b^{8} d^{3}}{b^{8}}\right )} x^{2} + \frac {80 \, b^{9} c^{2} d + 90 \, a b^{8} c d^{2} + a^{2} b^{7} d^{3}}{b^{8}}\right )} x^{2} + \frac {5 \, {\left (32 \, b^{9} c^{3} + 112 \, a b^{8} c^{2} d + 6 \, a^{2} b^{7} c d^{2} - a^{3} b^{6} d^{3}\right )}}{b^{8}}\right )} x^{2} + \frac {5 \, {\left (160 \, a b^{8} c^{3} + 48 \, a^{2} b^{7} c^{2} d - 18 \, a^{3} b^{6} c d^{2} + 3 \, a^{4} b^{5} d^{3}\right )}}{b^{8}}\right )} \sqrt {b x^{2} + a} x - \frac {3 \, {\left (32 \, a^{2} b^{3} c^{3} - 16 \, a^{3} b^{2} c^{2} d + 6 \, a^{4} b c d^{2} - a^{5} d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{256 \, b^{\frac {7}{2}}} \]
1/1280*(2*(4*(2*(8*b*d^3*x^2 + (30*b^9*c*d^2 + 11*a*b^8*d^3)/b^8)*x^2 + (8 0*b^9*c^2*d + 90*a*b^8*c*d^2 + a^2*b^7*d^3)/b^8)*x^2 + 5*(32*b^9*c^3 + 112 *a*b^8*c^2*d + 6*a^2*b^7*c*d^2 - a^3*b^6*d^3)/b^8)*x^2 + 5*(160*a*b^8*c^3 + 48*a^2*b^7*c^2*d - 18*a^3*b^6*c*d^2 + 3*a^4*b^5*d^3)/b^8)*sqrt(b*x^2 + a )*x - 3/256*(32*a^2*b^3*c^3 - 16*a^3*b^2*c^2*d + 6*a^4*b*c*d^2 - a^5*d^3)* log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(7/2)
Timed out. \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^3 \, dx=\int {\left (b\,x^2+a\right )}^{3/2}\,{\left (d\,x^2+c\right )}^3 \,d x \]