Integrand size = 20, antiderivative size = 173 \[ \int x^3 (A+B x) \left (a+b x^2\right )^{5/2} \, dx=\frac {3 a^4 B x \sqrt {a+b x^2}}{256 b^2}+\frac {a^3 B x \left (a+b x^2\right )^{3/2}}{128 b^2}+\frac {a^2 B x \left (a+b x^2\right )^{5/2}}{160 b^2}+\frac {A x^2 \left (a+b x^2\right )^{7/2}}{9 b}+\frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {a (160 A+189 B x) \left (a+b x^2\right )^{7/2}}{5040 b^2}+\frac {3 a^5 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{256 b^{5/2}} \]
1/128*a^3*B*x*(b*x^2+a)^(3/2)/b^2+1/160*a^2*B*x*(b*x^2+a)^(5/2)/b^2+1/9*A* x^2*(b*x^2+a)^(7/2)/b+1/10*B*x^3*(b*x^2+a)^(7/2)/b-1/5040*a*(189*B*x+160*A )*(b*x^2+a)^(7/2)/b^2+3/256*a^5*B*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(5/ 2)+3/256*a^4*B*x*(b*x^2+a)^(1/2)/b^2
Time = 0.29 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.79 \[ \int x^3 (A+B x) \left (a+b x^2\right )^{5/2} \, dx=\frac {\sqrt {b} \sqrt {a+b x^2} \left (896 b^4 x^8 (10 A+9 B x)+10 a^3 b x^2 (128 A+63 B x)-5 a^4 (512 A+189 B x)+24 a^2 b^2 x^4 (800 A+651 B x)+16 a b^3 x^6 (1520 A+1323 B x)\right )-945 a^5 B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{80640 b^{5/2}} \]
(Sqrt[b]*Sqrt[a + b*x^2]*(896*b^4*x^8*(10*A + 9*B*x) + 10*a^3*b*x^2*(128*A + 63*B*x) - 5*a^4*(512*A + 189*B*x) + 24*a^2*b^2*x^4*(800*A + 651*B*x) + 16*a*b^3*x^6*(1520*A + 1323*B*x)) - 945*a^5*B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(80640*b^(5/2))
Time = 0.29 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.12, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.550, Rules used = {533, 533, 25, 27, 533, 455, 211, 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.
| \(\displaystyle \int x^3 \left (a+b x^2\right )^{5/2} (A+B x) \, dx\) |
| \(\Big \downarrow \) 533 |
| \(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {\int x^2 (3 a B-10 A b x) \left (b x^2+a\right )^{5/2}dx}{10 b}\) |
| \(\Big \downarrow \) 533 |
| \(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {-\frac {\int -a b x (20 A+27 B x) \left (b x^2+a\right )^{5/2}dx}{9 b}-\frac {10}{9} A x^2 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {\frac {\int a b x (20 A+27 B x) \left (b x^2+a\right )^{5/2}dx}{9 b}-\frac {10}{9} A x^2 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {\frac {1}{9} a \int x (20 A+27 B x) \left (b x^2+a\right )^{5/2}dx-\frac {10}{9} A x^2 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 533 |
| \(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {\frac {1}{9} a \left (\frac {27 B x \left (a+b x^2\right )^{7/2}}{8 b}-\frac {\int (27 a B-160 A b x) \left (b x^2+a\right )^{5/2}dx}{8 b}\right )-\frac {10}{9} A x^2 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {\frac {1}{9} a \left (\frac {27 B x \left (a+b x^2\right )^{7/2}}{8 b}-\frac {27 a B \int \left (b x^2+a\right )^{5/2}dx-\frac {160}{7} A \left (a+b x^2\right )^{7/2}}{8 b}\right )-\frac {10}{9} A x^2 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {\frac {1}{9} a \left (\frac {27 B x \left (a+b x^2\right )^{7/2}}{8 b}-\frac {27 a B \left (\frac {5}{6} a \int \left (b x^2+a\right )^{3/2}dx+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )-\frac {160}{7} A \left (a+b x^2\right )^{7/2}}{8 b}\right )-\frac {10}{9} A x^2 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {\frac {1}{9} a \left (\frac {27 B x \left (a+b x^2\right )^{7/2}}{8 b}-\frac {27 a B \left (\frac {5}{6} a \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )-\frac {160}{7} A \left (a+b x^2\right )^{7/2}}{8 b}\right )-\frac {10}{9} A x^2 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {\frac {1}{9} a \left (\frac {27 B x \left (a+b x^2\right )^{7/2}}{8 b}-\frac {27 a B \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )-\frac {160}{7} A \left (a+b x^2\right )^{7/2}}{8 b}\right )-\frac {10}{9} A x^2 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {\frac {1}{9} a \left (\frac {27 B x \left (a+b x^2\right )^{7/2}}{8 b}-\frac {27 a B \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )-\frac {160}{7} A \left (a+b x^2\right )^{7/2}}{8 b}\right )-\frac {10}{9} A x^2 \left (a+b x^2\right )^{7/2}}{10 b}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{7/2}}{10 b}-\frac {\frac {1}{9} a \left (\frac {27 B x \left (a+b x^2\right )^{7/2}}{8 b}-\frac {27 a B \left (\frac {5}{6} a \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )+\frac {1}{6} x \left (a+b x^2\right )^{5/2}\right )-\frac {160}{7} A \left (a+b x^2\right )^{7/2}}{8 b}\right )-\frac {10}{9} A x^2 \left (a+b x^2\right )^{7/2}}{10 b}\) |
(B*x^3*(a + b*x^2)^(7/2))/(10*b) - ((-10*A*x^2*(a + b*x^2)^(7/2))/9 + (a*( (27*B*x*(a + b*x^2)^(7/2))/(8*b) - ((-160*A*(a + b*x^2)^(7/2))/7 + 27*a*B* ((x*(a + b*x^2)^(5/2))/6 + (5*a*((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))/(8*b)))/9)/(10*b)
3.1.15.3.1 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)^(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[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, p}, x] && !LeQ[p, -1]
Int[(x_)^(m_.)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[d*x^m*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 2))), x] - Simp[1/(b*(m + 2* p + 2)) Int[x^(m - 1)*(a + b*x^2)^p*Simp[a*d*m - b*c*(m + 2*p + 2)*x, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && IGtQ[m, 0] && GtQ[p, -1] && Integer Q[2*p]
Time = 3.43 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.79
| method | result | size |
| risch | \(-\frac {\left (-8064 B \,b^{4} x^{9}-8960 A \,b^{4} x^{8}-21168 B a \,b^{3} x^{7}-24320 A a \,b^{3} x^{6}-15624 B \,a^{2} b^{2} x^{5}-19200 A \,a^{2} b^{2} x^{4}-630 B \,a^{3} b \,x^{3}-1280 A \,a^{3} b \,x^{2}+945 B \,a^{4} x +2560 A \,a^{4}\right ) \sqrt {b \,x^{2}+a}}{80640 b^{2}}+\frac {3 a^{5} B \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{256 b^{\frac {5}{2}}}\) | \(137\) |
| default | \(B \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{10 b}-\frac {3 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{8 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\right )}{8 b}\right )}{10 b}\right )+A \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{9 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {7}{2}}}{63 b^{2}}\right )\) | \(152\) |
-1/80640*(-8064*B*b^4*x^9-8960*A*b^4*x^8-21168*B*a*b^3*x^7-24320*A*a*b^3*x ^6-15624*B*a^2*b^2*x^5-19200*A*a^2*b^2*x^4-630*B*a^3*b*x^3-1280*A*a^3*b*x^ 2+945*B*a^4*x+2560*A*a^4)/b^2*(b*x^2+a)^(1/2)+3/256*a^5*B/b^(5/2)*ln(x*b^( 1/2)+(b*x^2+a)^(1/2))
Time = 0.29 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.75 \[ \int x^3 (A+B x) \left (a+b x^2\right )^{5/2} \, dx=\left [\frac {945 \, B a^{5} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8064 \, B b^{5} x^{9} + 8960 \, A b^{5} x^{8} + 21168 \, B a b^{4} x^{7} + 24320 \, A a b^{4} x^{6} + 15624 \, B a^{2} b^{3} x^{5} + 19200 \, A a^{2} b^{3} x^{4} + 630 \, B a^{3} b^{2} x^{3} + 1280 \, A a^{3} b^{2} x^{2} - 945 \, B a^{4} b x - 2560 \, A a^{4} b\right )} \sqrt {b x^{2} + a}}{161280 \, b^{3}}, -\frac {945 \, B a^{5} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8064 \, B b^{5} x^{9} + 8960 \, A b^{5} x^{8} + 21168 \, B a b^{4} x^{7} + 24320 \, A a b^{4} x^{6} + 15624 \, B a^{2} b^{3} x^{5} + 19200 \, A a^{2} b^{3} x^{4} + 630 \, B a^{3} b^{2} x^{3} + 1280 \, A a^{3} b^{2} x^{2} - 945 \, B a^{4} b x - 2560 \, A a^{4} b\right )} \sqrt {b x^{2} + a}}{80640 \, b^{3}}\right ] \]
[1/161280*(945*B*a^5*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(8064*B*b^5*x^9 + 8960*A*b^5*x^8 + 21168*B*a*b^4*x^7 + 24320*A*a*b^ 4*x^6 + 15624*B*a^2*b^3*x^5 + 19200*A*a^2*b^3*x^4 + 630*B*a^3*b^2*x^3 + 12 80*A*a^3*b^2*x^2 - 945*B*a^4*b*x - 2560*A*a^4*b)*sqrt(b*x^2 + a))/b^3, -1/ 80640*(945*B*a^5*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8064*B*b^5 *x^9 + 8960*A*b^5*x^8 + 21168*B*a*b^4*x^7 + 24320*A*a*b^4*x^6 + 15624*B*a^ 2*b^3*x^5 + 19200*A*a^2*b^3*x^4 + 630*B*a^3*b^2*x^3 + 1280*A*a^3*b^2*x^2 - 945*B*a^4*b*x - 2560*A*a^4*b)*sqrt(b*x^2 + a))/b^3]
Time = 0.63 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.15 \[ \int x^3 (A+B x) \left (a+b x^2\right )^{5/2} \, dx=\begin {cases} \frac {3 B a^{5} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{256 b^{2}} + \sqrt {a + b x^{2}} \left (- \frac {2 A a^{4}}{63 b^{2}} + \frac {A a^{3} x^{2}}{63 b} + \frac {5 A a^{2} x^{4}}{21} + \frac {19 A a b x^{6}}{63} + \frac {A b^{2} x^{8}}{9} - \frac {3 B a^{4} x}{256 b^{2}} + \frac {B a^{3} x^{3}}{128 b} + \frac {31 B a^{2} x^{5}}{160} + \frac {21 B a b x^{7}}{80} + \frac {B b^{2} x^{9}}{10}\right ) & \text {for}\: b \neq 0 \\a^{\frac {5}{2}} \left (\frac {A x^{4}}{4} + \frac {B x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]
Piecewise((3*B*a**5*Piecewise((log(2*sqrt(b)*sqrt(a + b*x**2) + 2*b*x)/sqr t(b), Ne(a, 0)), (x*log(x)/sqrt(b*x**2), True))/(256*b**2) + sqrt(a + b*x* *2)*(-2*A*a**4/(63*b**2) + A*a**3*x**2/(63*b) + 5*A*a**2*x**4/21 + 19*A*a* b*x**6/63 + A*b**2*x**8/9 - 3*B*a**4*x/(256*b**2) + B*a**3*x**3/(128*b) + 31*B*a**2*x**5/160 + 21*B*a*b*x**7/80 + B*b**2*x**9/10), Ne(b, 0)), (a**(5 /2)*(A*x**4/4 + B*x**5/5), True))
Time = 0.19 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.84 \[ \int x^3 (A+B x) \left (a+b x^2\right )^{5/2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} B x^{3}}{10 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {7}{2}} A x^{2}}{9 \, b} - \frac {3 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} B a x}{80 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B a^{2} x}{160 \, b^{2}} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a^{3} x}{128 \, b^{2}} + \frac {3 \, \sqrt {b x^{2} + a} B a^{4} x}{256 \, b^{2}} + \frac {3 \, B a^{5} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{256 \, b^{\frac {5}{2}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {7}{2}} A a}{63 \, b^{2}} \]
1/10*(b*x^2 + a)^(7/2)*B*x^3/b + 1/9*(b*x^2 + a)^(7/2)*A*x^2/b - 3/80*(b*x ^2 + a)^(7/2)*B*a*x/b^2 + 1/160*(b*x^2 + a)^(5/2)*B*a^2*x/b^2 + 1/128*(b*x ^2 + a)^(3/2)*B*a^3*x/b^2 + 3/256*sqrt(b*x^2 + a)*B*a^4*x/b^2 + 3/256*B*a^ 5*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 2/63*(b*x^2 + a)^(7/2)*A*a/b^2
Time = 0.29 (sec) , antiderivative size = 140, normalized size of antiderivative = 0.81 \[ \int x^3 (A+B x) \left (a+b x^2\right )^{5/2} \, dx=-\frac {3 \, B a^{5} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{256 \, b^{\frac {5}{2}}} - \frac {1}{80640} \, {\left (\frac {2560 \, A a^{4}}{b^{2}} + {\left (\frac {945 \, B a^{4}}{b^{2}} - 2 \, {\left (\frac {640 \, A a^{3}}{b} + {\left (\frac {315 \, B a^{3}}{b} + 4 \, {\left (2400 \, A a^{2} + {\left (1953 \, B a^{2} + 2 \, {\left (1520 \, A a b + 7 \, {\left (189 \, B a b + 8 \, {\left (9 \, B b^{2} x + 10 \, A b^{2}\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} x\right )} \sqrt {b x^{2} + a} \]
-3/256*B*a^5*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2) - 1/80640*(256 0*A*a^4/b^2 + (945*B*a^4/b^2 - 2*(640*A*a^3/b + (315*B*a^3/b + 4*(2400*A*a ^2 + (1953*B*a^2 + 2*(1520*A*a*b + 7*(189*B*a*b + 8*(9*B*b^2*x + 10*A*b^2) *x)*x)*x)*x)*x)*x)*x)*x)*sqrt(b*x^2 + a)
Timed out. \[ \int x^3 (A+B x) \left (a+b x^2\right )^{5/2} \, dx=\int x^3\,{\left (b\,x^2+a\right )}^{5/2}\,\left (A+B\,x\right ) \,d x \]