Integrand size = 20, antiderivative size = 127 \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=\frac {a^2 B x \sqrt {a+b x^2}}{16 b^2}+\frac {A x^2 \left (a+b x^2\right )^{3/2}}{5 b}+\frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {a (16 A+15 B x) \left (a+b x^2\right )^{3/2}}{120 b^2}+\frac {a^3 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{5/2}} \]
1/5*A*x^2*(b*x^2+a)^(3/2)/b+1/6*B*x^3*(b*x^2+a)^(3/2)/b-1/120*a*(15*B*x+16 *A)*(b*x^2+a)^(3/2)/b^2+1/16*a^3*B*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(5 /2)+1/16*a^2*B*x*(b*x^2+a)^(1/2)/b^2
Time = 0.17 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.80 \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=\frac {\sqrt {a+b x^2} \left (-32 a^2 A-15 a^2 B x+16 a A b x^2+10 a b B x^3+48 A b^2 x^4+40 b^2 B x^5\right )}{240 b^2}-\frac {a^3 B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 b^{5/2}} \]
(Sqrt[a + b*x^2]*(-32*a^2*A - 15*a^2*B*x + 16*a*A*b*x^2 + 10*a*b*B*x^3 + 4 8*A*b^2*x^4 + 40*b^2*B*x^5))/(240*b^2) - (a^3*B*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(16*b^(5/2))
Time = 0.26 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.18, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {533, 27, 533, 25, 27, 533, 455, 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.
\(\displaystyle \int x^3 \sqrt {a+b x^2} (A+B x) \, dx\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {\int 3 x^2 (a B-2 A b x) \sqrt {b x^2+a}dx}{6 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {\int x^2 (a B-2 A b x) \sqrt {b x^2+a}dx}{2 b}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {-\frac {\int -a b x (4 A+5 B x) \sqrt {b x^2+a}dx}{5 b}-\frac {2}{5} A x^2 \left (a+b x^2\right )^{3/2}}{2 b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {\frac {\int a b x (4 A+5 B x) \sqrt {b x^2+a}dx}{5 b}-\frac {2}{5} A x^2 \left (a+b x^2\right )^{3/2}}{2 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {\frac {1}{5} a \int x (4 A+5 B x) \sqrt {b x^2+a}dx-\frac {2}{5} A x^2 \left (a+b x^2\right )^{3/2}}{2 b}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {\frac {1}{5} a \left (\frac {5 B x \left (a+b x^2\right )^{3/2}}{4 b}-\frac {\int (5 a B-16 A b x) \sqrt {b x^2+a}dx}{4 b}\right )-\frac {2}{5} A x^2 \left (a+b x^2\right )^{3/2}}{2 b}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {\frac {1}{5} a \left (\frac {5 B x \left (a+b x^2\right )^{3/2}}{4 b}-\frac {5 a B \int \sqrt {b x^2+a}dx-\frac {16}{3} A \left (a+b x^2\right )^{3/2}}{4 b}\right )-\frac {2}{5} A x^2 \left (a+b x^2\right )^{3/2}}{2 b}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {\frac {1}{5} a \left (\frac {5 B x \left (a+b x^2\right )^{3/2}}{4 b}-\frac {5 a B \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )-\frac {16}{3} A \left (a+b x^2\right )^{3/2}}{4 b}\right )-\frac {2}{5} A x^2 \left (a+b x^2\right )^{3/2}}{2 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {\frac {1}{5} a \left (\frac {5 B x \left (a+b x^2\right )^{3/2}}{4 b}-\frac {5 a B \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )-\frac {16}{3} A \left (a+b x^2\right )^{3/2}}{4 b}\right )-\frac {2}{5} A x^2 \left (a+b x^2\right )^{3/2}}{2 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {B x^3 \left (a+b x^2\right )^{3/2}}{6 b}-\frac {\frac {1}{5} a \left (\frac {5 B x \left (a+b x^2\right )^{3/2}}{4 b}-\frac {5 a B \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )-\frac {16}{3} A \left (a+b x^2\right )^{3/2}}{4 b}\right )-\frac {2}{5} A x^2 \left (a+b x^2\right )^{3/2}}{2 b}\) |
(B*x^3*(a + b*x^2)^(3/2))/(6*b) - ((-2*A*x^2*(a + b*x^2)^(3/2))/5 + (a*((5 *B*x*(a + b*x^2)^(3/2))/(4*b) - ((-16*A*(a + b*x^2)^(3/2))/3 + 5*a*B*((x*S qrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*Sqrt[b]))) /(4*b)))/5)/(2*b)
3.1.1.3.1 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 = 4.33 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.70
method | result | size |
risch | \(-\frac {\left (-40 b^{2} B \,x^{5}-48 A \,b^{2} x^{4}-10 B a b \,x^{3}-16 a A b \,x^{2}+15 a^{2} B x +32 a^{2} A \right ) \sqrt {b \,x^{2}+a}}{240 b^{2}}+\frac {a^{3} B \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {5}{2}}}\) | \(89\) |
default | \(B \left (\frac {x^{3} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4 b}-\frac {a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4 b}\right )}{2 b}\right )+A \left (\frac {x^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{5 b}-\frac {2 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{15 b^{2}}\right )\) | \(120\) |
-1/240*(-40*B*b^2*x^5-48*A*b^2*x^4-10*B*a*b*x^3-16*A*a*b*x^2+15*B*a^2*x+32 *A*a^2)/b^2*(b*x^2+a)^(1/2)+1/16*a^3*B/b^(5/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2 ))
Time = 0.31 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.62 \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=\left [\frac {15 \, B a^{3} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 10 \, B a b^{2} x^{3} + 16 \, A a b^{2} x^{2} - 15 \, B a^{2} b x - 32 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{480 \, b^{3}}, -\frac {15 \, B a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 10 \, B a b^{2} x^{3} + 16 \, A a b^{2} x^{2} - 15 \, B a^{2} b x - 32 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{240 \, b^{3}}\right ] \]
[1/480*(15*B*a^3*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(40*B*b^3*x^5 + 48*A*b^3*x^4 + 10*B*a*b^2*x^3 + 16*A*a*b^2*x^2 - 15*B*a ^2*b*x - 32*A*a^2*b)*sqrt(b*x^2 + a))/b^3, -1/240*(15*B*a^3*sqrt(-b)*arcta n(sqrt(-b)*x/sqrt(b*x^2 + a)) - (40*B*b^3*x^5 + 48*A*b^3*x^4 + 10*B*a*b^2* x^3 + 16*A*a*b^2*x^2 - 15*B*a^2*b*x - 32*A*a^2*b)*sqrt(b*x^2 + a))/b^3]
Time = 0.44 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.09 \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=\begin {cases} \frac {B a^{3} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{16 b^{2}} + \sqrt {a + b x^{2}} \left (- \frac {2 A a^{2}}{15 b^{2}} + \frac {A a x^{2}}{15 b} + \frac {A x^{4}}{5} - \frac {B a^{2} x}{16 b^{2}} + \frac {B a x^{3}}{24 b} + \frac {B x^{5}}{6}\right ) & \text {for}\: b \neq 0 \\\sqrt {a} \left (\frac {A x^{4}}{4} + \frac {B x^{5}}{5}\right ) & \text {otherwise} \end {cases} \]
Piecewise((B*a**3*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))/(16*b**2) + sqrt(a + b*x**2) *(-2*A*a**2/(15*b**2) + A*a*x**2/(15*b) + A*x**4/5 - B*a**2*x/(16*b**2) + B*a*x**3/(24*b) + B*x**5/6), Ne(b, 0)), (sqrt(a)*(A*x**4/4 + B*x**5/5), Tr ue))
Time = 0.20 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.84 \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B x^{3}}{6 \, b} + \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} A x^{2}}{5 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x}{8 \, b^{2}} + \frac {\sqrt {b x^{2} + a} B a^{2} x}{16 \, b^{2}} + \frac {B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {5}{2}}} - \frac {2 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A a}{15 \, b^{2}} \]
1/6*(b*x^2 + a)^(3/2)*B*x^3/b + 1/5*(b*x^2 + a)^(3/2)*A*x^2/b - 1/8*(b*x^2 + a)^(3/2)*B*a*x/b^2 + 1/16*sqrt(b*x^2 + a)*B*a^2*x/b^2 + 1/16*B*a^3*arcs inh(b*x/sqrt(a*b))/b^(5/2) - 2/15*(b*x^2 + a)^(3/2)*A*a/b^2
Time = 0.31 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.73 \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=-\frac {B a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {5}{2}}} + \frac {1}{240} \, \sqrt {b x^{2} + a} {\left ({\left (2 \, {\left ({\left (4 \, {\left (5 \, B x + 6 \, A\right )} x + \frac {5 \, B a}{b}\right )} x + \frac {8 \, A a}{b}\right )} x - \frac {15 \, B a^{2}}{b^{2}}\right )} x - \frac {32 \, A a^{2}}{b^{2}}\right )} \]
-1/16*B*a^3*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2) + 1/240*sqrt(b* x^2 + a)*((2*((4*(5*B*x + 6*A)*x + 5*B*a/b)*x + 8*A*a/b)*x - 15*B*a^2/b^2) *x - 32*A*a^2/b^2)
Timed out. \[ \int x^3 (A+B x) \sqrt {a+b x^2} \, dx=\int x^3\,\sqrt {b\,x^2+a}\,\left (A+B\,x\right ) \,d x \]