Integrand size = 18, antiderivative size = 103 \[ \int x (A+B x) \left (a+b x^2\right )^{3/2} \, dx=-\frac {a^2 B x \sqrt {a+b x^2}}{16 b}-\frac {a B x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {(6 A+5 B x) \left (a+b x^2\right )^{5/2}}{30 b}-\frac {a^3 B \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}} \]
-1/24*a*B*x*(b*x^2+a)^(3/2)/b+1/30*(5*B*x+6*A)*(b*x^2+a)^(5/2)/b-1/16*a^3* B*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(3/2)-1/16*a^2*B*x*(b*x^2+a)^(1/2)/ b
Time = 0.21 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.98 \[ \int x (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\frac {\sqrt {a+b x^2} \left (48 a^2 A+15 a^2 B x+96 a A b x^2+70 a b B x^3+48 A b^2 x^4+40 b^2 B x^5\right )}{240 b}+\frac {a^3 B \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 b^{3/2}} \]
(Sqrt[a + b*x^2]*(48*a^2*A + 15*a^2*B*x + 96*a*A*b*x^2 + 70*a*b*B*x^3 + 48 *A*b^2*x^4 + 40*b^2*B*x^5))/(240*b) + (a^3*B*Log[-(Sqrt[b]*x) + Sqrt[a + b *x^2]])/(16*b^(3/2))
Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {533, 455, 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.
\(\displaystyle \int x \left (a+b x^2\right )^{3/2} (A+B x) \, dx\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {B x \left (a+b x^2\right )^{5/2}}{6 b}-\frac {\int (a B-6 A b x) \left (b x^2+a\right )^{3/2}dx}{6 b}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {B x \left (a+b x^2\right )^{5/2}}{6 b}-\frac {a B \int \left (b x^2+a\right )^{3/2}dx-\frac {6}{5} A \left (a+b x^2\right )^{5/2}}{6 b}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {B x \left (a+b x^2\right )^{5/2}}{6 b}-\frac {a B \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )-\frac {6}{5} A \left (a+b x^2\right )^{5/2}}{6 b}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {B x \left (a+b x^2\right )^{5/2}}{6 b}-\frac {a B \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )-\frac {6}{5} A \left (a+b x^2\right )^{5/2}}{6 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {B x \left (a+b x^2\right )^{5/2}}{6 b}-\frac {a B \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )-\frac {6}{5} A \left (a+b x^2\right )^{5/2}}{6 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {B x \left (a+b x^2\right )^{5/2}}{6 b}-\frac {a B \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )-\frac {6}{5} A \left (a+b x^2\right )^{5/2}}{6 b}\) |
(B*x*(a + b*x^2)^(5/2))/(6*b) - ((-6*A*(a + b*x^2)^(5/2))/5 + a*B*((x*(a + b*x^2)^(3/2))/4 + (3*a*((x*Sqrt[a + b*x^2])/2 + (a*ArcTanh[(Sqrt[b]*x)/Sq rt[a + b*x^2]])/(2*Sqrt[b])))/4))/(6*b)
3.1.10.3.1 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[((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 = 3.41 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.86
method | result | size |
risch | \(\frac {\left (40 b^{2} B \,x^{5}+48 A \,b^{2} x^{4}+70 B a b \,x^{3}+96 a A b \,x^{2}+15 a^{2} B x +48 a^{2} A \right ) \sqrt {b \,x^{2}+a}}{240 b}-\frac {B \,a^{3} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {3}{2}}}\) | \(89\) |
default | \(B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )+\frac {A \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{5 b}\) | \(92\) |
1/240*(40*B*b^2*x^5+48*A*b^2*x^4+70*B*a*b*x^3+96*A*a*b*x^2+15*B*a^2*x+48*A *a^2)/b*(b*x^2+a)^(1/2)-1/16*B*a^3/b^(3/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))
Time = 0.34 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.99 \[ \int x (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\left [\frac {15 \, B a^{3} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 70 \, B a b^{2} x^{3} + 96 \, A a b^{2} x^{2} + 15 \, B a^{2} b x + 48 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{480 \, b^{2}}, \frac {15 \, B a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (40 \, B b^{3} x^{5} + 48 \, A b^{3} x^{4} + 70 \, B a b^{2} x^{3} + 96 \, A a b^{2} x^{2} + 15 \, B a^{2} b x + 48 \, A a^{2} b\right )} \sqrt {b x^{2} + a}}{240 \, b^{2}}\right ] \]
[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 + 70*B*a*b^2*x^3 + 96*A*a*b^2*x^2 + 15*B*a ^2*b*x + 48*A*a^2*b)*sqrt(b*x^2 + a))/b^2, 1/240*(15*B*a^3*sqrt(-b)*arctan (sqrt(-b)*x/sqrt(b*x^2 + a)) + (40*B*b^3*x^5 + 48*A*b^3*x^4 + 70*B*a*b^2*x ^3 + 96*A*a*b^2*x^2 + 15*B*a^2*b*x + 48*A*a^2*b)*sqrt(b*x^2 + a))/b^2]
Time = 0.45 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.30 \[ \int x (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\begin {cases} - \frac {B a^{3} \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right )}{16 b} + \sqrt {a + b x^{2}} \left (\frac {A a^{2}}{5 b} + \frac {2 A a x^{2}}{5} + \frac {A b x^{4}}{5} + \frac {B a^{2} x}{16 b} + \frac {7 B a x^{3}}{24} + \frac {B b x^{5}}{6}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (\frac {A x^{2}}{2} + \frac {B x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]
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) + sqrt(a + b*x**2)*( A*a**2/(5*b) + 2*A*a*x**2/5 + A*b*x**4/5 + B*a**2*x/(16*b) + 7*B*a*x**3/24 + B*b*x**5/6), Ne(b, 0)), (a**(3/2)*(A*x**2/2 + B*x**3/3), True))
Time = 0.22 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.83 \[ \int x (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x}{24 \, b} - \frac {\sqrt {b x^{2} + a} B a^{2} x}{16 \, b} - \frac {B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} A}{5 \, b} \]
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 - 1/16*B*a^3*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 1/5*(b*x ^2 + a)^(5/2)*A/b
Time = 0.30 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.86 \[ \int x (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\frac {B a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {3}{2}}} + \frac {1}{240} \, \sqrt {b x^{2} + a} {\left (\frac {48 \, A a^{2}}{b} + {\left (\frac {15 \, B a^{2}}{b} + 2 \, {\left (48 \, A a + {\left (35 \, B a + 4 \, {\left (5 \, B b x + 6 \, A b\right )} x\right )} x\right )} x\right )} x\right )} \]
1/16*B*a^3*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(3/2) + 1/240*sqrt(b*x ^2 + a)*(48*A*a^2/b + (15*B*a^2/b + 2*(48*A*a + (35*B*a + 4*(5*B*b*x + 6*A *b)*x)*x)*x)*x)
Timed out. \[ \int x (A+B x) \left (a+b x^2\right )^{3/2} \, dx=\int x\,{\left (b\,x^2+a\right )}^{3/2}\,\left (A+B\,x\right ) \,d x \]