Integrand size = 19, antiderivative size = 118 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {a (6 A b-a B) x \sqrt {a+b x^2}}{16 b}+\frac {(6 A b-a B) x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {B x \left (a+b x^2\right )^{5/2}}{6 b}+\frac {a^2 (6 A b-a B) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}} \] Output:
1/16*a*(6*A*b-B*a)*x*(b*x^2+a)^(1/2)/b+1/24*(6*A*b-B*a)*x*(b*x^2+a)^(3/2)/ b+1/6*B*x*(b*x^2+a)^(5/2)/b+1/16*a^2*(6*A*b-B*a)*arctanh(b^(1/2)*x/(b*x^2+ a)^(1/2))/b^(3/2)
Time = 0.14 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.84 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {x \sqrt {a+b x^2} \left (30 a A b+3 a^2 B+12 A b^2 x^2+14 a b B x^2+8 b^2 B x^4\right )}{48 b}+\frac {a^2 (-6 A b+a B) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 b^{3/2}} \] Input:
Integrate[(a + b*x^2)^(3/2)*(A + B*x^2),x]
Output:
(x*Sqrt[a + b*x^2]*(30*a*A*b + 3*a^2*B + 12*A*b^2*x^2 + 14*a*b*B*x^2 + 8*b ^2*B*x^4))/(48*b) + (a^2*(-6*A*b + a*B)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2] ])/(16*b^(3/2))
Time = 0.20 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.89, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {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.
\(\displaystyle \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {(6 A b-a B) \int \left (b x^2+a\right )^{3/2}dx}{6 b}+\frac {B x \left (a+b x^2\right )^{5/2}}{6 b}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {(6 A b-a B) \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {B x \left (a+b x^2\right )^{5/2}}{6 b}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {(6 A b-a B) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {B x \left (a+b x^2\right )^{5/2}}{6 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {(6 A b-a B) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {B x \left (a+b x^2\right )^{5/2}}{6 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {(6 A b-a B) \left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {B x \left (a+b x^2\right )^{5/2}}{6 b}\) |
Input:
Int[(a + b*x^2)^(3/2)*(A + B*x^2),x]
Output:
(B*x*(a + b*x^2)^(5/2))/(6*b) + ((6*A*b - a*B)*((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)
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]
Time = 0.54 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.75
method | result | size |
risch | \(\frac {x \left (8 b^{2} B \,x^{4}+12 A \,b^{2} x^{2}+14 B a b \,x^{2}+30 a b A +3 a^{2} B \right ) \sqrt {b \,x^{2}+a}}{48 b}+\frac {a^{2} \left (6 A b -B a \right ) \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {3}{2}}}\) | \(88\) |
pseudoelliptic | \(\frac {\left (\frac {3}{2} a^{2} b A -\frac {1}{4} a^{3} B \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )+\left (\frac {5 \left (\frac {7 x^{2} B}{15}+A \right ) a \,b^{\frac {3}{2}}}{2}+x^{2} \left (\frac {2 x^{2} B}{3}+A \right ) b^{\frac {5}{2}}+\frac {B \,a^{2} \sqrt {b}}{4}\right ) \sqrt {b \,x^{2}+a}\, x}{4 b^{\frac {3}{2}}}\) | \(89\) |
default | \(A \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+B \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )\) | \(130\) |
Input:
int((b*x^2+a)^(3/2)*(B*x^2+A),x,method=_RETURNVERBOSE)
Output:
1/48/b*x*(8*B*b^2*x^4+12*A*b^2*x^2+14*B*a*b*x^2+30*A*a*b+3*B*a^2)*(b*x^2+a )^(1/2)+1/16*a^2*(6*A*b-B*a)/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))
Time = 0.09 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.75 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\left [-\frac {3 \, {\left (B a^{3} - 6 \, A a^{2} b\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, B b^{3} x^{5} + 2 \, {\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} x^{3} + 3 \, {\left (B a^{2} b + 10 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{2}}, \frac {3 \, {\left (B a^{3} - 6 \, A a^{2} b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (8 \, B b^{3} x^{5} + 2 \, {\left (7 \, B a b^{2} + 6 \, A b^{3}\right )} x^{3} + 3 \, {\left (B a^{2} b + 10 \, A a b^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{2}}\right ] \] Input:
integrate((b*x^2+a)^(3/2)*(B*x^2+A),x, algorithm="fricas")
Output:
[-1/96*(3*(B*a^3 - 6*A*a^2*b)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqr t(b)*x - a) - 2*(8*B*b^3*x^5 + 2*(7*B*a*b^2 + 6*A*b^3)*x^3 + 3*(B*a^2*b + 10*A*a*b^2)*x)*sqrt(b*x^2 + a))/b^2, 1/48*(3*(B*a^3 - 6*A*a^2*b)*sqrt(-b)* arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (8*B*b^3*x^5 + 2*(7*B*a*b^2 + 6*A*b^3 )*x^3 + 3*(B*a^2*b + 10*A*a*b^2)*x)*sqrt(b*x^2 + a))/b^2]
Time = 0.34 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.48 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {B b x^{5}}{6} + \frac {x^{3} \left (A b^{2} + \frac {7 B a b}{6}\right )}{4 b} + \frac {x \left (2 A a b + B a^{2} - \frac {3 a \left (A b^{2} + \frac {7 B a b}{6}\right )}{4 b}\right )}{2 b}\right ) + \left (A a^{2} - \frac {a \left (2 A a b + B a^{2} - \frac {3 a \left (A b^{2} + \frac {7 B a b}{6}\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (A x + \frac {B x^{3}}{3}\right ) & \text {otherwise} \end {cases} \] Input:
integrate((b*x**2+a)**(3/2)*(B*x**2+A),x)
Output:
Piecewise((sqrt(a + b*x**2)*(B*b*x**5/6 + x**3*(A*b**2 + 7*B*a*b/6)/(4*b) + x*(2*A*a*b + B*a**2 - 3*a*(A*b**2 + 7*B*a*b/6)/(4*b))/(2*b)) + (A*a**2 - a*(2*A*a*b + B*a**2 - 3*a*(A*b**2 + 7*B*a*b/6)/(4*b))/(2*b))*Piecewise((l og(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)*(A*x + B*x**3/3), True))
Time = 0.04 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.98 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} A x + \frac {3}{8} \, \sqrt {b x^{2} + a} A a x + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} B x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} B a x}{24 \, b} - \frac {\sqrt {b x^{2} + a} B a^{2} x}{16 \, b} - \frac {B a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} + \frac {3 \, A a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} \] Input:
integrate((b*x^2+a)^(3/2)*(B*x^2+A),x, algorithm="maxima")
Output:
1/4*(b*x^2 + a)^(3/2)*A*x + 3/8*sqrt(b*x^2 + a)*A*a*x + 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) + 3/8*A*a^2*arcsinh(b*x/sqrt (a*b))/sqrt(b)
Time = 0.12 (sec) , antiderivative size = 102, normalized size of antiderivative = 0.86 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, B b x^{2} + \frac {7 \, B a b^{4} + 6 \, A b^{5}}{b^{4}}\right )} x^{2} + \frac {3 \, {\left (B a^{2} b^{3} + 10 \, A a b^{4}\right )}}{b^{4}}\right )} \sqrt {b x^{2} + a} x + \frac {{\left (B a^{3} - 6 \, A a^{2} b\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {3}{2}}} \] Input:
integrate((b*x^2+a)^(3/2)*(B*x^2+A),x, algorithm="giac")
Output:
1/48*(2*(4*B*b*x^2 + (7*B*a*b^4 + 6*A*b^5)/b^4)*x^2 + 3*(B*a^2*b^3 + 10*A* a*b^4)/b^4)*sqrt(b*x^2 + a)*x + 1/16*(B*a^3 - 6*A*a^2*b)*log(abs(-sqrt(b)* x + sqrt(b*x^2 + a)))/b^(3/2)
Timed out. \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\int \left (B\,x^2+A\right )\,{\left (b\,x^2+a\right )}^{3/2} \,d x \] Input:
int((A + B*x^2)*(a + b*x^2)^(3/2),x)
Output:
int((A + B*x^2)*(a + b*x^2)^(3/2), x)
Time = 0.16 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.68 \[ \int \left (a+b x^2\right )^{3/2} \left (A+B x^2\right ) \, dx=\frac {33 \sqrt {b \,x^{2}+a}\, a^{2} b x +26 \sqrt {b \,x^{2}+a}\, a \,b^{2} x^{3}+8 \sqrt {b \,x^{2}+a}\, b^{3} x^{5}+15 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3}}{48 b} \] Input:
int((b*x^2+a)^(3/2)*(B*x^2+A),x)
Output:
(33*sqrt(a + b*x**2)*a**2*b*x + 26*sqrt(a + b*x**2)*a*b**2*x**3 + 8*sqrt(a + b*x**2)*b**3*x**5 + 15*sqrt(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt( a))*a**3)/(48*b)