Integrand size = 19, antiderivative size = 118 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \, dx=\frac {a (6 b c-a d) x \sqrt {a+b x^2}}{16 b}+\frac {(6 b c-a d) x \left (a+b x^2\right )^{3/2}}{24 b}+\frac {d x \left (a+b x^2\right )^{5/2}}{6 b}+\frac {a^2 (6 b c-a d) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 b^{3/2}} \]
1/24*(-a*d+6*b*c)*x*(b*x^2+a)^(3/2)/b+1/6*d*x*(b*x^2+a)^(5/2)/b+1/16*a^2*( -a*d+6*b*c)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(3/2)+1/16*a*(-a*d+6*b*c) *x*(b*x^2+a)^(1/2)/b
Time = 0.14 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.84 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \, dx=\frac {x \sqrt {a+b x^2} \left (30 a b c+3 a^2 d+12 b^2 c x^2+14 a b d x^2+8 b^2 d x^4\right )}{48 b}+\frac {a^2 (-6 b c+a d) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 b^{3/2}} \]
(x*Sqrt[a + b*x^2]*(30*a*b*c + 3*a^2*d + 12*b^2*c*x^2 + 14*a*b*d*x^2 + 8*b ^2*d*x^4))/(48*b) + (a^2*(-6*b*c + a*d)*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.
\(\displaystyle \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \, dx\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {(6 b c-a d) \int \left (b x^2+a\right )^{3/2}dx}{6 b}+\frac {d x \left (a+b x^2\right )^{5/2}}{6 b}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {(6 b c-a d) \left (\frac {3}{4} a \int \sqrt {b x^2+a}dx+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {d x \left (a+b x^2\right )^{5/2}}{6 b}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {(6 b c-a d) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{\sqrt {b x^2+a}}dx+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {d x \left (a+b x^2\right )^{5/2}}{6 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {(6 b c-a d) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right )}{6 b}+\frac {d x \left (a+b x^2\right )^{5/2}}{6 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\left (\frac {3}{4} a \left (\frac {a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 \sqrt {b}}+\frac {1}{2} x \sqrt {a+b x^2}\right )+\frac {1}{4} x \left (a+b x^2\right )^{3/2}\right ) (6 b c-a d)}{6 b}+\frac {d x \left (a+b x^2\right )^{5/2}}{6 b}\) |
(d*x*(a + b*x^2)^(5/2))/(6*b) + ((6*b*c - a*d)*((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)
3.1.55.3.1 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]
Time = 2.35 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.74
method | result | size |
risch | \(\frac {x \left (8 b^{2} d \,x^{4}+14 x^{2} a b d +12 b^{2} c \,x^{2}+3 a^{2} d +30 a b c \right ) \sqrt {b \,x^{2}+a}}{48 b}-\frac {a^{2} \left (a d -6 b c \right ) \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 b^{\frac {3}{2}}}\) | \(87\) |
pseudoelliptic | \(-\frac {\left (a^{3} d -6 a^{2} b c \right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )-x \sqrt {b \,x^{2}+a}\, \left (10 \left (\frac {7 d \,x^{2}}{15}+c \right ) a \,b^{\frac {3}{2}}+\left (\frac {8}{3} d \,x^{4}+4 c \,x^{2}\right ) b^{\frac {5}{2}}+a^{2} d \sqrt {b}\right )}{16 b^{\frac {3}{2}}}\) | \(90\) |
default | \(c \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )+d \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6 b}-\frac {a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6 b}\right )\) | \(130\) |
1/48/b*x*(8*b^2*d*x^4+14*a*b*d*x^2+12*b^2*c*x^2+3*a^2*d+30*a*b*c)*(b*x^2+a )^(1/2)-1/16*a^2*(a*d-6*b*c)/b^(3/2)*ln(x*b^(1/2)+(b*x^2+a)^(1/2))
Time = 0.28 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.78 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \, dx=\left [-\frac {3 \, {\left (6 \, a^{2} b c - a^{3} d\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (8 \, b^{3} d x^{5} + 2 \, {\left (6 \, b^{3} c + 7 \, a b^{2} d\right )} x^{3} + 3 \, {\left (10 \, a b^{2} c + a^{2} b d\right )} x\right )} \sqrt {b x^{2} + a}}{96 \, b^{2}}, -\frac {3 \, {\left (6 \, a^{2} b c - a^{3} d\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, b^{3} d x^{5} + 2 \, {\left (6 \, b^{3} c + 7 \, a b^{2} d\right )} x^{3} + 3 \, {\left (10 \, a b^{2} c + a^{2} b d\right )} x\right )} \sqrt {b x^{2} + a}}{48 \, b^{2}}\right ] \]
[-1/96*(3*(6*a^2*b*c - a^3*d)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 + a)*sqr t(b)*x - a) - 2*(8*b^3*d*x^5 + 2*(6*b^3*c + 7*a*b^2*d)*x^3 + 3*(10*a*b^2*c + a^2*b*d)*x)*sqrt(b*x^2 + a))/b^2, -1/48*(3*(6*a^2*b*c - a^3*d)*sqrt(-b) *arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*b^3*d*x^5 + 2*(6*b^3*c + 7*a*b^2* d)*x^3 + 3*(10*a*b^2*c + a^2*b*d)*x)*sqrt(b*x^2 + a))/b^2]
Time = 0.34 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.48 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \, dx=\begin {cases} \sqrt {a + b x^{2}} \left (\frac {b d x^{5}}{6} + \frac {x^{3} \cdot \left (\frac {7 a b d}{6} + b^{2} c\right )}{4 b} + \frac {x \left (a^{2} d + 2 a b c - \frac {3 a \left (\frac {7 a b d}{6} + b^{2} c\right )}{4 b}\right )}{2 b}\right ) + \left (a^{2} c - \frac {a \left (a^{2} d + 2 a b c - \frac {3 a \left (\frac {7 a b d}{6} + b^{2} c\right )}{4 b}\right )}{2 b}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {b} \sqrt {a + b x^{2}} + 2 b x \right )}}{\sqrt {b}} & \text {for}\: a \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {b x^{2}}} & \text {otherwise} \end {cases}\right ) & \text {for}\: b \neq 0 \\a^{\frac {3}{2}} \left (c x + \frac {d x^{3}}{3}\right ) & \text {otherwise} \end {cases} \]
Piecewise((sqrt(a + b*x**2)*(b*d*x**5/6 + x**3*(7*a*b*d/6 + b**2*c)/(4*b) + x*(a**2*d + 2*a*b*c - 3*a*(7*a*b*d/6 + b**2*c)/(4*b))/(2*b)) + (a**2*c - a*(a**2*d + 2*a*b*c - 3*a*(7*a*b*d/6 + b**2*c)/(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)*(c*x + d*x**3/3), True))
Time = 0.21 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.98 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \, dx=\frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} c x + \frac {3}{8} \, \sqrt {b x^{2} + a} a c x + \frac {{\left (b x^{2} + a\right )}^{\frac {5}{2}} d x}{6 \, b} - \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} a d x}{24 \, b} - \frac {\sqrt {b x^{2} + a} a^{2} d x}{16 \, b} + \frac {3 \, a^{2} c \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} - \frac {a^{3} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, b^{\frac {3}{2}}} \]
1/4*(b*x^2 + a)^(3/2)*c*x + 3/8*sqrt(b*x^2 + a)*a*c*x + 1/6*(b*x^2 + a)^(5 /2)*d*x/b - 1/24*(b*x^2 + a)^(3/2)*a*d*x/b - 1/16*sqrt(b*x^2 + a)*a^2*d*x/ b + 3/8*a^2*c*arcsinh(b*x/sqrt(a*b))/sqrt(b) - 1/16*a^3*d*arcsinh(b*x/sqrt (a*b))/b^(3/2)
Time = 0.28 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.87 \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \, dx=\frac {1}{48} \, {\left (2 \, {\left (4 \, b d x^{2} + \frac {6 \, b^{5} c + 7 \, a b^{4} d}{b^{4}}\right )} x^{2} + \frac {3 \, {\left (10 \, a b^{4} c + a^{2} b^{3} d\right )}}{b^{4}}\right )} \sqrt {b x^{2} + a} x - \frac {{\left (6 \, a^{2} b c - a^{3} d\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, b^{\frac {3}{2}}} \]
1/48*(2*(4*b*d*x^2 + (6*b^5*c + 7*a*b^4*d)/b^4)*x^2 + 3*(10*a*b^4*c + a^2* b^3*d)/b^4)*sqrt(b*x^2 + a)*x - 1/16*(6*a^2*b*c - a^3*d)*log(abs(-sqrt(b)* x + sqrt(b*x^2 + a)))/b^(3/2)
Timed out. \[ \int \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) \, dx=\int {\left (b\,x^2+a\right )}^{3/2}\,\left (d\,x^2+c\right ) \,d x \]