Integrand size = 35, antiderivative size = 212 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=-\frac {a c x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{8 d (a+b x)}+\frac {b x^2 \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{5 d (a+b x)}-\frac {(8 b c-15 a d x) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{60 d^2 (a+b x)}-\frac {a c^2 \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{3/2} (a+b x)} \]
1/5*b*x^2*(d*x^2+c)^(3/2)*((b*x+a)^2)^(1/2)/d/(b*x+a)-1/60*(-15*a*d*x+8*b* c)*(d*x^2+c)^(3/2)*((b*x+a)^2)^(1/2)/d^2/(b*x+a)-1/8*a*c^2*arctanh(x*d^(1/ 2)/(d*x^2+c)^(1/2))*((b*x+a)^2)^(1/2)/d^(3/2)/(b*x+a)-1/8*a*c*x*((b*x+a)^2 )^(1/2)*(d*x^2+c)^(1/2)/d/(b*x+a)
Time = 0.17 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.50 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (\sqrt {c+d x^2} \left (15 a d x \left (c+2 d x^2\right )+8 b \left (-2 c^2+c d x^2+3 d^2 x^4\right )\right )+15 a c^2 \sqrt {d} \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )\right )}{120 d^2 (a+b x)} \]
(Sqrt[(a + b*x)^2]*(Sqrt[c + d*x^2]*(15*a*d*x*(c + 2*d*x^2) + 8*b*(-2*c^2 + c*d*x^2 + 3*d^2*x^4)) + 15*a*c^2*Sqrt[d]*Log[-(Sqrt[d]*x) + Sqrt[c + d*x ^2]]))/(120*d^2*(a + b*x))
Time = 0.29 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.71, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1334, 27, 533, 533, 25, 27, 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.
| \(\displaystyle \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx\) |
| \(\Big \downarrow \) 1334 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int 2 b x^2 (a+b x) \sqrt {d x^2+c}dx}{2 b (a+b x)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x^2 (a+b x) \sqrt {d x^2+c}dx}{a+b x}\) |
| \(\Big \downarrow \) 533 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x^2 \left (c+d x^2\right )^{3/2}}{5 d}-\frac {\int x (2 b c-5 a d x) \sqrt {d x^2+c}dx}{5 d}\right )}{a+b x}\) |
| \(\Big \downarrow \) 533 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x^2 \left (c+d x^2\right )^{3/2}}{5 d}-\frac {-\frac {\int -c d (5 a+8 b x) \sqrt {d x^2+c}dx}{4 d}-\frac {5}{4} a x \left (c+d x^2\right )^{3/2}}{5 d}\right )}{a+b x}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x^2 \left (c+d x^2\right )^{3/2}}{5 d}-\frac {\frac {\int c d (5 a+8 b x) \sqrt {d x^2+c}dx}{4 d}-\frac {5}{4} a x \left (c+d x^2\right )^{3/2}}{5 d}\right )}{a+b x}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x^2 \left (c+d x^2\right )^{3/2}}{5 d}-\frac {\frac {1}{4} c \int (5 a+8 b x) \sqrt {d x^2+c}dx-\frac {5}{4} a x \left (c+d x^2\right )^{3/2}}{5 d}\right )}{a+b x}\) |
| \(\Big \downarrow \) 455 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x^2 \left (c+d x^2\right )^{3/2}}{5 d}-\frac {\frac {1}{4} c \left (5 a \int \sqrt {d x^2+c}dx+\frac {8 b \left (c+d x^2\right )^{3/2}}{3 d}\right )-\frac {5}{4} a x \left (c+d x^2\right )^{3/2}}{5 d}\right )}{a+b x}\) |
| \(\Big \downarrow \) 211 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x^2 \left (c+d x^2\right )^{3/2}}{5 d}-\frac {\frac {1}{4} c \left (5 a \left (\frac {1}{2} c \int \frac {1}{\sqrt {d x^2+c}}dx+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {8 b \left (c+d x^2\right )^{3/2}}{3 d}\right )-\frac {5}{4} a x \left (c+d x^2\right )^{3/2}}{5 d}\right )}{a+b x}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x^2 \left (c+d x^2\right )^{3/2}}{5 d}-\frac {\frac {1}{4} c \left (5 a \left (\frac {1}{2} c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {8 b \left (c+d x^2\right )^{3/2}}{3 d}\right )-\frac {5}{4} a x \left (c+d x^2\right )^{3/2}}{5 d}\right )}{a+b x}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x^2 \left (c+d x^2\right )^{3/2}}{5 d}-\frac {\frac {1}{4} c \left (5 a \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}}+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {8 b \left (c+d x^2\right )^{3/2}}{3 d}\right )-\frac {5}{4} a x \left (c+d x^2\right )^{3/2}}{5 d}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((b*x^2*(c + d*x^2)^(3/2))/(5*d) - ((-5*a*x *(c + d*x^2)^(3/2))/4 + (c*((8*b*(c + d*x^2)^(3/2))/(3*d) + 5*a*((x*Sqrt[c + d*x^2])/2 + (c*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(2*Sqrt[d]))))/4)/ (5*d)))/(a + b*x)
3.1.40.3.1 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]
Int[((g_.) + (h_.)*(x_))^(m_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_)*((d_ ) + (f_.)*(x_)^2)^(q_), x_Symbol] :> Simp[(a + b*x + c*x^2)^FracPart[p]/((4 *c)^IntPart[p]*(b + 2*c*x)^(2*FracPart[p])) Int[(g + h*x)^m*(b + 2*c*x)^( 2*p)*(d + f*x^2)^q, x], x] /; FreeQ[{a, b, c, d, f, g, h, m, p, q}, x] && E qQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.45 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.49
| method | result | size |
| default | \(\frac {\operatorname {csgn}\left (b x +a \right ) \left (24 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d^{\frac {3}{2}} b \,x^{2}+30 \left (d \,x^{2}+c \right )^{\frac {3}{2}} d^{\frac {3}{2}} a x -16 \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}\, b c -15 \sqrt {d \,x^{2}+c}\, d^{\frac {3}{2}} a c x -15 \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) a \,c^{2} d \right )}{120 d^{\frac {5}{2}}}\) | \(103\) |
| risch | \(\frac {\left (24 b \,x^{4} d^{2}+30 a \,x^{3} d^{2}+8 b c \,x^{2} d +15 a c x d -16 b \,c^{2}\right ) \sqrt {d \,x^{2}+c}\, \sqrt {\left (b x +a \right )^{2}}}{120 d^{2} \left (b x +a \right )}-\frac {c^{2} a \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) \sqrt {\left (b x +a \right )^{2}}}{8 d^{\frac {3}{2}} \left (b x +a \right )}\) | \(112\) |
1/120*csgn(b*x+a)*(24*(d*x^2+c)^(3/2)*d^(3/2)*b*x^2+30*(d*x^2+c)^(3/2)*d^( 3/2)*a*x-16*(d*x^2+c)^(3/2)*d^(1/2)*b*c-15*(d*x^2+c)^(1/2)*d^(3/2)*a*c*x-1 5*ln(d^(1/2)*x+(d*x^2+c)^(1/2))*a*c^2*d)/d^(5/2)
Time = 0.28 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.83 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\left [\frac {15 \, a c^{2} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (24 \, b d^{2} x^{4} + 30 \, a d^{2} x^{3} + 8 \, b c d x^{2} + 15 \, a c d x - 16 \, b c^{2}\right )} \sqrt {d x^{2} + c}}{240 \, d^{2}}, \frac {15 \, a c^{2} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (24 \, b d^{2} x^{4} + 30 \, a d^{2} x^{3} + 8 \, b c d x^{2} + 15 \, a c d x - 16 \, b c^{2}\right )} \sqrt {d x^{2} + c}}{120 \, d^{2}}\right ] \]
[1/240*(15*a*c^2*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(24*b*d^2*x^4 + 30*a*d^2*x^3 + 8*b*c*d*x^2 + 15*a*c*d*x - 16*b*c^2)*sqr t(d*x^2 + c))/d^2, 1/120*(15*a*c^2*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (24*b*d^2*x^4 + 30*a*d^2*x^3 + 8*b*c*d*x^2 + 15*a*c*d*x - 16*b*c^2) *sqrt(d*x^2 + c))/d^2]
\[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\int x^{2} \sqrt {c + d x^{2}} \sqrt {\left (a + b x\right )^{2}}\, dx \]
\[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\int { \sqrt {d x^{2} + c} \sqrt {{\left (b x + a\right )}^{2}} x^{2} \,d x } \]
Time = 0.29 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.55 \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\frac {a c^{2} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{8 \, d^{\frac {3}{2}}} + \frac {1}{120} \, \sqrt {d x^{2} + c} {\left ({\left (2 \, {\left (3 \, {\left (4 \, b x \mathrm {sgn}\left (b x + a\right ) + 5 \, a \mathrm {sgn}\left (b x + a\right )\right )} x + \frac {4 \, b c \mathrm {sgn}\left (b x + a\right )}{d}\right )} x + \frac {15 \, a c \mathrm {sgn}\left (b x + a\right )}{d}\right )} x - \frac {16 \, b c^{2} \mathrm {sgn}\left (b x + a\right )}{d^{2}}\right )} \]
1/8*a*c^2*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))*sgn(b*x + a)/d^(3/2) + 1/ 120*sqrt(d*x^2 + c)*((2*(3*(4*b*x*sgn(b*x + a) + 5*a*sgn(b*x + a))*x + 4*b *c*sgn(b*x + a)/d)*x + 15*a*c*sgn(b*x + a)/d)*x - 16*b*c^2*sgn(b*x + a)/d^ 2)
Timed out. \[ \int x^2 \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\int x^2\,\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+c} \,d x \]