Integrand size = 33, antiderivative size = 161 \[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=-\frac {b c x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{8 d (a+b x)}+\frac {(4 a+3 b x) \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{12 d (a+b x)}-\frac {b c^2 \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{3/2} (a+b x)} \]
1/12*(3*b*x+4*a)*(d*x^2+c)^(3/2)*((b*x+a)^2)^(1/2)/d/(b*x+a)-1/8*b*c^2*arc tanh(x*d^(1/2)/(d*x^2+c)^(1/2))*((b*x+a)^2)^(1/2)/d^(3/2)/(b*x+a)-1/8*b*c* x*((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2)/d/(b*x+a)
Time = 0.13 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.59 \[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (\sqrt {d} \sqrt {c+d x^2} \left (8 a \left (c+d x^2\right )+3 b x \left (c+2 d x^2\right )\right )+3 b c^2 \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )\right )}{24 d^{3/2} (a+b x)} \]
(Sqrt[(a + b*x)^2]*(Sqrt[d]*Sqrt[c + d*x^2]*(8*a*(c + d*x^2) + 3*b*x*(c + 2*d*x^2)) + 3*b*c^2*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]]))/(24*d^(3/2)*(a + b*x))
Time = 0.24 (sec) , antiderivative size = 122, normalized size of antiderivative = 0.76, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {1334, 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.
\(\displaystyle \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx\) |
\(\Big \downarrow \) 1334 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int 2 b x (a+b x) \sqrt {d x^2+c}dx}{2 b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int x (a+b x) \sqrt {d x^2+c}dx}{a+b x}\) |
\(\Big \downarrow \) 533 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {\int (b c-4 a d x) \sqrt {d x^2+c}dx}{4 d}\right )}{a+b x}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {b c \int \sqrt {d x^2+c}dx-\frac {4}{3} a \left (c+d x^2\right )^{3/2}}{4 d}\right )}{a+b x}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {b c \left (\frac {1}{2} c \int \frac {1}{\sqrt {d x^2+c}}dx+\frac {1}{2} x \sqrt {c+d x^2}\right )-\frac {4}{3} a \left (c+d x^2\right )^{3/2}}{4 d}\right )}{a+b x}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {b c \left (\frac {1}{2} c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}+\frac {1}{2} x \sqrt {c+d x^2}\right )-\frac {4}{3} a \left (c+d x^2\right )^{3/2}}{4 d}\right )}{a+b x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (\frac {b x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {b c \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}}+\frac {1}{2} x \sqrt {c+d x^2}\right )-\frac {4}{3} a \left (c+d x^2\right )^{3/2}}{4 d}\right )}{a+b x}\) |
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((b*x*(c + d*x^2)^(3/2))/(4*d) - ((-4*a*(c + d*x^2)^(3/2))/3 + b*c*((x*Sqrt[c + d*x^2])/2 + (c*ArcTanh[(Sqrt[d]*x)/Sq rt[c + d*x^2]])/(2*Sqrt[d])))/(4*d)))/(a + b*x)
3.1.41.3.1 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 = 0.51 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.52
method | result | size |
default | \(\frac {\operatorname {csgn}\left (b x +a \right ) \left (6 \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}\, b x +8 a \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}-3 \sqrt {d \,x^{2}+c}\, \sqrt {d}\, b c x -3 \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) b \,c^{2}\right )}{24 d^{\frac {3}{2}}}\) | \(83\) |
risch | \(\frac {\left (6 b d \,x^{3}+8 a d \,x^{2}+3 b c x +8 a c \right ) \sqrt {d \,x^{2}+c}\, \sqrt {\left (b x +a \right )^{2}}}{24 d \left (b x +a \right )}-\frac {c^{2} b \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) \sqrt {\left (b x +a \right )^{2}}}{8 d^{\frac {3}{2}} \left (b x +a \right )}\) | \(97\) |
1/24*csgn(b*x+a)*(6*(d*x^2+c)^(3/2)*d^(1/2)*b*x+8*a*(d*x^2+c)^(3/2)*d^(1/2 )-3*(d*x^2+c)^(1/2)*d^(1/2)*b*c*x-3*ln(d^(1/2)*x+(d*x^2+c)^(1/2))*b*c^2)/d ^(3/2)
Time = 0.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.98 \[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\left [\frac {3 \, b c^{2} \sqrt {d} \log \left (-2 \, d x^{2} + 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (6 \, b d^{2} x^{3} + 8 \, a d^{2} x^{2} + 3 \, b c d x + 8 \, a c d\right )} \sqrt {d x^{2} + c}}{48 \, d^{2}}, \frac {3 \, b c^{2} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (6 \, b d^{2} x^{3} + 8 \, a d^{2} x^{2} + 3 \, b c d x + 8 \, a c d\right )} \sqrt {d x^{2} + c}}{24 \, d^{2}}\right ] \]
[1/48*(3*b*c^2*sqrt(d)*log(-2*d*x^2 + 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2 *(6*b*d^2*x^3 + 8*a*d^2*x^2 + 3*b*c*d*x + 8*a*c*d)*sqrt(d*x^2 + c))/d^2, 1 /24*(3*b*c^2*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (6*b*d^2*x^3 + 8*a*d^2*x^2 + 3*b*c*d*x + 8*a*c*d)*sqrt(d*x^2 + c))/d^2]
\[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\int x \sqrt {c + d x^{2}} \sqrt {\left (a + b x\right )^{2}}\, dx \]
\[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\int { \sqrt {d x^{2} + c} \sqrt {{\left (b x + a\right )}^{2}} x \,d x } \]
Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.61 \[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\frac {b c^{2} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{8 \, d^{\frac {3}{2}}} + \frac {1}{24} \, \sqrt {d x^{2} + c} {\left ({\left (2 \, {\left (3 \, b x \mathrm {sgn}\left (b x + a\right ) + 4 \, a \mathrm {sgn}\left (b x + a\right )\right )} x + \frac {3 \, b c \mathrm {sgn}\left (b x + a\right )}{d}\right )} x + \frac {8 \, a c \mathrm {sgn}\left (b x + a\right )}{d}\right )} \]
1/8*b*c^2*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))*sgn(b*x + a)/d^(3/2) + 1/ 24*sqrt(d*x^2 + c)*((2*(3*b*x*sgn(b*x + a) + 4*a*sgn(b*x + a))*x + 3*b*c*s gn(b*x + a)/d)*x + 8*a*c*sgn(b*x + a)/d)
Timed out. \[ \int x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\int x\,\sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+c} \,d x \]