Integrand size = 34, antiderivative size = 178 \[ \int \sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=-\frac {(b c-4 a d) x \sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{8 d \left (a+b x^2\right )}+\frac {b x \left (c+d x^2\right )^{3/2} \sqrt {a^2+2 a b x^2+b^2 x^4}}{4 d \left (a+b x^2\right )}-\frac {c (b c-4 a d) \sqrt {a^2+2 a b x^2+b^2 x^4} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{3/2} \left (a+b x^2\right )} \]
1/4*b*x*(d*x^2+c)^(3/2)*((b*x^2+a)^2)^(1/2)/d/(b*x^2+a)-1/8*c*(-4*a*d+b*c) *arctanh(x*d^(1/2)/(d*x^2+c)^(1/2))*((b*x^2+a)^2)^(1/2)/d^(3/2)/(b*x^2+a)- 1/8*(-4*a*d+b*c)*x*(d*x^2+c)^(1/2)*((b*x^2+a)^2)^(1/2)/d/(b*x^2+a)
Time = 0.12 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.54 \[ \int \sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {\sqrt {\left (a+b x^2\right )^2} \left (\sqrt {d} x \sqrt {c+d x^2} \left (4 a d+b \left (c+2 d x^2\right )\right )+c (b c-4 a d) \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )\right )}{8 d^{3/2} \left (a+b x^2\right )} \]
(Sqrt[(a + b*x^2)^2]*(Sqrt[d]*x*Sqrt[c + d*x^2]*(4*a*d + b*(c + 2*d*x^2)) + c*(b*c - 4*a*d)*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]]))/(8*d^(3/2)*(a + b* x^2))
Time = 0.24 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.64, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {1384, 27, 299, 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.
\(\displaystyle \int \sqrt {a^2+2 a b x^2+b^2 x^4} \sqrt {c+d x^2} \, dx\) |
\(\Big \downarrow \) 1384 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int b \left (b x^2+a\right ) \sqrt {d x^2+c}dx}{b \left (a+b x^2\right )}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \int \left (b x^2+a\right ) \sqrt {d x^2+c}dx}{a+b x^2}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {b x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {(b c-4 a d) \int \sqrt {d x^2+c}dx}{4 d}\right )}{a+b x^2}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {b x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {(b c-4 a d) \left (\frac {1}{2} c \int \frac {1}{\sqrt {d x^2+c}}dx+\frac {1}{2} x \sqrt {c+d x^2}\right )}{4 d}\right )}{a+b x^2}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {b x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {(b c-4 a d) \left (\frac {1}{2} c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}+\frac {1}{2} x \sqrt {c+d x^2}\right )}{4 d}\right )}{a+b x^2}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x^2+b^2 x^4} \left (\frac {b x \left (c+d x^2\right )^{3/2}}{4 d}-\frac {(b c-4 a d) \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}}+\frac {1}{2} x \sqrt {c+d x^2}\right )}{4 d}\right )}{a+b x^2}\) |
(Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4]*((b*x*(c + d*x^2)^(3/2))/(4*d) - ((b*c - 4*a*d)*((x*Sqrt[c + d*x^2])/2 + (c*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/( 2*Sqrt[d])))/(4*d)))/(a + b*x^2)
3.3.69.3.1 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[((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]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> S imp[(a + b*x^n + c*x^(2*n))^FracPart[p]/(c^IntPart[p]*(b/2 + c*x^n)^(2*Frac Part[p])) Int[u*(b/2 + c*x^n)^(2*p), x], x] /; FreeQ[{a, b, c, n, p}, x] && EqQ[n2, 2*n] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2] && NeQ[u, x^(n - 1)] && NeQ[u, x^(2*n - 1)] && !(EqQ[p, 1/2] && EqQ[u, x^(-2*n - 1)])
Time = 0.32 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.58
method | result | size |
risch | \(\frac {x \left (2 b d \,x^{2}+4 d a +b c \right ) \sqrt {d \,x^{2}+c}\, \sqrt {\left (b \,x^{2}+a \right )^{2}}}{8 d \left (b \,x^{2}+a \right )}+\frac {c \left (4 d a -b c \right ) \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) \sqrt {\left (b \,x^{2}+a \right )^{2}}}{8 d^{\frac {3}{2}} \left (b \,x^{2}+a \right )}\) | \(103\) |
default | \(\frac {\sqrt {\left (b \,x^{2}+a \right )^{2}}\, \left (2 \sqrt {d}\, \left (d \,x^{2}+c \right )^{\frac {3}{2}} b x +4 d^{\frac {3}{2}} \sqrt {d \,x^{2}+c}\, a x -\sqrt {d}\, \sqrt {d \,x^{2}+c}\, b c x +4 \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) a c d -\ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right ) b \,c^{2}\right )}{8 \left (b \,x^{2}+a \right ) d^{\frac {3}{2}}}\) | \(119\) |
1/8*x*(2*b*d*x^2+4*a*d+b*c)*(d*x^2+c)^(1/2)/d*((b*x^2+a)^2)^(1/2)/(b*x^2+a )+1/8*c*(4*a*d-b*c)/d^(3/2)*ln(d^(1/2)*x+(d*x^2+c)^(1/2))*((b*x^2+a)^2)^(1 /2)/(b*x^2+a)
Time = 0.27 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.87 \[ \int \sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\left [-\frac {{\left (b c^{2} - 4 \, a c d\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) - 2 \, {\left (2 \, b d^{2} x^{3} + {\left (b c d + 4 \, a d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{16 \, d^{2}}, \frac {{\left (b c^{2} - 4 \, a c d\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) + {\left (2 \, b d^{2} x^{3} + {\left (b c d + 4 \, a d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{8 \, d^{2}}\right ] \]
[-1/16*((b*c^2 - 4*a*c*d)*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d) *x - c) - 2*(2*b*d^2*x^3 + (b*c*d + 4*a*d^2)*x)*sqrt(d*x^2 + c))/d^2, 1/8* ((b*c^2 - 4*a*c*d)*sqrt(-d)*arctan(sqrt(-d)*x/sqrt(d*x^2 + c)) + (2*b*d^2* x^3 + (b*c*d + 4*a*d^2)*x)*sqrt(d*x^2 + c))/d^2]
\[ \int \sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\int \sqrt {c + d x^{2}} \sqrt {\left (a + b x^{2}\right )^{2}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.46 \[ \int \sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{2} \, \sqrt {d x^{2} + c} a x + \frac {{\left (d x^{2} + c\right )}^{\frac {3}{2}} b x}{4 \, d} - \frac {\sqrt {d x^{2} + c} b c x}{8 \, d} - \frac {b c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {3}{2}}} + \frac {a c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{2 \, \sqrt {d}} \]
1/2*sqrt(d*x^2 + c)*a*x + 1/4*(d*x^2 + c)^(3/2)*b*x/d - 1/8*sqrt(d*x^2 + c )*b*c*x/d - 1/8*b*c^2*arcsinh(d*x/sqrt(c*d))/d^(3/2) + 1/2*a*c*arcsinh(d*x /sqrt(c*d))/sqrt(d)
Time = 0.28 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.61 \[ \int \sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\frac {1}{8} \, {\left (2 \, b x^{2} \mathrm {sgn}\left (b x^{2} + a\right ) + \frac {b c d \mathrm {sgn}\left (b x^{2} + a\right ) + 4 \, a d^{2} \mathrm {sgn}\left (b x^{2} + a\right )}{d^{2}}\right )} \sqrt {d x^{2} + c} x + \frac {{\left (b c^{2} \mathrm {sgn}\left (b x^{2} + a\right ) - 4 \, a c d \mathrm {sgn}\left (b x^{2} + a\right )\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{8 \, d^{\frac {3}{2}}} \]
1/8*(2*b*x^2*sgn(b*x^2 + a) + (b*c*d*sgn(b*x^2 + a) + 4*a*d^2*sgn(b*x^2 + a))/d^2)*sqrt(d*x^2 + c)*x + 1/8*(b*c^2*sgn(b*x^2 + a) - 4*a*c*d*sgn(b*x^2 + a))*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(3/2)
Timed out. \[ \int \sqrt {c+d x^2} \sqrt {a^2+2 a b x^2+b^2 x^4} \, dx=\int \sqrt {d\,x^2+c}\,\sqrt {{\left (b\,x^2+a\right )}^2} \,d x \]