Integrand size = 32, antiderivative size = 148 \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\frac {a x \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2}}{2 (a+b x)}+\frac {b \sqrt {a^2+2 a b x+b^2 x^2} \left (c+d x^2\right )^{3/2}}{3 d (a+b x)}+\frac {a c \sqrt {a^2+2 a b x+b^2 x^2} \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d} (a+b x)} \] Output:
a*x*((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2)/(2*b*x+2*a)+1/3*b*((b*x+a)^2)^(1/2)* (d*x^2+c)^(3/2)/d/(b*x+a)+1/2*a*c*((b*x+a)^2)^(1/2)*arctanh(d^(1/2)*x/(d*x ^2+c)^(1/2))/d^(1/2)/(b*x+a)
Time = 0.23 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.57 \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\frac {\sqrt {(a+b x)^2} \left (\sqrt {c+d x^2} \left (3 a d x+2 b \left (c+d x^2\right )\right )-3 a c \sqrt {d} \log \left (-\sqrt {d} x+\sqrt {c+d x^2}\right )\right )}{6 d (a+b x)} \] Input:
Integrate[Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + d*x^2],x]
Output:
(Sqrt[(a + b*x)^2]*(Sqrt[c + d*x^2]*(3*a*d*x + 2*b*(c + d*x^2)) - 3*a*c*Sq rt[d]*Log[-(Sqrt[d]*x) + Sqrt[c + d*x^2]]))/(6*d*(a + b*x))
Time = 0.21 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.65, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {1298, 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.
\(\displaystyle \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx\) |
\(\Big \downarrow \) 1298 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int 2 b (a+b x) \sqrt {d x^2+c}dx}{2 b (a+b x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \int (a+b x) \sqrt {d x^2+c}dx}{a+b x}\) |
\(\Big \downarrow \) 455 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (a \int \sqrt {d x^2+c}dx+\frac {b \left (c+d x^2\right )^{3/2}}{3 d}\right )}{a+b x}\) |
\(\Big \downarrow \) 211 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (a \left (\frac {1}{2} c \int \frac {1}{\sqrt {d x^2+c}}dx+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {b \left (c+d x^2\right )^{3/2}}{3 d}\right )}{a+b x}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (a \left (\frac {1}{2} c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {b \left (c+d x^2\right )^{3/2}}{3 d}\right )}{a+b x}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (a \left (\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 \sqrt {d}}+\frac {1}{2} x \sqrt {c+d x^2}\right )+\frac {b \left (c+d x^2\right )^{3/2}}{3 d}\right )}{a+b x}\) |
Input:
Int[Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Sqrt[c + d*x^2],x]
Output:
(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*((b*(c + d*x^2)^(3/2))/(3*d) + a*((x*Sqrt[c + d*x^2])/2 + (c*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d*x^2]])/(2*Sqrt[d]))))/(a + b*x)
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[((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[(b + 2*c*x)^(2*p)*(d + f*x^2)^q, x], x] /; FreeQ[{a , b, c, d, f, p, q}, x] && EqQ[b^2 - 4*a*c, 0] && !IntegerQ[p]
Result contains higher order function than in optimal. Order 9 vs. order 3.
Time = 0.92 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.44
method | result | size |
default | \(\frac {\operatorname {csgn}\left (b x +a \right ) \left (2 \left (d \,x^{2}+c \right )^{\frac {3}{2}} \sqrt {d}\, b +3 a \sqrt {d \,x^{2}+c}\, d^{\frac {3}{2}} x +3 \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right ) a c d \right )}{6 d^{\frac {3}{2}}}\) | \(65\) |
risch | \(\frac {\left (2 b d \,x^{2}+3 a d x +2 b c \right ) \sqrt {d \,x^{2}+c}\, \sqrt {\left (b x +a \right )^{2}}}{6 d \left (b x +a \right )}+\frac {a c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right ) \sqrt {\left (b x +a \right )^{2}}}{2 \sqrt {d}\, \left (b x +a \right )}\) | \(88\) |
Input:
int(((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/6*csgn(b*x+a)*(2*(d*x^2+c)^(3/2)*d^(1/2)*b+3*a*(d*x^2+c)^(1/2)*d^(3/2)*x +3*ln(x*d^(1/2)+(d*x^2+c)^(1/2))*a*c*d)/d^(3/2)
Time = 0.07 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.86 \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\left [\frac {3 \, a c \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (2 \, b d x^{2} + 3 \, a d x + 2 \, b c\right )} \sqrt {d x^{2} + c}}{12 \, d}, -\frac {3 \, a c \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, b d x^{2} + 3 \, a d x + 2 \, b c\right )} \sqrt {d x^{2} + c}}{6 \, d}\right ] \] Input:
integrate(((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2),x, algorithm="fricas")
Output:
[1/12*(3*a*c*sqrt(d)*log(-2*d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*( 2*b*d*x^2 + 3*a*d*x + 2*b*c)*sqrt(d*x^2 + c))/d, -1/6*(3*a*c*sqrt(-d)*arct an(sqrt(-d)*x/sqrt(d*x^2 + c)) - (2*b*d*x^2 + 3*a*d*x + 2*b*c)*sqrt(d*x^2 + c))/d]
\[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\int \sqrt {c + d x^{2}} \sqrt {\left (a + b x\right )^{2}}\, dx \] Input:
integrate(((b*x+a)**2)**(1/2)*(d*x**2+c)**(1/2),x)
Output:
Integral(sqrt(c + d*x**2)*sqrt((a + b*x)**2), x)
\[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\int { \sqrt {d x^{2} + c} \sqrt {{\left (b x + a\right )}^{2}} \,d x } \] Input:
integrate(((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(d*x^2 + c)*sqrt((b*x + a)^2), x)
Time = 0.39 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.53 \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=-\frac {a c \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right ) \mathrm {sgn}\left (b x + a\right )}{2 \, \sqrt {d}} + \frac {1}{6} \, \sqrt {d x^{2} + c} {\left ({\left (2 \, b x \mathrm {sgn}\left (b x + a\right ) + 3 \, a \mathrm {sgn}\left (b x + a\right )\right )} x + \frac {2 \, b c \mathrm {sgn}\left (b x + a\right )}{d}\right )} \] Input:
integrate(((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2),x, algorithm="giac")
Output:
-1/2*a*c*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))*sgn(b*x + a)/sqrt(d) + 1/6 *sqrt(d*x^2 + c)*((2*b*x*sgn(b*x + a) + 3*a*sgn(b*x + a))*x + 2*b*c*sgn(b* x + a)/d)
Timed out. \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\int \sqrt {{\left (a+b\,x\right )}^2}\,\sqrt {d\,x^2+c} \,d x \] Input:
int(((a + b*x)^2)^(1/2)*(c + d*x^2)^(1/2),x)
Output:
int(((a + b*x)^2)^(1/2)*(c + d*x^2)^(1/2), x)
Time = 0.19 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.48 \[ \int \sqrt {a^2+2 a b x+b^2 x^2} \sqrt {c+d x^2} \, dx=\frac {3 \sqrt {d \,x^{2}+c}\, a d x +2 \sqrt {d \,x^{2}+c}\, b c +2 \sqrt {d \,x^{2}+c}\, b d \,x^{2}+3 \sqrt {d}\, \mathrm {log}\left (\frac {\sqrt {d \,x^{2}+c}+\sqrt {d}\, x}{\sqrt {c}}\right ) a c}{6 d} \] Input:
int(((b*x+a)^2)^(1/2)*(d*x^2+c)^(1/2),x)
Output:
(3*sqrt(c + d*x**2)*a*d*x + 2*sqrt(c + d*x**2)*b*c + 2*sqrt(c + d*x**2)*b* d*x**2 + 3*sqrt(d)*log((sqrt(c + d*x**2) + sqrt(d)*x)/sqrt(c))*a*c)/(6*d)