Integrand size = 27, antiderivative size = 276 \[ \int \frac {1}{c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\frac {2 b^2 c^3+2 a b^2 c x+4 a^2 c^3 x^2+\left (-b^2 c^2+3 a b^2 x+2 a c^4 x\right ) \sqrt {a x+\sqrt {b^2+a^2 x^2}}+\sqrt {b^2+a^2 x^2} \left (2 b^2 c+4 a c^3 x+\left (3 b^2+2 c^4\right ) \sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}{2 a c^3 \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}+\frac {b^2 \log \left (a x+\sqrt {b^2+a^2 x^2}\right )}{2 a c^3}-\frac {b^2 \log \left (c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}{a c^3}-\frac {c \log \left (c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}{a} \]
1/2*(2*b^2*c^3+2*a*b^2*c*x+4*a^2*c^3*x^2+(2*a*c^4*x+3*a*b^2*x-b^2*c^2)*(a* x+(a^2*x^2+b^2)^(1/2))^(1/2)+(a^2*x^2+b^2)^(1/2)*(2*b^2*c+4*a*c^3*x+(2*c^4 +3*b^2)*(a*x+(a^2*x^2+b^2)^(1/2))^(1/2)))/a/c^3/(a*x+(a^2*x^2+b^2)^(1/2))^ (3/2)+1/2*b^2*ln(a*x+(a^2*x^2+b^2)^(1/2))/a/c^3-b^2*ln(c+(a*x+(a^2*x^2+b^2 )^(1/2))^(1/2))/a/c^3-c*ln(c+(a*x+(a^2*x^2+b^2)^(1/2))^(1/2))/a
Time = 0.67 (sec) , antiderivative size = 259, normalized size of antiderivative = 0.94 \[ \int \frac {1}{c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\frac {\frac {2 b^2 c^3+2 a b^2 c x+4 a^2 c^3 x^2+\left (2 a c^4 x-b^2 \left (c^2-3 a x\right )\right ) \sqrt {a x+\sqrt {b^2+a^2 x^2}}+\sqrt {b^2+a^2 x^2} \left (2 b^2 c+4 a c^3 x+\left (3 b^2+2 c^4\right ) \sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}{\left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}+b^2 \log \left (a x+\sqrt {b^2+a^2 x^2}\right )-2 b^2 \log \left (c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )-2 c^4 \log \left (c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}{2 a c^3} \]
((2*b^2*c^3 + 2*a*b^2*c*x + 4*a^2*c^3*x^2 + (2*a*c^4*x - b^2*(c^2 - 3*a*x) )*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]] + Sqrt[b^2 + a^2*x^2]*(2*b^2*c + 4*a*c^3 *x + (3*b^2 + 2*c^4)*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]]))/(a*x + Sqrt[b^2 + a ^2*x^2])^(3/2) + b^2*Log[a*x + Sqrt[b^2 + a^2*x^2]] - 2*b^2*Log[c + Sqrt[a *x + Sqrt[b^2 + a^2*x^2]]] - 2*c^4*Log[c + Sqrt[a*x + Sqrt[b^2 + a^2*x^2]] ])/(2*a*c^3)
Time = 0.33 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.53, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2542, 2361, 2123, 2009}
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.
\(\displaystyle \int \frac {1}{\sqrt {\sqrt {a^2 x^2+b^2}+a x}+c} \, dx\) |
\(\Big \downarrow \) 2542 |
\(\displaystyle \frac {\int \frac {b^2+\left (a x+\sqrt {b^2+a^2 x^2}\right )^2}{\left (a x+\sqrt {b^2+a^2 x^2}\right )^2 \left (c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}d\left (a x+\sqrt {b^2+a^2 x^2}\right )}{2 a}\) |
\(\Big \downarrow \) 2361 |
\(\displaystyle \frac {\int \frac {b^2+\left (a x+\sqrt {b^2+a^2 x^2}\right )^2}{\left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2} \left (c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}d\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{a}\) |
\(\Big \downarrow \) 2123 |
\(\displaystyle \frac {\int \left (\frac {b^2}{c^3 \sqrt {a x+\sqrt {b^2+a^2 x^2}}}-\frac {b^2}{c^2 \left (a x+\sqrt {b^2+a^2 x^2}\right )}+\frac {b^2}{c \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}+\frac {-c^4-b^2}{c^3 \left (c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}\right )}+1\right )d\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {b^2 \log \left (\sqrt {a^2 x^2+b^2}+a x\right )}{c^3}+\frac {b^2}{c^2 \left (\sqrt {a^2 x^2+b^2}+a x\right )}-\frac {\left (b^2+c^4\right ) \log \left (\sqrt {a^2 x^2+b^2}+a x+c\right )}{c^3}-\frac {b^2}{2 c \left (\sqrt {a^2 x^2+b^2}+a x\right )^2}+\sqrt {\sqrt {a^2 x^2+b^2}+a x}}{a}\) |
(-1/2*b^2/(c*(a*x + Sqrt[b^2 + a^2*x^2])^2) + b^2/(c^2*(a*x + Sqrt[b^2 + a ^2*x^2])) + Sqrt[a*x + Sqrt[b^2 + a^2*x^2]] + (b^2*Log[a*x + Sqrt[b^2 + a^ 2*x^2]])/c^3 - ((b^2 + c^4)*Log[c + a*x + Sqrt[b^2 + a^2*x^2]])/c^3)/a
3.29.14.3.1 Defintions of rubi rules used
Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c , d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2])
Int[(Pq_)*(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*SubstFor[x^n, Pq, x]*(a + b*x)^p, x ], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && PolyQ[Pq, x^n] && IntegerQ[S implify[(m + 1)/n]]
Int[((g_.) + (h_.)*((d_.) + (e_.)*(x_) + (f_.)*Sqrt[(a_) + (c_.)*(x_)^2])^( n_))^(p_.), x_Symbol] :> Simp[1/(2*e) Subst[Int[(g + h*x^n)^p*((d^2 + a*f ^2 - 2*d*x + x^2)/(d - x)^2), x], x, d + e*x + f*Sqrt[a + c*x^2]], x] /; Fr eeQ[{a, c, d, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && IntegerQ[p]
\[\int \frac {1}{c +\sqrt {a x +\sqrt {a^{2} x^{2}+b^{2}}}}d x\]
Time = 0.27 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.49 \[ \int \frac {1}{c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\frac {a c^{2} x + 2 \, b^{2} \log \left (\sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}\right ) - \sqrt {a^{2} x^{2} + b^{2}} c^{2} - 2 \, {\left (c^{4} + b^{2}\right )} \log \left (c + \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}\right ) + 2 \, {\left (c^{3} - a c x + \sqrt {a^{2} x^{2} + b^{2}} c\right )} \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}{2 \, a c^{3}} \]
1/2*(a*c^2*x + 2*b^2*log(sqrt(a*x + sqrt(a^2*x^2 + b^2))) - sqrt(a^2*x^2 + b^2)*c^2 - 2*(c^4 + b^2)*log(c + sqrt(a*x + sqrt(a^2*x^2 + b^2))) + 2*(c^ 3 - a*c*x + sqrt(a^2*x^2 + b^2)*c)*sqrt(a*x + sqrt(a^2*x^2 + b^2)))/(a*c^3 )
\[ \int \frac {1}{c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\int \frac {1}{c + \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}\, dx \]
\[ \int \frac {1}{c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\int { \frac {1}{c + \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}} \,d x } \]
\[ \int \frac {1}{c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\int { \frac {1}{c + \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}} \,d x } \]
Timed out. \[ \int \frac {1}{c+\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\int \frac {1}{c+\sqrt {a\,x+\sqrt {a^2\,x^2+b^2}}} \,d x \]