Integrand size = 29, antiderivative size = 128 \[ \int \frac {d+c x}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\frac {b^4 c}{10 a^2 \left (a x+\sqrt {b^2+a^2 x^2}\right )^{5/2}}-\frac {b^2 d}{3 a \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}+\frac {d \sqrt {a x+\sqrt {b^2+a^2 x^2}}}{a}+\frac {c \left (a x+\sqrt {b^2+a^2 x^2}\right )^{3/2}}{6 a^2} \]
1/10*b^4*c/a^2/(a*x+(a^2*x^2+b^2)^(1/2))^(5/2)-1/3*b^2*d/a/(a*x+(a^2*x^2+b ^2)^(1/2))^(3/2)+d*(a*x+(a^2*x^2+b^2)^(1/2))^(1/2)/a+1/6*c*(a*x+(a^2*x^2+b ^2)^(1/2))^(3/2)/a^2
Time = 0.19 (sec) , antiderivative size = 110, normalized size of antiderivative = 0.86 \[ \int \frac {d+c x}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\frac {4 b^4 c+20 a^3 x^2 (3 d+c x) \left (a x+\sqrt {b^2+a^2 x^2}\right )+10 a b^2 \left (2 a x (2 d+c x)+(d+c x) \sqrt {b^2+a^2 x^2}\right )}{15 a^2 \left (a x+\sqrt {b^2+a^2 x^2}\right )^{5/2}} \]
(4*b^4*c + 20*a^3*x^2*(3*d + c*x)*(a*x + Sqrt[b^2 + a^2*x^2]) + 10*a*b^2*( 2*a*x*(2*d + c*x) + (d + c*x)*Sqrt[b^2 + a^2*x^2]))/(15*a^2*(a*x + Sqrt[b^ 2 + a^2*x^2])^(5/2))
Time = 0.40 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.98, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2544, 25, 2035, 2195, 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.
\(\displaystyle \int \frac {c x+d}{\sqrt {\sqrt {a^2 x^2+b^2}+a x}} \, dx\) |
\(\Big \downarrow \) 2544 |
\(\displaystyle \frac {\int -\frac {\left (b^2+\left (a x+\sqrt {b^2+a^2 x^2}\right )^2\right ) \left (c b^2-c \left (a x+\sqrt {b^2+a^2 x^2}\right )^2-2 a d \left (a x+\sqrt {b^2+a^2 x^2}\right )\right )}{\left (a x+\sqrt {b^2+a^2 x^2}\right )^{7/2}}d\left (a x+\sqrt {b^2+a^2 x^2}\right )}{4 a^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\left (b^2+\left (a x+\sqrt {b^2+a^2 x^2}\right )^2\right ) \left (c b^2-c \left (a x+\sqrt {b^2+a^2 x^2}\right )^2-2 a d \left (a x+\sqrt {b^2+a^2 x^2}\right )\right )}{\left (a x+\sqrt {b^2+a^2 x^2}\right )^{7/2}}d\left (a x+\sqrt {b^2+a^2 x^2}\right )}{4 a^2}\) |
\(\Big \downarrow \) 2035 |
\(\displaystyle -\frac {\int \frac {\left (b^2+\left (a x+\sqrt {b^2+a^2 x^2}\right )^2\right ) \left (c b^2-c \left (a x+\sqrt {b^2+a^2 x^2}\right )^2-2 a d \left (a x+\sqrt {b^2+a^2 x^2}\right )\right )}{\left (a x+\sqrt {b^2+a^2 x^2}\right )^3}d\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{2 a^2}\) |
\(\Big \downarrow \) 2195 |
\(\displaystyle -\frac {\int \left (\frac {c b^4}{\left (a x+\sqrt {b^2+a^2 x^2}\right )^3}-\frac {2 a d b^2}{\left (a x+\sqrt {b^2+a^2 x^2}\right )^2}-2 a d-c \left (a x+\sqrt {b^2+a^2 x^2}\right )\right )d\sqrt {a x+\sqrt {b^2+a^2 x^2}}}{2 a^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {-\frac {1}{3} c \left (\sqrt {a^2 x^2+b^2}+a x\right )^{3/2}+\frac {2 a b^2 d}{3 \left (\sqrt {a^2 x^2+b^2}+a x\right )^{3/2}}-2 a d \sqrt {\sqrt {a^2 x^2+b^2}+a x}-\frac {b^4 c}{5 \left (\sqrt {a^2 x^2+b^2}+a x\right )^{5/2}}}{2 a^2}\) |
-1/2*(-1/5*(b^4*c)/(a*x + Sqrt[b^2 + a^2*x^2])^(5/2) + (2*a*b^2*d)/(3*(a*x + Sqrt[b^2 + a^2*x^2])^(3/2)) - 2*a*d*Sqrt[a*x + Sqrt[b^2 + a^2*x^2]] - ( c*(a*x + Sqrt[b^2 + a^2*x^2])^(3/2))/3)/a^2
3.19.69.3.1 Defintions of rubi rules used
Int[(Fx_)*(x_)^(m_), x_Symbol] :> With[{k = Denominator[m]}, Simp[k Subst [Int[x^(k*(m + 1) - 1)*SubstPower[Fx, x, k], x], x, x^(1/k)], x]] /; Fracti onQ[m] && AlgebraicFunctionQ[Fx, x]
Int[(Pq_)*((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_.), x_ Symbol] :> Int[ExpandIntegrand[(d*x)^m*Pq*(a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, m}, x] && PolyQ[Pq, x^2] && IGtQ[p, -2]
Int[((g_.) + (h_.)*(x_))^(m_.)*((e_.)*(x_) + (f_.)*Sqrt[(a_.) + (c_.)*(x_)^ 2])^(n_.), x_Symbol] :> Simp[1/(2^(m + 1)*e^(m + 1)) Subst[Int[x^(n - m - 2)*(a*f^2 + x^2)*((-a)*f^2*h + 2*e*g*x + h*x^2)^m, x], x, e*x + f*Sqrt[a + c*x^2]], x] /; FreeQ[{a, c, e, f, g, h, n}, x] && EqQ[e^2 - c*f^2, 0] && I ntegerQ[m]
\[\int \frac {c x +d}{\sqrt {a x +\sqrt {a^{2} x^{2}+b^{2}}}}d x\]
Time = 0.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.77 \[ \int \frac {d+c x}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=-\frac {2 \, {\left (3 \, a^{3} c x^{3} + 5 \, a^{3} d x^{2} + a b^{2} c x - 5 \, a b^{2} d - {\left (3 \, a^{2} c x^{2} + 5 \, a^{2} d x + 2 \, b^{2} c\right )} \sqrt {a^{2} x^{2} + b^{2}}\right )} \sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}{15 \, a^{2} b^{2}} \]
-2/15*(3*a^3*c*x^3 + 5*a^3*d*x^2 + a*b^2*c*x - 5*a*b^2*d - (3*a^2*c*x^2 + 5*a^2*d*x + 2*b^2*c)*sqrt(a^2*x^2 + b^2))*sqrt(a*x + sqrt(a^2*x^2 + b^2))/ (a^2*b^2)
\[ \int \frac {d+c x}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\int \frac {c x + d}{\sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}}\, dx \]
\[ \int \frac {d+c x}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\int { \frac {c x + d}{\sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}} \,d x } \]
\[ \int \frac {d+c x}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\int { \frac {c x + d}{\sqrt {a x + \sqrt {a^{2} x^{2} + b^{2}}}} \,d x } \]
Timed out. \[ \int \frac {d+c x}{\sqrt {a x+\sqrt {b^2+a^2 x^2}}} \, dx=\int \frac {d+c\,x}{\sqrt {a\,x+\sqrt {a^2\,x^2+b^2}}} \,d x \]