Integrand size = 28, antiderivative size = 120 \[ \int \frac {(f+g x) (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=-\frac {b g x \sqrt {1+c^2 x^2}}{c \sqrt {d+c^2 d x^2}}+\frac {g \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{c^2 \sqrt {d+c^2 d x^2}}+\frac {f \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2}{2 b c \sqrt {d+c^2 d x^2}} \]
g*(c^2*x^2+1)*(a+b*arcsinh(c*x))/c^2/(c^2*d*x^2+d)^(1/2)-b*g*x*(c^2*x^2+1) ^(1/2)/c/(c^2*d*x^2+d)^(1/2)+1/2*f*(a+b*arcsinh(c*x))^2*(c^2*x^2+1)^(1/2)/ b/c/(c^2*d*x^2+d)^(1/2)
Time = 0.35 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.32 \[ \int \frac {(f+g x) (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\frac {2 \sqrt {d} g \left (a+a c^2 x^2-b c x \sqrt {1+c^2 x^2}\right )+2 b \sqrt {d} g \left (1+c^2 x^2\right ) \text {arcsinh}(c x)+b c \sqrt {d} f \sqrt {1+c^2 x^2} \text {arcsinh}(c x)^2+2 a c f \sqrt {d+c^2 d x^2} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )}{2 c^2 \sqrt {d} \sqrt {d+c^2 d x^2}} \]
(2*Sqrt[d]*g*(a + a*c^2*x^2 - b*c*x*Sqrt[1 + c^2*x^2]) + 2*b*Sqrt[d]*g*(1 + c^2*x^2)*ArcSinh[c*x] + b*c*Sqrt[d]*f*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]^2 + 2*a*c*f*Sqrt[d + c^2*d*x^2]*Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]])/(2* c^2*Sqrt[d]*Sqrt[d + c^2*d*x^2])
Time = 0.47 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.70, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.107, Rules used = {6260, 6253, 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.
\(\displaystyle \int \frac {(f+g x) (a+b \text {arcsinh}(c x))}{\sqrt {c^2 d x^2+d}} \, dx\) |
\(\Big \downarrow \) 6260 |
\(\displaystyle \frac {\sqrt {c^2 x^2+1} \int \frac {(f+g x) (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}dx}{\sqrt {c^2 d x^2+d}}\) |
\(\Big \downarrow \) 6253 |
\(\displaystyle \frac {\sqrt {c^2 x^2+1} \int \left (\frac {f (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}+\frac {g x (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}\right )dx}{\sqrt {c^2 d x^2+d}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\sqrt {c^2 x^2+1} \left (\frac {g \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))}{c^2}+\frac {f (a+b \text {arcsinh}(c x))^2}{2 b c}-\frac {b g x}{c}\right )}{\sqrt {c^2 d x^2+d}}\) |
(Sqrt[1 + c^2*x^2]*(-((b*g*x)/c) + (g*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x ]))/c^2 + (f*(a + b*ArcSinh[c*x])^2)/(2*b*c)))/Sqrt[d + c^2*d*x^2]
3.1.49.3.1 Defintions of rubi rules used
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d _) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a + b*ArcSinh[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n , 0] && ((EqQ[n, 1] && GtQ[p, -1]) || GtQ[p, 0] || EqQ[m, 1] || (EqQ[m, 2] && LtQ[p, -2]))
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d _) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Simp[Simp[(d + e*x^2)^p/(1 + c^2*x^2) ^p] Int[(f + g*x)^m*(1 + c^2*x^2)^p*(a + b*ArcSinh[c*x])^n, x], x] /; Fre eQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e, c^2*d] && IntegerQ[m] && IntegerQ [p - 1/2] && !GtQ[d, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(226\) vs. \(2(108)=216\).
Time = 0.83 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.89
method | result | size |
default | \(\frac {a f \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{\sqrt {c^{2} d}}+\frac {a g \sqrt {c^{2} d \,x^{2}+d}}{c^{2} d}+b \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, f \operatorname {arcsinh}\left (c x \right )^{2}}{2 \sqrt {c^{2} x^{2}+1}\, c d}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}+c x \sqrt {c^{2} x^{2}+1}+1\right ) g \left (-1+\operatorname {arcsinh}\left (c x \right )\right )}{2 c^{2} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}-c x \sqrt {c^{2} x^{2}+1}+1\right ) g \left (\operatorname {arcsinh}\left (c x \right )+1\right )}{2 c^{2} d \left (c^{2} x^{2}+1\right )}\right )\) | \(227\) |
parts | \(\frac {a f \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{\sqrt {c^{2} d}}+\frac {a g \sqrt {c^{2} d \,x^{2}+d}}{c^{2} d}+b \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, f \operatorname {arcsinh}\left (c x \right )^{2}}{2 \sqrt {c^{2} x^{2}+1}\, c d}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}+c x \sqrt {c^{2} x^{2}+1}+1\right ) g \left (-1+\operatorname {arcsinh}\left (c x \right )\right )}{2 c^{2} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}-c x \sqrt {c^{2} x^{2}+1}+1\right ) g \left (\operatorname {arcsinh}\left (c x \right )+1\right )}{2 c^{2} d \left (c^{2} x^{2}+1\right )}\right )\) | \(227\) |
a*f*ln(c^2*d*x/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)+a*g/c^2/d* (c^2*d*x^2+d)^(1/2)+b*(1/2*(d*(c^2*x^2+1))^(1/2)/(c^2*x^2+1)^(1/2)/c/d*f*a rcsinh(c*x)^2+1/2*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2+c*x*(c^2*x^2+1)^(1/2)+1)* g*(-1+arcsinh(c*x))/c^2/d/(c^2*x^2+1)+1/2*(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-c *x*(c^2*x^2+1)^(1/2)+1)*g*(arcsinh(c*x)+1)/c^2/d/(c^2*x^2+1))
\[ \int \frac {(f+g x) (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\int { \frac {{\left (g x + f\right )} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{\sqrt {c^{2} d x^{2} + d}} \,d x } \]
\[ \int \frac {(f+g x) (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\int \frac {\left (a + b \operatorname {asinh}{\left (c x \right )}\right ) \left (f + g x\right )}{\sqrt {d \left (c^{2} x^{2} + 1\right )}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.72 \[ \int \frac {(f+g x) (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\frac {b f \operatorname {arsinh}\left (c x\right )^{2}}{2 \, c \sqrt {d}} - \frac {b g x}{c \sqrt {d}} + \frac {a f \operatorname {arsinh}\left (c x\right )}{c \sqrt {d}} + \frac {\sqrt {c^{2} d x^{2} + d} b g \operatorname {arsinh}\left (c x\right )}{c^{2} d} + \frac {\sqrt {c^{2} d x^{2} + d} a g}{c^{2} d} \]
1/2*b*f*arcsinh(c*x)^2/(c*sqrt(d)) - b*g*x/(c*sqrt(d)) + a*f*arcsinh(c*x)/ (c*sqrt(d)) + sqrt(c^2*d*x^2 + d)*b*g*arcsinh(c*x)/(c^2*d) + sqrt(c^2*d*x^ 2 + d)*a*g/(c^2*d)
\[ \int \frac {(f+g x) (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\int { \frac {{\left (g x + f\right )} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{\sqrt {c^{2} d x^{2} + d}} \,d x } \]
Timed out. \[ \int \frac {(f+g x) (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\int \frac {\left (f+g\,x\right )\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{\sqrt {d\,c^2\,x^2+d}} \,d x \]