Integrand size = 30, antiderivative size = 258 \[ \int \frac {(f+g x)^2 (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=-\frac {2 b f g x \sqrt {1+c^2 x^2}}{c \sqrt {d+c^2 d x^2}}-\frac {b g^2 x^2 \sqrt {1+c^2 x^2}}{4 c \sqrt {d+c^2 d x^2}}+\frac {2 f g \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{c^2 \sqrt {d+c^2 d x^2}}+\frac {g^2 x \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{2 c^2 \sqrt {d+c^2 d x^2}}+\frac {f^2 \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2}{2 b c \sqrt {d+c^2 d x^2}}-\frac {g^2 \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2}{4 b c^3 \sqrt {d+c^2 d x^2}} \]
2*f*g*(c^2*x^2+1)*(a+b*arcsinh(c*x))/c^2/(c^2*d*x^2+d)^(1/2)+1/2*g^2*x*(c^ 2*x^2+1)*(a+b*arcsinh(c*x))/c^2/(c^2*d*x^2+d)^(1/2)-2*b*f*g*x*(c^2*x^2+1)^ (1/2)/c/(c^2*d*x^2+d)^(1/2)-1/4*b*g^2*x^2*(c^2*x^2+1)^(1/2)/c/(c^2*d*x^2+d )^(1/2)+1/2*f^2*(a+b*arcsinh(c*x))^2*(c^2*x^2+1)^(1/2)/b/c/(c^2*d*x^2+d)^( 1/2)-1/4*g^2*(a+b*arcsinh(c*x))^2*(c^2*x^2+1)^(1/2)/b/c^3/(c^2*d*x^2+d)^(1 /2)
Time = 0.71 (sec) , antiderivative size = 233, normalized size of antiderivative = 0.90 \[ \int \frac {(f+g x)^2 (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\frac {4 c \sqrt {d} g \left (-4 b c f x \sqrt {1+c^2 x^2}+a (4 f+g x) \left (1+c^2 x^2\right )\right )+4 b c \sqrt {d} g (4 f+g x) \left (1+c^2 x^2\right ) \text {arcsinh}(c x)+2 b \sqrt {d} \left (2 c^2 f^2-g^2\right ) \sqrt {1+c^2 x^2} \text {arcsinh}(c x)^2-b \sqrt {d} g^2 \sqrt {1+c^2 x^2} \cosh (2 \text {arcsinh}(c x))+4 a \left (2 c^2 f^2-g^2\right ) \sqrt {d+c^2 d x^2} \log \left (c d x+\sqrt {d} \sqrt {d+c^2 d x^2}\right )}{8 c^3 \sqrt {d} \sqrt {d+c^2 d x^2}} \]
(4*c*Sqrt[d]*g*(-4*b*c*f*x*Sqrt[1 + c^2*x^2] + a*(4*f + g*x)*(1 + c^2*x^2) ) + 4*b*c*Sqrt[d]*g*(4*f + g*x)*(1 + c^2*x^2)*ArcSinh[c*x] + 2*b*Sqrt[d]*( 2*c^2*f^2 - g^2)*Sqrt[1 + c^2*x^2]*ArcSinh[c*x]^2 - b*Sqrt[d]*g^2*Sqrt[1 + c^2*x^2]*Cosh[2*ArcSinh[c*x]] + 4*a*(2*c^2*f^2 - g^2)*Sqrt[d + c^2*d*x^2] *Log[c*d*x + Sqrt[d]*Sqrt[d + c^2*d*x^2]])/(8*c^3*Sqrt[d]*Sqrt[d + c^2*d*x ^2])
Time = 0.69 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.61, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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.
| \(\displaystyle \int \frac {(f+g x)^2 (a+b \text {arcsinh}(c x))}{\sqrt {c^2 d x^2+d}} \, dx\) |
| \(\Big \downarrow \) 6260 |
| \(\displaystyle \frac {\sqrt {c^2 x^2+1} \int \frac {(f+g x)^2 (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}dx}{\sqrt {c^2 d x^2+d}}\) |
| \(\Big \downarrow \) 6253 |
| \(\displaystyle \frac {\sqrt {c^2 x^2+1} \int \left (\frac {(a+b \text {arcsinh}(c x)) f^2}{\sqrt {c^2 x^2+1}}+\frac {2 g x (a+b \text {arcsinh}(c x)) f}{\sqrt {c^2 x^2+1}}+\frac {g^2 x^2 (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}\right )dx}{\sqrt {c^2 d x^2+d}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt {c^2 x^2+1} \left (-\frac {g^2 (a+b \text {arcsinh}(c x))^2}{4 b c^3}+\frac {2 f g \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))}{c^2}+\frac {g^2 x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))}{2 c^2}+\frac {f^2 (a+b \text {arcsinh}(c x))^2}{2 b c}-\frac {2 b f g x}{c}-\frac {b g^2 x^2}{4 c}\right )}{\sqrt {c^2 d x^2+d}}\) |
(Sqrt[1 + c^2*x^2]*((-2*b*f*g*x)/c - (b*g^2*x^2)/(4*c) + (2*f*g*Sqrt[1 + c ^2*x^2]*(a + b*ArcSinh[c*x]))/c^2 + (g^2*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSin h[c*x]))/(2*c^2) + (f^2*(a + b*ArcSinh[c*x])^2)/(2*b*c) - (g^2*(a + b*ArcS inh[c*x])^2)/(4*b*c^3)))/Sqrt[d + c^2*d*x^2]
3.1.48.3.1 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. \(483\) vs. \(2(230)=460\).
Time = 0.76 (sec) , antiderivative size = 484, normalized size of antiderivative = 1.88
| method | result | size |
| default | \(a \left (\frac {f^{2} \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{\sqrt {c^{2} d}}+g^{2} \left (\frac {x \sqrt {c^{2} d \,x^{2}+d}}{2 c^{2} d}-\frac {\ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{2 c^{2} \sqrt {c^{2} d}}\right )+\frac {2 f g \sqrt {c^{2} d \,x^{2}+d}}{c^{2} d}\right )+b \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (c x \right )^{2} \left (2 c^{2} f^{2}-g^{2}\right )}{4 \sqrt {c^{2} x^{2}+1}\, c^{3} d}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}+2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x +\sqrt {c^{2} x^{2}+1}\right ) g^{2} \left (-1+2 \,\operatorname {arcsinh}\left (c x \right )\right )}{16 c^{3} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}+c x \sqrt {c^{2} x^{2}+1}+1\right ) f g \left (-1+\operatorname {arcsinh}\left (c x \right )\right )}{c^{2} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}-c x \sqrt {c^{2} x^{2}+1}+1\right ) f g \left (\operatorname {arcsinh}\left (c x \right )+1\right )}{c^{2} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}-2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x -\sqrt {c^{2} x^{2}+1}\right ) g^{2} \left (1+2 \,\operatorname {arcsinh}\left (c x \right )\right )}{16 c^{3} d \left (c^{2} x^{2}+1\right )}\right )\) | \(484\) |
| parts | \(a \left (\frac {f^{2} \ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{\sqrt {c^{2} d}}+g^{2} \left (\frac {x \sqrt {c^{2} d \,x^{2}+d}}{2 c^{2} d}-\frac {\ln \left (\frac {c^{2} d x}{\sqrt {c^{2} d}}+\sqrt {c^{2} d \,x^{2}+d}\right )}{2 c^{2} \sqrt {c^{2} d}}\right )+\frac {2 f g \sqrt {c^{2} d \,x^{2}+d}}{c^{2} d}\right )+b \left (\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \operatorname {arcsinh}\left (c x \right )^{2} \left (2 c^{2} f^{2}-g^{2}\right )}{4 \sqrt {c^{2} x^{2}+1}\, c^{3} d}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}+2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x +\sqrt {c^{2} x^{2}+1}\right ) g^{2} \left (-1+2 \,\operatorname {arcsinh}\left (c x \right )\right )}{16 c^{3} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}+c x \sqrt {c^{2} x^{2}+1}+1\right ) f g \left (-1+\operatorname {arcsinh}\left (c x \right )\right )}{c^{2} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (c^{2} x^{2}-c x \sqrt {c^{2} x^{2}+1}+1\right ) f g \left (\operatorname {arcsinh}\left (c x \right )+1\right )}{c^{2} d \left (c^{2} x^{2}+1\right )}+\frac {\sqrt {d \left (c^{2} x^{2}+1\right )}\, \left (2 c^{3} x^{3}-2 c^{2} x^{2} \sqrt {c^{2} x^{2}+1}+2 c x -\sqrt {c^{2} x^{2}+1}\right ) g^{2} \left (1+2 \,\operatorname {arcsinh}\left (c x \right )\right )}{16 c^{3} d \left (c^{2} x^{2}+1\right )}\right )\) | \(484\) |
a*(f^2*ln(c^2*d*x/(c^2*d)^(1/2)+(c^2*d*x^2+d)^(1/2))/(c^2*d)^(1/2)+g^2*(1/ 2*x/c^2/d*(c^2*d*x^2+d)^(1/2)-1/2/c^2*ln(c^2*d*x/(c^2*d)^(1/2)+(c^2*d*x^2+ d)^(1/2))/(c^2*d)^(1/2))+2*f*g/c^2/d*(c^2*d*x^2+d)^(1/2))+b*(1/4*(d*(c^2*x ^2+1))^(1/2)*arcsinh(c*x)^2*(2*c^2*f^2-g^2)/(c^2*x^2+1)^(1/2)/c^3/d+1/16*( d*(c^2*x^2+1))^(1/2)*(2*c^3*x^3+2*c^2*x^2*(c^2*x^2+1)^(1/2)+2*c*x+(c^2*x^2 +1)^(1/2))*g^2*(-1+2*arcsinh(c*x))/c^3/d/(c^2*x^2+1)+(d*(c^2*x^2+1))^(1/2) *(c^2*x^2+c*x*(c^2*x^2+1)^(1/2)+1)*f*g*(-1+arcsinh(c*x))/c^2/d/(c^2*x^2+1) +(d*(c^2*x^2+1))^(1/2)*(c^2*x^2-c*x*(c^2*x^2+1)^(1/2)+1)*f*g*(arcsinh(c*x) +1)/c^2/d/(c^2*x^2+1)+1/16*(d*(c^2*x^2+1))^(1/2)*(2*c^3*x^3-2*c^2*x^2*(c^2 *x^2+1)^(1/2)+2*c*x-(c^2*x^2+1)^(1/2))*g^2*(1+2*arcsinh(c*x))/c^3/d/(c^2*x ^2+1))
\[ \int \frac {(f+g x)^2 (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\int { \frac {{\left (g x + f\right )}^{2} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{\sqrt {c^{2} d x^{2} + d}} \,d x } \]
integral((a*g^2*x^2 + 2*a*f*g*x + a*f^2 + (b*g^2*x^2 + 2*b*f*g*x + b*f^2)* arcsinh(c*x))/sqrt(c^2*d*x^2 + d), x)
\[ \int \frac {(f+g x)^2 (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\int \frac {\left (a + b \operatorname {asinh}{\left (c x \right )}\right ) \left (f + g x\right )^{2}}{\sqrt {d \left (c^{2} x^{2} + 1\right )}}\, dx \]
Exception generated. \[ \int \frac {(f+g x)^2 (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\text {Exception raised: RuntimeError} \]
\[ \int \frac {(f+g x)^2 (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\int { \frac {{\left (g x + f\right )}^{2} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{\sqrt {c^{2} d x^{2} + d}} \,d x } \]
Timed out. \[ \int \frac {(f+g x)^2 (a+b \text {arcsinh}(c x))}{\sqrt {d+c^2 d x^2}} \, dx=\int \frac {{\left (f+g\,x\right )}^2\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{\sqrt {d\,c^2\,x^2+d}} \,d x \]