Integrand size = 35, antiderivative size = 416 \[ \int (d+i c d x)^{5/2} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x)) \, dx=-\frac {2 i b d^2 x \sqrt {d+i c d x} \sqrt {f-i c f x}}{3 \sqrt {1+c^2 x^2}}-\frac {3 b c d^2 x^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}{16 \sqrt {1+c^2 x^2}}-\frac {2 i b c^2 d^2 x^3 \sqrt {d+i c d x} \sqrt {f-i c f x}}{9 \sqrt {1+c^2 x^2}}+\frac {b c^3 d^2 x^4 \sqrt {d+i c d x} \sqrt {f-i c f x}}{16 \sqrt {1+c^2 x^2}}+\frac {3}{8} d^2 x \sqrt {d+i c d x} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x))-\frac {1}{4} c^2 d^2 x^3 \sqrt {d+i c d x} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x))+\frac {2 i d^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{3 c}+\frac {5 d^2 \sqrt {d+i c d x} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x))^2}{16 b c \sqrt {1+c^2 x^2}} \]
3/8*d^2*x*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)-1/4*c^2*d ^2*x^3*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)+2/3*I*d^2*(c ^2*x^2+1)*(a+b*arcsinh(c*x))*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/c-2/3*I*b *d^2*x*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c^2*x^2+1)^(1/2)-3/16*b*c*d^2* x^2*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c^2*x^2+1)^(1/2)-2/9*I*b*c^2*d^2* x^3*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c^2*x^2+1)^(1/2)+1/16*b*c^3*d^2*x ^4*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/(c^2*x^2+1)^(1/2)+5/16*d^2*(a+b*arc sinh(c*x))^2*(d+I*c*d*x)^(1/2)*(f-I*c*f*x)^(1/2)/b/c/(c^2*x^2+1)^(1/2)
Time = 2.79 (sec) , antiderivative size = 361, normalized size of antiderivative = 0.87 \[ \int (d+i c d x)^{5/2} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x)) \, dx=\frac {48 a d^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2} \left (16 i+9 c x+16 i c^2 x^2-6 c^3 x^3\right )+720 a d^{5/2} \sqrt {f} \sqrt {1+c^2 x^2} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )+144 b d^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (2 \text {arcsinh}(c x)^2-\cosh (2 \text {arcsinh}(c x))+2 \text {arcsinh}(c x) \sinh (2 \text {arcsinh}(c x))\right )-64 i b d^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (9 c x-3 \text {arcsinh}(c x) \left (3 \sqrt {1+c^2 x^2}+\cosh (3 \text {arcsinh}(c x))\right )+\sinh (3 \text {arcsinh}(c x))\right )+9 b d^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (8 \text {arcsinh}(c x)^2+\cosh (4 \text {arcsinh}(c x))-4 \text {arcsinh}(c x) \sinh (4 \text {arcsinh}(c x))\right )}{1152 c \sqrt {1+c^2 x^2}} \]
(48*a*d^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2]*(16*I + 9* c*x + (16*I)*c^2*x^2 - 6*c^3*x^3) + 720*a*d^(5/2)*Sqrt[f]*Sqrt[1 + c^2*x^2 ]*Log[c*d*f*x + Sqrt[d]*Sqrt[f]*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]] + 144 *b*d^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(2*ArcSinh[c*x]^2 - Cosh[2*ArcS inh[c*x]] + 2*ArcSinh[c*x]*Sinh[2*ArcSinh[c*x]]) - (64*I)*b*d^2*Sqrt[d + I *c*d*x]*Sqrt[f - I*c*f*x]*(9*c*x - 3*ArcSinh[c*x]*(3*Sqrt[1 + c^2*x^2] + C osh[3*ArcSinh[c*x]]) + Sinh[3*ArcSinh[c*x]]) + 9*b*d^2*Sqrt[d + I*c*d*x]*S qrt[f - I*c*f*x]*(8*ArcSinh[c*x]^2 + Cosh[4*ArcSinh[c*x]] - 4*ArcSinh[c*x] *Sinh[4*ArcSinh[c*x]]))/(1152*c*Sqrt[1 + c^2*x^2])
Time = 0.83 (sec) , antiderivative size = 192, normalized size of antiderivative = 0.46, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {6211, 27, 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.
\(\displaystyle \int (d+i c d x)^{5/2} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x)) \, dx\) |
\(\Big \downarrow \) 6211 |
\(\displaystyle \frac {\sqrt {d+i c d x} \sqrt {f-i c f x} \int d^2 (i c x+1)^2 \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))dx}{\sqrt {c^2 x^2+1}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {d^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \int (i c x+1)^2 \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))dx}{\sqrt {c^2 x^2+1}}\) |
\(\Big \downarrow \) 6253 |
\(\displaystyle \frac {d^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \int \left (-c^2 \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x)) x^2+2 i c \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x)) x+\sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))\right )dx}{\sqrt {c^2 x^2+1}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {d^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \left (\frac {3}{8} x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))+\frac {2 i \left (c^2 x^2+1\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c}-\frac {1}{4} c^2 x^3 \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))+\frac {5 (a+b \text {arcsinh}(c x))^2}{16 b c}+\frac {1}{16} b c^3 x^4-\frac {2}{9} i b c^2 x^3-\frac {3}{16} b c x^2-\frac {2 i b x}{3}\right )}{\sqrt {c^2 x^2+1}}\) |
(d^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(((-2*I)/3)*b*x - (3*b*c*x^2)/16 - ((2*I)/9)*b*c^2*x^3 + (b*c^3*x^4)/16 + (3*x*Sqrt[1 + c^2*x^2]*(a + b*Arc Sinh[c*x]))/8 - (c^2*x^3*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/4 + (((2* I)/3)*(1 + c^2*x^2)^(3/2)*(a + b*ArcSinh[c*x]))/c + (5*(a + b*ArcSinh[c*x] )^2)/(16*b*c)))/Sqrt[1 + c^2*x^2]
3.6.34.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_))^(p_)*((f_ ) + (g_.)*(x_))^(q_), x_Symbol] :> Simp[(d + e*x)^q*((f + g*x)^q/(1 + c^2*x ^2)^q) Int[(d + e*x)^(p - q)*(1 + c^2*x^2)^q*(a + b*ArcSinh[c*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && EqQ[e*f + d*g, 0] && EqQ[c^2*d^ 2 + e^2, 0] && HalfIntegerQ[p, q] && GeQ[p - q, 0]
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]))
\[\int \left (i c d x +d \right )^{\frac {5}{2}} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right ) \sqrt {-i c f x +f}d x\]
\[ \int (d+i c d x)^{5/2} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x)) \, dx=\int { {\left (i \, c d x + d\right )}^{\frac {5}{2}} \sqrt {-i \, c f x + f} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} \,d x } \]
integral(-(b*c^2*d^2*x^2 - 2*I*b*c*d^2*x - b*d^2)*sqrt(I*c*d*x + d)*sqrt(- I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1)) - (a*c^2*d^2*x^2 - 2*I*a*c*d^2*x - a*d^2)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f), x)
Timed out. \[ \int (d+i c d x)^{5/2} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x)) \, dx=\text {Timed out} \]
Exception generated. \[ \int (d+i c d x)^{5/2} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int (d+i c d x)^{5/2} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x)) \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int (d+i c d x)^{5/2} \sqrt {f-i c f x} (a+b \text {arcsinh}(c x)) \, dx=\int \left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{5/2}\,\sqrt {f-c\,f\,x\,1{}\mathrm {i}} \,d x \]