Integrand size = 35, antiderivative size = 203 \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \, dx=\frac {b \left (1+c^2 x^2\right )^{3/2}}{6 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {x \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 x \left (1+c^2 x^2\right )^2 (a+b \text {arcsinh}(c x))}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b \left (1+c^2 x^2\right )^{5/2} \log \left (1+c^2 x^2\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
1/6*b*(c^2*x^2+1)^(3/2)/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+1/3*x*(c^2*x ^2+1)*(a+b*arcsinh(c*x))/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+2/3*x*(c^2*x^ 2+1)^2*(a+b*arcsinh(c*x))/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-1/3*b*(c^2*x ^2+1)^(5/2)*ln(c^2*x^2+1)/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)
Time = 1.26 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.95 \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \, dx=\frac {i \sqrt {f-i c f x} \left (6 a c x+4 a c^3 x^3+b \sqrt {1+c^2 x^2}+2 b c x \left (3+2 c^2 x^2\right ) \text {arcsinh}(c x)-2 b \left (1+c^2 x^2\right )^{3/2} \log (d (-1+i c x))-2 b \sqrt {1+c^2 x^2} \log (d+i c d x)-2 b c^2 x^2 \sqrt {1+c^2 x^2} \log (d+i c d x)\right )}{6 c d^2 f^3 (-i+c x) (i+c x)^2 \sqrt {d+i c d x}} \]
((I/6)*Sqrt[f - I*c*f*x]*(6*a*c*x + 4*a*c^3*x^3 + b*Sqrt[1 + c^2*x^2] + 2* b*c*x*(3 + 2*c^2*x^2)*ArcSinh[c*x] - 2*b*(1 + c^2*x^2)^(3/2)*Log[d*(-1 + I *c*x)] - 2*b*Sqrt[1 + c^2*x^2]*Log[d + I*c*d*x] - 2*b*c^2*x^2*Sqrt[1 + c^2 *x^2]*Log[d + I*c*d*x]))/(c*d^2*f^3*(-I + c*x)*(I + c*x)^2*Sqrt[d + I*c*d* x])
Time = 0.57 (sec) , antiderivative size = 132, normalized size of antiderivative = 0.65, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6211, 6203, 241, 6202, 240}
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.
| \(\displaystyle \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6211 |
| \(\displaystyle \frac {\left (c^2 x^2+1\right )^{5/2} \int \frac {a+b \text {arcsinh}(c x)}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 6203 |
| \(\displaystyle \frac {\left (c^2 x^2+1\right )^{5/2} \left (\frac {2}{3} \int \frac {a+b \text {arcsinh}(c x)}{\left (c^2 x^2+1\right )^{3/2}}dx-\frac {1}{3} b c \int \frac {x}{\left (c^2 x^2+1\right )^2}dx+\frac {x (a+b \text {arcsinh}(c x))}{3 \left (c^2 x^2+1\right )^{3/2}}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 241 |
| \(\displaystyle \frac {\left (c^2 x^2+1\right )^{5/2} \left (\frac {2}{3} \int \frac {a+b \text {arcsinh}(c x)}{\left (c^2 x^2+1\right )^{3/2}}dx+\frac {x (a+b \text {arcsinh}(c x))}{3 \left (c^2 x^2+1\right )^{3/2}}+\frac {b}{6 c \left (c^2 x^2+1\right )}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 6202 |
| \(\displaystyle \frac {\left (c^2 x^2+1\right )^{5/2} \left (\frac {2}{3} \left (\frac {x (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}-b c \int \frac {x}{c^2 x^2+1}dx\right )+\frac {x (a+b \text {arcsinh}(c x))}{3 \left (c^2 x^2+1\right )^{3/2}}+\frac {b}{6 c \left (c^2 x^2+1\right )}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 240 |
| \(\displaystyle \frac {\left (c^2 x^2+1\right )^{5/2} \left (\frac {x (a+b \text {arcsinh}(c x))}{3 \left (c^2 x^2+1\right )^{3/2}}+\frac {2}{3} \left (\frac {x (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}-\frac {b \log \left (c^2 x^2+1\right )}{2 c}\right )+\frac {b}{6 c \left (c^2 x^2+1\right )}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
((1 + c^2*x^2)^(5/2)*(b/(6*c*(1 + c^2*x^2)) + (x*(a + b*ArcSinh[c*x]))/(3* (1 + c^2*x^2)^(3/2)) + (2*((x*(a + b*ArcSinh[c*x]))/Sqrt[1 + c^2*x^2] - (b *Log[1 + c^2*x^2])/(2*c)))/3))/((d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))
3.6.69.3.1 Defintions of rubi rules used
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[x*((a + b*ArcSinh[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Simp [b*c*(n/d)*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]] Int[x*((a + b*ArcSinh[ c*x])^(n - 1)/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((d_) + (e_.)*(x_)^2)^(p_), x _Symbol] :> Simp[(-x)*(d + e*x^2)^(p + 1)*((a + b*ArcSinh[c*x])^n/(2*d*(p + 1))), x] + (Simp[(2*p + 3)/(2*d*(p + 1)) Int[(d + e*x^2)^(p + 1)*(a + b* ArcSinh[c*x])^n, x], x] + Simp[b*c*(n/(2*(p + 1)))*Simp[(d + e*x^2)^p/(1 + c^2*x^2)^p] Int[x*(1 + c^2*x^2)^(p + 1/2)*(a + b*ArcSinh[c*x])^(n - 1), x ], x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 0] && LtQ[p, -1] && NeQ[p, -3/2]
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]
\[\int \frac {a +b \,\operatorname {arcsinh}\left (c x \right )}{\left (i c d x +d \right )^{\frac {5}{2}} \left (-i c f x +f \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (i \, c d x + d\right )}^{\frac {5}{2}} {\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
-1/6*(sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*c*x^2 - 2*( 2*b*c^2*x^3 + 3*b*x)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sqrt(c ^2*x^2 + 1)) - (c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*sqrt(b^2/(c ^2*d^5*f^5))*log((sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c *d^2*f^2*x^2*sqrt(b^2/(c^2*d^5*f^5)) + b*c^2*x^4 + b*x^2)/(b*c^4*x^4 + 2*b *c^2*x^2 + b)) + (c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*sqrt(b^2/ (c^2*d^5*f^5))*log(-(sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f )*c*d^2*f^2*x^2*sqrt(b^2/(c^2*d^5*f^5)) - b*c^2*x^4 - b*x^2)/(b*c^4*x^4 + 2*b*c^2*x^2 + b)) + 2*(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*sqrt (b^2/(c^2*d^5*f^5))*log((sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f^2*x*sqrt(b^2/(c^2*d^5*f^5)) + b*c^2*x^3 + b*x)/(b*c^2*x^2 + b)) - 2*(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*sqrt(b^2/(c^2*d^5* f^5))*log(-(sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d^2*f ^2*x*sqrt(b^2/(c^2*d^5*f^5)) - b*c^2*x^3 - b*x)/(b*c^2*x^2 + b)) - 2*(2*a* c^2*x^3 + 3*a*x)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f) - 6*(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)*integral(-2/3*sqrt(c^2*x^2 + 1)*sqrt(I*c*d *x + d)*sqrt(-I*c*f*x + f)*b*c*x/(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^ 3*f^3), x))/(c^4*d^3*f^3*x^4 + 2*c^2*d^3*f^3*x^2 + d^3*f^3)
Timed out. \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \, dx=\text {Timed out} \]
Time = 0.23 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.78 \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \, dx=\frac {1}{6} \, b c {\left (\frac {1}{c^{4} d^{\frac {5}{2}} f^{\frac {5}{2}} x^{2} + c^{2} d^{\frac {5}{2}} f^{\frac {5}{2}}} - \frac {2 \, \log \left (c^{2} x^{2} + 1\right )}{c^{2} d^{\frac {5}{2}} f^{\frac {5}{2}}}\right )} + \frac {1}{3} \, b {\left (\frac {x}{{\left (c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}} d f} + \frac {2 \, x}{\sqrt {c^{2} d f x^{2} + d f} d^{2} f^{2}}\right )} \operatorname {arsinh}\left (c x\right ) + \frac {1}{3} \, a {\left (\frac {x}{{\left (c^{2} d f x^{2} + d f\right )}^{\frac {3}{2}} d f} + \frac {2 \, x}{\sqrt {c^{2} d f x^{2} + d f} d^{2} f^{2}}\right )} \]
1/6*b*c*(1/(c^4*d^(5/2)*f^(5/2)*x^2 + c^2*d^(5/2)*f^(5/2)) - 2*log(c^2*x^2 + 1)/(c^2*d^(5/2)*f^(5/2))) + 1/3*b*(x/((c^2*d*f*x^2 + d*f)^(3/2)*d*f) + 2*x/(sqrt(c^2*d*f*x^2 + d*f)*d^2*f^2))*arcsinh(c*x) + 1/3*a*(x/((c^2*d*f*x ^2 + d*f)^(3/2)*d*f) + 2*x/(sqrt(c^2*d*f*x^2 + d*f)*d^2*f^2))
Exception generated. \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:The choice was done assuming 0=[0,0 ,0,0]ext_reduce Error: Bad Argument TypeThe choice was done assuming 0=[0, 0,0,0]ext
Timed out. \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{5/2}\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]