Integrand size = 37, antiderivative size = 942 \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\frac {2 i b^2 d^2 \left (1+c^2 x^2\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 d^2 x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b^2 d^2 \left (1+c^2 x^2\right )^{5/2} \text {arcsinh}(c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b d^2 \left (1+c^2 x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b d^2 x \left (1+c^2 x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b c d^2 x^2 \left (1+c^2 x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i d^2 \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {d^2 x \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {c^2 d^2 x^3 \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 d^2 x \left (1+c^2 x^2\right )^2 (a+b \text {arcsinh}(c x))^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {d^2 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 i b d^2 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b d^2 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \log \left (1+e^{2 \text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b^2 d^2 \left (1+c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 d^2 \left (1+c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b^2 d^2 \left (1+c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
2/3*I*b*d^2*x*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))/(d+I*c*d*x)^(5/2)/(f-I* c*f*x)^(5/2)-2/3*b^2*d^2*x*(c^2*x^2+1)^2/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/ 2)+1/3*b^2*d^2*(c^2*x^2+1)^(5/2)*arcsinh(c*x)/c/(d+I*c*d*x)^(5/2)/(f-I*c*f *x)^(5/2)+1/3*b*d^2*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))/c/(d+I*c*d*x)^(5/ 2)/(f-I*c*f*x)^(5/2)-2/3*I*d^2*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/c/(d+I*c*d *x)^(5/2)/(f-I*c*f*x)^(5/2)-1/3*b*c*d^2*x^2*(c^2*x^2+1)^(3/2)*(a+b*arcsinh (c*x))/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+4/3*I*b*d^2*(c^2*x^2+1)^(5/2)*( a+b*arcsinh(c*x))*arctan(c*x+(c^2*x^2+1)^(1/2))/c/(d+I*c*d*x)^(5/2)/(f-I*c *f*x)^(5/2)+1/3*d^2*x*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(5/2)/( f-I*c*f*x)^(5/2)-1/3*c^2*d^2*x^3*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/(d+I*c*d *x)^(5/2)/(f-I*c*f*x)^(5/2)+2/3*d^2*x*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))^2/( d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+1/3*d^2*(c^2*x^2+1)^(5/2)*(a+b*arcsinh( c*x))^2/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+2/3*I*b^2*d^2*(c^2*x^2+1)^2/ c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-2/3*b*d^2*(c^2*x^2+1)^(5/2)*(a+b*arc sinh(c*x))*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x) ^(5/2)+2/3*b^2*d^2*(c^2*x^2+1)^(5/2)*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2))) /c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-2/3*b^2*d^2*(c^2*x^2+1)^(5/2)*polyl og(2,I*(c*x+(c^2*x^2+1)^(1/2)))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-1/3* b^2*d^2*(c^2*x^2+1)^(5/2)*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/c/(d+I*c*d *x)^(5/2)/(f-I*c*f*x)^(5/2)
Time = 6.96 (sec) , antiderivative size = 528, normalized size of antiderivative = 0.56 \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\frac {\sqrt {d+i c d x} \sqrt {f-i c f x} \left (\frac {a^2 (2 i+c x)}{(i+c x)^2}-\frac {a b \left (i \cosh \left (\frac {3}{2} \text {arcsinh}(c x)\right ) \left (\text {arcsinh}(c x)-2 \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+\frac {1}{2} i \log \left (1+c^2 x^2\right )\right )+\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right ) \left (-2+3 i \text {arcsinh}(c x)+6 i \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+\frac {3}{2} \log \left (1+c^2 x^2\right )\right )+2 \left (i+\left (-1+\sqrt {1+c^2 x^2}\right ) \text {arcsinh}(c x)+2 \left (2+\sqrt {1+c^2 x^2}\right ) \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )-\frac {1}{2} i \left (2+\sqrt {1+c^2 x^2}\right ) \log \left (1+c^2 x^2\right )\right ) \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )}{\sqrt {1+c^2 x^2} \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )-i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )^3}-\frac {b^2 \left ((1+i) \text {arcsinh}(c x)^2-\frac {\text {arcsinh}(c x) (2 i+\text {arcsinh}(c x))}{i+c x}+2 (i \pi +2 \text {arcsinh}(c x)) \log \left (1+i e^{-\text {arcsinh}(c x)}\right )+i \pi \left (3 \text {arcsinh}(c x)-4 \log \left (1+e^{\text {arcsinh}(c x)}\right )-2 \log \left (-\cos \left (\frac {1}{4} (\pi +2 i \text {arcsinh}(c x))\right )\right )+4 \log \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )\right )-4 \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}(c x)}\right )-\frac {2 \text {arcsinh}(c x)^2 \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )}{\left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )-i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )^3}-\frac {2 \left (-2+\text {arcsinh}(c x)^2\right ) \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )}{\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )-i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )}\right )}{\sqrt {1+c^2 x^2}}\right )}{3 c d f^3} \]
(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*((a^2*(2*I + c*x))/(I + c*x)^2 - (a*b *(I*Cosh[(3*ArcSinh[c*x])/2]*(ArcSinh[c*x] - 2*ArcTan[Coth[ArcSinh[c*x]/2] ] + (I/2)*Log[1 + c^2*x^2]) + Cosh[ArcSinh[c*x]/2]*(-2 + (3*I)*ArcSinh[c*x ] + (6*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + (3*Log[1 + c^2*x^2])/2) + 2*(I + (-1 + Sqrt[1 + c^2*x^2])*ArcSinh[c*x] + 2*(2 + Sqrt[1 + c^2*x^2])*ArcTan[C oth[ArcSinh[c*x]/2]] - (I/2)*(2 + Sqrt[1 + c^2*x^2])*Log[1 + c^2*x^2])*Sin h[ArcSinh[c*x]/2]))/(Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcS inh[c*x]/2])^3) - (b^2*((1 + I)*ArcSinh[c*x]^2 - (ArcSinh[c*x]*(2*I + ArcS inh[c*x]))/(I + c*x) + 2*(I*Pi + 2*ArcSinh[c*x])*Log[1 + I/E^ArcSinh[c*x]] + I*Pi*(3*ArcSinh[c*x] - 4*Log[1 + E^ArcSinh[c*x]] - 2*Log[-Cos[(Pi + (2* I)*ArcSinh[c*x])/4]] + 4*Log[Cosh[ArcSinh[c*x]/2]]) - 4*PolyLog[2, (-I)/E^ ArcSinh[c*x]] - (2*ArcSinh[c*x]^2*Sinh[ArcSinh[c*x]/2])/(Cosh[ArcSinh[c*x] /2] - I*Sinh[ArcSinh[c*x]/2])^3 - (2*(-2 + ArcSinh[c*x]^2)*Sinh[ArcSinh[c* x]/2])/(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])))/Sqrt[1 + c^2*x^2] ))/(3*c*d*f^3)
Time = 1.47 (sec) , antiderivative size = 442, normalized size of antiderivative = 0.47, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, 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.
| \(\displaystyle \int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6211 |
| \(\displaystyle \frac {\left (c^2 x^2+1\right )^{5/2} \int \frac {d^2 (i c x+1)^2 (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^2 \left (c^2 x^2+1\right )^{5/2} \int \frac {(i c x+1)^2 (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 6253 |
| \(\displaystyle \frac {d^2 \left (c^2 x^2+1\right )^{5/2} \int \left (-\frac {c^2 x^2 (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}+\frac {2 i c x (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}+\frac {(a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}\right )dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^2 \left (c^2 x^2+1\right )^{5/2} \left (\frac {4 i b \arctan \left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{3 c}-\frac {b c x^2 (a+b \text {arcsinh}(c x))}{3 \left (c^2 x^2+1\right )}+\frac {2 x (a+b \text {arcsinh}(c x))^2}{3 \sqrt {c^2 x^2+1}}+\frac {x (a+b \text {arcsinh}(c x))^2}{3 \left (c^2 x^2+1\right )^{3/2}}+\frac {2 i b x (a+b \text {arcsinh}(c x))}{3 \left (c^2 x^2+1\right )}-\frac {2 i (a+b \text {arcsinh}(c x))^2}{3 c \left (c^2 x^2+1\right )^{3/2}}+\frac {b (a+b \text {arcsinh}(c x))}{3 c \left (c^2 x^2+1\right )}-\frac {c^2 x^3 (a+b \text {arcsinh}(c x))^2}{3 \left (c^2 x^2+1\right )^{3/2}}+\frac {(a+b \text {arcsinh}(c x))^2}{3 c}-\frac {2 b \log \left (e^{2 \text {arcsinh}(c x)}+1\right ) (a+b \text {arcsinh}(c x))}{3 c}+\frac {2 b^2 \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{3 c}-\frac {2 b^2 \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{3 c}-\frac {b^2 \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(c x)}\right )}{3 c}+\frac {b^2 \text {arcsinh}(c x)}{3 c}-\frac {2 b^2 x}{3 \sqrt {c^2 x^2+1}}+\frac {2 i b^2}{3 c \sqrt {c^2 x^2+1}}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
(d^2*(1 + c^2*x^2)^(5/2)*((((2*I)/3)*b^2)/(c*Sqrt[1 + c^2*x^2]) - (2*b^2*x )/(3*Sqrt[1 + c^2*x^2]) + (b^2*ArcSinh[c*x])/(3*c) + (b*(a + b*ArcSinh[c*x ]))/(3*c*(1 + c^2*x^2)) + (((2*I)/3)*b*x*(a + b*ArcSinh[c*x]))/(1 + c^2*x^ 2) - (b*c*x^2*(a + b*ArcSinh[c*x]))/(3*(1 + c^2*x^2)) + (a + b*ArcSinh[c*x ])^2/(3*c) - (((2*I)/3)*(a + b*ArcSinh[c*x])^2)/(c*(1 + c^2*x^2)^(3/2)) + (x*(a + b*ArcSinh[c*x])^2)/(3*(1 + c^2*x^2)^(3/2)) - (c^2*x^3*(a + b*ArcSi nh[c*x])^2)/(3*(1 + c^2*x^2)^(3/2)) + (2*x*(a + b*ArcSinh[c*x])^2)/(3*Sqrt [1 + c^2*x^2]) + (((4*I)/3)*b*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]]) /c - (2*b*(a + b*ArcSinh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/(3*c) + (2*b^2 *PolyLog[2, (-I)*E^ArcSinh[c*x]])/(3*c) - (2*b^2*PolyLog[2, I*E^ArcSinh[c* x]])/(3*c) - (b^2*PolyLog[2, -E^(2*ArcSinh[c*x])])/(3*c)))/((d + I*c*d*x)^ (5/2)*(f - I*c*f*x)^(5/2))
3.7.3.3.1 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]))
\[\int \frac {\left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{2}}{\left (-i c f x +f \right )^{\frac {5}{2}} \sqrt {i c d x +d}}d x\]
\[ \int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{\sqrt {i \, c d x + d} {\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
1/3*((b^2*c*x + 2*I*b^2)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sq rt(c^2*x^2 + 1))^2 + 3*(c^3*d*f^3*x^2 + 2*I*c^2*d*f^3*x - c*d*f^3)*integra l(-1/3*(3*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a^2 + 2*(3*sqrt(I*c*d*x + d )*sqrt(-I*c*f*x + f)*a*b + (b^2*c*x + 2*I*b^2)*sqrt(c^2*x^2 + 1)*sqrt(I*c* d*x + d)*sqrt(-I*c*f*x + f))*log(c*x + sqrt(c^2*x^2 + 1)))/(c^4*d*f^3*x^4 + 2*I*c^3*d*f^3*x^3 + 2*I*c*d*f^3*x - d*f^3), x))/(c^3*d*f^3*x^2 + 2*I*c^2 *d*f^3*x - c*d*f^3)
\[ \int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\int \frac {\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\sqrt {i d \left (c x - i\right )} \left (- i f \left (c x + i\right )\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{\sqrt {i \, c d x + d} {\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x} (f-i c f x)^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{\sqrt {d+c\,d\,x\,1{}\mathrm {i}}\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]