Integrand size = 37, antiderivative size = 464 \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\frac {i f \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {f x \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {f \left (1+c^2 x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {4 i b f \left (1+c^2 x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b f \left (1+c^2 x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \log \left (1+e^{2 \text {arcsinh}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 b^2 f \left (1+c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {2 b^2 f \left (1+c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {b^2 f \left (1+c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]
I*f*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2) +f*x*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)+ f*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))^2/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^( 3/2)-4*I*b*f*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))*arctan(c*x+(c^2*x^2+1)^( 1/2))/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)-2*b*f*(c^2*x^2+1)^(3/2)*(a+b*a rcsinh(c*x))*ln(1+(c*x+(c^2*x^2+1)^(1/2))^2)/c/(d+I*c*d*x)^(3/2)/(f-I*c*f* x)^(3/2)-2*b^2*f*(c^2*x^2+1)^(3/2)*polylog(2,-I*(c*x+(c^2*x^2+1)^(1/2)))/c /(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)+2*b^2*f*(c^2*x^2+1)^(3/2)*polylog(2,I *(c*x+(c^2*x^2+1)^(1/2)))/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)-b^2*f*(c^2 *x^2+1)^(3/2)*polylog(2,-(c*x+(c^2*x^2+1)^(1/2))^2)/c/(d+I*c*d*x)^(3/2)/(f -I*c*f*x)^(3/2)
Time = 3.26 (sec) , antiderivative size = 508, normalized size of antiderivative = 1.09 \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\frac {\sqrt {d+i c d x} \sqrt {f-i c f x} \left ((-1+i) b^2 \sqrt {1+c^2 x^2} \text {arcsinh}(c x)^2 \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )-\sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+\left (i a^2+a^2 c x-4 i a b \sqrt {1+c^2 x^2} \arctan \left (\tanh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+2 i b^2 \pi \sqrt {1+c^2 x^2} \log \left (1-i e^{-\text {arcsinh}(c x)}\right )+4 i b^2 \pi \sqrt {1+c^2 x^2} \log \left (1+e^{\text {arcsinh}(c x)}\right )-a b \sqrt {1+c^2 x^2} \log \left (1+c^2 x^2\right )-4 i b^2 \pi \sqrt {1+c^2 x^2} \log \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )-2 i b^2 \pi \sqrt {1+c^2 x^2} \log \left (\sin \left (\frac {1}{4} (\pi +2 i \text {arcsinh}(c x))\right )\right )\right ) \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )+i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+4 b^2 \sqrt {1+c^2 x^2} \operatorname {PolyLog}\left (2,i e^{-\text {arcsinh}(c x)}\right ) \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )+i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+b \sqrt {1+c^2 x^2} \text {arcsinh}(c x) \left (i \cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right ) \left (2 a-b \pi +4 i b \log \left (1-i e^{-\text {arcsinh}(c x)}\right )\right )+\left (2 a+b \pi -4 i b \log \left (1-i e^{-\text {arcsinh}(c x)}\right )\right ) \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )\right )}{c d^2 f (-i+c x) (i+c x) \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )+i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )} \]
(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*((-1 + I)*b^2*Sqrt[1 + c^2*x^2]*ArcSi nh[c*x]^2*(Cosh[ArcSinh[c*x]/2] - Sinh[ArcSinh[c*x]/2]) + (I*a^2 + a^2*c*x - (4*I)*a*b*Sqrt[1 + c^2*x^2]*ArcTan[Tanh[ArcSinh[c*x]/2]] + (2*I)*b^2*Pi *Sqrt[1 + c^2*x^2]*Log[1 - I/E^ArcSinh[c*x]] + (4*I)*b^2*Pi*Sqrt[1 + c^2*x ^2]*Log[1 + E^ArcSinh[c*x]] - a*b*Sqrt[1 + c^2*x^2]*Log[1 + c^2*x^2] - (4* I)*b^2*Pi*Sqrt[1 + c^2*x^2]*Log[Cosh[ArcSinh[c*x]/2]] - (2*I)*b^2*Pi*Sqrt[ 1 + c^2*x^2]*Log[Sin[(Pi + (2*I)*ArcSinh[c*x])/4]])*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2]) + 4*b^2*Sqrt[1 + c^2*x^2]*PolyLog[2, I/E^ArcSinh [c*x]]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2]) + b*Sqrt[1 + c^2*x^ 2]*ArcSinh[c*x]*(I*Cosh[ArcSinh[c*x]/2]*(2*a - b*Pi + (4*I)*b*Log[1 - I/E^ ArcSinh[c*x]]) + (2*a + b*Pi - (4*I)*b*Log[1 - I/E^ArcSinh[c*x]])*Sinh[Arc Sinh[c*x]/2])))/(c*d^2*f*(-I + c*x)*(I + c*x)*(Cosh[ArcSinh[c*x]/2] + I*Si nh[ArcSinh[c*x]/2]))
Time = 0.94 (sec) , antiderivative size = 219, 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.
| \(\displaystyle \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx\) |
| \(\Big \downarrow \) 6211 |
| \(\displaystyle \frac {\left (c^2 x^2+1\right )^{3/2} \int \frac {f (1-i c x) (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{3/2}}dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {f \left (c^2 x^2+1\right )^{3/2} \int \frac {(1-i c x) (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{3/2}}dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\) |
| \(\Big \downarrow \) 6253 |
| \(\displaystyle \frac {f \left (c^2 x^2+1\right )^{3/2} \int \left (\frac {(a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{3/2}}-\frac {i c x (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{3/2}}\right )dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f \left (c^2 x^2+1\right )^{3/2} \left (-\frac {4 i b \arctan \left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{c}+\frac {x (a+b \text {arcsinh}(c x))^2}{\sqrt {c^2 x^2+1}}+\frac {i (a+b \text {arcsinh}(c x))^2}{c \sqrt {c^2 x^2+1}}+\frac {(a+b \text {arcsinh}(c x))^2}{c}-\frac {2 b \log \left (e^{2 \text {arcsinh}(c x)}+1\right ) (a+b \text {arcsinh}(c x))}{c}-\frac {2 b^2 \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{c}+\frac {2 b^2 \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{c}-\frac {b^2 \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(c x)}\right )}{c}\right )}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\) |
(f*(1 + c^2*x^2)^(3/2)*((a + b*ArcSinh[c*x])^2/c + (I*(a + b*ArcSinh[c*x]) ^2)/(c*Sqrt[1 + c^2*x^2]) + (x*(a + b*ArcSinh[c*x])^2)/Sqrt[1 + c^2*x^2] - ((4*I)*b*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]])/c - (2*b*(a + b*Arc Sinh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/c - (2*b^2*PolyLog[2, (-I)*E^ArcSi nh[c*x]])/c + (2*b^2*PolyLog[2, I*E^ArcSinh[c*x]])/c - (b^2*PolyLog[2, -E^ (2*ArcSinh[c*x])])/c))/((d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))
3.6.92.3.1 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]))
\[\int \frac {\left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{2}}{\left (i c d x +d \right )^{\frac {3}{2}} \sqrt {-i c f x +f}}d x\]
\[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )}^{\frac {3}{2}} \sqrt {-i \, c f x + f}} \,d x } \]
(sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b^2*log(c*x + sqrt(c^2*x^2 + 1))^2 + (c^2*d^2*f*x - I*c*d^2*f)*integral((-I*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a^2 - 2*(sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b^2 + I *sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a*b)*log(c*x + sqrt(c^2*x^2 + 1)))/( c^3*d^2*f*x^3 - I*c^2*d^2*f*x^2 + c*d^2*f*x - I*d^2*f), x))/(c^2*d^2*f*x - I*c*d^2*f)
\[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\int \frac {\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\left (i d \left (c x - i\right )\right )^{\frac {3}{2}} \sqrt {- i f \left (c x + i\right )}}\, dx \]
\[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )}^{\frac {3}{2}} \sqrt {-i \, c f x + f}} \,d x } \]
b^2*integrate(log(c*x + sqrt(c^2*x^2 + 1))^2/((I*c*d*x + d)^(3/2)*sqrt(-I* c*f*x + f)), x) + 2*I*sqrt(c^2*d*f*x^2 + d*f)*a*b*arcsinh(c*x)/(I*c^2*d^2* f*x + c*d^2*f) + I*sqrt(c^2*d*f*x^2 + d*f)*a^2/(I*c^2*d^2*f*x + c*d^2*f) - 2*a*b*log(I*c*x + 1)/(c*d^(3/2)*sqrt(f))
\[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )}^{\frac {3}{2}} \sqrt {-i \, c f x + f}} \,d x } \]
Timed out. \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2} \sqrt {f-i c f x}} \, dx=\int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{3/2}\,\sqrt {f-c\,f\,x\,1{}\mathrm {i}}} \,d x \]