Integrand size = 37, antiderivative size = 522 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\frac {d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \log \left (1+i e^{-\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 b^2 d^3 \left (1+c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 b d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 i b^2 d^3 \left (1+c^2 x^2\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i d^3 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
1/3*d^3*(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)+4/3*b*d^3*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))*ln(1+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)-4/3*b^2*d^3*(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*d^3*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))*sec(1/4*Pi+1/2* I*arcsinh(c*x))^2/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+4/3*I*b^2*d^3*(c^2 *x^2+1)^(5/2)*tan(1/4*Pi+1/2*I*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f* x)^(5/2)+1/3*I*d^3*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))^2*tan(1/4*Pi+1/2*I *arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-1/3*I*d^3*(c^2*x^2+1) ^(5/2)*(a+b*arcsinh(c*x))^2*sec(1/4*Pi+1/2*I*arcsinh(c*x))^2*tan(1/4*Pi+1/ 2*I*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)
Time = 10.07 (sec) , antiderivative size = 788, normalized size of antiderivative = 1.51 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\frac {\sqrt {i d (-i+c x)} \sqrt {-i f (i+c x)} \left (\frac {2 i a^2}{3 f^3 (i+c x)^2}-\frac {a^2}{3 f^3 (i+c x)}\right )}{c}-\frac {i a b \sqrt {i (-i d+c d x)} \sqrt {-i (i f+c f x)} \sqrt {-d f \left (1+c^2 x^2\right )} \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )+i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right ) \left (-\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 )+i \log \left (\sqrt {1+c^2 x^2}\right )\right )+\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right ) \left (4 i+3 \text {arcsinh}(c x)-6 \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+3 i \log \left (\sqrt {1+c^2 x^2}\right )\right )+2 \left (\sqrt {1+c^2 x^2} \left (i \text {arcsinh}(c x)+2 i \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+\log \left (\sqrt {1+c^2 x^2}\right )\right )+2 \left (1+i \text {arcsinh}(c x)+2 i \arctan \left (\coth \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+\log \left (\sqrt {1+c^2 x^2}\right )\right )\right ) \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )}{3 c f^3 (1+i c x) \sqrt {-((-i d+c d x) (i f+c f x))} \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )-i \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )^4}-\frac {i b^2 (-i+c x) \sqrt {i (-i d+c d x)} \sqrt {-i (i f+c f x)} \sqrt {-d f \left (1+c^2 x^2\right )} \left ((-1-i) \text {arcsinh}(c x)^2-\frac {2 \text {arcsinh}(c x) (2 i+\text {arcsinh}(c x))}{i+c x}-2 i (\pi -2 i \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 {4 \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 (4+\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 )}{3 c f^3 \sqrt {-((-i d+c d x) (i f+c f x))} \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 )^2} \]
(Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)]*((((2*I)/3)*a^2)/(f^3*(I + c* x)^2) - a^2/(3*f^3*(I + c*x))))/c - ((I/3)*a*b*Sqrt[I*((-I)*d + c*d*x)]*Sq rt[(-I)*(I*f + c*f*x)]*Sqrt[-(d*f*(1 + c^2*x^2))]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])*(-(Cosh[(3*ArcSinh[c*x])/2]*(ArcSinh[c*x] - 2*ArcT an[Coth[ArcSinh[c*x]/2]] + I*Log[Sqrt[1 + c^2*x^2]])) + Cosh[ArcSinh[c*x]/ 2]*(4*I + 3*ArcSinh[c*x] - 6*ArcTan[Coth[ArcSinh[c*x]/2]] + (3*I)*Log[Sqrt [1 + c^2*x^2]]) + 2*(Sqrt[1 + c^2*x^2]*(I*ArcSinh[c*x] + (2*I)*ArcTan[Coth [ArcSinh[c*x]/2]] + Log[Sqrt[1 + c^2*x^2]]) + 2*(1 + I*ArcSinh[c*x] + (2*I )*ArcTan[Coth[ArcSinh[c*x]/2]] + Log[Sqrt[1 + c^2*x^2]]))*Sinh[ArcSinh[c*x ]/2]))/(c*f^3*(1 + I*c*x)*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*(Cosh[Ar cSinh[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^4) - ((I/3)*b^2*(-I + c*x)*Sqrt[I* ((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[-(d*f*(1 + c^2*x^2))]*((-1 - I)*ArcSinh[c*x]^2 - (2*ArcSinh[c*x]*(2*I + ArcSinh[c*x]))/(I + c*x) - ( 2*I)*(Pi - (2*I)*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]] - (4 *ArcSinh[c*x]^2*Sinh[ArcSinh[c*x]/2])/(Cosh[ArcSinh[c*x]/2] - I*Sinh[ArcSi nh[c*x]/2])^3 + (2*(4 + ArcSinh[c*x]^2)*Sinh[ArcSinh[c*x]/2])/(Cosh[ArcSin h[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])))/(c*f^3*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x...
Time = 1.33 (sec) , antiderivative size = 271, normalized size of antiderivative = 0.52, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, Rules used = {6211, 27, 6259, 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.
| \(\displaystyle \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6211 |
| \(\displaystyle \frac {\left (c^2 x^2+1\right )^{5/2} \int \frac {d^3 (i c x+1)^3 (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^3 \left (c^2 x^2+1\right )^{5/2} \int \frac {(i c x+1)^3 (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 6259 |
| \(\displaystyle \frac {d^3 \left (c^2 x^2+1\right )^{5/2} \int \left (-\frac {i (a+b \text {arcsinh}(c x))^2}{(c x+i) \sqrt {c^2 x^2+1}}-\frac {2 (a+b \text {arcsinh}(c x))^2}{(c x+i)^2 \sqrt {c^2 x^2+1}}\right )dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^3 \left (c^2 x^2+1\right )^{5/2} \left (\frac {(a+b \text {arcsinh}(c x))^2}{3 c}+\frac {4 b \log \left (1+i e^{-\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{3 c}+\frac {i \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))^2}{3 c}+\frac {2 b \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))}{3 c}-\frac {i \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))^2}{3 c}-\frac {4 b^2 \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}(c x)}\right )}{3 c}+\frac {4 i b^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
(d^3*(1 + c^2*x^2)^(5/2)*((a + b*ArcSinh[c*x])^2/(3*c) + (4*b*(a + b*ArcSi nh[c*x])*Log[1 + I/E^ArcSinh[c*x]])/(3*c) - (4*b^2*PolyLog[2, (-I)/E^ArcSi nh[c*x]])/(3*c) + (2*b*(a + b*ArcSinh[c*x])*Sec[Pi/4 + (I/2)*ArcSinh[c*x]] ^2)/(3*c) + (((4*I)/3)*b^2*Tan[Pi/4 + (I/2)*ArcSinh[c*x]])/c + ((I/3)*(a + b*ArcSinh[c*x])^2*Tan[Pi/4 + (I/2)*ArcSinh[c*x]])/c - ((I/3)*(a + b*ArcSi nh[c*x])^2*Sec[Pi/4 + (I/2)*ArcSinh[c*x]]^2*Tan[Pi/4 + (I/2)*ArcSinh[c*x]] )/c))/((d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))
3.7.2.3.1 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[(a + b*ArcSinh[c* x])^n/Sqrt[d + e*x^2], (f + g*x)^m*(d + e*x^2)^(p + 1/2), x], x] /; FreeQ[{ a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IntegerQ[m] && ILtQ[p + 1/2, 0 ] && GtQ[d, 0] && IGtQ[n, 0]
\[\int \frac {\left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{2} \sqrt {i c d x +d}}{\left (-i c f x +f \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\int { \frac {\sqrt {i \, c d x + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
-1/3*((b^2*c*x - I*b^2)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sqr t(c^2*x^2 + 1))^2 - 3*(c^3*f^3*x^2 + 2*I*c^2*f^3*x - c*f^3)*integral(1/3*( -3*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 - 3*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*f^3*x^3 + 3*I*c^2*f^3*x^2 - 3 *c*f^3*x - I*f^3), x))/(c^3*f^3*x^2 + 2*I*c^2*f^3*x - c*f^3)
\[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\int \frac {\sqrt {i d \left (c x - i\right )} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\left (- i f \left (c x + i\right )\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\int { \frac {\sqrt {i \, c d x + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,\sqrt {d+c\,d\,x\,1{}\mathrm {i}}}{{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]