Integrand size = 37, antiderivative size = 743 \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2} (f-i c f x)^{3/2}} \, dx=-\frac {i b^2 f \left (1+c^2 x^2\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b^2 f x \left (1+c^2 x^2\right )^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b f \left (1+c^2 x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i b f x \left (1+c^2 x^2\right )^{3/2} (a+b \text {arcsinh}(c x))}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i f \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {f x \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 f x \left (1+c^2 x^2\right )^2 (a+b \text {arcsinh}(c x))^2}{3 (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 f \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b f \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 b f \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \log \left (1+e^{2 \text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {b^2 f \left (1+c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b^2 f \left (1+c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 b^2 f \left (1+c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
-1/3*I*b^2*f*(c^2*x^2+1)^2/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-1/3*b^2*f *x*(c^2*x^2+1)^2/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+1/3*b*f*(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)-1/3*I*b*f*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)+1 /3*I*f*(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*f*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)+2/3*f*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)+2/3*f*(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*f*(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)-4/3*b*f*(c^2 *x^2+1)^(5/2)*(a+b*arcsinh(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)-1/3*b^2*f*(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)+1/3*b^2*f*(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*f*(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 = 9.91 (sec) , antiderivative size = 754, normalized size of antiderivative = 1.01 \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2} (f-i c f x)^{3/2}} \, dx=\frac {\sqrt {i d (-i+c x)} \sqrt {-i f (i+c x)} \left (-\frac {i a^2}{6 d^3 f^2 (-i+c x)^2}+\frac {5 a^2}{12 d^3 f^2 (-i+c x)}+\frac {a^2}{4 d^3 f^2 (i+c x)}\right )}{c}+\frac {i a b \sqrt {i (-i d+c d x)} \sqrt {-i (i f+c f x)} \left (4 c x \text {arcsinh}(c x)+2 i \text {arcsinh}(c x) \cosh (2 \text {arcsinh}(c x))+\sqrt {1+c^2 x^2} \left (1-2 i \arctan \left (\tanh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+2 c x \left (\arctan \left (\tanh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )-2 i \log \left (\sqrt {1+c^2 x^2}\right )\right )-4 \log \left (\sqrt {1+c^2 x^2}\right )\right )\right )}{3 c d^2 f (-i+c x) \sqrt {-((-i d+c d x) (i f+c f x))} \sqrt {-d f \left (1+c^2 x^2\right )}}+\frac {i b^2 \sqrt {i (-i d+c d x)} \sqrt {-i (i f+c f x)} \sqrt {1+c^2 x^2} \left (7 \pi \text {arcsinh}(c x)+\frac {(2+i \text {arcsinh}(c x)) \text {arcsinh}(c x)}{-i+c x}-(1+4 i) \text {arcsinh}(c x)^2-5 (\pi +2 i \text {arcsinh}(c x)) \log \left (1-i e^{-\text {arcsinh}(c x)}\right )+3 (\pi -2 i \text {arcsinh}(c x)) \log \left (1+i e^{-\text {arcsinh}(c x)}\right )-16 \pi \log \left (1+e^{\text {arcsinh}(c x)}\right )-3 \pi \log \left (-\cos \left (\frac {1}{4} (\pi +2 i \text {arcsinh}(c x))\right )\right )+16 \pi \log \left (\cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )\right )+5 \pi \log \left (\sin \left (\frac {1}{4} (\pi +2 i \text {arcsinh}(c x))\right )\right )+6 i \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}(c x)}\right )+10 i \operatorname {PolyLog}\left (2,i e^{-\text {arcsinh}(c x)}\right )+\frac {3 i \text {arcsinh}(c x)^2 \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 )}+\frac {2 i \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 {\left (-4+5 \text {arcsinh}(c x)^2\right ) \sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )}{-i \cosh \left (\frac {1}{2} \text {arcsinh}(c x)\right )+\sinh \left (\frac {1}{2} \text {arcsinh}(c x)\right )}\right )}{6 c d^2 f \sqrt {-((-i d+c d x) (i f+c f x))} \sqrt {-d f \left (1+c^2 x^2\right )}} \]
(Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)]*(((-1/6*I)*a^2)/(d^3*f^2*(-I + c*x)^2) + (5*a^2)/(12*d^3*f^2*(-I + c*x)) + a^2/(4*d^3*f^2*(I + c*x))))/ c + ((I/3)*a*b*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*(4*c*x*Ar cSinh[c*x] + (2*I)*ArcSinh[c*x]*Cosh[2*ArcSinh[c*x]] + Sqrt[1 + c^2*x^2]*( 1 - (2*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 2*c*x*(ArcTan[Tanh[ArcSinh[c*x]/2 ]] - (2*I)*Log[Sqrt[1 + c^2*x^2]]) - 4*Log[Sqrt[1 + c^2*x^2]])))/(c*d^2*f* (-I + c*x)*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*Sqrt[-(d*f*(1 + c^2*x^2 ))]) + ((I/6)*b^2*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-I)*(I*f + c*f*x)]*Sqrt[1 + c^2*x^2]*(7*Pi*ArcSinh[c*x] + ((2 + I*ArcSinh[c*x])*ArcSinh[c*x])/(-I + c*x) - (1 + 4*I)*ArcSinh[c*x]^2 - 5*(Pi + (2*I)*ArcSinh[c*x])*Log[1 - I/E ^ArcSinh[c*x]] + 3*(Pi - (2*I)*ArcSinh[c*x])*Log[1 + I/E^ArcSinh[c*x]] - 1 6*Pi*Log[1 + E^ArcSinh[c*x]] - 3*Pi*Log[-Cos[(Pi + (2*I)*ArcSinh[c*x])/4]] + 16*Pi*Log[Cosh[ArcSinh[c*x]/2]] + 5*Pi*Log[Sin[(Pi + (2*I)*ArcSinh[c*x] )/4]] + (6*I)*PolyLog[2, (-I)/E^ArcSinh[c*x]] + (10*I)*PolyLog[2, I/E^ArcS inh[c*x]] + ((3*I)*ArcSinh[c*x]^2*Sinh[ArcSinh[c*x]/2])/(Cosh[ArcSinh[c*x] /2] - I*Sinh[ArcSinh[c*x]/2]) + ((2*I)*ArcSinh[c*x]^2*Sinh[ArcSinh[c*x]/2] )/(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])^3 + ((-4 + 5*ArcSinh[c*x ]^2)*Sinh[ArcSinh[c*x]/2])/((-I)*Cosh[ArcSinh[c*x]/2] + Sinh[ArcSinh[c*x]/ 2])))/(c*d^2*f*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*Sqrt[-(d*f*(1 + c^2 *x^2))])
Time = 1.15 (sec) , antiderivative size = 365, normalized size of antiderivative = 0.49, 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.
| \(\displaystyle \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2} (f-i c f x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 6211 |
| \(\displaystyle \frac {\left (c^2 x^2+1\right )^{5/2} \int \frac {f (1-i c x) (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {f \left (c^2 x^2+1\right )^{5/2} \int \frac {(1-i c x) (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 6253 |
| \(\displaystyle \frac {f \left (c^2 x^2+1\right )^{5/2} \int \left (\frac {(a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}-\frac {i c x (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}\right )dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f \left (c^2 x^2+1\right )^{5/2} \left (-\frac {2 i b \arctan \left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{3 c}-\frac {i b x (a+b \text {arcsinh}(c x))}{3 \left (c^2 x^2+1\right )}+\frac {b (a+b \text {arcsinh}(c x))}{3 c \left (c^2 x^2+1\right )}+\frac {2 x (a+b \text {arcsinh}(c x))^2}{3 \sqrt {c^2 x^2+1}}+\frac {x (a+b \text {arcsinh}(c x))^2}{3 \left (c^2 x^2+1\right )^{3/2}}+\frac {i (a+b \text {arcsinh}(c x))^2}{3 c \left (c^2 x^2+1\right )^{3/2}}+\frac {2 (a+b \text {arcsinh}(c x))^2}{3 c}-\frac {4 b \log \left (e^{2 \text {arcsinh}(c x)}+1\right ) (a+b \text {arcsinh}(c x))}{3 c}-\frac {b^2 \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{3 c}+\frac {b^2 \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{3 c}-\frac {2 b^2 \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(c x)}\right )}{3 c}-\frac {b^2 x}{3 \sqrt {c^2 x^2+1}}-\frac {i b^2}{3 c \sqrt {c^2 x^2+1}}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
(f*(1 + c^2*x^2)^(5/2)*(((-1/3*I)*b^2)/(c*Sqrt[1 + c^2*x^2]) - (b^2*x)/(3* Sqrt[1 + c^2*x^2]) + (b*(a + b*ArcSinh[c*x]))/(3*c*(1 + c^2*x^2)) - ((I/3) *b*x*(a + b*ArcSinh[c*x]))/(1 + c^2*x^2) + (2*(a + b*ArcSinh[c*x])^2)/(3*c ) + ((I/3)*(a + b*ArcSinh[c*x])^2)/(c*(1 + c^2*x^2)^(3/2)) + (x*(a + b*Arc Sinh[c*x])^2)/(3*(1 + c^2*x^2)^(3/2)) + (2*x*(a + b*ArcSinh[c*x])^2)/(3*Sq rt[1 + c^2*x^2]) - (((2*I)/3)*b*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x] ])/c - (4*b*(a + b*ArcSinh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/(3*c) - (b^2 *PolyLog[2, (-I)*E^ArcSinh[c*x]])/(3*c) + (b^2*PolyLog[2, I*E^ArcSinh[c*x] ])/(3*c) - (2*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))
3.6.99.3.1 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]))
\[\int \frac {\left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{2}}{\left (i c d x +d \right )^{\frac {5}{2}} \left (-i c f x +f \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2} (f-i c f x)^{3/2}} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )}^{\frac {5}{2}} {\left (-i \, c f x + f\right )}^{\frac {3}{2}}} \,d x } \]
1/3*((2*b^2*c^2*x^2 - 2*I*b^2*c*x + b^2)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1))^2 + 3*(c^4*d^3*f^2*x^3 - I*c^3*d^3*f^2*x^ 2 + c^2*d^3*f^2*x - I*c*d^3*f^2)*integral(1/3*(-3*I*sqrt(I*c*d*x + d)*sqrt (-I*c*f*x + f)*a^2 - 2*(3*I*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a*b + (2* b^2*c^2*x^2 - 2*I*b^2*c*x + 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^5*d^3*f^2*x^5 - I*c^4*d^3* f^2*x^4 + 2*c^3*d^3*f^2*x^3 - 2*I*c^2*d^3*f^2*x^2 + c*d^3*f^2*x - I*d^3*f^ 2), x))/(c^4*d^3*f^2*x^3 - I*c^3*d^3*f^2*x^2 + c^2*d^3*f^2*x - I*c*d^3*f^2 )
Timed out. \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2} (f-i c f x)^{3/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2} (f-i c f x)^{3/2}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2} (f-i c f x)^{3/2}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {(a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2} (f-i c f x)^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{5/2}\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]