Integrand size = 37, antiderivative size = 584 \[ \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\frac {8 d^4 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {d^4 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {32 b d^4 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \log \left (1+i e^{-\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {32 b^2 d^4 \left (1+c^2 x^2\right )^{5/2} \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {4 b d^4 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x)) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 i b^2 d^4 \left (1+c^2 x^2\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 i d^4 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i d^4 \left (1+c^2 x^2\right )^{5/2} (a+b \text {arcsinh}(c x))^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
8/3*d^4*(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)+1/3*d^4*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))^3/b/c/(d+I*c*d*x)^ (5/2)/(f-I*c*f*x)^(5/2)+32/3*b*d^4*(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)-32/3*b ^2*d^4*(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)+4/3*b*d^4*(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)+8/3 *I*b^2*d^4*(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)+8/3*I*d^4*(c^2*x^2+1)^(5/2)*(a+b*arcsinh(c*x))^2*t an(1/4*Pi+1/2*I*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-2/3*I* d^4*(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)
Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1617\) vs. \(2(584)=1168\).
Time = 17.78 (sec) , antiderivative size = 1617, normalized size of antiderivative = 2.77 \[ \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx =\text {Too large to display} \]
(Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)]*((((4*I)/3)*a^2*d)/(f^3*(I + c*x)^2) - (8*a^2*d)/(3*f^3*(I + c*x))))/c + (a^2*d^(3/2)*Log[c*d*f*x + Sqr t[d]*Sqrt[f]*Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)]])/(c*f^(5/2)) - ( (I/3)*a*b*d*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-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*ArcTan[Coth[ArcSinh[c*x]/2]] + I*Log[Sq rt[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]] + Lo g[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[ArcSinh[c*x]/2] - I*Sinh[ArcSinh[c*x] /2])^4) + (a*b*d*Sqrt[I*((-I)*d + c*d*x)]*Sqrt[(-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]*((14*I - 3*ArcSinh[c*x])*ArcSinh[c*x] + (28*I)*ArcTan[ Tanh[ArcSinh[c*x]/2]] - 14*Log[Sqrt[1 + c^2*x^2]]) + Cosh[ArcSinh[c*x]/2]* (8 + (6*I)*ArcSinh[c*x] + 9*ArcSinh[c*x]^2 - (84*I)*ArcTan[Tanh[ArcSinh[c* x]/2]] + 42*Log[Sqrt[1 + c^2*x^2]]) - (2*I)*(4 + (4*I)*ArcSinh[c*x] + 6*Ar cSinh[c*x]^2 - (56*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 28*Log[Sqrt[1 + c^2*x ^2]] + Sqrt[1 + c^2*x^2]*(ArcSinh[c*x]*(14*I + 3*ArcSinh[c*x]) - (28*I)...
Time = 1.37 (sec) , antiderivative size = 291, normalized size of antiderivative = 0.50, 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.
| \(\displaystyle \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6211 |
| \(\displaystyle \frac {\left (c^2 x^2+1\right )^{5/2} \int \frac {d^4 (i c x+1)^4 (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^4 \left (c^2 x^2+1\right )^{5/2} \int \frac {(i c x+1)^4 (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 6259 |
| \(\displaystyle \frac {d^4 \left (c^2 x^2+1\right )^{5/2} \int \left (-\frac {4 i (a+b \text {arcsinh}(c x))^2}{(c x+i) \sqrt {c^2 x^2+1}}-\frac {4 (a+b \text {arcsinh}(c x))^2}{(c x+i)^2 \sqrt {c^2 x^2+1}}+\frac {(a+b \text {arcsinh}(c x))^2}{\sqrt {c^2 x^2+1}}\right )dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^4 \left (c^2 x^2+1\right )^{5/2} \left (\frac {(a+b \text {arcsinh}(c x))^3}{3 b c}+\frac {8 (a+b \text {arcsinh}(c x))^2}{3 c}+\frac {32 b \log \left (1+i e^{-\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{3 c}+\frac {8 i \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))^2}{3 c}+\frac {4 b \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))}{3 c}-\frac {2 i \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))^2}{3 c}-\frac {32 b^2 \operatorname {PolyLog}\left (2,-i e^{-\text {arcsinh}(c x)}\right )}{3 c}+\frac {8 i b^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
(d^4*(1 + c^2*x^2)^(5/2)*((8*(a + b*ArcSinh[c*x])^2)/(3*c) + (a + b*ArcSin h[c*x])^3/(3*b*c) + (32*b*(a + b*ArcSinh[c*x])*Log[1 + I/E^ArcSinh[c*x]])/ (3*c) - (32*b^2*PolyLog[2, (-I)/E^ArcSinh[c*x]])/(3*c) + (4*b*(a + b*ArcSi nh[c*x])*Sec[Pi/4 + (I/2)*ArcSinh[c*x]]^2)/(3*c) + (((8*I)/3)*b^2*Tan[Pi/4 + (I/2)*ArcSinh[c*x]])/c + (((8*I)/3)*(a + b*ArcSinh[c*x])^2*Tan[Pi/4 + ( I/2)*ArcSinh[c*x]])/c - (((2*I)/3)*(a + b*ArcSinh[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))
3.7.1.3.1 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]
\[\int \frac {\left (i c d x +d \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{2}}{\left (-i c f x +f \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
integral(((b^2*c*d*x - I*b^2*d)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c *x + sqrt(c^2*x^2 + 1))^2 + 2*(a*b*c*d*x - I*a*b*d)*sqrt(I*c*d*x + d)*sqrt (-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1)) + (a^2*c*d*x - I*a^2*d)*sqrt(I *c*d*x + d)*sqrt(-I*c*f*x + f))/(c^3*f^3*x^3 + 3*I*c^2*f^3*x^2 - 3*c*f^3*x - I*f^3), x)
\[ \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\int \frac {\left (i d \left (c x - i\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\left (- i f \left (c x + i\right )\right )^{\frac {5}{2}}}\, dx \]
Timed out. \[ \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {(d+i c d x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(f-i c f x)^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{3/2}}{{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]