Integrand size = 35, antiderivative size = 470 \[ \int \frac {(d+i c d x)^{5/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=-\frac {i b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 i b d^5 \left (1+c^2 x^2\right )^{5/2}}{3 c (i+c x) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {5 b d^5 \left (1+c^2 x^2\right )^{5/2} \text {arcsinh}(c x)^2}{2 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i d^5 (1+i c x)^4 \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {10 i d^5 (1+i c x)^2 \left (1+c^2 x^2\right )^2 (a+b \text {arcsinh}(c x))}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 i d^5 \left (1+c^2 x^2\right )^3 (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \text {arcsinh}(c x) (a+b \text {arcsinh}(c x))}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 b d^5 \left (1+c^2 x^2\right )^{5/2} \log (i+c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
-I*b*d^5*x*(c^2*x^2+1)^(5/2)/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+8/3*I*b*d ^5*(c^2*x^2+1)^(5/2)/c/(c*x+I)/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)-5/2*b*d ^5*(c^2*x^2+1)^(5/2)*arcsinh(c*x)^2/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)- 2/3*I*d^5*(1+I*c*x)^4*(c^2*x^2+1)*(a+b*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/( f-I*c*f*x)^(5/2)+10/3*I*d^5*(1+I*c*x)^2*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))/c /(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+5*I*d^5*(c^2*x^2+1)^3*(a+b*arcsinh(c* x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+5*d^5*(c^2*x^2+1)^(5/2)*arcsinh( c*x)*(a+b*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2)+28/3*b*d^5*( c^2*x^2+1)^(5/2)*ln(c*x+I)/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. \(1083\) vs. \(2(470)=940\).
Time = 15.42 (sec) , antiderivative size = 1083, normalized size of antiderivative = 2.30 \[ \int \frac {(d+i c d x)^{5/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx =\text {Too large to display} \]
(((4*I)*a*d^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(-23 + (34*I)*c*x + 3*c^ 2*x^2))/(f^3*(I + c*x)^2) + (60*a*d^(5/2)*Log[c*d*f*x + Sqrt[d]*Sqrt[f]*Sq rt[d + I*c*d*x]*Sqrt[f - I*c*f*x]])/f^(5/2) - ((2*I)*b*d^2*Sqrt[d + I*c*d* x]*Sqrt[f - I*c*f*x]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])*(-(Co sh[(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]*(4*I + 3*ArcSinh[c*x] - 6*Ar cTan[Coth[ArcSinh[c*x]/2]] + ((3*I)/2)*Log[1 + c^2*x^2]) + 2*(2 + (2*I)*Ar cSinh[c*x] + (4*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + Log[1 + c^2*x^2] + (Sqrt [1 + c^2*x^2]*((2*I)*ArcSinh[c*x] + (4*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + L og[1 + c^2*x^2]))/2)*Sinh[ArcSinh[c*x]/2]))/(f^3*(1 + I*c*x)*(Cosh[ArcSinh [c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^4) + (2*b*d^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2])*(Cosh[(3*ArcSin h[c*x])/2]*((14*I - 3*ArcSinh[c*x])*ArcSinh[c*x] + (28*I)*ArcTan[Tanh[ArcS inh[c*x]/2]] - 7*Log[1 + c^2*x^2]) + Cosh[ArcSinh[c*x]/2]*(8 + (6*I)*ArcSi nh[c*x] + 9*ArcSinh[c*x]^2 - (84*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 21*Log[ 1 + c^2*x^2]) - (2*I)*(4 + (4*I)*ArcSinh[c*x] + 6*ArcSinh[c*x]^2 - (56*I)* ArcTan[Tanh[ArcSinh[c*x]/2]] + 14*Log[1 + c^2*x^2] + Sqrt[1 + c^2*x^2]*(Ar cSinh[c*x]*(14*I + 3*ArcSinh[c*x]) - (28*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 7*Log[1 + c^2*x^2]))*Sinh[ArcSinh[c*x]/2]))/(f^3*(1 + I*c*x)*(Cosh[ArcSin h[c*x]/2] - I*Sinh[ArcSinh[c*x]/2])^4) - (I*b*d^2*Sqrt[d + I*c*d*x]*Sqr...
Time = 0.67 (sec) , antiderivative size = 228, normalized size of antiderivative = 0.49, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {6211, 27, 6252, 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.
| \(\displaystyle \int \frac {(d+i c d x)^{5/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6211 |
| \(\displaystyle \frac {\left (c^2 x^2+1\right )^{5/2} \int \frac {d^5 (i c x+1)^5 (a+b \text {arcsinh}(c x))}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^5 \left (c^2 x^2+1\right )^{5/2} \int \frac {(i c x+1)^5 (a+b \text {arcsinh}(c x))}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 6252 |
| \(\displaystyle \frac {d^5 \left (c^2 x^2+1\right )^{5/2} \left (-b c \int \left (-\frac {2 i (i c x+1)^4}{3 c \left (c^2 x^2+1\right )^2}+\frac {20 i (i c x+1)}{3 c \left (c^2 x^2+1\right )}+\frac {5 \text {arcsinh}(c x)}{c \sqrt {c^2 x^2+1}}+\frac {5 i}{3 c}\right )dx-\frac {2 i (1+i c x)^4 (a+b \text {arcsinh}(c x))}{3 c \left (c^2 x^2+1\right )^{3/2}}+\frac {20 i (1+i c x) (a+b \text {arcsinh}(c x))}{3 c \sqrt {c^2 x^2+1}}+\frac {5 i \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))}{3 c}+\frac {5 \text {arcsinh}(c x) (a+b \text {arcsinh}(c x))}{c}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^5 \left (c^2 x^2+1\right )^{5/2} \left (-\frac {2 i (1+i c x)^4 (a+b \text {arcsinh}(c x))}{3 c \left (c^2 x^2+1\right )^{3/2}}+\frac {20 i (1+i c x) (a+b \text {arcsinh}(c x))}{3 c \sqrt {c^2 x^2+1}}+\frac {5 i \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))}{3 c}+\frac {5 \text {arcsinh}(c x) (a+b \text {arcsinh}(c x))}{c}-b c \left (\frac {5 \text {arcsinh}(c x)^2}{2 c^2}-\frac {8 i}{3 c^2 (c x+i)}-\frac {28 \log (c x+i)}{3 c^2}+\frac {i x}{c}\right )\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
(d^5*(1 + c^2*x^2)^(5/2)*((((-2*I)/3)*(1 + I*c*x)^4*(a + b*ArcSinh[c*x]))/ (c*(1 + c^2*x^2)^(3/2)) + (((20*I)/3)*(1 + I*c*x)*(a + b*ArcSinh[c*x]))/(c *Sqrt[1 + c^2*x^2]) + (((5*I)/3)*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/c + (5*ArcSinh[c*x]*(a + b*ArcSinh[c*x]))/c - b*c*((I*x)/c - ((8*I)/3)/(c^2 *(I + c*x)) + (5*ArcSinh[c*x]^2)/(2*c^2) - (28*Log[I + c*x])/(3*c^2))))/(( d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))
3.6.64.3.1 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_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + ( e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p , x]}, Simp[(a + b*ArcSinh[c*x]) u, x] - Simp[b*c Int[1/Sqrt[1 + c^2*x^ 2] u, x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IGtQ [m, 0] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3])
\[\int \frac {\left (i c d x +d \right )^{\frac {5}{2}} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )}{\left (-i c f x +f \right )^{\frac {5}{2}}}d x\]
\[ \int \frac {(d+i c d x)^{5/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{\frac {5}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{{\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
integral(((I*b*c^2*d^2*x^2 + 2*b*c*d^2*x - I*b*d^2)*sqrt(I*c*d*x + d)*sqrt (-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1)) + (I*a*c^2*d^2*x^2 + 2*a*c*d^2 *x - I*a*d^2)*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)
Timed out. \[ \int \frac {(d+i c d x)^{5/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {(d+i c d x)^{5/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\int { \frac {{\left (i \, c d x + d\right )}^{\frac {5}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{{\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
-1/3*(-3*I*(c^2*d*f*x^2 + d*f)^(5/2)/(c^5*f^5*x^4 + 4*I*c^4*f^5*x^3 - 6*c^ 3*f^5*x^2 - 4*I*c^2*f^5*x + c*f^5) + 15*I*(c^2*d*f*x^2 + d*f)^(3/2)*d/(3*I *c^4*f^4*x^3 - 9*c^3*f^4*x^2 - 9*I*c^2*f^4*x + 3*c*f^4) - 10*I*sqrt(c^2*d* f*x^2 + d*f)*d^2/(c^3*f^3*x^2 + 2*I*c^2*f^3*x - c*f^3) - 105*I*sqrt(c^2*d* f*x^2 + d*f)*d^2/(-3*I*c^2*f^3*x + 3*c*f^3) - 15*d^3*arcsinh(c*x)/(c*f^3*s qrt(d/f)))*a + b*integrate((I*c*d*x + d)^(5/2)*log(c*x + sqrt(c^2*x^2 + 1) )/(-I*c*f*x + f)^(5/2), x)
Exception generated. \[ \int \frac {(d+i c d x)^{5/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {(d+i c d x)^{5/2} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{5/2}}{{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]