Integrand size = 37, antiderivative size = 752 \[ \int \frac {(f-i c f x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2}} \, dx=-\frac {2 i a b f^3 x \left (1+c^2 x^2\right )^{3/2}}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {2 i b^2 f^3 \left (1+c^2 x^2\right )^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {2 i b^2 f^3 x \left (1+c^2 x^2\right )^{3/2} \text {arcsinh}(c x)}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {4 i f^3 \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {4 f^3 x \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {4 f^3 \left (1+c^2 x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {i f^3 \left (1+c^2 x^2\right )^2 (a+b \text {arcsinh}(c x))^2}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {f^3 \left (1+c^2 x^2\right )^{3/2} (a+b \text {arcsinh}(c x))^3}{b c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {16 i b f^3 \left (1+c^2 x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {8 b f^3 \left (1+c^2 x^2\right )^{3/2} (a+b \text {arcsinh}(c x)) \log \left (1+e^{2 \text {arcsinh}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {8 b^2 f^3 \left (1+c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}+\frac {8 b^2 f^3 \left (1+c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {4 b^2 f^3 \left (1+c^2 x^2\right )^{3/2} \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(c x)}\right )}{c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]
-2*I*a*b*f^3*x*(c^2*x^2+1)^(3/2)/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)+2*I*b ^2*f^3*(c^2*x^2+1)^2/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)-2*I*b^2*f^3*x*( c^2*x^2+1)^(3/2)*arcsinh(c*x)/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)+4*I*f^3* (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)+4*f ^3*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)+ 4*f^3*(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)+I*f^3*(c^2*x^2+1)^2*(a+b*arcsinh(c*x))^2/c/(d+I*c*d*x)^(3/2)/(f-I *c*f*x)^(3/2)-f^3*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))^3/b/c/(d+I*c*d*x)^( 3/2)/(f-I*c*f*x)^(3/2)-16*I*b*f^3*(c^2*x^2+1)^(3/2)*(a+b*arcsinh(c*x))*arc tan(c*x+(c^2*x^2+1)^(1/2))/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)-8*b*f^3*( c^2*x^2+1)^(3/2)*(a+b*arcsinh(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)-8*b^2*f^3*(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)+8*b^2*f^3*(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)-4*b^2*f^3*(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 = 14.85 (sec) , antiderivative size = 1174, normalized size of antiderivative = 1.56 \[ \int \frac {(f-i c f x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2}} \, dx =\text {Too large to display} \]
((I/3)*f*(-3*a^2*(-5*I + c*x)*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2]) + 9*a^2*Sqrt[d]* Sqrt[f]*(-I + c*x)*Sqrt[1 + c^2*x^2]*Log[c*d*f*x + Sqrt[d]*Sqrt[f]*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]]*((-I)*Cosh[ArcSinh[c*x]/2] + Sinh[ArcSinh[c* x]/2]) + 6*a*b*(I - c*x)*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(Cosh[ArcSinh [c*x]/2]*(-(c*x) + (2 + Sqrt[1 + c^2*x^2])*ArcSinh[c*x] + I*ArcSinh[c*x]^2 + 4*ArcTan[Coth[ArcSinh[c*x]/2]] + I*Log[1 + c^2*x^2]) + I*(-(c*x) + (-2 + Sqrt[1 + c^2*x^2])*ArcSinh[c*x] + I*ArcSinh[c*x]^2 + 4*ArcTan[Coth[ArcSi nh[c*x]/2]] + I*Log[1 + c^2*x^2])*Sinh[ArcSinh[c*x]/2]) + (3*I)*a*b*(I - c *x)*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(Cosh[ArcSinh[c*x]/2]*(ArcSinh[c*x ]*(-4*I + ArcSinh[c*x]) + (8*I)*ArcTan[Tanh[ArcSinh[c*x]/2]] + 2*Log[1 + c ^2*x^2]) + I*(ArcSinh[c*x]*(4*I + ArcSinh[c*x]) + (8*I)*ArcTan[Tanh[ArcSin h[c*x]/2]] + 2*Log[1 + c^2*x^2])*Sinh[ArcSinh[c*x]/2]) + I*b^2*(I - c*x)*S qrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*((6 - 6*I)*ArcSinh[c*x]^2*(Cosh[ArcSinh [c*x]/2] - Sinh[ArcSinh[c*x]/2]) + ArcSinh[c*x]^3*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2]) + 6*ArcSinh[c*x]*(I*Pi + 4*Log[1 - I/E^ArcSinh[c*x ]])*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2]) - 24*PolyLog[2, I/E^Ar cSinh[c*x]]*(Cosh[ArcSinh[c*x]/2] + I*Sinh[ArcSinh[c*x]/2]) + 12*Pi*(Log[1 - I/E^ArcSinh[c*x]] + 2*Log[1 + E^ArcSinh[c*x]] - 2*Log[Cosh[ArcSinh[c*x] /2]] - Log[Sin[(Pi + (2*I)*ArcSinh[c*x])/4]])*((-I)*Cosh[ArcSinh[c*x]/2...
Time = 1.27 (sec) , antiderivative size = 313, normalized size of antiderivative = 0.42, 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.
| \(\displaystyle \int \frac {(f-i c f x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2}} \, dx\) |
| \(\Big \downarrow \) 6211 |
| \(\displaystyle \frac {\left (c^2 x^2+1\right )^{3/2} \int \frac {f^3 (1-i c x)^3 (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{3/2}}dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {f^3 \left (c^2 x^2+1\right )^{3/2} \int \frac {(1-i c x)^3 (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{3/2}}dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\) |
| \(\Big \downarrow \) 6259 |
| \(\displaystyle \frac {f^3 \left (c^2 x^2+1\right )^{3/2} \int \left (\frac {i c x (a+b \text {arcsinh}(c x))^2}{\sqrt {c^2 x^2+1}}-\frac {3 (a+b \text {arcsinh}(c x))^2}{\sqrt {c^2 x^2+1}}-\frac {4 i (c x+i) (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{3/2}}\right )dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f^3 \left (c^2 x^2+1\right )^{3/2} \left (-\frac {16 i b \arctan \left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{c}+\frac {i \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^2}{c}+\frac {4 x (a+b \text {arcsinh}(c x))^2}{\sqrt {c^2 x^2+1}}+\frac {4 i (a+b \text {arcsinh}(c x))^2}{c \sqrt {c^2 x^2+1}}-\frac {(a+b \text {arcsinh}(c x))^3}{b c}+\frac {4 (a+b \text {arcsinh}(c x))^2}{c}-\frac {8 b \log \left (e^{2 \text {arcsinh}(c x)}+1\right ) (a+b \text {arcsinh}(c x))}{c}-2 i a b x-\frac {8 b^2 \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{c}+\frac {8 b^2 \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{c}-\frac {4 b^2 \operatorname {PolyLog}\left (2,-e^{2 \text {arcsinh}(c x)}\right )}{c}-2 i b^2 x \text {arcsinh}(c x)+\frac {2 i b^2 \sqrt {c^2 x^2+1}}{c}\right )}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\) |
(f^3*(1 + c^2*x^2)^(3/2)*((-2*I)*a*b*x + ((2*I)*b^2*Sqrt[1 + c^2*x^2])/c - (2*I)*b^2*x*ArcSinh[c*x] + (4*(a + b*ArcSinh[c*x])^2)/c + ((4*I)*(a + b*A rcSinh[c*x])^2)/(c*Sqrt[1 + c^2*x^2]) + (4*x*(a + b*ArcSinh[c*x])^2)/Sqrt[ 1 + c^2*x^2] + (I*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/c - (a + b*Arc Sinh[c*x])^3/(b*c) - ((16*I)*b*(a + b*ArcSinh[c*x])*ArcTan[E^ArcSinh[c*x]] )/c - (8*b*(a + b*ArcSinh[c*x])*Log[1 + E^(2*ArcSinh[c*x])])/c - (8*b^2*Po lyLog[2, (-I)*E^ArcSinh[c*x]])/c + (8*b^2*PolyLog[2, I*E^ArcSinh[c*x]])/c - (4*b^2*PolyLog[2, -E^(2*ArcSinh[c*x])])/c))/((d + I*c*d*x)^(3/2)*(f - I* c*f*x)^(3/2))
3.6.80.3.1 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]
\[\int \frac {\left (-i c f x +f \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{2}}{\left (i c d x +d \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(f-i c f x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2}} \, dx=\int { \frac {{\left (-i \, c f x + f\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )}^{\frac {3}{2}}} \,d x } \]
integral(((I*b^2*c*f*x - b^2*f)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c *x + sqrt(c^2*x^2 + 1))^2 - 2*(-I*a*b*c*f*x + a*b*f)*sqrt(I*c*d*x + d)*sqr t(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1)) + (I*a^2*c*f*x - a^2*f)*sqrt( I*c*d*x + d)*sqrt(-I*c*f*x + f))/(c^2*d^2*x^2 - 2*I*c*d^2*x - d^2), x)
\[ \int \frac {(f-i c f x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2}} \, dx=\int \frac {\left (- i f \left (c x + i\right )\right )^{\frac {3}{2}} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\left (i d \left (c x - i\right )\right )^{\frac {3}{2}}}\, dx \]
\[ \int \frac {(f-i c f x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2}} \, dx=\int { \frac {{\left (-i \, c f x + f\right )}^{\frac {3}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )}^{\frac {3}{2}}} \,d x } \]
a^2*(I*(c^2*d*f*x^2 + d*f)^(3/2)/(c^3*d^3*x^2 - 2*I*c^2*d^3*x - c*d^3) + 6 *I*sqrt(c^2*d*f*x^2 + d*f)*f/(I*c^2*d^2*x + c*d^2) - 3*f^2*arcsinh(c*x)/(c *d^2*sqrt(f/d))) + integrate((-I*c*f*x + f)^(3/2)*b^2*log(c*x + sqrt(c^2*x ^2 + 1))^2/(I*c*d*x + d)^(3/2) + 2*(-I*c*f*x + f)^(3/2)*a*b*log(c*x + sqrt (c^2*x^2 + 1))/(I*c*d*x + d)^(3/2), x)
Timed out. \[ \int \frac {(f-i c f x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(f-i c f x)^{3/2} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{3/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{3/2}}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]