Integrand size = 37, antiderivative size = 525 \[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2}} \, dx=-\frac {f \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2}{3 c d^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {4 i b^2 f \sqrt {1+c^2 x^2} \cot \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c d^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {i f \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c d^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {2 b f \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x)) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c d^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i f \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c d^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {4 b f \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x)) \log \left (1+i e^{\text {arcsinh}(c x)}\right )}{3 c d^2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {4 b^2 f \sqrt {1+c^2 x^2} \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{3 c d^2 \sqrt {d+i c d x} \sqrt {f-i c f x}} \] Output:
-1/3*f*(c^2*x^2+1)^(1/2)*(a+b*arcsinh(c*x))^2/c/d^2/(d+I*c*d*x)^(1/2)/(f-I *c*f*x)^(1/2)-4/3*I*b^2*f*(c^2*x^2+1)^(1/2)*cot(1/4*Pi+1/2*I*arcsinh(c*x)) /c/d^2/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-1/3*I*f*(c^2*x^2+1)^(1/2)*(a+b* arcsinh(c*x))^2*cot(1/4*Pi+1/2*I*arcsinh(c*x))/c/d^2/(d+I*c*d*x)^(1/2)/(f- I*c*f*x)^(1/2)+2/3*b*f*(c^2*x^2+1)^(1/2)*(a+b*arcsinh(c*x))*csc(1/4*Pi+1/2 *I*arcsinh(c*x))^2/c/d^2/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+1/3*I*f*(c^2* x^2+1)^(1/2)*(a+b*arcsinh(c*x))^2*cot(1/4*Pi+1/2*I*arcsinh(c*x))*csc(1/4*P i+1/2*I*arcsinh(c*x))^2/c/d^2/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+4/3*b*f* (c^2*x^2+1)^(1/2)*(a+b*arcsinh(c*x))*ln(1+I*(c*x+(c^2*x^2+1)^(1/2)))/c/d^2 /(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+4/3*b^2*f*(c^2*x^2+1)^(1/2)*polylog(2 ,-I*(c*x+(c^2*x^2+1)^(1/2)))/c/d^2/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)
Time = 9.31 (sec) , antiderivative size = 783, normalized size of antiderivative = 1.49 \[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2}} \, dx =\text {Too large to display} \] Input:
Integrate[(Sqrt[f - I*c*f*x]*(a + b*ArcSinh[c*x])^2)/(d + I*c*d*x)^(5/2),x ]
Output:
(Sqrt[I*d*(-I + c*x)]*Sqrt[(-I)*f*(I + c*x)]*((((-2*I)/3)*a^2)/(d^3*(-I + c*x)^2) - a^2/(3*d^3*(-I + c*x))))/c + ((I/3)*a*b*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])*((-I)*Cosh[(3*ArcSinh[c*x])/2]*(ArcSinh[c*x] - 2*ArcTan[Coth[ArcSinh[c*x]/2]] - I*Log[Sqrt[1 + c^2*x^2]]) + Cosh[ArcSinh[ c*x]/2]*(4 + (3*I)*ArcSinh[c*x] - (6*I)*ArcTan[Coth[ArcSinh[c*x]/2]] + 3*L og[Sqrt[1 + c^2*x^2]]) + 2*(Sqrt[1 + c^2*x^2]*(ArcSinh[c*x] + 2*ArcTan[Cot h[ArcSinh[c*x]/2]] + I*Log[Sqrt[1 + c^2*x^2]]) + 2*(I + ArcSinh[c*x] + 2*A rcTan[Coth[ArcSinh[c*x]/2]] + I*Log[Sqrt[1 + c^2*x^2]]))*Sinh[ArcSinh[c*x] /2]))/(c*d^3*(I + c*x)*Sqrt[-(((-I)*d + c*d*x)*(I*f + c*f*x))]*(Cosh[ArcSi nh[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*(ArcSinh[c*x ] - 4*Log[1 + E^ArcSinh[c*x]] + 4*Log[Cosh[ArcSinh[c*x]/2]] + 2*Log[Sin[(P i + (2*I)*ArcSinh[c*x])/4]]) + 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[ArcSinh[c*x]/ 2])^3 + (2*(4 + ArcSinh[c*x]^2)*Sinh[ArcSinh[c*x]/2])/(Cosh[ArcSinh[c*x]/2 ] + I*Sinh[ArcSinh[c*x]/2])))/(c*d^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]/2])^2)
Time = 2.19 (sec) , antiderivative size = 267, normalized size of antiderivative = 0.51, 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.
\(\displaystyle \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2}} \, dx\) |
\(\Big \downarrow \) 6211 |
\(\displaystyle \frac {\left (c^2 x^2+1\right )^{5/2} \int \frac {f^3 (1-i c x)^3 (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {f^3 \left (c^2 x^2+1\right )^{5/2} \int \frac {(1-i c x)^3 (a+b \text {arcsinh}(c x))^2}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
\(\Big \downarrow \) 6259 |
\(\displaystyle \frac {f^3 \left (c^2 x^2+1\right )^{5/2} \int \left (\frac {i (a+b \text {arcsinh}(c x))^2}{(c x-i) \sqrt {c^2 x^2+1}}-\frac {2 (a+b \text {arcsinh}(c x))^2}{(c x-i)^2 \sqrt {c^2 x^2+1}}\right )dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {f^3 \left (c^2 x^2+1\right )^{5/2} \left (-\frac {(a+b \text {arcsinh}(c x))^2}{3 c}+\frac {4 b \log \left (1+i e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{3 c}-\frac {i \cot \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))^2}{3 c}+\frac {2 b \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))}{3 c}+\frac {i \cot \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) \csc ^2\left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right ) (a+b \text {arcsinh}(c x))^2}{3 c}+\frac {4 b^2 \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{3 c}-\frac {4 i b^2 \cot \left (\frac {\pi }{4}+\frac {1}{2} i \text {arcsinh}(c x)\right )}{3 c}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
Input:
Int[(Sqrt[f - I*c*f*x]*(a + b*ArcSinh[c*x])^2)/(d + I*c*d*x)^(5/2),x]
Output:
(f^3*(1 + c^2*x^2)^(5/2)*(-1/3*(a + b*ArcSinh[c*x])^2/c - (((4*I)/3)*b^2*C ot[Pi/4 + (I/2)*ArcSinh[c*x]])/c - ((I/3)*(a + b*ArcSinh[c*x])^2*Cot[Pi/4 + (I/2)*ArcSinh[c*x]])/c + (2*b*(a + b*ArcSinh[c*x])*Csc[Pi/4 + (I/2)*ArcS inh[c*x]]^2)/(3*c) + ((I/3)*(a + b*ArcSinh[c*x])^2*Cot[Pi/4 + (I/2)*ArcSin h[c*x]]*Csc[Pi/4 + (I/2)*ArcSinh[c*x]]^2)/c + (4*b*(a + b*ArcSinh[c*x])*Lo g[1 + I*E^ArcSinh[c*x]])/(3*c) + (4*b^2*PolyLog[2, (-I)*E^ArcSinh[c*x]])/( 3*c)))/((d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))
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]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2756 vs. \(2 (443 ) = 886\).
Time = 7.90 (sec) , antiderivative size = 2757, normalized size of antiderivative = 5.25
method | result | size |
default | \(\text {Expression too large to display}\) | \(2757\) |
parts | \(\text {Expression too large to display}\) | \(2757\) |
Input:
int((f-I*c*f*x)^(1/2)*(a+b*arcsinh(x*c))^2/(d+I*c*d*x)^(5/2),x,method=_RET URNVERBOSE)
Output:
-2/3*b^2*(-I*(I+x*c)*f)^(1/2)*(I*(x*c-I)*d)^(1/2)/d^3/(3*c^2*x^2-1)/(c^2*x ^2+1)^2*arcsinh(x*c)*x+4*b^2*(-I*(I+x*c)*f)^(1/2)*(I*(x*c-I)*d)^(1/2)/d^3/ (3*c^2*x^2-1)/(c^2*x^2+1)^(3/2)*c^3*x^4-4/3*b^2*(-I*(I+x*c)*f)^(1/2)*(I*(x *c-I)*d)^(1/2)/d^3/(3*c^2*x^2-1)/(c^2*x^2+1)*c^2*x^3-1/3*b^2*(-I*(I+x*c)*f )^(1/2)*(I*(x*c-I)*d)^(1/2)/d^3/(3*c^2*x^2-1)/(c^2*x^2+1)^(3/2)/c*arcsinh( x*c)^2+2/3*b^2*(-I*(I+x*c)*f)^(1/2)*(I*(x*c-I)*d)^(1/2)/d^3/(3*c^2*x^2-1)/ (c^2*x^2+1)*arcsinh(x*c)*x-8/3*b^2*(-I*(I+x*c)*f)^(1/2)*(I*(x*c-I)*d)^(1/2 )/d^3/(3*c^2*x^2-1)/(c^2*x^2+1)^2*c^4*x^5+4/3*b^2/(c^2*x^2+1)^(1/2)*(I*(x* c-I)*d)^(1/2)*(-I*(I+x*c)*f)^(1/2)/c/d^3*arcsinh(x*c)*ln(1+I*(x*c+(c^2*x^2 +1)^(1/2)))+4/3*I*b^2*(-I*(I+x*c)*f)^(1/2)*(I*(x*c-I)*d)^(1/2)/d^3/(3*c^2* x^2-1)/(c^2*x^2+1)/c-b^2*(-I*(I+x*c)*f)^(1/2)*(I*(x*c-I)*d)^(1/2)/d^3/(3*c ^2*x^2-1)/(c^2*x^2+1)^2*arcsinh(x*c)^2*x+8/3*b^2*(-I*(I+x*c)*f)^(1/2)*(I*( x*c-I)*d)^(1/2)/d^3/(3*c^2*x^2-1)/(c^2*x^2+1)^(3/2)*c*x^2-4/3*b^2*(-I*(I+x *c)*f)^(1/2)*(I*(x*c-I)*d)^(1/2)/d^3/(3*c^2*x^2-1)/(c^2*x^2+1)^(3/2)/c*arc sinh(x*c)-4/3*b^2*(-I*(I+x*c)*f)^(1/2)*(I*(x*c-I)*d)^(1/2)/d^3/(3*c^2*x^2- 1)/(c^2*x^2+1)*x+8/3*b^2*(-I*(I+x*c)*f)^(1/2)*(I*(x*c-I)*d)^(1/2)/d^3/(3*c ^2*x^2-1)/(c^2*x^2+1)^2*x-2/3*b^2/(c^2*x^2+1)^(1/2)*(I*(x*c-I)*d)^(1/2)*(- I*(I+x*c)*f)^(1/2)/c/d^3*arcsinh(x*c)^2+4/3*b^2/(c^2*x^2+1)^(1/2)*(I*(x*c- I)*d)^(1/2)*(-I*(I+x*c)*f)^(1/2)/c/d^3*polylog(2,-I*(x*c+(c^2*x^2+1)^(1/2) ))-4/3*b^2*(-I*(I+x*c)*f)^(1/2)*(I*(x*c-I)*d)^(1/2)/d^3/(3*c^2*x^2-1)/(...
\[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2}} \, dx=\int { \frac {\sqrt {-i \, c f x + f} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((f-I*c*f*x)^(1/2)*(a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(5/2),x, algo rithm="fricas")
Output:
-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*d^3*x^2 - 2*I*c^2*d^3*x - c*d^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*d^3*x^3 - 3*I*c^2*d^3*x^2 - 3* c*d^3*x + I*d^3), x))/(c^3*d^3*x^2 - 2*I*c^2*d^3*x - c*d^3)
\[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2}} \, dx=\int \frac {\sqrt {- i f \left (c x + i\right )} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}{\left (i d \left (c x - i\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((f-I*c*f*x)**(1/2)*(a+b*asinh(c*x))**2/(d+I*c*d*x)**(5/2),x)
Output:
Integral(sqrt(-I*f*(c*x + I))*(a + b*asinh(c*x))**2/(I*d*(c*x - I))**(5/2) , x)
Timed out. \[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2}} \, dx=\text {Timed out} \] Input:
integrate((f-I*c*f*x)^(1/2)*(a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(5/2),x, algo rithm="maxima")
Output:
Timed out
\[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2}} \, dx=\int { \frac {\sqrt {-i \, c f x + f} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{{\left (i \, c d x + d\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((f-I*c*f*x)^(1/2)*(a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(5/2),x, algo rithm="giac")
Output:
integrate(sqrt(-I*c*f*x + f)*(b*arcsinh(c*x) + a)^2/(I*c*d*x + d)^(5/2), x )
Timed out. \[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2}} \, dx=\int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,\sqrt {f-c\,f\,x\,1{}\mathrm {i}}}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \] Input:
int(((a + b*asinh(c*x))^2*(f - c*f*x*1i)^(1/2))/(d + c*d*x*1i)^(5/2),x)
Output:
int(((a + b*asinh(c*x))^2*(f - c*f*x*1i)^(1/2))/(d + c*d*x*1i)^(5/2), x)
\[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))^2}{(d+i c d x)^{5/2}} \, dx=\frac {\sqrt {f}\, \left (-6 \sqrt {c i x +1}\, \sqrt {-c i x +1}\, \left (\int \frac {\sqrt {-c i x +1}\, \mathit {asinh} \left (c x \right )}{\sqrt {c i x +1}\, c^{2} x^{2}-2 \sqrt {c i x +1}\, c i x -\sqrt {c i x +1}}d x \right ) a b \,c^{2} i x -6 \sqrt {c i x +1}\, \sqrt {-c i x +1}\, \left (\int \frac {\sqrt {-c i x +1}\, \mathit {asinh} \left (c x \right )}{\sqrt {c i x +1}\, c^{2} x^{2}-2 \sqrt {c i x +1}\, c i x -\sqrt {c i x +1}}d x \right ) a b c -3 \sqrt {c i x +1}\, \sqrt {-c i x +1}\, \left (\int \frac {\sqrt {-c i x +1}\, \mathit {asinh} \left (c x \right )^{2}}{\sqrt {c i x +1}\, c^{2} x^{2}-2 \sqrt {c i x +1}\, c i x -\sqrt {c i x +1}}d x \right ) b^{2} c^{2} i x -3 \sqrt {c i x +1}\, \sqrt {-c i x +1}\, \left (\int \frac {\sqrt {-c i x +1}\, \mathit {asinh} \left (c x \right )^{2}}{\sqrt {c i x +1}\, c^{2} x^{2}-2 \sqrt {c i x +1}\, c i x -\sqrt {c i x +1}}d x \right ) b^{2} c -a^{2} c^{2} i \,x^{2}+2 a^{2} c x +a^{2} i \right )}{3 \sqrt {d}\, \sqrt {c i x +1}\, \sqrt {-c i x +1}\, c \,d^{2} \left (c i x +1\right )} \] Input:
int((f-I*c*f*x)^(1/2)*(a+b*asinh(c*x))^2/(d+I*c*d*x)^(5/2),x)
Output:
(sqrt(f)*( - 6*sqrt(c*i*x + 1)*sqrt( - c*i*x + 1)*int((sqrt( - c*i*x + 1)* asinh(c*x))/(sqrt(c*i*x + 1)*c**2*x**2 - 2*sqrt(c*i*x + 1)*c*i*x - sqrt(c* i*x + 1)),x)*a*b*c**2*i*x - 6*sqrt(c*i*x + 1)*sqrt( - c*i*x + 1)*int((sqrt ( - c*i*x + 1)*asinh(c*x))/(sqrt(c*i*x + 1)*c**2*x**2 - 2*sqrt(c*i*x + 1)* c*i*x - sqrt(c*i*x + 1)),x)*a*b*c - 3*sqrt(c*i*x + 1)*sqrt( - c*i*x + 1)*i nt((sqrt( - c*i*x + 1)*asinh(c*x)**2)/(sqrt(c*i*x + 1)*c**2*x**2 - 2*sqrt( c*i*x + 1)*c*i*x - sqrt(c*i*x + 1)),x)*b**2*c**2*i*x - 3*sqrt(c*i*x + 1)*s qrt( - c*i*x + 1)*int((sqrt( - c*i*x + 1)*asinh(c*x)**2)/(sqrt(c*i*x + 1)* c**2*x**2 - 2*sqrt(c*i*x + 1)*c*i*x - sqrt(c*i*x + 1)),x)*b**2*c - a**2*c* *2*i*x**2 + 2*a**2*c*x + a**2*i))/(3*sqrt(d)*sqrt(c*i*x + 1)*sqrt( - c*i*x + 1)*c*d**2*(c*i*x + 1))