Integrand size = 37, antiderivative size = 615 \[ \int \frac {(f-i c f x)^{5/2} (a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x}} \, dx=-\frac {68 i b^2 f^3 \left (1+c^2 x^2\right )}{9 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 b^2 f^3 x \left (1+c^2 x^2\right )}{4 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {2 i b^2 f^3 \left (1+c^2 x^2\right )^2}{27 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b^2 f^3 \sqrt {1+c^2 x^2} \text {arcsinh}(c x)}{4 c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {22 i b f^3 x \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {3 b c f^3 x^2 \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {2 i b c^2 f^3 x^3 \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))}{9 \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {11 i f^3 \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))^2}{3 c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {3 f^3 x \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))^2}{2 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i c f^3 x^2 \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))^2}{3 \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {5 f^3 \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^3}{6 b c \sqrt {d+i c d x} \sqrt {f-i c f x}} \]
-68/9*I*b^2*f^3*(c^2*x^2+1)/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-3/4*b^2* f^3*x*(c^2*x^2+1)/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+2/27*I*b^2*f^3*(c^2* x^2+1)^2/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-11/3*I*f^3*(c^2*x^2+1)*(a+b *arcsinh(c*x))^2/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-3/2*f^3*x*(c^2*x^2+ 1)*(a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+1/3*I*c*f^3*x^ 2*(c^2*x^2+1)*(a+b*arcsinh(c*x))^2/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+3/4 *b^2*f^3*arcsinh(c*x)*(c^2*x^2+1)^(1/2)/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1 /2)+22/3*I*b*f^3*x*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/(d+I*c*d*x)^(1/2)/ (f-I*c*f*x)^(1/2)+3/2*b*c*f^3*x^2*(a+b*arcsinh(c*x))*(c^2*x^2+1)^(1/2)/(d+ I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-2/9*I*b*c^2*f^3*x^3*(a+b*arcsinh(c*x))*(c ^2*x^2+1)^(1/2)/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+5/6*f^3*(a+b*arcsinh(c *x))^3*(c^2*x^2+1)^(1/2)/b/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)
Time = 14.05 (sec) , antiderivative size = 723, normalized size of antiderivative = 1.18 \[ \int \frac {(f-i c f x)^{5/2} (a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x}} \, dx=\frac {1620 i a b c f^2 x \sqrt {d+i c d x} \sqrt {f-i c f x}-792 i a^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}-1620 i b^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}-324 a^2 c f^2 x \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}+72 i a^2 c^2 f^2 x^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2}+180 b^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \text {arcsinh}(c x)^3+162 a b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \cosh (2 \text {arcsinh}(c x))+4 i b^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \cosh (3 \text {arcsinh}(c x))+6 b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \text {arcsinh}(c x) \left (27 b \cosh (2 \text {arcsinh}(c x))+2 i \left (-4 b c x \left (-33+c^2 x^2\right )+27 a (-5+2 i c x) \sqrt {1+c^2 x^2}+3 a \cosh (3 \text {arcsinh}(c x))\right )\right )+540 a^2 \sqrt {d} f^{5/2} \sqrt {1+c^2 x^2} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )-81 b^2 f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh (2 \text {arcsinh}(c x))+18 b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \text {arcsinh}(c x)^2 \left (30 a-45 i b \sqrt {1+c^2 x^2}+i b \cosh (3 \text {arcsinh}(c x))-9 b \sinh (2 \text {arcsinh}(c x))\right )-12 i a b f^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \sinh (3 \text {arcsinh}(c x))}{216 c d \sqrt {1+c^2 x^2}} \]
((1620*I)*a*b*c*f^2*x*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x] - (792*I)*a^2*f^ 2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] - (1620*I)*b^2*f^2 *Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] - 324*a^2*c*f^2*x*S qrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] + (72*I)*a^2*c^2*f^2* x^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2] + 180*b^2*f^2*Sq rt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^3 + 162*a*b*f^2*Sqrt[d + I* c*d*x]*Sqrt[f - I*c*f*x]*Cosh[2*ArcSinh[c*x]] + (4*I)*b^2*f^2*Sqrt[d + I*c *d*x]*Sqrt[f - I*c*f*x]*Cosh[3*ArcSinh[c*x]] + 6*b*f^2*Sqrt[d + I*c*d*x]*S qrt[f - I*c*f*x]*ArcSinh[c*x]*(27*b*Cosh[2*ArcSinh[c*x]] + (2*I)*(-4*b*c*x *(-33 + c^2*x^2) + 27*a*(-5 + (2*I)*c*x)*Sqrt[1 + c^2*x^2] + 3*a*Cosh[3*Ar cSinh[c*x]])) + 540*a^2*Sqrt[d]*f^(5/2)*Sqrt[1 + c^2*x^2]*Log[c*d*f*x + Sq rt[d]*Sqrt[f]*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]] - 81*b^2*f^2*Sqrt[d + I *c*d*x]*Sqrt[f - I*c*f*x]*Sinh[2*ArcSinh[c*x]] + 18*b*f^2*Sqrt[d + I*c*d*x ]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^2*(30*a - (45*I)*b*Sqrt[1 + c^2*x^2] + I* b*Cosh[3*ArcSinh[c*x]] - 9*b*Sinh[2*ArcSinh[c*x]]) - (12*I)*a*b*f^2*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sinh[3*ArcSinh[c*x]])/(216*c*d*Sqrt[1 + c^2* x^2])
Time = 1.00 (sec) , antiderivative size = 312, normalized size of antiderivative = 0.51, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {6211, 27, 6258, 3042, 3798, 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.
| \(\displaystyle \int \frac {(f-i c f x)^{5/2} (a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x}} \, dx\) |
| \(\Big \downarrow \) 6211 |
| \(\displaystyle \frac {\sqrt {c^2 x^2+1} \int \frac {f^3 (1-i c x)^3 (a+b \text {arcsinh}(c x))^2}{\sqrt {c^2 x^2+1}}dx}{\sqrt {d+i c d x} \sqrt {f-i c f x}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {f^3 \sqrt {c^2 x^2+1} \int \frac {(1-i c x)^3 (a+b \text {arcsinh}(c x))^2}{\sqrt {c^2 x^2+1}}dx}{\sqrt {d+i c d x} \sqrt {f-i c f x}}\) |
| \(\Big \downarrow \) 6258 |
| \(\displaystyle \frac {f^3 \sqrt {c^2 x^2+1} \int \left (c-i c^2 x\right )^3 (a+b \text {arcsinh}(c x))^2d\text {arcsinh}(c x)}{c^4 \sqrt {d+i c d x} \sqrt {f-i c f x}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {f^3 \sqrt {c^2 x^2+1} \int (a+b \text {arcsinh}(c x))^2 (c-c \sin (i \text {arcsinh}(c x)))^3d\text {arcsinh}(c x)}{c^4 \sqrt {d+i c d x} \sqrt {f-i c f x}}\) |
| \(\Big \downarrow \) 3798 |
| \(\displaystyle \frac {f^3 \sqrt {c^2 x^2+1} \int \left (i x^3 (a+b \text {arcsinh}(c x))^2 c^6-3 x^2 (a+b \text {arcsinh}(c x))^2 c^5-3 i x (a+b \text {arcsinh}(c x))^2 c^4+(a+b \text {arcsinh}(c x))^2 c^3\right )d\text {arcsinh}(c x)}{c^4 \sqrt {d+i c d x} \sqrt {f-i c f x}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {f^3 \sqrt {c^2 x^2+1} \left (-\frac {2}{9} i b c^6 x^3 (a+b \text {arcsinh}(c x))+\frac {3}{2} b c^5 x^2 (a+b \text {arcsinh}(c x))+\frac {22}{3} i b c^4 x (a+b \text {arcsinh}(c x))+\frac {5 c^3 (a+b \text {arcsinh}(c x))^3}{6 b}+\frac {1}{3} i c^5 x^2 \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^2-\frac {3}{2} c^4 x \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^2-\frac {11}{3} i c^3 \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^2+\frac {3}{4} b^2 c^3 \text {arcsinh}(c x)-\frac {3}{4} b^2 c^4 x \sqrt {c^2 x^2+1}+\frac {2}{27} i b^2 c^3 \left (c^2 x^2+1\right )^{3/2}-\frac {68}{9} i b^2 c^3 \sqrt {c^2 x^2+1}\right )}{c^4 \sqrt {d+i c d x} \sqrt {f-i c f x}}\) |
(f^3*Sqrt[1 + c^2*x^2]*(((-68*I)/9)*b^2*c^3*Sqrt[1 + c^2*x^2] - (3*b^2*c^4 *x*Sqrt[1 + c^2*x^2])/4 + ((2*I)/27)*b^2*c^3*(1 + c^2*x^2)^(3/2) + (3*b^2* c^3*ArcSinh[c*x])/4 + ((22*I)/3)*b*c^4*x*(a + b*ArcSinh[c*x]) + (3*b*c^5*x ^2*(a + b*ArcSinh[c*x]))/2 - ((2*I)/9)*b*c^6*x^3*(a + b*ArcSinh[c*x]) - (( 11*I)/3)*c^3*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2 - (3*c^4*x*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/2 + (I/3)*c^5*x^2*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2 + (5*c^3*(a + b*ArcSinh[c*x])^3)/(6*b)))/(c^4*Sqrt[d + I *c*d*x]*Sqrt[f - I*c*f*x])
3.6.85.3.1 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[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_.) , x_Symbol] :> Int[ExpandIntegrand[(c + d*x)^m, (a + b*Sin[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[n, 0] && (EqQ[n, 1] || IGtQ[ m, 0] || NeQ[a^2 - b^2, 0])
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_.))/S qrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[1/(c^(m + 1)*Sqrt[d]) Subst[I nt[(a + b*x)^n*(c*f + g*Sinh[x])^m, x], x, ArcSinh[c*x]], x] /; FreeQ[{a, b , c, d, e, f, g, n}, x] && EqQ[e, c^2*d] && IntegerQ[m] && GtQ[d, 0] && (Gt Q[m, 0] || IGtQ[n, 0])
\[\int \frac {\left (-i c f x +f \right )^{\frac {5}{2}} \left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{2}}{\sqrt {i c d x +d}}d x\]
\[ \int \frac {(f-i c f x)^{5/2} (a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x}} \, dx=\int { \frac {{\left (-i \, c f x + f\right )}^{\frac {5}{2}} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{\sqrt {i \, c d x + d}} \,d x } \]
integral(((I*b^2*c^2*f^2*x^2 - 2*b^2*c*f^2*x - I*b^2*f^2)*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^2*f^2*x^ 2 + 2*a*b*c*f^2*x + I*a*b*f^2)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c* x + sqrt(c^2*x^2 + 1)) + (I*a^2*c^2*f^2*x^2 - 2*a^2*c*f^2*x - I*a^2*f^2)*s qrt(I*c*d*x + d)*sqrt(-I*c*f*x + f))/(c*d*x - I*d), x)
Timed out. \[ \int \frac {(f-i c f x)^{5/2} (a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x}} \, dx=\text {Timed out} \]
Exception generated. \[ \int \frac {(f-i c f x)^{5/2} (a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x}} \, dx=\text {Exception raised: RuntimeError} \]
Exception generated. \[ \int \frac {(f-i c f x)^{5/2} (a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x}} \, dx=\text {Exception raised: TypeError} \]
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. \[ \int \frac {(f-i c f x)^{5/2} (a+b \text {arcsinh}(c x))^2}{\sqrt {d+i c d x}} \, dx=\int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}}{\sqrt {d+c\,d\,x\,1{}\mathrm {i}}} \,d x \]