\(\int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))}{\sqrt {d+i c d x}} \, dx\) [204]

Optimal result

Integrand size = 35, antiderivative size = 158 \[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))}{\sqrt {d+i c d x}} \, dx=\frac {i b f x \sqrt {1+c^2 x^2}}{\sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {i f \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {f \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^2}{2 b c \sqrt {d+i c d x} \sqrt {f-i c f x}} \] Output:

I*b*f*x*(c^2*x^2+1)^(1/2)/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-I*f*(c^2*x^2 
+1)*(a+b*arcsinh(c*x))/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+1/2*f*(c^2*x^ 
2+1)^(1/2)*(a+b*arcsinh(c*x))^2/b/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)
 

Mathematica [A] (verified)

Time = 1.30 (sec) , antiderivative size = 227, normalized size of antiderivative = 1.44 \[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))}{\sqrt {d+i c d x}} \, dx=\frac {2 i \sqrt {d+i c d x} \sqrt {f-i c f x} \left (b c x-a \sqrt {1+c^2 x^2}\right )-2 i b \sqrt {d+i c d x} \sqrt {f-i c f x} \sqrt {1+c^2 x^2} \text {arcsinh}(c x)+b \sqrt {d+i c d x} \sqrt {f-i c f x} \text {arcsinh}(c x)^2+2 a \sqrt {d} \sqrt {f} \sqrt {1+c^2 x^2} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )}{2 c d \sqrt {1+c^2 x^2}} \] Input:

Integrate[(Sqrt[f - I*c*f*x]*(a + b*ArcSinh[c*x]))/Sqrt[d + I*c*d*x],x]
 

Output:

((2*I)*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(b*c*x - a*Sqrt[1 + c^2*x^2]) - 
 (2*I)*b*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*Sqrt[1 + c^2*x^2]*ArcSinh[c*x 
] + b*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^2 + 2*a*Sqrt[d]*Sqr 
t[f]*Sqrt[1 + c^2*x^2]*Log[c*d*f*x + Sqrt[d]*Sqrt[f]*Sqrt[d + I*c*d*x]*Sqr 
t[f - I*c*f*x]])/(2*c*d*Sqrt[1 + c^2*x^2])
 

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.61, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {6211, 27, 6253, 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.

\(\displaystyle \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))}{\sqrt {d+i c d x}} \, dx\)

\(\Big \downarrow \) 6211

\(\displaystyle \frac {\sqrt {c^2 x^2+1} \int \frac {f (1-i c x) (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}dx}{\sqrt {d+i c d x} \sqrt {f-i c f x}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {f \sqrt {c^2 x^2+1} \int \frac {(1-i c x) (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}dx}{\sqrt {d+i c d x} \sqrt {f-i c f x}}\)

\(\Big \downarrow \) 6253

\(\displaystyle \frac {f \sqrt {c^2 x^2+1} \int \left (\frac {a+b \text {arcsinh}(c x)}{\sqrt {c^2 x^2+1}}-\frac {i c x (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}\right )dx}{\sqrt {d+i c d x} \sqrt {f-i c f x}}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {f \sqrt {c^2 x^2+1} \left (-\frac {i \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))}{c}+\frac {(a+b \text {arcsinh}(c x))^2}{2 b c}+i b x\right )}{\sqrt {d+i c d x} \sqrt {f-i c f x}}\)

Input:

Int[(Sqrt[f - I*c*f*x]*(a + b*ArcSinh[c*x]))/Sqrt[d + I*c*d*x],x]
 

Output:

(f*Sqrt[1 + c^2*x^2]*(I*b*x - (I*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x]))/c 
 + (a + b*ArcSinh[c*x])^2/(2*b*c)))/(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 6211

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]
 

rule 6253

Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)*((f_) + (g_.)*(x_))^(m_.)*((d 
_) + (e_.)*(x_)^2)^(p_), x_Symbol] :> Int[ExpandIntegrand[(d + e*x^2)^p*(a 
+ b*ArcSinh[c*x])^n, (f + g*x)^m, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] 
 && EqQ[e, c^2*d] && IGtQ[m, 0] && IntegerQ[p + 1/2] && GtQ[d, 0] && IGtQ[n 
, 0] && ((EqQ[n, 1] && GtQ[p, -1]) || GtQ[p, 0] || EqQ[m, 1] || (EqQ[m, 2] 
&& LtQ[p, -2]))
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 312 vs. \(2 (132 ) = 264\).

Time = 4.17 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.98

method	result	size
default	\(-\frac {i a \sqrt {-i c f x +f}\, \sqrt {i c d x +d}}{c d}+\frac {a f \sqrt {\left (-i c f x +f \right ) \left (i c d x +d \right )}\, \ln \left (\frac {c^{2} d f x}{\sqrt {c^{2} d f}}+\sqrt {c^{2} d f \,x^{2}+d f}\right )}{\sqrt {-i c f x +f}\, \sqrt {i c d x +d}\, \sqrt {c^{2} d f}}+b \left (\frac {\operatorname {arcsinh}\left (x c \right )^{2} \sqrt {i \left (x c -i\right ) d}\, \sqrt {-i \left (x c +i\right ) f}}{2 \sqrt {c^{2} x^{2}+1}\, c d}-\frac {i \left (\operatorname {arcsinh}\left (x c \right )-1\right ) \left (c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}\, x c +1\right ) \sqrt {i \left (x c -i\right ) d}\, \sqrt {-i \left (x c +i\right ) f}}{2 c \left (c^{2} x^{2}+1\right ) d}-\frac {i \left (\operatorname {arcsinh}\left (x c \right )+1\right ) \left (c^{2} x^{2}-\sqrt {c^{2} x^{2}+1}\, x c +1\right ) \sqrt {i \left (x c -i\right ) d}\, \sqrt {-i \left (x c +i\right ) f}}{2 c \left (c^{2} x^{2}+1\right ) d}\right )\)	\(313\)
parts	\(-\frac {i a \sqrt {-i c f x +f}\, \sqrt {i c d x +d}}{c d}+\frac {a f \sqrt {\left (-i c f x +f \right ) \left (i c d x +d \right )}\, \ln \left (\frac {c^{2} d f x}{\sqrt {c^{2} d f}}+\sqrt {c^{2} d f \,x^{2}+d f}\right )}{\sqrt {-i c f x +f}\, \sqrt {i c d x +d}\, \sqrt {c^{2} d f}}+b \left (\frac {\operatorname {arcsinh}\left (x c \right )^{2} \sqrt {i \left (x c -i\right ) d}\, \sqrt {-i \left (x c +i\right ) f}}{2 \sqrt {c^{2} x^{2}+1}\, c d}-\frac {i \left (\operatorname {arcsinh}\left (x c \right )-1\right ) \left (c^{2} x^{2}+\sqrt {c^{2} x^{2}+1}\, x c +1\right ) \sqrt {i \left (x c -i\right ) d}\, \sqrt {-i \left (x c +i\right ) f}}{2 c \left (c^{2} x^{2}+1\right ) d}-\frac {i \left (\operatorname {arcsinh}\left (x c \right )+1\right ) \left (c^{2} x^{2}-\sqrt {c^{2} x^{2}+1}\, x c +1\right ) \sqrt {i \left (x c -i\right ) d}\, \sqrt {-i \left (x c +i\right ) f}}{2 c \left (c^{2} x^{2}+1\right ) d}\right )\)	\(313\)

Input:

int((f-I*c*f*x)^(1/2)*(a+b*arcsinh(x*c))/(d+I*c*d*x)^(1/2),x,method=_RETUR 
NVERBOSE)
 

Output:

-I*a/c/d*(f-I*c*f*x)^(1/2)*(d+I*c*d*x)^(1/2)+a*f*((f-I*c*f*x)*(d+I*c*d*x)) 
^(1/2)/(f-I*c*f*x)^(1/2)/(d+I*c*d*x)^(1/2)*ln(c^2*d*f*x/(c^2*d*f)^(1/2)+(c 
^2*d*f*x^2+d*f)^(1/2))/(c^2*d*f)^(1/2)+b*(1/2*arcsinh(x*c)^2*(I*(x*c-I)*d) 
^(1/2)*(-I*(I+x*c)*f)^(1/2)/(c^2*x^2+1)^(1/2)/c/d-1/2*I*(arcsinh(x*c)-1)*( 
c^2*x^2+(c^2*x^2+1)^(1/2)*x*c+1)*(I*(x*c-I)*d)^(1/2)*(-I*(I+x*c)*f)^(1/2)/ 
c/(c^2*x^2+1)/d-1/2*I*(arcsinh(x*c)+1)*(c^2*x^2-(c^2*x^2+1)^(1/2)*x*c+1)*( 
I*(x*c-I)*d)^(1/2)*(-I*(I+x*c)*f)^(1/2)/c/(c^2*x^2+1)/d)
 

Fricas [F]

\[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))}{\sqrt {d+i c d x}} \, dx=\int { \frac {\sqrt {-i \, c f x + f} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{\sqrt {i \, c d x + d}} \,d x } \] Input:

integrate((f-I*c*f*x)^(1/2)*(a+b*arcsinh(c*x))/(d+I*c*d*x)^(1/2),x, algori 
thm="fricas")
 

Output:

integral((-I*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*log(c*x + sqrt(c^2*x^2 
 + 1)) - I*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a)/(c*d*x - I*d), x)
 

Sympy [F]

\[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))}{\sqrt {d+i c d x}} \, dx=\int \frac {\sqrt {- i f \left (c x + i\right )} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\sqrt {i d \left (c x - i\right )}}\, dx \] Input:

integrate((f-I*c*f*x)**(1/2)*(a+b*asinh(c*x))/(d+I*c*d*x)**(1/2),x)
 

Output:

Integral(sqrt(-I*f*(c*x + I))*(a + b*asinh(c*x))/sqrt(I*d*(c*x - I)), x)
 

Maxima [F]

\[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))}{\sqrt {d+i c d x}} \, dx=\int { \frac {\sqrt {-i \, c f x + f} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{\sqrt {i \, c d x + d}} \,d x } \] Input:

integrate((f-I*c*f*x)^(1/2)*(a+b*arcsinh(c*x))/(d+I*c*d*x)^(1/2),x, algori 
thm="maxima")
 

Output:

a*(f*arcsinh(c*x)/(c*d*sqrt(f/d)) - I*sqrt(c^2*d*f*x^2 + d*f)/(c*d)) + b*i 
ntegrate(sqrt(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1))/sqrt(I*c*d*x + d) 
, x)
 

Giac [F]

\[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))}{\sqrt {d+i c d x}} \, dx=\int { \frac {\sqrt {-i \, c f x + f} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{\sqrt {i \, c d x + d}} \,d x } \] Input:

integrate((f-I*c*f*x)^(1/2)*(a+b*arcsinh(c*x))/(d+I*c*d*x)^(1/2),x, algori 
thm="giac")
 

Output:

integrate(sqrt(-I*c*f*x + f)*(b*arcsinh(c*x) + a)/sqrt(I*c*d*x + d), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))}{\sqrt {d+i c d x}} \, dx=\int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {f-c\,f\,x\,1{}\mathrm {i}}}{\sqrt {d+c\,d\,x\,1{}\mathrm {i}}} \,d x \] Input:

int(((a + b*asinh(c*x))*(f - c*f*x*1i)^(1/2))/(d + c*d*x*1i)^(1/2),x)
 

Output:

int(((a + b*asinh(c*x))*(f - c*f*x*1i)^(1/2))/(d + c*d*x*1i)^(1/2), x)
 

Reduce [F]

\[ \int \frac {\sqrt {f-i c f x} (a+b \text {arcsinh}(c x))}{\sqrt {d+i c d x}} \, dx=\frac {\sqrt {f}\, \left (2 \mathit {asin} \left (\frac {\sqrt {-c i x +1}}{\sqrt {2}}\right ) a i -\sqrt {c i x +1}\, \sqrt {-c i x +1}\, a i +\left (\int \frac {\sqrt {-c i x +1}\, \mathit {asinh} \left (c x \right )}{\sqrt {c i x +1}}d x \right ) b c \right )}{\sqrt {d}\, c} \] Input:

int((f-I*c*f*x)^(1/2)*(a+b*asinh(c*x))/(d+I*c*d*x)^(1/2),x)
                                                                                    
                                                                                    
 

Output:

(sqrt(f)*(2*asin(sqrt( - c*i*x + 1)/sqrt(2))*a*i - sqrt(c*i*x + 1)*sqrt( - 
 c*i*x + 1)*a*i + int((sqrt( - c*i*x + 1)*asinh(c*x))/sqrt(c*i*x + 1),x)*b 
*c))/(sqrt(d)*c)