3.6.90 \(\int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{\sqrt {f-i c f x}} \, dx\) [590]

3.6.90.1 Optimal result

Integrand size = 37, antiderivative size = 259 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{\sqrt {f-i c f x}} \, dx=-\frac {2 i a b d x \sqrt {1+c^2 x^2}}{\sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {2 i b^2 d \left (1+c^2 x^2\right )}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}-\frac {2 i b^2 d x \sqrt {1+c^2 x^2} \text {arcsinh}(c x)}{\sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {i d \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))^2}{c \sqrt {d+i c d x} \sqrt {f-i c f x}}+\frac {d \sqrt {1+c^2 x^2} (a+b \text {arcsinh}(c x))^3}{3 b c \sqrt {d+i c d x} \sqrt {f-i c f x}} \]

output

2*I*b^2*d*(c^2*x^2+1)/c/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+I*d*(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)-2*I*a*b*d*x*( 
c^2*x^2+1)^(1/2)/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)-2*I*b^2*d*x*arcsinh(c 
*x)*(c^2*x^2+1)^(1/2)/(d+I*c*d*x)^(1/2)/(f-I*c*f*x)^(1/2)+1/3*d*(a+b*arcsi 
nh(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)
 

3.6.90.2 Mathematica [A] (verified)

Time = 2.64 (sec) , antiderivative size = 315, normalized size of antiderivative = 1.22 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{\sqrt {f-i c f x}} \, dx=\frac {3 i \sqrt {d+i c d x} \sqrt {f-i c f x} \left (-2 a b c x+a^2 \sqrt {1+c^2 x^2}+2 b^2 \sqrt {1+c^2 x^2}\right )-6 i b \sqrt {d+i c d x} \sqrt {f-i c f x} \left (b c x-a \sqrt {1+c^2 x^2}\right ) \text {arcsinh}(c x)+3 b \sqrt {d+i c d x} \sqrt {f-i c f x} \left (a+i b \sqrt {1+c^2 x^2}\right ) \text {arcsinh}(c x)^2+b^2 \sqrt {d+i c d x} \sqrt {f-i c f x} \text {arcsinh}(c x)^3+3 a^2 \sqrt {d} \sqrt {f} \sqrt {1+c^2 x^2} \log \left (c d f x+\sqrt {d} \sqrt {f} \sqrt {d+i c d x} \sqrt {f-i c f x}\right )}{3 c f \sqrt {1+c^2 x^2}} \]

input

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

output

((3*I)*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x]*(-2*a*b*c*x + a^2*Sqrt[1 + c^2* 
x^2] + 2*b^2*Sqrt[1 + c^2*x^2]) - (6*I)*b*Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f 
*x]*(b*c*x - a*Sqrt[1 + c^2*x^2])*ArcSinh[c*x] + 3*b*Sqrt[d + I*c*d*x]*Sqr 
t[f - I*c*f*x]*(a + I*b*Sqrt[1 + c^2*x^2])*ArcSinh[c*x]^2 + b^2*Sqrt[d + I 
*c*d*x]*Sqrt[f - I*c*f*x]*ArcSinh[c*x]^3 + 3*a^2*Sqrt[d]*Sqrt[f]*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] 
])/(3*c*f*Sqrt[1 + c^2*x^2])
 

3.6.90.3 Rubi [A] (verified)

Time = 0.72 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.52, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.108, 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.

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

\(\Big \downarrow \) 6211

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 6253

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {d \sqrt {c^2 x^2+1} \left (\frac {i \sqrt {c^2 x^2+1} (a+b \text {arcsinh}(c x))^2}{c}+\frac {(a+b \text {arcsinh}(c x))^3}{3 b c}-2 i a b x-2 i b^2 x \text {arcsinh}(c x)+\frac {2 i b^2 \sqrt {c^2 x^2+1}}{c}\right )}{\sqrt {d+i c d x} \sqrt {f-i c f x}}\)

input

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

output

(d*Sqrt[1 + c^2*x^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] + (I*Sqrt[1 + c^2*x^2]*(a + b*ArcSinh[c*x])^2)/c + ( 
a + b*ArcSinh[c*x])^3/(3*b*c)))/(Sqrt[d + I*c*d*x]*Sqrt[f - I*c*f*x])
 

3.6.90.3.1 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]))
 

3.6.90.4 Maple [F]

\[\int \frac {\left (a +b \,\operatorname {arcsinh}\left (c x \right )\right )^{2} \sqrt {i c d x +d}}{\sqrt {-i c f x +f}}d x\]

input

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

output

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

3.6.90.5 Fricas [F]

\[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{\sqrt {f-i c f x}} \, dx=\int { \frac {\sqrt {i \, c d x + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{\sqrt {-i \, c f x + f}} \,d x } \]

input

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

output

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

3.6.90.6 Sympy [F]

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

input

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

output

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

3.6.90.7 Maxima [F]

\[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{\sqrt {f-i c f x}} \, dx=\int { \frac {\sqrt {i \, c d x + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{\sqrt {-i \, c f x + f}} \,d x } \]

input

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

output

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

3.6.90.8 Giac [F]

\[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{\sqrt {f-i c f x}} \, dx=\int { \frac {\sqrt {i \, c d x + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}{\sqrt {-i \, c f x + f}} \,d x } \]

input

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

output

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

3.6.90.9 Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))^2}{\sqrt {f-i c f x}} \, dx=\int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,\sqrt {d+c\,d\,x\,1{}\mathrm {i}}}{\sqrt {f-c\,f\,x\,1{}\mathrm {i}}} \,d x \]

input

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

output

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