Integrand size = 35, antiderivative size = 103 \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}} \, dx=\frac {x \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}-\frac {b \left (1+c^2 x^2\right )^{3/2} \log \left (1+c^2 x^2\right )}{2 c (d+i c d x)^{3/2} (f-i c f x)^{3/2}} \]
x*(c^2*x^2+1)*(a+b*arcsinh(c*x))/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)-1/2*b *(c^2*x^2+1)^(3/2)*ln(c^2*x^2+1)/c/(d+I*c*d*x)^(3/2)/(f-I*c*f*x)^(3/2)
Time = 1.19 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.15 \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}} \, dx=\frac {i \sqrt {f-i c f x} \left (2 a c x+2 b c x \text {arcsinh}(c x)-b \sqrt {1+c^2 x^2} \log (d (-1+i c x))-b \sqrt {1+c^2 x^2} \log (d+i c d x)\right )}{2 c d f^2 (i+c x) \sqrt {d+i c d x}} \]
((I/2)*Sqrt[f - I*c*f*x]*(2*a*c*x + 2*b*c*x*ArcSinh[c*x] - b*Sqrt[1 + c^2* x^2]*Log[d*(-1 + I*c*x)] - b*Sqrt[1 + c^2*x^2]*Log[d + I*c*d*x]))/(c*d*f^2 *(I + c*x)*Sqrt[d + I*c*d*x])
Time = 0.45 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.80, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {6211, 6202, 240}
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.
\(\displaystyle \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}} \, dx\) |
\(\Big \downarrow \) 6211 |
\(\displaystyle \frac {\left (c^2 x^2+1\right )^{3/2} \int \frac {a+b \text {arcsinh}(c x)}{\left (c^2 x^2+1\right )^{3/2}}dx}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\) |
\(\Big \downarrow \) 6202 |
\(\displaystyle \frac {\left (c^2 x^2+1\right )^{3/2} \left (\frac {x (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}-b c \int \frac {x}{c^2 x^2+1}dx\right )}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {\left (c^2 x^2+1\right )^{3/2} \left (\frac {x (a+b \text {arcsinh}(c x))}{\sqrt {c^2 x^2+1}}-\frac {b \log \left (c^2 x^2+1\right )}{2 c}\right )}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}}\) |
((1 + c^2*x^2)^(3/2)*((x*(a + b*ArcSinh[c*x]))/Sqrt[1 + c^2*x^2] - (b*Log[ 1 + c^2*x^2])/(2*c)))/((d + I*c*d*x)^(3/2)*(f - I*c*f*x)^(3/2))
3.6.62.3.1 Defintions of rubi rules used
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((a_.) + ArcSinh[(c_.)*(x_)]*(b_.))^(n_.)/((d_) + (e_.)*(x_)^2)^(3/2), x_Symbol] :> Simp[x*((a + b*ArcSinh[c*x])^n/(d*Sqrt[d + e*x^2])), x] - Simp [b*c*(n/d)*Simp[Sqrt[1 + c^2*x^2]/Sqrt[d + e*x^2]] Int[x*((a + b*ArcSinh[ c*x])^(n - 1)/(1 + c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[e, c^2*d] && GtQ[n, 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]
\[\int \frac {a +b \,\operatorname {arcsinh}\left (c x \right )}{\left (i c d x +d \right )^{\frac {3}{2}} \left (-i c f x +f \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (i \, c d x + d\right )}^{\frac {3}{2}} {\left (-i \, c f x + f\right )}^{\frac {3}{2}}} \,d x } \]
1/4*(4*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*x*log(c*x + sqrt(c^2*x^2 + 1 )) + 4*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*a*x + (c^2*d^2*f^2*x^2 + d^2*f ^2)*sqrt(b^2/(c^2*d^3*f^3))*log((b*c^2*x^4 + sqrt(c^2*x^2 + 1)*sqrt(I*c*d* x + d)*sqrt(-I*c*f*x + f)*c*d*f*x^2*sqrt(b^2/(c^2*d^3*f^3)) + b*x^2)/(b*c^ 4*x^4 + 2*b*c^2*x^2 + b)) - (c^2*d^2*f^2*x^2 + d^2*f^2)*sqrt(b^2/(c^2*d^3* f^3))*log((b*c^2*x^4 - sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d*f*x^2*sqrt(b^2/(c^2*d^3*f^3)) + b*x^2)/(b*c^4*x^4 + 2*b*c^2*x^2 + b)) - 2*(c^2*d^2*f^2*x^2 + d^2*f^2)*sqrt(b^2/(c^2*d^3*f^3))*log((b*c^2*x^3 + sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*c*d*f*x*sqrt(b^2 /(c^2*d^3*f^3)) + b*x)/(b*c^2*x^2 + b)) + 2*(c^2*d^2*f^2*x^2 + d^2*f^2)*sq rt(b^2/(c^2*d^3*f^3))*log((b*c^2*x^3 - sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d) *sqrt(-I*c*f*x + f)*c*d*f*x*sqrt(b^2/(c^2*d^3*f^3)) + b*x)/(b*c^2*x^2 + b) ) + 4*(c^2*d^2*f^2*x^2 + d^2*f^2)*integral(-sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*c*x/(c^4*d^2*f^2*x^4 + 2*c^2*d^2*f^2*x^2 + d^2* f^2), x))/(c^2*d^2*f^2*x^2 + d^2*f^2)
\[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}} \, dx=\int \frac {a + b \operatorname {asinh}{\left (c x \right )}}{\left (i d \left (c x - i\right )\right )^{\frac {3}{2}} \left (- i f \left (c x + i\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.80 \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}} \, dx=\frac {b x \operatorname {arsinh}\left (c x\right )}{\sqrt {c^{2} d f x^{2} + d f} d f} + \frac {a x}{\sqrt {c^{2} d f x^{2} + d f} d f} - \frac {b \sqrt {\frac {1}{d f}} \log \left (x^{2} + \frac {1}{c^{2}}\right )}{2 \, c d f} \]
b*x*arcsinh(c*x)/(sqrt(c^2*d*f*x^2 + d*f)*d*f) + a*x/(sqrt(c^2*d*f*x^2 + d *f)*d*f) - 1/2*b*sqrt(1/(d*f))*log(x^2 + 1/c^2)/(c*d*f)
\[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}} \, dx=\int { \frac {b \operatorname {arsinh}\left (c x\right ) + a}{{\left (i \, c d x + d\right )}^{\frac {3}{2}} {\left (-i \, c f x + f\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {a+b \text {arcsinh}(c x)}{(d+i c d x)^{3/2} (f-i c f x)^{3/2}} \, dx=\int \frac {a+b\,\mathrm {asinh}\left (c\,x\right )}{{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{3/2}\,{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{3/2}} \,d x \]