Integrand size = 35, antiderivative size = 185 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\frac {2 i b d^3 \left (1+c^2 x^2\right )^{5/2}}{3 c (i+c x) (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {i d^3 (1+i c x)^3 \left (1+c^2 x^2\right ) (a+b \text {arcsinh}(c x))}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {b d^3 \left (1+c^2 x^2\right )^{5/2} \log (i+c x)}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
2/3*I*b*d^3*(c^2*x^2+1)^(5/2)/c/(c*x+I)/(d+I*c*d*x)^(5/2)/(f-I*c*f*x)^(5/2 )-1/3*I*d^3*(1+I*c*x)^3*(c^2*x^2+1)*(a+b*arcsinh(c*x))/c/(d+I*c*d*x)^(5/2) /(f-I*c*f*x)^(5/2)+1/3*b*d^3*(c^2*x^2+1)^(5/2)*ln(c*x+I)/c/(d+I*c*d*x)^(5/ 2)/(f-I*c*f*x)^(5/2)
Time = 1.75 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.71 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=-\frac {i d \sqrt {f-i c f x} \left ((-i+c x) \left (-i a+a c x+b \sqrt {1+c^2 x^2}\right )+b (-i+c x)^2 \text {arcsinh}(c x)-b (i+c x) \sqrt {1+c^2 x^2} \log (d (-1+i c x))\right )}{3 c f^3 (i+c x)^2 \sqrt {d+i c d x}} \]
((-1/3*I)*d*Sqrt[f - I*c*f*x]*((-I + c*x)*((-I)*a + a*c*x + b*Sqrt[1 + c^2 *x^2]) + b*(-I + c*x)^2*ArcSinh[c*x] - b*(I + c*x)*Sqrt[1 + c^2*x^2]*Log[d *(-1 + I*c*x)]))/(c*f^3*(I + c*x)^2*Sqrt[d + I*c*d*x])
Time = 0.56 (sec) , antiderivative size = 121, normalized size of antiderivative = 0.65, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6211, 27, 6252, 27, 456, 49, 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.
| \(\displaystyle \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6211 |
| \(\displaystyle \frac {\left (c^2 x^2+1\right )^{5/2} \int \frac {d^3 (i c x+1)^3 (a+b \text {arcsinh}(c x))}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^3 \left (c^2 x^2+1\right )^{5/2} \int \frac {(i c x+1)^3 (a+b \text {arcsinh}(c x))}{\left (c^2 x^2+1\right )^{5/2}}dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 6252 |
| \(\displaystyle \frac {d^3 \left (c^2 x^2+1\right )^{5/2} \left (-b c \int -\frac {i (i c x+1)^3}{3 c \left (c^2 x^2+1\right )^2}dx-\frac {i (1+i c x)^3 (a+b \text {arcsinh}(c x))}{3 c \left (c^2 x^2+1\right )^{3/2}}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d^3 \left (c^2 x^2+1\right )^{5/2} \left (\frac {1}{3} i b \int \frac {(i c x+1)^3}{\left (c^2 x^2+1\right )^2}dx-\frac {i (1+i c x)^3 (a+b \text {arcsinh}(c x))}{3 c \left (c^2 x^2+1\right )^{3/2}}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 456 |
| \(\displaystyle \frac {d^3 \left (c^2 x^2+1\right )^{5/2} \left (\frac {1}{3} i b \int \frac {i c x+1}{(1-i c x)^2}dx-\frac {i (1+i c x)^3 (a+b \text {arcsinh}(c x))}{3 c \left (c^2 x^2+1\right )^{3/2}}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 49 |
| \(\displaystyle \frac {d^3 \left (c^2 x^2+1\right )^{5/2} \left (\frac {1}{3} i b \int \left (-\frac {i}{c x+i}-\frac {2}{(c x+i)^2}\right )dx-\frac {i (1+i c x)^3 (a+b \text {arcsinh}(c x))}{3 c \left (c^2 x^2+1\right )^{3/2}}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {d^3 \left (c^2 x^2+1\right )^{5/2} \left (\frac {1}{3} i b \left (\frac {2}{c (c x+i)}-\frac {i \log (c x+i)}{c}\right )-\frac {i (1+i c x)^3 (a+b \text {arcsinh}(c x))}{3 c \left (c^2 x^2+1\right )^{3/2}}\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\) |
(d^3*(1 + c^2*x^2)^(5/2)*(((-1/3*I)*(1 + I*c*x)^3*(a + b*ArcSinh[c*x]))/(c *(1 + c^2*x^2)^(3/2)) + (I/3)*b*(2/(c*(I + c*x)) - (I*Log[I + c*x])/c)))/( (d + I*c*d*x)^(5/2)*(f - I*c*f*x)^(5/2))
3.6.66.3.1 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[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ (c + d*x)^(n + p)*(a/c + (b/d)*x)^p, x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[b*c^2 + a*d^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[c, 0] && !Integ erQ[n]))
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_.))*((f_) + (g_.)*(x_))^(m_.)*((d_) + ( e_.)*(x_)^2)^(p_), x_Symbol] :> With[{u = IntHide[(f + g*x)^m*(d + e*x^2)^p , x]}, Simp[(a + b*ArcSinh[c*x]) u, x] - Simp[b*c Int[1/Sqrt[1 + c^2*x^ 2] u, x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[e, c^2*d] && IGtQ [m, 0] && ILtQ[p + 1/2, 0] && GtQ[d, 0] && (LtQ[m, -2*p - 1] || GtQ[m, 3])
\[\int \frac {\left (a +b \,\operatorname {arcsinh}\left (c x \right )\right ) \sqrt {i c d x +d}}{\left (-i c f x +f \right )^{\frac {5}{2}}}d x\]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 548 vs. \(2 (141) = 282\).
Time = 0.33 (sec) , antiderivative size = 548, normalized size of antiderivative = 2.96 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=-\frac {4 \, \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} b c x + 2 \, {\left (b c^{2} x^{2} - 2 i \, b c x - b\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + {\left (c^{4} f^{3} x^{3} + i \, c^{3} f^{3} x^{2} + c^{2} f^{3} x + i \, c f^{3}\right )} \sqrt {\frac {b^{2} d}{c^{2} f^{5}}} \log \left (-\frac {{\left (-i \, b c^{6} x^{2} + 2 \, b c^{5} x + 2 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + {\left (i \, c^{9} f^{3} x^{4} - 2 \, c^{8} f^{3} x^{3} + i \, c^{7} f^{3} x^{2} - 2 \, c^{6} f^{3} x\right )} \sqrt {\frac {b^{2} d}{c^{2} f^{5}}}}{8 \, {\left (b c^{3} x^{3} + i \, b c^{2} x^{2} + b c x + i \, b\right )}}\right ) - {\left (c^{4} f^{3} x^{3} + i \, c^{3} f^{3} x^{2} + c^{2} f^{3} x + i \, c f^{3}\right )} \sqrt {\frac {b^{2} d}{c^{2} f^{5}}} \log \left (-\frac {{\left (-i \, b c^{6} x^{2} + 2 \, b c^{5} x + 2 i \, b c^{4}\right )} \sqrt {c^{2} x^{2} + 1} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f} + {\left (-i \, c^{9} f^{3} x^{4} + 2 \, c^{8} f^{3} x^{3} - i \, c^{7} f^{3} x^{2} + 2 \, c^{6} f^{3} x\right )} \sqrt {\frac {b^{2} d}{c^{2} f^{5}}}}{8 \, {\left (b c^{3} x^{3} + i \, b c^{2} x^{2} + b c x + i \, b\right )}}\right ) + 2 \, {\left (a c^{2} x^{2} - 2 i \, a c x - a\right )} \sqrt {i \, c d x + d} \sqrt {-i \, c f x + f}}{6 \, {\left (c^{4} f^{3} x^{3} + i \, c^{3} f^{3} x^{2} + c^{2} f^{3} x + i \, c f^{3}\right )}} \]
-1/6*(4*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*b*c*x + 2*( b*c^2*x^2 - 2*I*b*c*x - b)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f)*log(c*x + sqrt(c^2*x^2 + 1)) + (c^4*f^3*x^3 + I*c^3*f^3*x^2 + c^2*f^3*x + I*c*f^3)*s qrt(b^2*d/(c^2*f^5))*log(-1/8*((-I*b*c^6*x^2 + 2*b*c^5*x + 2*I*b*c^4)*sqrt (c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f) + (I*c^9*f^3*x^4 - 2*c^ 8*f^3*x^3 + I*c^7*f^3*x^2 - 2*c^6*f^3*x)*sqrt(b^2*d/(c^2*f^5)))/(b*c^3*x^3 + I*b*c^2*x^2 + b*c*x + I*b)) - (c^4*f^3*x^3 + I*c^3*f^3*x^2 + c^2*f^3*x + I*c*f^3)*sqrt(b^2*d/(c^2*f^5))*log(-1/8*((-I*b*c^6*x^2 + 2*b*c^5*x + 2*I *b*c^4)*sqrt(c^2*x^2 + 1)*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f) + (-I*c^9*f ^3*x^4 + 2*c^8*f^3*x^3 - I*c^7*f^3*x^2 + 2*c^6*f^3*x)*sqrt(b^2*d/(c^2*f^5) ))/(b*c^3*x^3 + I*b*c^2*x^2 + b*c*x + I*b)) + 2*(a*c^2*x^2 - 2*I*a*c*x - a )*sqrt(I*c*d*x + d)*sqrt(-I*c*f*x + f))/(c^4*f^3*x^3 + I*c^3*f^3*x^2 + c^2 *f^3*x + I*c*f^3)
\[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\int \frac {\sqrt {i d \left (c x - i\right )} \left (a + b \operatorname {asinh}{\left (c x \right )}\right )}{\left (- i f \left (c x + i\right )\right )^{\frac {5}{2}}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.19 \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=-\frac {1}{3} \, b c {\left (\frac {6 \, \sqrt {d}}{3 i \, c^{3} f^{\frac {5}{2}} x - 3 \, c^{2} f^{\frac {5}{2}}} - \frac {\sqrt {d} \log \left (c x + i\right )}{c^{2} f^{\frac {5}{2}}}\right )} - \frac {1}{3} \, b {\left (-\frac {2 i \, \sqrt {c^{2} d f x^{2} + d f}}{c^{3} f^{3} x^{2} + 2 i \, c^{2} f^{3} x - c f^{3}} - \frac {3 i \, \sqrt {c^{2} d f x^{2} + d f}}{-3 i \, c^{2} f^{3} x + 3 \, c f^{3}}\right )} \operatorname {arsinh}\left (c x\right ) - \frac {1}{3} \, a {\left (-\frac {2 i \, \sqrt {c^{2} d f x^{2} + d f}}{c^{3} f^{3} x^{2} + 2 i \, c^{2} f^{3} x - c f^{3}} - \frac {3 i \, \sqrt {c^{2} d f x^{2} + d f}}{-3 i \, c^{2} f^{3} x + 3 \, c f^{3}}\right )} \]
-1/3*b*c*(6*sqrt(d)/(3*I*c^3*f^(5/2)*x - 3*c^2*f^(5/2)) - sqrt(d)*log(c*x + I)/(c^2*f^(5/2))) - 1/3*b*(-2*I*sqrt(c^2*d*f*x^2 + d*f)/(c^3*f^3*x^2 + 2 *I*c^2*f^3*x - c*f^3) - 3*I*sqrt(c^2*d*f*x^2 + d*f)/(-3*I*c^2*f^3*x + 3*c* f^3))*arcsinh(c*x) - 1/3*a*(-2*I*sqrt(c^2*d*f*x^2 + d*f)/(c^3*f^3*x^2 + 2* I*c^2*f^3*x - c*f^3) - 3*I*sqrt(c^2*d*f*x^2 + d*f)/(-3*I*c^2*f^3*x + 3*c*f ^3))
\[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\int { \frac {\sqrt {i \, c d x + d} {\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}}{{\left (-i \, c f x + f\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {d+i c d x} (a+b \text {arcsinh}(c x))}{(f-i c f x)^{5/2}} \, dx=\int \frac {\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )\,\sqrt {d+c\,d\,x\,1{}\mathrm {i}}}{{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]