Optimal. Leaf size=794 \[ -\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (1+c^2 x^2\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {112 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {112 b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \text {PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {16 i b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
[Out]
________________________________________________________________________________________
Rubi [A]
time = 0.93, antiderivative size = 794, normalized size of antiderivative = 1.00, number of steps
used = 25, number of rules used = 16, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.432, Rules used = {5796,
5844, 5783, 5798, 5772, 267, 5843, 3399, 4271, 3852, 8, 4269, 3797, 2221, 2317, 2438}
\begin {gather*} -\frac {2 i a b d^5 x \left (c^2 x^2+1\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 d^5 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {112 b d^5 \left (c^2 x^2+1\right )^{5/2} \log \left (1+i e^{-\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {112 b^2 d^5 \left (c^2 x^2+1\right )^{5/2} \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (c^2 x^2+1\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {16 i b^2 d^5 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
[Out]
Rule 8
Rule 267
Rule 2221
Rule 2317
Rule 2438
Rule 3399
Rule 3797
Rule 3852
Rule 4269
Rule 4271
Rule 5772
Rule 5783
Rule 5796
Rule 5798
Rule 5843
Rule 5844
Rubi steps
\begin {align*} \int \frac {(d+i c d x)^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{(f-i c f x)^{5/2}} \, dx &=\frac {\left (1+c^2 x^2\right )^{5/2} \int \frac {(d+i c d x)^5 \left (a+b \sinh ^{-1}(c x)\right )^2}{\left (1+c^2 x^2\right )^{5/2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {\left (1+c^2 x^2\right )^{5/2} \int \left (\frac {5 d^5 \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}}+\frac {i c d^5 x \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}}-\frac {8 d^5 \left (a+b \sinh ^{-1}(c x)\right )^2}{(i+c x)^2 \sqrt {1+c^2 x^2}}-\frac {12 i d^5 \left (a+b \sinh ^{-1}(c x)\right )^2}{(i+c x) \sqrt {1+c^2 x^2}}\right ) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {\left (12 i d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{(i+c x) \sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (5 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (8 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {\left (a+b \sinh ^{-1}(c x)\right )^2}{(i+c x)^2 \sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (i c d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x \left (a+b \sinh ^{-1}(c x)\right )^2}{\sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (12 i d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {(a+b x)^2}{i c+c \sinh (x)} \, dx,x,\sinh ^{-1}(c x)\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \left (a+b \sinh ^{-1}(c x)\right ) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (8 c d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {(a+b x)^2}{(i c+c \sinh (x))^2} \, dx,x,\sinh ^{-1}(c x)\right )}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (2 i b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \sinh ^{-1}(c x) \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int (a+b x)^2 \csc ^4\left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (6 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int (a+b x)^2 \csc ^2\left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {12 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (4 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int (a+b x)^2 \csc ^2\left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (24 i b d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int (a+b x) \cot \left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (8 b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \csc ^2\left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (2 i b^2 c d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \int \frac {x}{\sqrt {1+c^2 x^2}} \, dx}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (1+c^2 x^2\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {12 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (16 i b d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int (a+b x) \cot \left (\frac {\pi }{4}-\frac {i x}{2}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (48 i b d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {e^{-x} (a+b x)}{1+i e^{-x}} \, dx,x,\sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (16 i b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int 1 \, dx,x,\cot \left (\frac {\pi }{4}-\frac {1}{2} i \sinh ^{-1}(c x)\right )\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (1+c^2 x^2\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {16 i b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}-\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {48 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (32 i b d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {e^{-x} (a+b x)}{1+i e^{-x}} \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (48 b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \log \left (1+i e^{-x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (1+c^2 x^2\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {16 i b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}-\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {112 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (32 b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \log \left (1+i e^{-x}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {\left (48 b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{-\sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (1+c^2 x^2\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {16 i b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}-\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {112 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {48 b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {\left (32 b^2 d^5 \left (1+c^2 x^2\right )^{5/2}\right ) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ &=-\frac {2 i a b d^5 x \left (1+c^2 x^2\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {2 i b^2 d^5 \left (1+c^2 x^2\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {2 i b^2 d^5 x \left (1+c^2 x^2\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {i d^5 \left (1+c^2 x^2\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {5 d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {16 i b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \cot \left (\frac {\pi }{4}-\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {112 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \log \left (1+i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {112 b^2 d^5 \left (1+c^2 x^2\right )^{5/2} \text {Li}_2\left (-i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {8 b d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right ) \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac {28 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac {4 i d^5 \left (1+c^2 x^2\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2 \sec ^2\left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac {\pi }{4}+\frac {1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}\\ \end {align*}
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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice
the leaf count of optimal. \(2552\) vs. \(2(794)=1588\).
time = 11.05, size = 2552, normalized size = 3.21 \begin {gather*} \text {Result too large to show} \end {gather*}
Warning: Unable to verify antiderivative.
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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (i c d x +d \right )^{\frac {5}{2}} \left (a +b \arcsinh \left (c x \right )\right )^{2}}{\left (-i c f x +f \right )^{\frac {5}{2}}}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2\,{\left (d+c\,d\,x\,1{}\mathrm {i}\right )}^{5/2}}{{\left (f-c\,f\,x\,1{}\mathrm {i}\right )}^{5/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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