Optimal. Leaf size=794 \[ -\frac{112 b^2 d^5 \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i a b d^5 x \left (c^2 x^2+1\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{5 d^5 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i d^5 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{28 d^5 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{112 b d^5 \left (c^2 x^2+1\right )^{5/2} \log \left (1+i e^{-\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 b d^5 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{28 i d^5 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{4 i d^5 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 i b^2 d^5 \left (c^2 x^2+1\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i b^2 d^5 x \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{16 i b^2 d^5 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 1.39589, antiderivative size = 794, normalized size of antiderivative = 1., 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 = {5712, 5833, 5675, 5717, 5653, 261, 5831, 3318, 4186, 3767, 8, 4184, 3716, 2190, 2279, 2391} \[ -\frac{112 b^2 d^5 \left (c^2 x^2+1\right )^{5/2} \text{PolyLog}\left (2,-i e^{-\sinh ^{-1}(c x)}\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i a b d^5 x \left (c^2 x^2+1\right )^{5/2}}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{5 d^5 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^3}{3 b c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{i d^5 \left (c^2 x^2+1\right )^3 \left (a+b \sinh ^{-1}(c x)\right )^2}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{28 d^5 \left (c^2 x^2+1\right )^{5/2} \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{112 b d^5 \left (c^2 x^2+1\right )^{5/2} \log \left (1+i e^{-\sinh ^{-1}(c x)}\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{8 b d^5 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{28 i d^5 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{4 i d^5 \left (c^2 x^2+1\right )^{5/2} \sec ^2\left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \tan \left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right ) \left (a+b \sinh ^{-1}(c x)\right )^2}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{2 i b^2 d^5 \left (c^2 x^2+1\right )^3}{c (d+i c d x)^{5/2} (f-i c f x)^{5/2}}-\frac{2 i b^2 d^5 x \left (c^2 x^2+1\right )^{5/2} \sinh ^{-1}(c x)}{(d+i c d x)^{5/2} (f-i c f x)^{5/2}}+\frac{16 i b^2 d^5 \left (c^2 x^2+1\right )^{5/2} \tan \left (\frac{\pi }{4}+\frac{1}{2} i \sinh ^{-1}(c x)\right )}{3 c (d+i c d x)^{5/2} (f-i c f x)^{5/2}} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 5712
Rule 5833
Rule 5675
Rule 5717
Rule 5653
Rule 261
Rule 5831
Rule 3318
Rule 4186
Rule 3767
Rule 8
Rule 4184
Rule 3716
Rule 2190
Rule 2279
Rule 2391
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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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 ) \operatorname{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*}
Mathematica [B] time = 13.6484, size = 2552, normalized size = 3.21 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
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Maple [F] time = 0.273, size = 0, normalized size = 0. \begin{align*} \int{ \left ( a+b{\it Arcsinh} \left ( cx \right ) \right ) ^{2} \left ( d+icdx \right ) ^{{\frac{5}{2}}} \left ( f-icfx \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (i \, b^{2} c^{2} d^{2} x^{2} + 2 \, b^{2} c d^{2} x - i \, b^{2} d^{2}\right )} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f} \log \left (c x + \sqrt{c^{2} x^{2} + 1}\right )^{2} +{\left (2 i \, a b c^{2} d^{2} x^{2} + 4 \, a b c d^{2} x - 2 i \, a b d^{2}\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 (i \, a^{2} c^{2} d^{2} x^{2} + 2 \, a^{2} c d^{2} x - i \, a^{2} d^{2}\right )} \sqrt{i \, c d x + d} \sqrt{-i \, c f x + f}}{c^{3} f^{3} x^{3} + 3 i \, c^{2} f^{3} x^{2} - 3 \, c f^{3} x - i \, f^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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