Optimal. Leaf size=180 \[ \frac{\left (2-5 a^2\right ) b^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a^2 \left (1-a^2\right )^2 x}-\frac{b^3 \csc ^{-1}(a+b x)}{3 a^3}-\frac{\left (6 a^4-5 a^2+2\right ) b^3 \tan ^{-1}\left (\frac{a-\tan \left (\frac{1}{2} \csc ^{-1}(a+b x)\right )}{\sqrt{1-a^2}}\right )}{3 a^3 \left (1-a^2\right )^{5/2}}-\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a \left (1-a^2\right ) x^2}-\frac{\csc ^{-1}(a+b x)}{3 x^3} \]
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Rubi [A] time = 0.297021, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.9, Rules used = {5259, 4427, 3785, 4060, 3919, 3831, 2660, 618, 204} \[ \frac{\left (2-5 a^2\right ) b^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a^2 \left (1-a^2\right )^2 x}-\frac{b^3 \csc ^{-1}(a+b x)}{3 a^3}-\frac{\left (6 a^4-5 a^2+2\right ) b^3 \tan ^{-1}\left (\frac{a-\tan \left (\frac{1}{2} \csc ^{-1}(a+b x)\right )}{\sqrt{1-a^2}}\right )}{3 a^3 \left (1-a^2\right )^{5/2}}-\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a \left (1-a^2\right ) x^2}-\frac{\csc ^{-1}(a+b x)}{3 x^3} \]
Antiderivative was successfully verified.
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Rule 5259
Rule 4427
Rule 3785
Rule 4060
Rule 3919
Rule 3831
Rule 2660
Rule 618
Rule 204
Rubi steps
\begin{align*} \int \frac{\csc ^{-1}(a+b x)}{x^4} \, dx &=-\left (b^3 \operatorname{Subst}\left (\int \frac{x \cot (x) \csc (x)}{(-a+\csc (x))^4} \, dx,x,\csc ^{-1}(a+b x)\right )\right )\\ &=-\frac{\csc ^{-1}(a+b x)}{3 x^3}+\frac{1}{3} b^3 \operatorname{Subst}\left (\int \frac{1}{(-a+\csc (x))^3} \, dx,x,\csc ^{-1}(a+b x)\right )\\ &=-\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a \left (1-a^2\right ) x^2}-\frac{\csc ^{-1}(a+b x)}{3 x^3}-\frac{b^3 \operatorname{Subst}\left (\int \frac{2 \left (1-a^2\right )-2 a \csc (x)-\csc ^2(x)}{(-a+\csc (x))^2} \, dx,x,\csc ^{-1}(a+b x)\right )}{6 a \left (1-a^2\right )}\\ &=-\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a \left (1-a^2\right ) x^2}+\frac{\left (2-5 a^2\right ) b^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a^2 \left (1-a^2\right )^2 x}-\frac{\csc ^{-1}(a+b x)}{3 x^3}+\frac{b^3 \operatorname{Subst}\left (\int \frac{2 \left (1-a^2\right )^2-a \left (1-4 a^2\right ) \csc (x)}{-a+\csc (x)} \, dx,x,\csc ^{-1}(a+b x)\right )}{6 a^2 \left (1-a^2\right )^2}\\ &=-\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a \left (1-a^2\right ) x^2}+\frac{\left (2-5 a^2\right ) b^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a^2 \left (1-a^2\right )^2 x}-\frac{b^3 \csc ^{-1}(a+b x)}{3 a^3}-\frac{\csc ^{-1}(a+b x)}{3 x^3}+\frac{\left (\left (2-5 a^2+6 a^4\right ) b^3\right ) \operatorname{Subst}\left (\int \frac{\csc (x)}{-a+\csc (x)} \, dx,x,\csc ^{-1}(a+b x)\right )}{6 a^3 \left (1-a^2\right )^2}\\ &=-\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a \left (1-a^2\right ) x^2}+\frac{\left (2-5 a^2\right ) b^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a^2 \left (1-a^2\right )^2 x}-\frac{b^3 \csc ^{-1}(a+b x)}{3 a^3}-\frac{\csc ^{-1}(a+b x)}{3 x^3}+\frac{\left (\left (2-5 a^2+6 a^4\right ) b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-a \sin (x)} \, dx,x,\csc ^{-1}(a+b x)\right )}{6 a^3 \left (1-a^2\right )^2}\\ &=-\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a \left (1-a^2\right ) x^2}+\frac{\left (2-5 a^2\right ) b^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a^2 \left (1-a^2\right )^2 x}-\frac{b^3 \csc ^{-1}(a+b x)}{3 a^3}-\frac{\csc ^{-1}(a+b x)}{3 x^3}+\frac{\left (\left (2-5 a^2+6 a^4\right ) b^3\right ) \operatorname{Subst}\left (\int \frac{1}{1-2 a x+x^2} \, dx,x,\tan \left (\frac{1}{2} \csc ^{-1}(a+b x)\right )\right )}{3 a^3 \left (1-a^2\right )^2}\\ &=-\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a \left (1-a^2\right ) x^2}+\frac{\left (2-5 a^2\right ) b^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a^2 \left (1-a^2\right )^2 x}-\frac{b^3 \csc ^{-1}(a+b x)}{3 a^3}-\frac{\csc ^{-1}(a+b x)}{3 x^3}-\frac{\left (2 \left (2-5 a^2+6 a^4\right ) b^3\right ) \operatorname{Subst}\left (\int \frac{1}{-4 \left (1-a^2\right )-x^2} \, dx,x,-2 a+2 \tan \left (\frac{1}{2} \csc ^{-1}(a+b x)\right )\right )}{3 a^3 \left (1-a^2\right )^2}\\ &=-\frac{b (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a \left (1-a^2\right ) x^2}+\frac{\left (2-5 a^2\right ) b^2 (a+b x) \sqrt{1-\frac{1}{(a+b x)^2}}}{6 a^2 \left (1-a^2\right )^2 x}-\frac{b^3 \csc ^{-1}(a+b x)}{3 a^3}-\frac{\csc ^{-1}(a+b x)}{3 x^3}-\frac{\left (2-5 a^2+6 a^4\right ) b^3 \tan ^{-1}\left (\frac{a-\tan \left (\frac{1}{2} \csc ^{-1}(a+b x)\right )}{\sqrt{1-a^2}}\right )}{3 a^3 \left (1-a^2\right )^{5/2}}\\ \end{align*}
Mathematica [C] time = 0.490824, size = 241, normalized size = 1.34 \[ \frac{1}{6} \left (\frac{b \sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} \left (-a^2 \left (5 b^2 x^2+1\right )-4 a^3 b x+a^4+a b x+2 b^2 x^2\right )}{a^2 \left (a^2-1\right )^2 x^2}+\frac{i \left (6 a^4-5 a^2+2\right ) b^3 \log \left (\frac{12 a^3 \left (a^2-1\right )^2 \left (-\sqrt{\frac{a^2+2 a b x+b^2 x^2-1}{(a+b x)^2}} (a+b x)-\frac{i \left (a^2+a b x-1\right )}{\sqrt{1-a^2}}\right )}{\left (6 a^4-5 a^2+2\right ) b^3 x}\right )}{a^3 \left (1-a^2\right )^{5/2}}-\frac{2 b^3 \sin ^{-1}\left (\frac{1}{a+b x}\right )}{a^3}-\frac{2 \csc ^{-1}(a+b x)}{x^3}\right ) \]
Antiderivative was successfully verified.
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Maple [B] time = 0.24, size = 759, normalized size = 4.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{x^{3} \int \frac{{\left (b^{2} x + a b\right )} e^{\left (\frac{1}{2} \, \log \left (b x + a + 1\right ) + \frac{1}{2} \, \log \left (b x + a - 1\right )\right )}}{b^{2} x^{5} + 2 \, a b x^{4} +{\left (a^{2} - 1\right )} x^{3} +{\left (b^{2} x^{5} + 2 \, a b x^{4} +{\left (a^{2} - 1\right )} x^{3}\right )}{\left (b x + a + 1\right )}{\left (b x + a - 1\right )}}\,{d x} + \arctan \left (1, \sqrt{b x + a + 1} \sqrt{b x + a - 1}\right )}{3 \, x^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A] time = 3.80911, size = 1242, normalized size = 6.9 \begin{align*} \left [\frac{{\left (6 \, a^{4} - 5 \, a^{2} + 2\right )} \sqrt{a^{2} - 1} b^{3} x^{3} \log \left (\frac{a^{2} b x + a^{3} + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}{\left (a^{2} + \sqrt{a^{2} - 1} a - 1\right )} +{\left (a b x + a^{2} - 1\right )} \sqrt{a^{2} - 1} - a}{x}\right ) + 4 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} b^{3} x^{3} \arctan \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) -{\left (5 \, a^{5} - 7 \, a^{3} + 2 \, a\right )} b^{3} x^{3} - 2 \,{\left (a^{9} - 3 \, a^{7} + 3 \, a^{5} - a^{3}\right )} \operatorname{arccsc}\left (b x + a\right ) -{\left ({\left (5 \, a^{5} - 7 \, a^{3} + 2 \, a\right )} b^{2} x^{2} -{\left (a^{6} - 2 \, a^{4} + a^{2}\right )} b x\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{6 \,{\left (a^{9} - 3 \, a^{7} + 3 \, a^{5} - a^{3}\right )} x^{3}}, -\frac{2 \,{\left (6 \, a^{4} - 5 \, a^{2} + 2\right )} \sqrt{-a^{2} + 1} b^{3} x^{3} \arctan \left (-\frac{\sqrt{-a^{2} + 1} b x - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1} \sqrt{-a^{2} + 1}}{a^{2} - 1}\right ) - 4 \,{\left (a^{6} - 3 \, a^{4} + 3 \, a^{2} - 1\right )} b^{3} x^{3} \arctan \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) +{\left (5 \, a^{5} - 7 \, a^{3} + 2 \, a\right )} b^{3} x^{3} + 2 \,{\left (a^{9} - 3 \, a^{7} + 3 \, a^{5} - a^{3}\right )} \operatorname{arccsc}\left (b x + a\right ) +{\left ({\left (5 \, a^{5} - 7 \, a^{3} + 2 \, a\right )} b^{2} x^{2} -{\left (a^{6} - 2 \, a^{4} + a^{2}\right )} b x\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{6 \,{\left (a^{9} - 3 \, a^{7} + 3 \, a^{5} - a^{3}\right )} x^{3}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{acsc}{\left (a + b x \right )}}{x^{4}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.72845, size = 792, normalized size = 4.4 \begin{align*} -\frac{1}{3} \, b{\left (\frac{{\left (6 \, a^{4} b^{8} - 5 \, a^{2} b^{8} + 2 \, b^{8}\right )} \arctan \left (-\frac{x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}}{\sqrt{-a^{2} + 1}}\right )}{{\left (a^{7} b^{6} \mathrm{sgn}\left (b x + a\right ) - 2 \, a^{5} b^{6} \mathrm{sgn}\left (b x + a\right ) + a^{3} b^{6} \mathrm{sgn}\left (b x + a\right )\right )} \sqrt{-a^{2} + 1}} - \frac{2 \, b^{2} \arctan \left (-\frac{{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} b + a{\left | b \right |}}{b}\right )}{a^{3} \mathrm{sgn}\left (b x + a\right )} - \frac{4 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{3} a^{3} b^{8} - 8 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} a^{5} b^{8} + 2 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{2} a^{4} b^{7}{\left | b \right |} - 6 \, a^{6} b^{7}{\left | b \right |} -{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{3} a b^{8} + 11 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} a^{3} b^{8} - 4 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{2} a^{2} b^{7}{\left | b \right |} + 14 \, a^{4} b^{7}{\left | b \right |} - 3 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )} a b^{8} + 2 \,{\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{2} b^{7}{\left | b \right |} - 10 \, a^{2} b^{7}{\left | b \right |} + 2 \, b^{7}{\left | b \right |}}{{\left (a^{6} b^{6} \mathrm{sgn}\left (b x + a\right ) - 2 \, a^{4} b^{6} \mathrm{sgn}\left (b x + a\right ) + a^{2} b^{6} \mathrm{sgn}\left (b x + a\right )\right )}{\left ({\left (x{\left | b \right |} - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right )}^{2} - a^{2} + 1\right )}^{2}}\right )} - \frac{\arcsin \left (\frac{1}{b x + a}\right )}{3 \, x^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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