Integrand size = 24, antiderivative size = 156 \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\frac {4 b \sqrt {1+c^2 x^2}}{3 c^5 d}-\frac {b \left (1+c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac {x (a+b \text {arcsinh}(c x))}{c^4 d}+\frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d}+\frac {2 (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{c^5 d}-\frac {i b \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{c^5 d}+\frac {i b \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{c^5 d} \]
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Time = 0.17 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5812, 5789, 4265, 2317, 2438, 267, 272, 45} \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\frac {2 \arctan \left (e^{\text {arcsinh}(c x)}\right ) (a+b \text {arcsinh}(c x))}{c^5 d}-\frac {x (a+b \text {arcsinh}(c x))}{c^4 d}+\frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d}-\frac {i b \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{c^5 d}+\frac {i b \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{c^5 d}-\frac {b \left (c^2 x^2+1\right )^{3/2}}{9 c^5 d}+\frac {4 b \sqrt {c^2 x^2+1}}{3 c^5 d} \]
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Rule 45
Rule 267
Rule 272
Rule 2317
Rule 2438
Rule 4265
Rule 5789
Rule 5812
Rubi steps \begin{align*} \text {integral}& = \frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d}-\frac {\int \frac {x^2 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx}{c^2}-\frac {b \int \frac {x^3}{\sqrt {1+c^2 x^2}} \, dx}{3 c d} \\ & = -\frac {x (a+b \text {arcsinh}(c x))}{c^4 d}+\frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d}+\frac {\int \frac {a+b \text {arcsinh}(c x)}{d+c^2 d x^2} \, dx}{c^4}+\frac {b \int \frac {x}{\sqrt {1+c^2 x^2}} \, dx}{c^3 d}-\frac {b \text {Subst}\left (\int \frac {x}{\sqrt {1+c^2 x}} \, dx,x,x^2\right )}{6 c d} \\ & = \frac {b \sqrt {1+c^2 x^2}}{c^5 d}-\frac {x (a+b \text {arcsinh}(c x))}{c^4 d}+\frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d}+\frac {\text {Subst}(\int (a+b x) \text {sech}(x) \, dx,x,\text {arcsinh}(c x))}{c^5 d}-\frac {b \text {Subst}\left (\int \left (-\frac {1}{c^2 \sqrt {1+c^2 x}}+\frac {\sqrt {1+c^2 x}}{c^2}\right ) \, dx,x,x^2\right )}{6 c d} \\ & = \frac {4 b \sqrt {1+c^2 x^2}}{3 c^5 d}-\frac {b \left (1+c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac {x (a+b \text {arcsinh}(c x))}{c^4 d}+\frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d}+\frac {2 (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{c^5 d}-\frac {(i b) \text {Subst}\left (\int \log \left (1-i e^x\right ) \, dx,x,\text {arcsinh}(c x)\right )}{c^5 d}+\frac {(i b) \text {Subst}\left (\int \log \left (1+i e^x\right ) \, dx,x,\text {arcsinh}(c x)\right )}{c^5 d} \\ & = \frac {4 b \sqrt {1+c^2 x^2}}{3 c^5 d}-\frac {b \left (1+c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac {x (a+b \text {arcsinh}(c x))}{c^4 d}+\frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d}+\frac {2 (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{c^5 d}-\frac {(i b) \text {Subst}\left (\int \frac {\log (1-i x)}{x} \, dx,x,e^{\text {arcsinh}(c x)}\right )}{c^5 d}+\frac {(i b) \text {Subst}\left (\int \frac {\log (1+i x)}{x} \, dx,x,e^{\text {arcsinh}(c x)}\right )}{c^5 d} \\ & = \frac {4 b \sqrt {1+c^2 x^2}}{3 c^5 d}-\frac {b \left (1+c^2 x^2\right )^{3/2}}{9 c^5 d}-\frac {x (a+b \text {arcsinh}(c x))}{c^4 d}+\frac {x^3 (a+b \text {arcsinh}(c x))}{3 c^2 d}+\frac {2 (a+b \text {arcsinh}(c x)) \arctan \left (e^{\text {arcsinh}(c x)}\right )}{c^5 d}-\frac {i b \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )}{c^5 d}+\frac {i b \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{c^5 d} \\ \end{align*}
Time = 0.17 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.09 \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\frac {-9 a c x+3 a c^3 x^3+11 b \sqrt {1+c^2 x^2}-b c^2 x^2 \sqrt {1+c^2 x^2}-9 b c x \text {arcsinh}(c x)+3 b c^3 x^3 \text {arcsinh}(c x)+9 a \arctan (c x)+9 i b \text {arcsinh}(c x) \log \left (1-i e^{\text {arcsinh}(c x)}\right )-9 i b \text {arcsinh}(c x) \log \left (1+i e^{\text {arcsinh}(c x)}\right )-9 i b \operatorname {PolyLog}\left (2,-i e^{\text {arcsinh}(c x)}\right )+9 i b \operatorname {PolyLog}\left (2,i e^{\text {arcsinh}(c x)}\right )}{9 c^5 d} \]
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Time = 0.35 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.33
method | result | size |
derivativedivides | \(\frac {\frac {a \left (\frac {c^{3} x^{3}}{3}-c x +\arctan \left (c x \right )\right )}{d}+\frac {b \left (\frac {\operatorname {arcsinh}\left (c x \right ) c^{3} x^{3}}{3}-\operatorname {arcsinh}\left (c x \right ) c x +\operatorname {arcsinh}\left (c x \right ) \arctan \left (c x \right )-\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{9}+\frac {11 \sqrt {c^{2} x^{2}+1}}{9}+\arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-\arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )\right )}{d}}{c^{5}}\) | \(208\) |
default | \(\frac {\frac {a \left (\frac {c^{3} x^{3}}{3}-c x +\arctan \left (c x \right )\right )}{d}+\frac {b \left (\frac {\operatorname {arcsinh}\left (c x \right ) c^{3} x^{3}}{3}-\operatorname {arcsinh}\left (c x \right ) c x +\operatorname {arcsinh}\left (c x \right ) \arctan \left (c x \right )-\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{9}+\frac {11 \sqrt {c^{2} x^{2}+1}}{9}+\arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-\arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )\right )}{d}}{c^{5}}\) | \(208\) |
parts | \(\frac {a \left (\frac {\frac {1}{3} x^{3} c^{2}-x}{c^{4}}+\frac {\arctan \left (c x \right )}{c^{5}}\right )}{d}+\frac {b \left (\frac {\operatorname {arcsinh}\left (c x \right ) c^{3} x^{3}}{3}-\operatorname {arcsinh}\left (c x \right ) c x +\operatorname {arcsinh}\left (c x \right ) \arctan \left (c x \right )-\frac {c^{2} x^{2} \sqrt {c^{2} x^{2}+1}}{9}+\frac {11 \sqrt {c^{2} x^{2}+1}}{9}+\arctan \left (c x \right ) \ln \left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-\arctan \left (c x \right ) \ln \left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )-i \operatorname {dilog}\left (1+\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )+i \operatorname {dilog}\left (1-\frac {i \left (i c x +1\right )}{\sqrt {c^{2} x^{2}+1}}\right )\right )}{d \,c^{5}}\) | \(215\) |
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\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{c^{2} d x^{2} + d} \,d x } \]
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\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\frac {\int \frac {a x^{4}}{c^{2} x^{2} + 1}\, dx + \int \frac {b x^{4} \operatorname {asinh}{\left (c x \right )}}{c^{2} x^{2} + 1}\, dx}{d} \]
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\[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\int { \frac {{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )} x^{4}}{c^{2} d x^{2} + d} \,d x } \]
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Exception generated. \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\text {Exception raised: TypeError} \]
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Timed out. \[ \int \frac {x^4 (a+b \text {arcsinh}(c x))}{d+c^2 d x^2} \, dx=\int \frac {x^4\,\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}{d\,c^2\,x^2+d} \,d x \]
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