Optimal. Leaf size=495 \[ -\frac {e^2 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2 c^5}+\frac {9 e^2 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^5}-\frac {5 e^2 \sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^5}+\frac {e^2 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{8 b^2 c^5}-\frac {9 e^2 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^5}+\frac {5 e^2 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{16 b^2 c^5}+\frac {d e \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{2 b^2 c^3}-\frac {3 d e \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b^2 c^3}-\frac {d e \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{2 b^2 c^3}+\frac {3 d e \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 \left (a+b \sinh ^{-1}(c x)\right )}{b}\right )}{2 b^2 c^3}-\frac {d^2 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b^2 c}+\frac {d^2 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a+b \sinh ^{-1}(c x)}{b}\right )}{b^2 c}-\frac {d^2 \sqrt {c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {2 d e x^2 \sqrt {c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {e^2 x^4 \sqrt {c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]
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Rubi [A] time = 0.87, antiderivative size = 483, normalized size of antiderivative = 0.98, number of steps used = 26, number of rules used = 7, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {5706, 5655, 5779, 3303, 3298, 3301, 5665} \[ \frac {d e \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{2 b^2 c^3}-\frac {3 d e \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{2 b^2 c^3}-\frac {e^2 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2 c^5}+\frac {9 e^2 \sinh \left (\frac {3 a}{b}\right ) \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2 c^5}-\frac {5 e^2 \sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2 c^5}-\frac {d e \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{2 b^2 c^3}+\frac {3 d e \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{2 b^2 c^3}+\frac {e^2 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2 c^5}-\frac {9 e^2 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2 c^5}+\frac {5 e^2 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2 c^5}-\frac {d^2 \sinh \left (\frac {a}{b}\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{b^2 c}+\frac {d^2 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{b^2 c}-\frac {d^2 \sqrt {c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {2 d e x^2 \sqrt {c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {e^2 x^4 \sqrt {c^2 x^2+1}}{b c \left (a+b \sinh ^{-1}(c x)\right )} \]
Antiderivative was successfully verified.
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Rule 3298
Rule 3301
Rule 3303
Rule 5655
Rule 5665
Rule 5706
Rule 5779
Rubi steps
\begin {align*} \int \frac {\left (d+e x^2\right )^2}{\left (a+b \sinh ^{-1}(c x)\right )^2} \, dx &=\int \left (\frac {d^2}{\left (a+b \sinh ^{-1}(c x)\right )^2}+\frac {2 d e x^2}{\left (a+b \sinh ^{-1}(c x)\right )^2}+\frac {e^2 x^4}{\left (a+b \sinh ^{-1}(c x)\right )^2}\right ) \, dx\\ &=d^2 \int \frac {1}{\left (a+b \sinh ^{-1}(c x)\right )^2} \, dx+(2 d e) \int \frac {x^2}{\left (a+b \sinh ^{-1}(c x)\right )^2} \, dx+e^2 \int \frac {x^4}{\left (a+b \sinh ^{-1}(c x)\right )^2} \, dx\\ &=-\frac {d^2 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {2 d e x^2 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {e^2 x^4 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {\left (c d^2\right ) \int \frac {x}{\sqrt {1+c^2 x^2} \left (a+b \sinh ^{-1}(c x)\right )} \, dx}{b}+\frac {(2 d e) \operatorname {Subst}\left (\int \left (-\frac {\sinh (x)}{4 (a+b x)}+\frac {3 \sinh (3 x)}{4 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^3}+\frac {e^2 \operatorname {Subst}\left (\int \left (\frac {\sinh (x)}{8 (a+b x)}-\frac {9 \sinh (3 x)}{16 (a+b x)}+\frac {5 \sinh (5 x)}{16 (a+b x)}\right ) \, dx,x,\sinh ^{-1}(c x)\right )}{b c^5}\\ &=-\frac {d^2 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {2 d e x^2 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {e^2 x^4 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {d^2 \operatorname {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c}-\frac {(d e) \operatorname {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}+\frac {(3 d e) \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}+\frac {e^2 \operatorname {Subst}\left (\int \frac {\sinh (x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^5}+\frac {\left (5 e^2\right ) \operatorname {Subst}\left (\int \frac {\sinh (5 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^5}-\frac {\left (9 e^2\right ) \operatorname {Subst}\left (\int \frac {\sinh (3 x)}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^5}\\ &=-\frac {d^2 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {2 d e x^2 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {e^2 x^4 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}+\frac {\left (d^2 \cosh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c}-\frac {\left (d e \cosh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}+\frac {\left (e^2 \cosh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^5}+\frac {\left (3 d e \cosh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}-\frac {\left (9 e^2 \cosh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^5}+\frac {\left (5 e^2 \cosh \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\sinh \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^5}-\frac {\left (d^2 \sinh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{b c}+\frac {\left (d e \sinh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}-\frac {\left (e^2 \sinh \left (\frac {a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {a}{b}+x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{8 b c^5}-\frac {\left (3 d e \sinh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{2 b c^3}+\frac {\left (9 e^2 \sinh \left (\frac {3 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {3 a}{b}+3 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^5}-\frac {\left (5 e^2 \sinh \left (\frac {5 a}{b}\right )\right ) \operatorname {Subst}\left (\int \frac {\cosh \left (\frac {5 a}{b}+5 x\right )}{a+b x} \, dx,x,\sinh ^{-1}(c x)\right )}{16 b c^5}\\ &=-\frac {d^2 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {2 d e x^2 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {e^2 x^4 \sqrt {1+c^2 x^2}}{b c \left (a+b \sinh ^{-1}(c x)\right )}-\frac {d^2 \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )}{b^2 c}+\frac {d e \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )}{2 b^2 c^3}-\frac {e^2 \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right ) \sinh \left (\frac {a}{b}\right )}{8 b^2 c^5}-\frac {3 d e \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {3 a}{b}\right )}{2 b^2 c^3}+\frac {9 e^2 \text {Chi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {3 a}{b}\right )}{16 b^2 c^5}-\frac {5 e^2 \text {Chi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c x)\right ) \sinh \left (\frac {5 a}{b}\right )}{16 b^2 c^5}+\frac {d^2 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{b^2 c}-\frac {d e \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{2 b^2 c^3}+\frac {e^2 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )}{8 b^2 c^5}+\frac {3 d e \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{2 b^2 c^3}-\frac {9 e^2 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (\frac {3 a}{b}+3 \sinh ^{-1}(c x)\right )}{16 b^2 c^5}+\frac {5 e^2 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (\frac {5 a}{b}+5 \sinh ^{-1}(c x)\right )}{16 b^2 c^5}\\ \end {align*}
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Mathematica [A] time = 2.08, size = 356, normalized size = 0.72 \[ -\frac {-16 c^4 d^2 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )+3 e \sinh \left (\frac {3 a}{b}\right ) \left (8 c^2 d-3 e\right ) \text {Chi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+8 c^2 d e \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )-24 c^2 d e \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )+2 \sinh \left (\frac {a}{b}\right ) \left (8 c^4 d^2-4 c^2 d e+e^2\right ) \text {Chi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )+\frac {16 b c^4 d^2 \sqrt {c^2 x^2+1}}{a+b \sinh ^{-1}(c x)}+\frac {32 b c^4 d e x^2 \sqrt {c^2 x^2+1}}{a+b \sinh ^{-1}(c x)}+\frac {16 b c^4 e^2 x^4 \sqrt {c^2 x^2+1}}{a+b \sinh ^{-1}(c x)}+5 e^2 \sinh \left (\frac {5 a}{b}\right ) \text {Chi}\left (5 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-2 e^2 \cosh \left (\frac {a}{b}\right ) \text {Shi}\left (\frac {a}{b}+\sinh ^{-1}(c x)\right )+9 e^2 \cosh \left (\frac {3 a}{b}\right ) \text {Shi}\left (3 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )-5 e^2 \cosh \left (\frac {5 a}{b}\right ) \text {Shi}\left (5 \left (\frac {a}{b}+\sinh ^{-1}(c x)\right )\right )}{16 b^2 c^5} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.60, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}{b^{2} \operatorname {arsinh}\left (c x\right )^{2} + 2 \, a b \operatorname {arsinh}\left (c x\right ) + a^{2}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x^{2} + d\right )}^{2}}{{\left (b \operatorname {arsinh}\left (c x\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.40, size = 1036, normalized size = 2.09 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ -\frac {c^{3} e^{2} x^{7} + {\left (2 \, c^{3} d e + c e^{2}\right )} x^{5} + c d^{2} x + {\left (c^{3} d^{2} + 2 \, c d e\right )} x^{3} + {\left (c^{2} e^{2} x^{6} + {\left (2 \, c^{2} d e + e^{2}\right )} x^{4} + {\left (c^{2} d^{2} + 2 \, d e\right )} x^{2} + d^{2}\right )} \sqrt {c^{2} x^{2} + 1}}{a b c^{3} x^{2} + \sqrt {c^{2} x^{2} + 1} a b c^{2} x + a b c + {\left (b^{2} c^{3} x^{2} + \sqrt {c^{2} x^{2} + 1} b^{2} c^{2} x + b^{2} c\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right )} + \int \frac {5 \, c^{5} e^{2} x^{8} + 2 \, {\left (3 \, c^{5} d e + 5 \, c^{3} e^{2}\right )} x^{6} + {\left (c^{5} d^{2} + 12 \, c^{3} d e + 5 \, c e^{2}\right )} x^{4} + c d^{2} + 2 \, {\left (c^{3} d^{2} + 3 \, c d e\right )} x^{2} + {\left (5 \, c^{3} e^{2} x^{6} + 3 \, {\left (2 \, c^{3} d e + c e^{2}\right )} x^{4} - c d^{2} + {\left (c^{3} d^{2} + 2 \, c d e\right )} x^{2}\right )} {\left (c^{2} x^{2} + 1\right )} + {\left (10 \, c^{4} e^{2} x^{7} + {\left (12 \, c^{4} d e + 13 \, c^{2} e^{2}\right )} x^{5} + 2 \, {\left (c^{4} d^{2} + 7 \, c^{2} d e + 2 \, e^{2}\right )} x^{3} + {\left (c^{2} d^{2} + 4 \, d e\right )} x\right )} \sqrt {c^{2} x^{2} + 1}}{a b c^{5} x^{4} + {\left (c^{2} x^{2} + 1\right )} a b c^{3} x^{2} + 2 \, a b c^{3} x^{2} + a b c + {\left (b^{2} c^{5} x^{4} + {\left (c^{2} x^{2} + 1\right )} b^{2} c^{3} x^{2} + 2 \, b^{2} c^{3} x^{2} + b^{2} c + 2 \, {\left (b^{2} c^{4} x^{3} + b^{2} c^{2} x\right )} \sqrt {c^{2} x^{2} + 1}\right )} \log \left (c x + \sqrt {c^{2} x^{2} + 1}\right ) + 2 \, {\left (a b c^{4} x^{3} + a b c^{2} x\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x} \]
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 \[ \int \frac {{\left (e\,x^2+d\right )}^2}{{\left (a+b\,\mathrm {asinh}\left (c\,x\right )\right )}^2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x^{2}\right )^{2}}{\left (a + b \operatorname {asinh}{\left (c x \right )}\right )^{2}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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