Optimal. Leaf size=179 \[ \frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {b c \left (2 c^2 d+3 e\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 d^{3/2}}-\frac {b \sqrt {c^2 d-e} \left (c^2 d+2 e\right ) \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d^2}-\frac {b c \sqrt {d+e x^2}}{6 d x^2} \]
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Rubi [A] time = 0.27, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {271, 264, 4976, 12, 573, 149, 156, 63, 208} \[ \frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {b c \left (2 c^2 d+3 e\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 d^{3/2}}-\frac {b \sqrt {c^2 d-e} \left (c^2 d+2 e\right ) \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d^2}-\frac {b c \sqrt {d+e x^2}}{6 d x^2} \]
Antiderivative was successfully verified.
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Rule 12
Rule 63
Rule 149
Rule 156
Rule 208
Rule 264
Rule 271
Rule 573
Rule 4976
Rubi steps
\begin {align*} \int \frac {a+b \tan ^{-1}(c x)}{x^4 \sqrt {d+e x^2}} \, dx &=-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}-(b c) \int \frac {\sqrt {d+e x^2} \left (-d+2 e x^2\right )}{3 d^2 x^3 \left (1+c^2 x^2\right )} \, dx\\ &=-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}-\frac {(b c) \int \frac {\sqrt {d+e x^2} \left (-d+2 e x^2\right )}{x^3 \left (1+c^2 x^2\right )} \, dx}{3 d^2}\\ &=-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}-\frac {(b c) \operatorname {Subst}\left (\int \frac {\sqrt {d+e x} (-d+2 e x)}{x^2 \left (1+c^2 x\right )} \, dx,x,x^2\right )}{6 d^2}\\ &=-\frac {b c \sqrt {d+e x^2}}{6 d x^2}-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}-\frac {(b c) \operatorname {Subst}\left (\int \frac {\frac {1}{2} d \left (2 c^2 d+3 e\right )+\frac {1}{2} e \left (c^2 d+4 e\right ) x}{x \left (1+c^2 x\right ) \sqrt {d+e x}} \, dx,x,x^2\right )}{6 d^2}\\ &=-\frac {b c \sqrt {d+e x^2}}{6 d x^2}-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}+\frac {\left (b c \left (c^2 d-e\right ) \left (c^2 d+2 e\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+c^2 x\right ) \sqrt {d+e x}} \, dx,x,x^2\right )}{6 d^2}-\frac {\left (b c \left (2 c^2 d+3 e\right )\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {d+e x}} \, dx,x,x^2\right )}{12 d}\\ &=-\frac {b c \sqrt {d+e x^2}}{6 d x^2}-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}+\frac {\left (b c \left (c^2 d-e\right ) \left (c^2 d+2 e\right )\right ) \operatorname {Subst}\left (\int \frac {1}{1-\frac {c^2 d}{e}+\frac {c^2 x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{3 d^2 e}-\frac {\left (b c \left (2 c^2 d+3 e\right )\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {d}{e}+\frac {x^2}{e}} \, dx,x,\sqrt {d+e x^2}\right )}{6 d e}\\ &=-\frac {b c \sqrt {d+e x^2}}{6 d x^2}-\frac {\sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d x^3}+\frac {2 e \sqrt {d+e x^2} \left (a+b \tan ^{-1}(c x)\right )}{3 d^2 x}+\frac {b c \left (2 c^2 d+3 e\right ) \tanh ^{-1}\left (\frac {\sqrt {d+e x^2}}{\sqrt {d}}\right )}{6 d^{3/2}}-\frac {b \sqrt {c^2 d-e} \left (c^2 d+2 e\right ) \tanh ^{-1}\left (\frac {c \sqrt {d+e x^2}}{\sqrt {c^2 d-e}}\right )}{3 d^2}\\ \end {align*}
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Mathematica [C] time = 0.53, size = 372, normalized size = 2.08 \[ -\frac {\frac {\sqrt {d+e x^2} \left (2 a \left (d-2 e x^2\right )+b c d x\right )}{x^3}-b c \sqrt {d} \left (2 c^2 d+3 e\right ) \log \left (\sqrt {d} \sqrt {d+e x^2}+d\right )+b c \sqrt {d} \log (x) \left (2 c^2 d+3 e\right )+\frac {b \left (c^4 d^2+c^2 d e-2 e^2\right ) \log \left (\frac {12 c d^2 \left (\sqrt {c^2 d-e} \sqrt {d+e x^2}+c d-i e x\right )}{b (c x+i) \sqrt {c^2 d-e} \left (c^4 d^2+c^2 d e-2 e^2\right )}\right )}{\sqrt {c^2 d-e}}+\frac {b \left (c^4 d^2+c^2 d e-2 e^2\right ) \log \left (\frac {12 c d^2 \left (\sqrt {c^2 d-e} \sqrt {d+e x^2}+c d+i e x\right )}{b (c x-i) \sqrt {c^2 d-e} \left (c^4 d^2+c^2 d e-2 e^2\right )}\right )}{\sqrt {c^2 d-e}}+\frac {2 b \tan ^{-1}(c x) \left (d-2 e x^2\right ) \sqrt {d+e x^2}}{x^3}}{6 d^2} \]
Antiderivative was successfully verified.
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fricas [A] time = 0.62, size = 868, normalized size = 4.85 \[ \left [\frac {{\left (b c^{2} d + 2 \, b e\right )} \sqrt {c^{2} d - e} x^{3} \log \left (\frac {c^{4} e^{2} x^{4} + 8 \, c^{4} d^{2} - 8 \, c^{2} d e + 2 \, {\left (4 \, c^{4} d e - 3 \, c^{2} e^{2}\right )} x^{2} - 4 \, {\left (c^{3} e x^{2} + 2 \, c^{3} d - c e\right )} \sqrt {c^{2} d - e} \sqrt {e x^{2} + d} + e^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) + {\left (2 \, b c^{3} d + 3 \, b c e\right )} \sqrt {d} x^{3} \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) - 2 \, {\left (b c d x - 4 \, a e x^{2} + 2 \, a d - 2 \, {\left (2 \, b e x^{2} - b d\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{12 \, d^{2} x^{3}}, -\frac {2 \, {\left (b c^{2} d + 2 \, b e\right )} \sqrt {-c^{2} d + e} x^{3} \arctan \left (-\frac {{\left (c^{2} e x^{2} + 2 \, c^{2} d - e\right )} \sqrt {-c^{2} d + e} \sqrt {e x^{2} + d}}{2 \, {\left (c^{3} d^{2} - c d e + {\left (c^{3} d e - c e^{2}\right )} x^{2}\right )}}\right ) - {\left (2 \, b c^{3} d + 3 \, b c e\right )} \sqrt {d} x^{3} \log \left (-\frac {e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {d} + 2 \, d}{x^{2}}\right ) + 2 \, {\left (b c d x - 4 \, a e x^{2} + 2 \, a d - 2 \, {\left (2 \, b e x^{2} - b d\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{12 \, d^{2} x^{3}}, -\frac {2 \, {\left (2 \, b c^{3} d + 3 \, b c e\right )} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) - {\left (b c^{2} d + 2 \, b e\right )} \sqrt {c^{2} d - e} x^{3} \log \left (\frac {c^{4} e^{2} x^{4} + 8 \, c^{4} d^{2} - 8 \, c^{2} d e + 2 \, {\left (4 \, c^{4} d e - 3 \, c^{2} e^{2}\right )} x^{2} - 4 \, {\left (c^{3} e x^{2} + 2 \, c^{3} d - c e\right )} \sqrt {c^{2} d - e} \sqrt {e x^{2} + d} + e^{2}}{c^{4} x^{4} + 2 \, c^{2} x^{2} + 1}\right ) + 2 \, {\left (b c d x - 4 \, a e x^{2} + 2 \, a d - 2 \, {\left (2 \, b e x^{2} - b d\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{12 \, d^{2} x^{3}}, -\frac {{\left (b c^{2} d + 2 \, b e\right )} \sqrt {-c^{2} d + e} x^{3} \arctan \left (-\frac {{\left (c^{2} e x^{2} + 2 \, c^{2} d - e\right )} \sqrt {-c^{2} d + e} \sqrt {e x^{2} + d}}{2 \, {\left (c^{3} d^{2} - c d e + {\left (c^{3} d e - c e^{2}\right )} x^{2}\right )}}\right ) + {\left (2 \, b c^{3} d + 3 \, b c e\right )} \sqrt {-d} x^{3} \arctan \left (\frac {\sqrt {-d}}{\sqrt {e x^{2} + d}}\right ) + {\left (b c d x - 4 \, a e x^{2} + 2 \, a d - 2 \, {\left (2 \, b e x^{2} - b d\right )} \arctan \left (c x\right )\right )} \sqrt {e x^{2} + d}}{6 \, d^{2} x^{3}}\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 \[ \mathit {sage}_{0} x \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.26, size = 0, normalized size = 0.00 \[ \int \frac {a +b \arctan \left (c x \right )}{x^{4} \sqrt {e \,x^{2}+d}}\, dx \]
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 \[ \frac {1}{3} \, a {\left (\frac {2 \, \sqrt {e x^{2} + d} e}{d^{2} x} - \frac {\sqrt {e x^{2} + d}}{d x^{3}}\right )} + b \int \frac {\arctan \left (c x\right )}{\sqrt {e x^{2} + d} x^{4}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {a+b\,\mathrm {atan}\left (c\,x\right )}{x^4\,\sqrt {e\,x^2+d}} \,d x \]
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 \[ \int \frac {a + b \operatorname {atan}{\left (c x \right )}}{x^{4} \sqrt {d + e x^{2}}}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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