Optimal. Leaf size=296 \[ -\frac {a f \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}+\frac {a f \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}-\frac {f \log (a+b \cos (c+d x))}{d^2 \left (a^2-b^2\right )}-\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac {b (e+f x) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
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Rubi [A] time = 0.52, antiderivative size = 296, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {3324, 3321, 2264, 2190, 2279, 2391, 2668, 31} \[ -\frac {a f \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}+\frac {a f \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d^2 \left (a^2-b^2\right )^{3/2}}-\frac {f \log (a+b \cos (c+d x))}{d^2 \left (a^2-b^2\right )}-\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{d \left (a^2-b^2\right )^{3/2}}+\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{d \left (a^2-b^2\right )^{3/2}}-\frac {b (e+f x) \sin (c+d x)}{d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]
Antiderivative was successfully verified.
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Rule 31
Rule 2190
Rule 2264
Rule 2279
Rule 2391
Rule 2668
Rule 3321
Rule 3324
Rubi steps
\begin {align*} \int \frac {e+f x}{(a+b \cos (c+d x))^2} \, dx &=-\frac {b (e+f x) \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {a \int \frac {e+f x}{a+b \cos (c+d x)} \, dx}{a^2-b^2}+\frac {(b f) \int \frac {\sin (c+d x)}{a+b \cos (c+d x)} \, dx}{\left (a^2-b^2\right ) d}\\ &=-\frac {b (e+f x) \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {(2 a) \int \frac {e^{i (c+d x)} (e+f x)}{b+2 a e^{i (c+d x)}+b e^{2 i (c+d x)}} \, dx}{a^2-b^2}-\frac {f \operatorname {Subst}\left (\int \frac {1}{a+x} \, dx,x,b \cos (c+d x)\right )}{\left (a^2-b^2\right ) d^2}\\ &=-\frac {f \log (a+b \cos (c+d x))}{\left (a^2-b^2\right ) d^2}-\frac {b (e+f x) \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {(2 a b) \int \frac {e^{i (c+d x)} (e+f x)}{2 a-2 \sqrt {a^2-b^2}+2 b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}-\frac {(2 a b) \int \frac {e^{i (c+d x)} (e+f x)}{2 a+2 \sqrt {a^2-b^2}+2 b e^{i (c+d x)}} \, dx}{\left (a^2-b^2\right )^{3/2}}\\ &=-\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {f \log (a+b \cos (c+d x))}{\left (a^2-b^2\right ) d^2}-\frac {b (e+f x) \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {(i a f) \int \log \left (1+\frac {2 b e^{i (c+d x)}}{2 a-2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d}-\frac {(i a f) \int \log \left (1+\frac {2 b e^{i (c+d x)}}{2 a+2 \sqrt {a^2-b^2}}\right ) \, dx}{\left (a^2-b^2\right )^{3/2} d}\\ &=-\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {f \log (a+b \cos (c+d x))}{\left (a^2-b^2\right ) d^2}-\frac {b (e+f x) \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac {(a f) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 b x}{2 a-2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {(a f) \operatorname {Subst}\left (\int \frac {\log \left (1+\frac {2 b x}{2 a+2 \sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{\left (a^2-b^2\right )^{3/2} d^2}\\ &=-\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}+\frac {i a (e+f x) \log \left (1+\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d}-\frac {f \log (a+b \cos (c+d x))}{\left (a^2-b^2\right ) d^2}-\frac {a f \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}+\frac {a f \text {Li}_2\left (-\frac {b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{\left (a^2-b^2\right )^{3/2} d^2}-\frac {b (e+f x) \sin (c+d x)}{\left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end {align*}
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Mathematica [B] time = 9.75, size = 933, normalized size = 3.15 \[ \frac {\left (\frac {2 a (d e-c f) \tan ^{-1}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b}}+f \log \left (\sec ^2\left (\frac {1}{2} (c+d x)\right )\right )-f \log \left ((a+b \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )\right )-\frac {i a f \left (\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {\sqrt {a+b}-\sqrt {b-a} \tan \left (\frac {1}{2} (c+d x)\right )}{i \sqrt {b-a}+\sqrt {a+b}}\right )+\text {Li}_2\left (\frac {\sqrt {b-a} \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {b-a}-i \sqrt {a+b}}\right )\right )}{\sqrt {b-a} \sqrt {a+b}}+\frac {i a f \left (\log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right ) \log \left (\frac {i \left (\sqrt {b-a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt {a+b}\right )}{\sqrt {b-a}+i \sqrt {a+b}}\right )+\text {Li}_2\left (\frac {\sqrt {b-a} \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {b-a}+i \sqrt {a+b}}\right )\right )}{\sqrt {b-a} \sqrt {a+b}}-\frac {i a f \left (\log \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right ) \log \left (\frac {\sqrt {b-a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt {a+b}}{i \sqrt {b-a}+\sqrt {a+b}}\right )+\text {Li}_2\left (\frac {\sqrt {b-a} \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right )}{\sqrt {b-a}-i \sqrt {a+b}}\right )\right )}{\sqrt {b-a} \sqrt {a+b}}+\frac {i a f \left (\log \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right ) \log \left (\frac {i \left (\sqrt {a+b}-\sqrt {b-a} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {b-a}+i \sqrt {a+b}}\right )+\text {Li}_2\left (\frac {\sqrt {b-a} \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right )}{\sqrt {b-a}+i \sqrt {a+b}}\right )\right )}{\sqrt {b-a} \sqrt {a+b}}\right ) (a d e+a d f x+b f \sin (c+d x)) \left (\sqrt {a+b}-\sqrt {b-a} \tan \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sqrt {b-a} \tan \left (\frac {1}{2} (c+d x)\right )+\sqrt {a+b}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )}{\left (a^2-b^2\right ) d^2 (a+b \cos (c+d x)) \left (a \left (d e-c f+i f \log \left (1-i \tan \left (\frac {1}{2} (c+d x)\right )\right )-i f \log \left (i \tan \left (\frac {1}{2} (c+d x)\right )+1\right )\right )+b f \sin (c+d x)\right )}+\frac {-b d e \sin (c+d x)+b c f \sin (c+d x)-b f (c+d x) \sin (c+d x)}{(a-b) (a+b) d^2 (a+b \cos (c+d x))} \]
Warning: Unable to verify antiderivative.
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fricas [B] time = 1.59, size = 1482, normalized size = 5.01 \[ \text {result too large to display} \]
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 \[ \int \frac {f x + e}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [B] time = 0.70, size = 674, normalized size = 2.28 \[ \frac {2 i \left (f x +e \right ) \left (a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{d \left (-a^{2}+b^{2}\right ) \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}-\frac {2 f \ln \left ({\mathrm e}^{i \left (d x +c \right )}\right )}{d^{2} \left (-a^{2}+b^{2}\right )}+\frac {f \ln \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{d^{2} \left (-a^{2}+b^{2}\right )}+\frac {2 i a e \arctan \left (\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d \left (-a^{2}+b^{2}\right )^{\frac {3}{2}}}+\frac {i a f \ln \left (\frac {-b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) x}{d \left (-a^{2}+b^{2}\right ) \sqrt {a^{2}-b^{2}}}+\frac {i a f \ln \left (\frac {-b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} \left (-a^{2}+b^{2}\right ) \sqrt {a^{2}-b^{2}}}-\frac {i a f \ln \left (\frac {b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) x}{d \left (-a^{2}+b^{2}\right ) \sqrt {a^{2}-b^{2}}}-\frac {i a f \ln \left (\frac {b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right ) c}{d^{2} \left (-a^{2}+b^{2}\right ) \sqrt {a^{2}-b^{2}}}+\frac {a f \dilog \left (\frac {-b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}-a}{-a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} \left (-a^{2}+b^{2}\right ) \sqrt {a^{2}-b^{2}}}-\frac {a f \dilog \left (\frac {b \,{\mathrm e}^{i \left (d x +c \right )}+\sqrt {a^{2}-b^{2}}+a}{a +\sqrt {a^{2}-b^{2}}}\right )}{d^{2} \left (-a^{2}+b^{2}\right ) \sqrt {a^{2}-b^{2}}}-\frac {2 i a f c \arctan \left (\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}+2 a}{2 \sqrt {-a^{2}+b^{2}}}\right )}{d^{2} \left (-a^{2}+b^{2}\right )^{\frac {3}{2}}} \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F(-1)] time = 0.00, size = -1, normalized size = -0.00 \[ \text {Hanged} \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]
Verification of antiderivative is not currently implemented for this CAS.
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