Integrand size = 27, antiderivative size = 177 \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=-\frac {b (a+3 b) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {(a-3 b) b \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {a b^4 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a b^2-b \left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \]
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Time = 0.19 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2916, 12, 837, 815} \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 d \left (a^2-b^2\right )}-\frac {\sec ^2(c+d x) \left (4 a b^2-b \left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 d \left (a^2-b^2\right )^2}+\frac {a b^4 \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^3}-\frac {b (a+3 b) \log (1-\sin (c+d x))}{16 d (a+b)^3}+\frac {b (a-3 b) \log (\sin (c+d x)+1)}{16 d (a-b)^3} \]
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Rule 12
Rule 815
Rule 837
Rule 2916
Rubi steps \begin{align*} \text {integral}& = \frac {b^5 \text {Subst}\left (\int \frac {x}{b (a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {b^4 \text {Subst}\left (\int \frac {x}{(a+x) \left (b^2-x^2\right )^3} \, dx,x,b \sin (c+d x)\right )}{d} \\ & = \frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {b^2 \text {Subst}\left (\int \frac {-a b^2+3 b^2 x}{(a+x) \left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{4 \left (a^2-b^2\right ) d} \\ & = \frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a b^2-b \left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac {\text {Subst}\left (\int \frac {a b^2 \left (a^2-5 b^2\right )+b^2 \left (a^2+3 b^2\right ) x}{(a+x) \left (b^2-x^2\right )} \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \\ & = \frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a b^2-b \left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac {\text {Subst}\left (\int \left (\frac {(a-b)^2 b (a+3 b)}{2 (a+b) (b-x)}+\frac {8 a b^4}{(a-b) (a+b) (a+x)}+\frac {(a-3 b) b (a+b)^2}{2 (a-b) (b+x)}\right ) \, dx,x,b \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \\ & = -\frac {b (a+3 b) \log (1-\sin (c+d x))}{16 (a+b)^3 d}+\frac {(a-3 b) b \log (1+\sin (c+d x))}{16 (a-b)^3 d}+\frac {a b^4 \log (a+b \sin (c+d x))}{\left (a^2-b^2\right )^3 d}+\frac {\sec ^4(c+d x) (a-b \sin (c+d x))}{4 \left (a^2-b^2\right ) d}-\frac {\sec ^2(c+d x) \left (4 a b^2-b \left (a^2+3 b^2\right ) \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d} \\ \end{align*}
Time = 4.20 (sec) , antiderivative size = 161, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {-(a-b)^3 b (a+3 b) \log (1-\sin (c+d x))+(a-3 b) b (a+b)^3 \log (1+\sin (c+d x))+16 a b^4 \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}+4 \sec ^4(c+d x) (a-b \sin (c+d x))+\frac {2 b \sec ^2(c+d x) \left (-4 a b+\left (a^2+3 b^2\right ) \sin (c+d x)\right )}{a^2-b^2}}{16 \left (a^2-b^2\right ) d} \]
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Time = 0.94 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-a +3 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a -3 b \right ) b \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {a +3 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {\left (a +3 b \right ) b \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}+\frac {b^{4} a \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\) | \(170\) |
default | \(\frac {\frac {1}{2 \left (8 a -8 b \right ) \left (1+\sin \left (d x +c \right )\right )^{2}}-\frac {-a +3 b}{16 \left (a -b \right )^{2} \left (1+\sin \left (d x +c \right )\right )}+\frac {\left (a -3 b \right ) b \ln \left (1+\sin \left (d x +c \right )\right )}{16 \left (a -b \right )^{3}}+\frac {1}{2 \left (8 a +8 b \right ) \left (\sin \left (d x +c \right )-1\right )^{2}}-\frac {a +3 b}{16 \left (a +b \right )^{2} \left (\sin \left (d x +c \right )-1\right )}-\frac {\left (a +3 b \right ) b \ln \left (\sin \left (d x +c \right )-1\right )}{16 \left (a +b \right )^{3}}+\frac {b^{4} a \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a +b \right )^{3} \left (a -b \right )^{3}}}{d}\) | \(170\) |
parallelrisch | \(\frac {8 b^{4} \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) a \ln \left (2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \left (\sec ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )-\left (a +3 b \right ) b \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a -b \right )^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-2 \left (-\frac {b \left (a -3 b \right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a +b \right )^{2} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\left (\left (a^{3}-a \,b^{2}\right ) \cos \left (2 d x +2 c \right )+\frac {\left (a^{3}-3 a \,b^{2}\right ) \cos \left (4 d x +4 c \right )}{4}+\frac {\left (-a^{2} b -3 b^{3}\right ) \sin \left (3 d x +3 c \right )}{4}+\frac {\left (7 a^{2} b -11 b^{3}\right ) \sin \left (d x +c \right )}{4}-\frac {5 a^{3}}{4}+\frac {7 a \,b^{2}}{4}\right ) \left (a -b \right )\right ) \left (a +b \right )}{2 \left (a -b \right )^{3} \left (a +b \right )^{3} d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\) | \(302\) |
norman | \(\frac {\frac {4 a \,b^{2} \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 \left (-a^{3}+2 a \,b^{2}\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {2 \left (-a^{3}+2 a \,b^{2}\right ) \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (7 a^{2}-3 b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {b \left (7 a^{2}-3 b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 \left (a^{4}-2 a^{2} b^{2}+b^{4}\right ) d}-\frac {b \left (a^{2}-5 b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}-\frac {b \left (a^{2}-5 b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d \left (a^{4}-2 a^{2} b^{2}+b^{4}\right )}}{\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {a \,b^{4} \ln \left (\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a +2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {\left (a -3 b \right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}-\frac {\left (a +3 b \right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}\) | \(473\) |
risch | \(-\frac {2 i a \,b^{4} c}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}+\frac {i a b c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {2 i a \,b^{4} x}{a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}}+\frac {3 i b^{2} x}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}+\frac {3 i b^{2} c}{8 d \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right )}-\frac {i a b c}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}+\frac {3 i b^{2} x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {3 i b^{2} c}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {i a b x}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right )}+\frac {i a b x}{8 a^{3}+24 a^{2} b +24 a \,b^{2}+8 b^{3}}+\frac {-8 a \,b^{2} {\mathrm e}^{6 i \left (d x +c \right )}-i a^{2} b \,{\mathrm e}^{7 i \left (d x +c \right )}-3 i b^{3} {\mathrm e}^{7 i \left (d x +c \right )}+16 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}-32 a \,b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+7 i a^{2} b \,{\mathrm e}^{5 i \left (d x +c \right )}-11 i b^{3} {\mathrm e}^{5 i \left (d x +c \right )}-8 a \,b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-7 i a^{2} b \,{\mathrm e}^{3 i \left (d x +c \right )}+11 i b^{3} {\mathrm e}^{3 i \left (d x +c \right )}+i a^{2} b \,{\mathrm e}^{i \left (d x +c \right )}+3 i b^{3} {\mathrm e}^{i \left (d x +c \right )}}{4 \left (a^{2}-b^{2}\right )^{2} d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) a b}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) b^{2}}{8 \left (a^{3}-3 a^{2} b +3 a \,b^{2}-b^{3}\right ) d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) a b}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) b^{2}}{8 \left (a^{3}+3 a^{2} b +3 a \,b^{2}+b^{3}\right ) d}+\frac {a \,b^{4} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-1+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}\right )}{d \left (a^{6}-3 a^{4} b^{2}+3 a^{2} b^{4}-b^{6}\right )}\) | \(761\) |
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Time = 0.50 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.44 \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {16 \, a b^{4} \cos \left (d x + c\right )^{4} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left (a^{4} b - 6 \, a^{2} b^{3} - 8 \, a b^{4} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{4} b - 6 \, a^{2} b^{3} + 8 \, a b^{4} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a^{5} - 8 \, a^{3} b^{2} + 4 \, a b^{4} - 8 \, {\left (a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (2 \, a^{4} b - 4 \, a^{2} b^{3} + 2 \, b^{5} - {\left (a^{4} b + 2 \, a^{2} b^{3} - 3 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{16 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} d \cos \left (d x + c\right )^{4}} \]
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\[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\int \frac {\sin {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]
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Time = 0.20 (sec) , antiderivative size = 267, normalized size of antiderivative = 1.51 \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, a b^{4} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}} + \frac {{\left (a b - 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (a b + 3 \, b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (4 \, a b^{2} \sin \left (d x + c\right )^{2} - {\left (a^{2} b + 3 \, b^{3}\right )} \sin \left (d x + c\right )^{3} + 2 \, a^{3} - 6 \, a b^{2} - {\left (a^{2} b - 5 \, b^{3}\right )} \sin \left (d x + c\right )\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{4} + a^{4} - 2 \, a^{2} b^{2} + b^{4} - 2 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \sin \left (d x + c\right )^{2}}}{16 \, d} \]
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Time = 0.49 (sec) , antiderivative size = 323, normalized size of antiderivative = 1.82 \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {\frac {16 \, a b^{5} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{6} b - 3 \, a^{4} b^{3} + 3 \, a^{2} b^{5} - b^{7}} + \frac {{\left (a b - 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}} - \frac {{\left (a b + 3 \, b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}} + \frac {2 \, {\left (6 \, a b^{4} \sin \left (d x + c\right )^{4} - a^{4} b \sin \left (d x + c\right )^{3} - 2 \, a^{2} b^{3} \sin \left (d x + c\right )^{3} + 3 \, b^{5} \sin \left (d x + c\right )^{3} + 4 \, a^{3} b^{2} \sin \left (d x + c\right )^{2} - 16 \, a b^{4} \sin \left (d x + c\right )^{2} - a^{4} b \sin \left (d x + c\right ) + 6 \, a^{2} b^{3} \sin \left (d x + c\right ) - 5 \, b^{5} \sin \left (d x + c\right ) + 2 \, a^{5} - 8 \, a^{3} b^{2} + 12 \, a b^{4}\right )}}{{\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \]
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Time = 12.41 (sec) , antiderivative size = 483, normalized size of antiderivative = 2.73 \[ \int \frac {\sec ^4(c+d x) \tan (c+d x)}{a+b \sin (c+d x)} \, dx=\frac {a\,b^4\,\ln \left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+a\right )}{d\,\left (a^6-3\,a^4\,b^2+3\,a^2\,b^4-b^6\right )}-\frac {\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (2\,a\,b^2-a^3\right )}{a^4-2\,a^2\,b^2+b^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (a^2\,b-5\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (7\,a^2\,b-3\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (7\,a^2\,b-3\,b^3\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}-\frac {4\,a\,b^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{a^4-2\,a^2\,b^2+b^4}+\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-5\,b^2\right )}{4\,\left (a^4-2\,a^2\,b^2+b^4\right )}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left (\frac {b}{8\,{\left (a+b\right )}^2}+\frac {b^2}{4\,{\left (a+b\right )}^3}\right )}{d}+\frac {b\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )\,\left (a-3\,b\right )}{8\,d\,{\left (a-b\right )}^3} \]
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