Integrand size = 18, antiderivative size = 128 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (4 e^2+b^2 c^2 \log ^2(F)\right )}-\frac {2 e F^{c (a+b x)} \cos (d+e x) \sin (d+e x)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {b c F^{c (a+b x)} \log (F) \sin ^2(d+e x)}{4 e^2+b^2 c^2 \log ^2(F)} \]
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Time = 0.06 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4519, 2225} \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=\frac {b c \log (F) \sin ^2(d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}-\frac {2 e \sin (d+e x) \cos (d+e x) F^{c (a+b x)}}{b^2 c^2 \log ^2(F)+4 e^2}+\frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (b^2 c^2 \log ^2(F)+4 e^2\right )} \]
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Rule 2225
Rule 4519
Rubi steps \begin{align*} \text {integral}& = -\frac {2 e F^{c (a+b x)} \cos (d+e x) \sin (d+e x)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {b c F^{c (a+b x)} \log (F) \sin ^2(d+e x)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {\left (2 e^2\right ) \int F^{c (a+b x)} \, dx}{4 e^2+b^2 c^2 \log ^2(F)} \\ & = \frac {2 e^2 F^{c (a+b x)}}{b c \log (F) \left (4 e^2+b^2 c^2 \log ^2(F)\right )}-\frac {2 e F^{c (a+b x)} \cos (d+e x) \sin (d+e x)}{4 e^2+b^2 c^2 \log ^2(F)}+\frac {b c F^{c (a+b x)} \log (F) \sin ^2(d+e x)}{4 e^2+b^2 c^2 \log ^2(F)} \\ \end{align*}
Time = 0.14 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.67 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=\frac {F^{c (a+b x)} \left (4 e^2+b^2 c^2 \log ^2(F)-b^2 c^2 \cos (2 (d+e x)) \log ^2(F)-2 b c e \log (F) \sin (2 (d+e x))\right )}{8 b c e^2 \log (F)+2 b^3 c^3 \log ^3(F)} \]
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Time = 0.77 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.74
method | result | size |
parallelrisch | \(-\frac {F^{c \left (x b +a \right )} \left (\frac {b^{2} c^{2} \ln \left (F \right )^{2} \cos \left (2 e x +2 d \right )}{2}-\frac {b^{2} c^{2} \ln \left (F \right )^{2}}{2}+e \sin \left (2 e x +2 d \right ) b c \ln \left (F \right )-2 e^{2}\right )}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}\) | \(95\) |
risch | \(\frac {F^{c \left (x b +a \right )}}{2 b c \ln \left (F \right )}-\frac {F^{c \left (x b +a \right )} b c \ln \left (F \right ) \cos \left (2 e x +2 d \right )}{2 \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}-\frac {e \,F^{c \left (x b +a \right )} \sin \left (2 e x +2 d \right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}\) | \(106\) |
norman | \(\frac {-\frac {4 e \,{\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {4 e \,{\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{3}}{4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}}+\frac {2 e^{2} {\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {2 e^{2} {\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{4}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}+\frac {4 \left (e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right ) {\mathrm e}^{c \left (x b +a \right ) \ln \left (F \right )} \tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}}{b c \ln \left (F \right ) \left (4 e^{2}+b^{2} c^{2} \ln \left (F \right )^{2}\right )}}{\left (1+\tan \left (\frac {e x}{2}+\frac {d}{2}\right )^{2}\right )^{2}}\) | \(268\) |
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Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.70 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=-\frac {{\left (2 \, b c e \cos \left (e x + d\right ) \log \left (F\right ) \sin \left (e x + d\right ) + {\left (b^{2} c^{2} \cos \left (e x + d\right )^{2} - b^{2} c^{2}\right )} \log \left (F\right )^{2} - 2 \, e^{2}\right )} F^{b c x + a c}}{b^{3} c^{3} \log \left (F\right )^{3} + 4 \, b c e^{2} \log \left (F\right )} \]
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Result contains complex when optimal does not.
Time = 1.06 (sec) , antiderivative size = 741, normalized size of antiderivative = 5.79 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=\begin {cases} x \sin ^{2}{\left (d \right )} & \text {for}\: F = 1 \wedge b = 0 \wedge c = 0 \wedge e = 0 \\\frac {x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {x \cos ^{2}{\left (d + e x \right )}}{2} - \frac {\sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} & \text {for}\: F = 1 \\F^{a c} \left (\frac {x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {x \cos ^{2}{\left (d + e x \right )}}{2} - \frac {\sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e}\right ) & \text {for}\: b = 0 \\\frac {x \sin ^{2}{\left (d + e x \right )}}{2} + \frac {x \cos ^{2}{\left (d + e x \right )}}{2} - \frac {\sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{2 e} & \text {for}\: c = 0 \\\frac {F^{a c + b c x} x \sin ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{4} - \frac {i F^{a c + b c x} x \sin {\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )} \cos {\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{2} - \frac {F^{a c + b c x} x \cos ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{4} + \frac {3 i F^{a c + b c x} \sin {\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )} \cos {\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{2 b c \log {\left (F \right )}} + \frac {F^{a c + b c x} \cos ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} - d \right )}}{b c \log {\left (F \right )}} & \text {for}\: e = - \frac {i b c \log {\left (F \right )}}{2} \\\frac {F^{a c + b c x} x \sin ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{4} - \frac {i F^{a c + b c x} x \sin {\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )} \cos {\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{2} - \frac {F^{a c + b c x} x \cos ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{4} + \frac {F^{a c + b c x} \sin ^{2}{\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{b c \log {\left (F \right )}} - \frac {i F^{a c + b c x} \sin {\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )} \cos {\left (\frac {i b c x \log {\left (F \right )}}{2} + d \right )}}{2 b c \log {\left (F \right )}} & \text {for}\: e = \frac {i b c \log {\left (F \right )}}{2} \\\frac {F^{a c + b c x} b^{2} c^{2} \log {\left (F \right )}^{2} \sin ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + 4 b c e^{2} \log {\left (F \right )}} - \frac {2 F^{a c + b c x} b c e \log {\left (F \right )} \sin {\left (d + e x \right )} \cos {\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + 4 b c e^{2} \log {\left (F \right )}} + \frac {2 F^{a c + b c x} e^{2} \sin ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + 4 b c e^{2} \log {\left (F \right )}} + \frac {2 F^{a c + b c x} e^{2} \cos ^{2}{\left (d + e x \right )}}{b^{3} c^{3} \log {\left (F \right )}^{3} + 4 b c e^{2} \log {\left (F \right )}} & \text {otherwise} \end {cases} \]
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Leaf count of result is larger than twice the leaf count of optimal. 356 vs. \(2 (128) = 256\).
Time = 0.22 (sec) , antiderivative size = 356, normalized size of antiderivative = 2.78 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=-\frac {{\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right ) \log \left (F\right )^{2} + 2 \, F^{a c} b c e \log \left (F\right ) \sin \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, e x\right ) + {\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right ) \log \left (F\right )^{2} - 2 \, F^{a c} b c e \log \left (F\right ) \sin \left (2 \, d\right )\right )} F^{b c x} \cos \left (2 \, e x + 4 \, d\right ) - {\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right ) - 2 \, F^{a c} b c e \cos \left (2 \, d\right ) \log \left (F\right )\right )} F^{b c x} \sin \left (2 \, e x\right ) + {\left (F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right ) + 2 \, F^{a c} b c e \cos \left (2 \, d\right ) \log \left (F\right )\right )} F^{b c x} \sin \left (2 \, e x + 4 \, d\right ) - 2 \, {\left (F^{a c} b^{2} c^{2} \cos \left (2 \, d\right )^{2} \log \left (F\right )^{2} + F^{a c} b^{2} c^{2} \log \left (F\right )^{2} \sin \left (2 \, d\right )^{2} + 4 \, {\left (F^{a c} \cos \left (2 \, d\right )^{2} + F^{a c} \sin \left (2 \, d\right )^{2}\right )} e^{2}\right )} F^{b c x}}{4 \, {\left (b^{3} c^{3} \cos \left (2 \, d\right )^{2} \log \left (F\right )^{3} + b^{3} c^{3} \log \left (F\right )^{3} \sin \left (2 \, d\right )^{2} + 4 \, {\left (b c \cos \left (2 \, d\right )^{2} \log \left (F\right ) + b c \log \left (F\right ) \sin \left (2 \, d\right )^{2}\right )} e^{2}\right )}} \]
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Result contains complex when optimal does not.
Time = 0.32 (sec) , antiderivative size = 915, normalized size of antiderivative = 7.15 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=\text {Too large to display} \]
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Time = 29.59 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.74 \[ \int F^{c (a+b x)} \sin ^2(d+e x) \, dx=\frac {F^{a\,c+b\,c\,x}\,\left (2\,e^2+\frac {b^2\,c^2\,{\ln \left (F\right )}^2}{2}-\frac {b^2\,c^2\,{\ln \left (F\right )}^2\,\cos \left (2\,d+2\,e\,x\right )}{2}-b\,c\,e\,\ln \left (F\right )\,\sin \left (2\,d+2\,e\,x\right )\right )}{b\,c\,\ln \left (F\right )\,\left (b^2\,c^2\,{\ln \left (F\right )}^2+4\,e^2\right )} \]
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