Integrand size = 23, antiderivative size = 245 \[ \int f^{a+b x+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{-2 i d+\frac {b^2 \log ^2(f)}{8 i f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {b \log (f)-2 x (2 i f-c \log (f))}{2 \sqrt {2 i f-c \log (f)}}\right )}{8 \sqrt {2 i f-c \log (f)}}-\frac {e^{2 i d-\frac {b^2 \log ^2(f)}{8 i f+4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {b \log (f)+2 x (2 i f+c \log (f))}{2 \sqrt {2 i f+c \log (f)}}\right )}{8 \sqrt {2 i f+c \log (f)}} \]
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Time = 0.50 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4560, 2266, 2235, 2325, 2236} \[ \int f^{a+b x+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\frac {\sqrt {\pi } f^a e^{\frac {b^2 \log ^2(f)}{-4 c \log (f)+8 i f}-2 i d} \text {erf}\left (\frac {b \log (f)-2 x (-c \log (f)+2 i f)}{2 \sqrt {-c \log (f)+2 i f}}\right )}{8 \sqrt {-c \log (f)+2 i f}}-\frac {\sqrt {\pi } f^a e^{2 i d-\frac {b^2 \log ^2(f)}{4 c \log (f)+8 i f}} \text {erfi}\left (\frac {b \log (f)+2 x (c \log (f)+2 i f)}{2 \sqrt {c \log (f)+2 i f}}\right )}{8 \sqrt {c \log (f)+2 i f}}+\frac {\sqrt {\pi } f^{a-\frac {b^2}{4 c}} \text {erfi}\left (\frac {\sqrt {\log (f)} (b+2 c x)}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}} \]
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Rule 2235
Rule 2236
Rule 2266
Rule 2325
Rule 4560
Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {1}{2} f^{a+b x+c x^2}-\frac {1}{4} e^{-2 i d-2 i f x^2} f^{a+b x+c x^2}-\frac {1}{4} e^{2 i d+2 i f x^2} f^{a+b x+c x^2}\right ) \, dx \\ & = -\left (\frac {1}{4} \int e^{-2 i d-2 i f x^2} f^{a+b x+c x^2} \, dx\right )-\frac {1}{4} \int e^{2 i d+2 i f x^2} f^{a+b x+c x^2} \, dx+\frac {1}{2} \int f^{a+b x+c x^2} \, dx \\ & = -\left (\frac {1}{4} \int \exp \left (-2 i d+a \log (f)+b x \log (f)-x^2 (2 i f-c \log (f))\right ) \, dx\right )-\frac {1}{4} \int \exp \left (2 i d+a \log (f)+b x \log (f)+x^2 (2 i f+c \log (f))\right ) \, dx+\frac {1}{2} f^{a-\frac {b^2}{4 c}} \int f^{\frac {(b+2 c x)^2}{4 c}} \, dx \\ & = \frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}-\frac {1}{4} \left (e^{-2 i d+\frac {b^2 \log ^2(f)}{8 i f-4 c \log (f)}} f^a\right ) \int \exp \left (\frac {(b \log (f)+2 x (-2 i f+c \log (f)))^2}{4 (-2 i f+c \log (f))}\right ) \, dx-\frac {1}{4} \left (e^{2 i d-\frac {b^2 \log ^2(f)}{8 i f+4 c \log (f)}} f^a\right ) \int \exp \left (\frac {(b \log (f)+2 x (2 i f+c \log (f)))^2}{4 (2 i f+c \log (f))}\right ) \, dx \\ & = \frac {f^{a-\frac {b^2}{4 c}} \sqrt {\pi } \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{4 \sqrt {c} \sqrt {\log (f)}}+\frac {e^{-2 i d+\frac {b^2 \log ^2(f)}{8 i f-4 c \log (f)}} f^a \sqrt {\pi } \text {erf}\left (\frac {b \log (f)-2 x (2 i f-c \log (f))}{2 \sqrt {2 i f-c \log (f)}}\right )}{8 \sqrt {2 i f-c \log (f)}}-\frac {e^{2 i d-\frac {b^2 \log ^2(f)}{8 i f+4 c \log (f)}} f^a \sqrt {\pi } \text {erfi}\left (\frac {b \log (f)+2 x (2 i f+c \log (f))}{2 \sqrt {2 i f+c \log (f)}}\right )}{8 \sqrt {2 i f+c \log (f)}} \\ \end{align*}
Time = 2.23 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.22 \[ \int f^{a+b x+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\frac {1}{8} f^a \sqrt {\pi } \left (\frac {2 f^{-\frac {b^2}{4 c}} \text {erfi}\left (\frac {(b+2 c x) \sqrt {\log (f)}}{2 \sqrt {c}}\right )}{\sqrt {c} \sqrt {\log (f)}}+\frac {\sqrt [4]{-1} e^{\frac {b^2 \log ^2(f)}{8 i f-4 c \log (f)}} \left (\text {erf}\left (\frac {\sqrt [4]{-1} (4 f x+i (b+2 c x) \log (f))}{2 \sqrt {2 f+i c \log (f)}}\right ) \sqrt {2 f+i c \log (f)} (2 i f+c \log (f)) (\cos (2 d)-i \sin (2 d))+e^{\frac {i b^2 f \log ^2(f)}{4 f^2+c^2 \log ^2(f)}} \text {erf}\left (\frac {(-1)^{3/4} (4 f x-i (b+2 c x) \log (f))}{2 \sqrt {2 f-i c \log (f)}}\right ) \sqrt {2 f-i c \log (f)} (2 f+i c \log (f)) (\cos (2 d)+i \sin (2 d))\right )}{4 f^2+c^2 \log ^2(f)}\right ) \]
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Time = 0.66 (sec) , antiderivative size = 227, normalized size of antiderivative = 0.93
method | result | size |
risch | \(\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}+8 i d \ln \left (f \right ) c +16 d f}{4 \left (c \ln \left (f \right )-2 i f \right )}} \operatorname {erf}\left (-x \sqrt {2 i f -c \ln \left (f \right )}+\frac {\ln \left (f \right ) b}{2 \sqrt {2 i f -c \ln \left (f \right )}}\right )}{8 \sqrt {2 i f -c \ln \left (f \right )}}+\frac {\sqrt {\pi }\, f^{a} {\mathrm e}^{-\frac {\ln \left (f \right )^{2} b^{2}-8 i d \ln \left (f \right ) c +16 d f}{4 \left (2 i f +c \ln \left (f \right )\right )}} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )-2 i f}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )-2 i f}}\right )}{8 \sqrt {-c \ln \left (f \right )-2 i f}}-\frac {\sqrt {\pi }\, f^{-\frac {b^{2}}{4 c}} f^{a} \operatorname {erf}\left (-\sqrt {-c \ln \left (f \right )}\, x +\frac {\ln \left (f \right ) b}{2 \sqrt {-c \ln \left (f \right )}}\right )}{4 \sqrt {-c \ln \left (f \right )}}\) | \(227\) |
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Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 402 vs. \(2 (185) = 370\).
Time = 0.27 (sec) , antiderivative size = 402, normalized size of antiderivative = 1.64 \[ \int f^{a+b x+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\frac {\sqrt {\pi } {\left (c^{2} \log \left (f\right )^{2} - 2 i \, c f \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) - 2 i \, f} \operatorname {erf}\left (\frac {{\left (8 \, f^{2} x - 2 i \, b f \log \left (f\right ) + {\left (2 \, c^{2} x + b c\right )} \log \left (f\right )^{2}\right )} \sqrt {-c \log \left (f\right ) - 2 i \, f}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )}}\right ) e^{\left (\frac {16 \, a f^{2} \log \left (f\right ) - {\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )^{3} + 32 i \, d f^{2} - 2 \, {\left (-4 i \, c^{2} d - i \, b^{2} f\right )} \log \left (f\right )^{2}}{4 \, {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )}}\right )} + \sqrt {\pi } {\left (c^{2} \log \left (f\right )^{2} + 2 i \, c f \log \left (f\right )\right )} \sqrt {-c \log \left (f\right ) + 2 i \, f} \operatorname {erf}\left (\frac {{\left (8 \, f^{2} x + 2 i \, b f \log \left (f\right ) + {\left (2 \, c^{2} x + b c\right )} \log \left (f\right )^{2}\right )} \sqrt {-c \log \left (f\right ) + 2 i \, f}}{2 \, {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )}}\right ) e^{\left (\frac {16 \, a f^{2} \log \left (f\right ) - {\left (b^{2} c - 4 \, a c^{2}\right )} \log \left (f\right )^{3} - 32 i \, d f^{2} - 2 \, {\left (4 i \, c^{2} d + i \, b^{2} f\right )} \log \left (f\right )^{2}}{4 \, {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )}}\right )} - \frac {2 \, \sqrt {\pi } {\left (c^{2} \log \left (f\right )^{2} + 4 \, f^{2}\right )} \sqrt {-c \log \left (f\right )} \operatorname {erf}\left (\frac {{\left (2 \, c x + b\right )} \sqrt {-c \log \left (f\right )}}{2 \, c}\right )}{f^{\frac {b^{2} - 4 \, a c}{4 \, c}}}}{8 \, {\left (c^{3} \log \left (f\right )^{3} + 4 \, c f^{2} \log \left (f\right )\right )}} \]
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\[ \int f^{a+b x+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\int f^{a + b x + c x^{2}} \sin ^{2}{\left (d + f x^{2} \right )}\, dx \]
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Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.24 (sec) , antiderivative size = 997, normalized size of antiderivative = 4.07 \[ \int f^{a+b x+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\text {Too large to display} \]
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\[ \int f^{a+b x+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\int { f^{c x^{2} + b x + a} \sin \left (f x^{2} + d\right )^{2} \,d x } \]
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Timed out. \[ \int f^{a+b x+c x^2} \sin ^2\left (d+f x^2\right ) \, dx=\int f^{c\,x^2+b\,x+a}\,{\sin \left (f\,x^2+d\right )}^2 \,d x \]
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