Integrand size = 16, antiderivative size = 21 \[ \int e^{c+b^2 x^2} \text {erfi}(b x) \, dx=\frac {e^c \sqrt {\pi } \text {erfi}(b x)^2}{4 b} \]
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Time = 0.01 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {6510, 30} \[ \int e^{c+b^2 x^2} \text {erfi}(b x) \, dx=\frac {\sqrt {\pi } e^c \text {erfi}(b x)^2}{4 b} \]
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Rule 30
Rule 6510
Rubi steps \begin{align*} \text {integral}& = \frac {\left (e^c \sqrt {\pi }\right ) \text {Subst}(\int x \, dx,x,\text {erfi}(b x))}{2 b} \\ & = \frac {e^c \sqrt {\pi } \text {erfi}(b x)^2}{4 b} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 21, normalized size of antiderivative = 1.00 \[ \int e^{c+b^2 x^2} \text {erfi}(b x) \, dx=\frac {e^c \sqrt {\pi } \text {erfi}(b x)^2}{4 b} \]
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\[\int {\mathrm e}^{b^{2} x^{2}+c} \operatorname {erfi}\left (b x \right )d x\]
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none
Time = 0.25 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.76 \[ \int e^{c+b^2 x^2} \text {erfi}(b x) \, dx=\frac {\sqrt {\pi } \operatorname {erfi}\left (b x\right )^{2} e^{c}}{4 \, b} \]
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Time = 0.15 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.90 \[ \int e^{c+b^2 x^2} \text {erfi}(b x) \, dx=\begin {cases} \frac {\sqrt {\pi } e^{c} \operatorname {erfi}^{2}{\left (b x \right )}}{4 b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]
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\[ \int e^{c+b^2 x^2} \text {erfi}(b x) \, dx=\int { \operatorname {erfi}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )} \,d x } \]
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\[ \int e^{c+b^2 x^2} \text {erfi}(b x) \, dx=\int { \operatorname {erfi}\left (b x\right ) e^{\left (b^{2} x^{2} + c\right )} \,d x } \]
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Time = 5.00 (sec) , antiderivative size = 91, normalized size of antiderivative = 4.33 \[ \int e^{c+b^2 x^2} \text {erfi}(b x) \, dx=\frac {\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b^2\,x}{\sqrt {b^2}}\right )\,{\mathrm {e}}^c\,\mathrm {erfi}\left (b\,x\right )}{2\,\sqrt {b^2}}-\frac {\sqrt {\pi }\,{\mathrm {e}}^c\,{\mathrm {erf}\left (x\,\sqrt {-b^2}\right )}^2}{4\,b}-\frac {b\,\sqrt {\pi }\,\mathrm {erfi}\left (\frac {b^2\,x}{\sqrt {b^2}}\right )\,{\mathrm {e}}^c\,\mathrm {erf}\left (x\,\sqrt {-b^2}\right )}{2\,\sqrt {b^2}\,\sqrt {-b^2}} \]
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