Integrand size = 15, antiderivative size = 18 \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi } \text {erf}(b x)^2}{4 b} \]
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Time = 0.01 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {6508, 30} \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi } \text {erf}(b x)^2}{4 b} \]
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Rule 30
Rule 6508
Rubi steps \begin{align*} \text {integral}& = \frac {\sqrt {\pi } \text {Subst}(\int x \, dx,x,\text {erf}(b x))}{2 b} \\ & = \frac {\sqrt {\pi } \text {erf}(b x)^2}{4 b} \\ \end{align*}
Time = 0.00 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.00 \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi } \text {erf}(b x)^2}{4 b} \]
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Time = 0.26 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83
method | result | size |
default | \(\frac {\operatorname {erf}\left (b x \right )^{2} \sqrt {\pi }}{4 b}\) | \(15\) |
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Time = 0.26 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi } \operatorname {erf}\left (b x\right )^{2}}{4 \, b} \]
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Time = 0.30 (sec) , antiderivative size = 15, normalized size of antiderivative = 0.83 \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\begin {cases} \frac {\sqrt {\pi } \operatorname {erf}^{2}{\left (b x \right )}}{4 b} & \text {for}\: b \neq 0 \\0 & \text {otherwise} \end {cases} \]
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Time = 0.19 (sec) , antiderivative size = 14, normalized size of antiderivative = 0.78 \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi } \operatorname {erf}\left (b x\right )^{2}}{4 \, b} \]
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\[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\int { \operatorname {erf}\left (b x\right ) e^{\left (-b^{2} x^{2}\right )} \,d x } \]
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Time = 0.20 (sec) , antiderivative size = 41, normalized size of antiderivative = 2.28 \[ \int e^{-b^2 x^2} \text {erf}(b x) \, dx=\frac {\sqrt {\pi }\,\mathrm {erf}\left (x\,\sqrt {b^2}\right )\,\mathrm {erf}\left (b\,x\right )}{2\,\sqrt {b^2}}-\frac {\sqrt {\pi }\,{\mathrm {erf}\left (x\,\sqrt {b^2}\right )}^2}{4\,b} \]
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